mm-2759 === . Finally notice a familiar limit inside the log limit(log(A)) = log(limit(A)) log10(e)=1/ln(10) --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Tough Limit? lim ( Log(1-1/(10^x)) * (10^x - 1)) = -1/Ln(10) as x approaches >infinity >Where Log is the Logarithm to the base 10 and Ln is the logarithm to >the base e. >I have tried manipulating it to 0/0 and infinity/infinity form so I can >use L'Hospitals rule but have had no luck with the resulting >derivatives. Any ideas? > Substitute n = 10^x to make it easier on the eyes. > Then move the (n-1) into the log() -- log(a)*b = log(a^b). > Finally notice a familiar limit inside the log > limit(log(A)) = log(limit(A)) > log10(e)=1/ln(10) > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. Angela === Subject: Re: Tough Limit? lim ( Log(1-1/(10^x)) * (10^x - 1)) = -1/Ln(10) as x approaches >infinity >Where Log is the Logarithm to the base 10 and Ln is the logarithm to >the base e. >I have tried manipulating it to 0/0 and infinity/infinity form so I can >use L'Hospitals rule but have had no luck with the resulting >derivatives. Any ideas? > Substitute n = 10^x to make it easier on the eyes. > Then move the (n-1) into the log() -- log(a)*b = log(a^b). > Finally notice a familiar limit inside the log > limit(log(A)) = log(limit(A)) > log10(e)=1/ln(10) > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. === Subject: Spanish-American Whoa The Spanish-American war began in 1898. To date it has cost American taxpayers over $300,000,000,000.00. That's over 300 billion for the non-mathemeticians. Tell your congresspersons and your senators it is time for you to stop paying for a war that ended over 100 years ago. Imperfect as I am if I had a choice I would choose to remain imperfect. I enjoy life a lot more than the perfect members of our species. Pax Vobiscum Michelangelo Merisi da Caravaggio . === Subject: Re: Spanish-American Whoa > The Spanish-American war began in 1898. > To date it has cost American taxpayers over $300,000,000,000.00. > That's over 300 billion for the non-mathemeticians. Actually it's only 0.3 of a billion here in Aus. Scott === Subject: Re: Spanish-American Whoa > The Spanish-American war began in 1898. > To date it has cost American taxpayers over $300,000,000,000.00. > That's over 300 billion for the non-mathemeticians. > Tell your congresspersons and your senators it is time for you to stop paying for a war that ended over 100 years ago. Could you provide a few details about current expenditures? === Subject: Re: Spanish-American Whoa :-)= Could you provide a few details about current expenditures? Expenditures for the war since it ended or at least by the end of 1899 = USD$zero.00. The telephone excise tax that was imposed to help pay for the war is still in effect. Paying for Guantanamo is not an excuse for a temporary tax imposed in 1898 for the S-A War. Imperfect as I am, if I had a choice I would remain imperfect. I enjoy life so much more than the perfect members of our species. Pax Vobiscum Michelangelo Merisi da Caravaggio === Subject: Re: Spanish-American Whoa >>The Spanish-American war began in 1898. >>To date it has cost American taxpayers over $300,000,000,000.00. >>That's over 300 billion for the non-mathemeticians. >>Tell your congresspersons and your senators it is time for you to stop > paying for a war that ended over 100 years ago. > Could you provide a few details about current expenditures? One word: Guantanamo. Ceded by the Spanish, it was never Cuban territory and remains a US possession despite what Neo-cons would have you beleive. James === Subject: Re: Spanish-American Whoa >The Spanish-American war began in 1898. >To date it has cost American taxpayers over $300,000,000,000.00. >That's over 300 billion for the non-mathemeticians. >Tell your congresspersons and your senators it is time for you to stop >> paying for a war that ended over 100 years ago. >> Could you provide a few details about current expenditures? >One word: Guantanamo. >Ceded by the Spanish, it was never Cuban territory and remains a US >possession despite what Neo-cons would have you beleive. Quite wrong. Gitmo is a leasehold from Cuba. The 1898 treaty with Spain treats the US presence as an occupation: Article I. Spain relinquishes all claim of sovereignty over and title to Cuba.And as the island is, upon its evacuation by Spain, to be occupied by the United States, the United States will, so long as such occupation shall last, assume and discharge the obligations that may under international law result from the fact of its occupation, for the protection of life and property. A treaty between the USA and the new Republic of Cuba was concluded in 1903 and modified in 1934: ARTICLE III Until the two contracting parties agree to the modification or abrogation of the stipulations of the agreement in regard to the lease to the United States of America of lands in Cuba for coaling and naval stations signed by the President of the Republic of Cuba on February 16, 1903, and by the President of the United States of America on the 23d day of the same month and year, the stipulations of that agreement with regard to the naval station of Guantanamo shall continue in effect. The supplementary agreement in regard to naval or coaling stations signed between the two Governments on July 2, 1903, also shall continue in effect in the same form and on the same conditions with respect to the naval station at Guantanamo. So long as the United States of America shall not abandon the said naval station of Guantanamo or the two Governments shall not agree to a modification of its present limits, the station shall continue to have the territorial area that it now has, with the limits that it has on the date of the signature of the present Treaty. According to the treaty of 1903: ARTICLE I The United States of America agrees and covenants to pay to the Republic of Cuba the annual sum of two thousand dollars, in gold coin of the United States, as long as the former shall occupy and use said areas of land by virtue of said agreement. I understand the USA still tries to pay the rent every year, and Castro duly refuses it. Cuba retains sovereignty over Guantanamo Bay, which is why the administration claims the US Constitution doesn't apply there, and the Supreme Court agrees. ************* DAVE HATUNEN (hatunen@cox.net) ************* * Tucson Arizona, out where the cacti grow * * My typos & mispellings are intentional copyright traps * === Subject: Re: Spanish-American Whoa > Cuba retains sovereignty over Guantanamo Bay, which is why the > administration claims the US Constitution doesn't apply there, > and the Supreme Court agrees. In what way does the Supreme Court agree? Quoting from their ruling: United States courts have jurisdiction to consider challenges to the legality of the detention of foreign nationals captured abroad in connection with hostilities and incarcerated at Guantanamo Bay. Question: Do the sailors and marines stationed at Gitmo still have the protection of the US Constitution and the Bill of Rights? === Subject: Re: Spanish-American Whoa On Fri, 1 Jul 2005 19:39:00 -0700, Richard Henry >> Cuba retains sovereignty over Guantanamo Bay, which is why the >> administration claims the US Constitution doesn't apply there, >> and the Supreme Court agrees. >In what way does the Supreme Court agree? Quoting from their ruling: >United States courts have jurisdiction to consider challenges to the >legality of the detention of foreign nationals captured abroad in connection >with hostilities and incarcerated at Guantanamo Bay. Note that I said that was the administration's claim, not mine. You've quoted from the summary of Rasul v Bush, but note the next paragraph: (a) The District Court has jurisdiction to hear petitioners.89 habeas challenges under 28 U.S. C. Û2241, which authorizes district courts, .8bwithin their respective jurisdictions,.8a to entertain habeas applications by persons claiming to be held .8bin custody in violation of the á laws á of the United States,.8a ÛÛ2241(a), (c)(3). Such jurisdiction extends to aliens held in a territory over which the United States exercises plenary and exclusive jurisdiction, but not .8bultimate sovereignty..8a Pp. 4Ö16. It's not too clear what the ultimate sovereignty phrase means here. But if you read the decision itself, you will see that Rasul v Bush is decided on federal *law* USC 3341, not the Constitution, which is also what it says above in (a). ************* DAVE HATUNEN (hatunen@cox.net) ************* * Tucson Arizona, out where the cacti grow * * My typos & mispellings are intentional copyright traps * === Subject: Re: Spanish-American Whoa resulting in: :-)= One word: Guantanamo. The telephone excise tax that was imposed to help pay for the war is still in effect. Paying for Guantanamo is not an excuse for a temporary tax imposed in 1898 for the S-A War. Imperfect as I am, if I had a choice I would remain imperfect. I enjoy life so much more than the perfect members of our species. Pax Vobiscum Michelangelo Merisi da Caravaggio === Subject: Re: Spanish-American Whoa Ascher let their cat run across the keyboard > resulting in: > :-)= One word: Guantanamo. > The telephone excise tax that was imposed to help pay for the war is still in effect. > Paying for Guantanamo is not an excuse for a temporary tax imposed in 1898 for the S-A War. That statement makes no sense. > Imperfect as I am, if I had a choice I would remain imperfect. Your clock is wrong, too. === Subject: Re: Spanish-American Whoa Approx. Fri, 1 Jul 2005 19:33:09 -0700, someone calling themselves Richard Henry let their cat run across the keyboard resulting in: :-)= That statement makes no sense. I must agree that a temporary tax imposed in 1898 is still in effect in 2005 makes no sense. Don't tell me. I already figured it out. The government is betting that your priority is to flame me and keep paying the tax rather than try to get Congress to repeal the tax. It makes absolutely no sense whatever as the tax was imposed to help defray the costs of a 6 month war that ended over 105 years ago. Imperfect as I am, if I had a choice I would remain imperfect. I enjoy life so much more than the perfect members of our species. Pax Vobiscum Michelangelo Merisi da Caravaggio === Subject: Question about exp (x) function Using the series expression for exp(x), how does one prove exp(x) * exp(y) = exp(x+y) === Subject: Re: Question about exp (x) function > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) Others suggested handling the series directly. Another way: With some standard calculus, we can use the series to verify (d/dx) exp(x) = exp(x) and directly exp(0)=1. Then forget the series and with given x and y, define g(t) = exp(x+y-t) * exp(t) . Verify that g'(t) = 0 for all t, so g is a constant (as a function of t), in particular g(y) = g(0). That is what you wanted. === Subject: Re: Question about exp (x) function > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) ln(exp(x) * exp(y)) = ln(exp(x+y)) => ln(exp(x)+ln(exp(y)) = x+y => x+y = x+y that is about it isn't it? === Subject: Re: Question about exp (x) function > ln(exp(x) * exp(y)) = ln(exp(x+y)) > => ln(exp(x)+ln(exp(y)) = x+y > => x+y = x+y > that is about it isn't it? No. First of all, you have deduced that *if* exp(x)*exp(y) = exp(x + y) *then* x + y = x + y. So? What do you think you get from the fact that the thing you want to prove implies a true statement? Furthermore, you've used that fact that log(a*b) = log(a) + log(b). This is at least as hard as the original problem. Also, the original problem makes sense over the complex numbers, but there you have no nice log function. Jose Carlos Santos === Subject: Re: Question about exp (x) function >> ln(exp(x) * exp(y)) = ln(exp(x+y)) >> => ln(exp(x)+ln(exp(y)) = x+y >> => x+y = x+y >> that is about it isn't it? > No. First of all, you have deduced that *if* exp(x)*exp(y) = exp(x + y) > *then* x + y = x + y. So? What do you think you get from the fact that > the thing you want to prove implies a true statement? > Furthermore, you've used that fact that log(a*b) = log(a) + log(b). This > is at least as hard as the original problem. > Also, the original problem makes sense over the complex numbers, but > there you have no nice log function. > Jose Carlos Santos May I also add that the original question required use of the series expansion, which the log solution clearly does not use. === Subject: Re: Question about exp (x) function >Using the series expression for exp(x), how does one prove >exp(x) * exp(y) = exp(x+y) Collect terms of like total power. I.e., x^3*y^2 is of 5th power. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Question about exp (x) function exp(x) * exp(y) = exp(x+y) > Collect terms of like total power. I.e., x^3*y^2 is of 5th power. === Subject: Re: Question about exp (x) function > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) You can find it written out in the beginning of Rudin's Real and Complex Analysis which we used to call Big Rudin back when I was studying these things. Achava === Subject: Re: Question about exp (x) function >> Using the series expression for exp(x), how does one prove >> exp(x) * exp(y) = exp(x+y) > You can find it written out in the beginning of Rudin's Real and > Complex Analysis which we used to call Big Rudin back when I was > studying these things. > Achava And in chapter 8 of principles... aka little or baby Rudin. === Subject: Re: Question about exp (x) function > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) It uses these two items: (1) rearrangement of absolutely convergent series (2) the binomial theorem -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Question about exp (x) function > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) exp(x+y) = sum(n=0,oo) [(x+y)^n]/n!. Expand (x+y)^n by the binomial theorem, then switch the order of summation. === Subject: Re: Question about exp (x) function agapito6314@aol.com a .8ecrit : > Using the series expression for exp(x), how does one prove > exp(x) * exp(y) = exp(x+y) with the Cauchy product of series === Subject: abstract algebra I am given that [G:H] = m. I am to show that for all g in G that g^m is in H. This is what I have so far. [G:H] = m means that there are m distinct cosets of H that partition G into m parts. One such coset is He = H itself. If g is in H then Hg = H (and gH = H). I am trying to conclude that g^m is not in any cosets of the form gH ( /= H). If I can show this then g^m is in H, since g^m is in one coset. Please do not give me the complete proof but rather a hint. === Subject: Re: abstract algebra > I am given that [G:H] = m. I am to show that for all g in G that g^m is in > H. > This is what I have so far. [G:H] = m means that there are m distinct cosets > of H that partition G into m parts. One such coset is He = H itself. If g > is in H then Hg = H (and gH = H). I am trying to conclude that g^m is not in > any cosets of the form gH ( /= H). If I can show this then g^m is in H, > since g^m is in one coset. > Please do not give me the complete proof but rather a hint. Let G=S_3, and H be the subgroup generated by (12). [G:H]=3, but (13)^3=(13) which is not in H. === Subject: Re: abstract algebra H must be a normal subgroup for this to be true. <(12)> is not normal in S_3. Van === Subject: Re: abstract algebra > H must be a normal subgroup for this to be true. > <(12)> is not normal in S_3. > Van I know. The OP had missed this condition, and I was providing him with a counterexample. === Subject: Re: abstract algebra I think that H must be normal. Consider the map f from G to G/H defined as f(g) = gH. g^m in H <=> f(g^m) = H <==> f(g)^m = H... === Subject: Re: abstract algebra if H is a NORMAL subgroup of G, than the cosets inherit a structure of a quotient group: G/H . what is the quotient group's order? what can you say about the order of any element in such a group? === Subject: Re: Cantor and the binary tree > There is no largest natural number in the set N. But we know that every > n e N is finite and has a finite initial segment. Why can't you grasp > the simple truth that a proof of the finity of n does not necessarily > need fix that n? > Card({1,2,3,...,omega}) = aleph_0 > Cantor says: Every number less than omega is a finite number and is > surpassed byother finite nunmbers. But omega is not in the set N!!!! > Card({1,2,3,...}) cannot be infinite, although there is no largest n. > Because we know that all naturals are finite. One can even use this > fact to state: A countable set, i.e. a set the elements of which can be > bijected with N, is *not* actually infinite. In particular N is not > actally infinite (though it is potentially infinite). Eh? Considere the sets A = {1,2,3,...,omega} and B = {1,2,3,...} and the bijection f: A -> B: f(omege) = 1, f(n) = n + 1 if n != omega. By what you state the cardinalities of these two sets are the same (and they are). But for one you state the cardinality is omega, and for the other you state it is finite. Aren't you a bit inconsistent? > Otherwise there was the following contradiction: > By the axiom of infinity guarantees, after von Neumann, the existence > of all the natural numbers {}, {{}}, {{},{{}}}, ... > The axiom of infinity guarantees infinite sets. > Both values and cardinality are created by this axiom. Why is the > cardinality infinite but not the values? Because the values are unbounded. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree There is no largest natural number in the set N. But we know that every > > n e N is finite and has a finite initial segment. Why can't you grasp > > the simple truth that a proof of the finity of n does not necessarily > > need fix that n? > > Card({1,2,3,...,omega}) = aleph_0 > > Cantor says: Every number less than omega is a finite number and is > > surpassed byother finite nunmbers. But omega is not in the set N!!!! > > Card({1,2,3,...}) cannot be infinite, although there is no largest n. > > Because we know that all naturals are finite. One can even use this > > fact to state: A countable set, i.e. a set the elements of which can be > > bijected with N, is *not* actually infinite. In particular N is not > > actally infinite (though it is potentially infinite). > Eh? Considere the sets A = {1,2,3,...,omega} and B = {1,2,3,...} and > the bijection f: A -> B: > f(omege) = 1, > f(n) = n + 1 if n != omega. > By what you state the cardinalities of these two sets are the same (and > they are). But for one you state the cardinality is omega, and for the > other you state it is finite. Aren't you a bit inconsistent? Your bijection with f(n) = n+1 would apply to the sets {0,1,2,3,...} and {1,2,3,...} although I believe with Bolzano that a bijection does not prove or express any equivalence of sets. But omega is not a single number, because omega = 1 + omega = 2 + omega = ...= omega + omega = omega * omega = 2^omega = omega^omega. Therefore no bijection can include it. > > Otherwise there was the following contradiction: > > By the axiom of infinity guarantees, after von Neumann, the existence > > of all the natural numbers {}, {{}}, {{},{{}}}, ... > > The axiom of infinity guarantees infinite sets. > > Both values and cardinality are created by this axiom. Why is the > > cardinality infinite but not the values? > Because the values are unbounded. Exactly. If an actually infinite set is created, then also an actually infinite number is created, simultaneously. The same is true for the initial segments. Therefore it is clear: Both, segments and values, are unbounded but always finite. That is potential infinity. === Subject: Re: Cantor and the binary tree ... > Eh? Considere the sets A = {1,2,3,...,omega} and B = {1,2,3,...} and > the bijection f: A -> B: > f(omege) = 1, > f(n) = n + 1 if n != omega. > By what you state the cardinalities of these two sets are the same (and > they are). But for one you state the cardinality is omega, and for the > other you state it is finite. Aren't you a bit inconsistent? > Your bijection with f(n) = n+1 would apply to the sets {0,1,2,3,...} > and {1,2,3,...} although I believe with Bolzano that a bijection does > not prove or express any equivalence of sets. Nothing is shown about equivalence at all. Only about the equality of the cardinal number of the set. > But omega is not a single > number, because omega = 1 + omega = 2 + omega = ...= omega + omega = > omega * omega = 2^omega = omega^omega. Therefore no bijection can > include it. And omega != omega + 1 != omega + 2, ... And you are wrong in stating that omega + omega = omega. You are confusing omega with oo, which is a concept, quite different from omega. > Therefore no bijection can > include it. I do not understand. You ar talking about the set {1, 2, 3, ..., omega}, so omega is an element of that set. In bijections we only talk about elements of a set, not about what they represent. > Because the values are unbounded. > Exactly. If an actually infinite set is created, then also an actually > infinite number is created, simultaneously. The same is true for the > initial segments. Therefore it is clear: Both, segments and values, are > unbounded but always finite. That is potential infinity. You do not consider limits. 1. What is lim{n -> oo} 1/n 2. What is lim{n -> oo} {1, 2, 3, ..., n} So in your opinion, the first limit is a potential 0, not an actual 0, if I understand you. Now I wonder in what way we should distinguish that potential 0 from an actual 0. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree Matt Gutting said: > Jan de Vos said: > >Um, you want me to agree that you have an infinite binary tree with >only finite paths, or infinite paths with only finite numbers of >nodes, or infinite numbers of nodes a finite distance from the root. >>The third of these is what I would like you to believe, the first two >>are obviously not true :-) Note that I don't mean there are >>infinitely many nodes with /a specific/ distance to the root, just >>that all nodes have /some/ finite distance to the root. > > > If there are finitely many branchings at each node (2), and finitely many > levels of branching, there is no way one can have infinitely many branches. > It's mathematically impossible. If you want an infinite number of branches or > nodes, then you either have to have infinite branchings at at least one point, > or at least one path that is infinitely long. If you have 2 branchins at each > node, and n (some finite number) of levels in the tree, then you have 2^n > branches, which is finite. > > True; but, as you yourself say, this is only true if n is some finite number. > If the levels in the tree cannot be described by a finite number (that is, if > one cannot say there are exactly n levels in the tree for some finite number > n), then your conclusion no longer holds. The number of levels must be infinite. If you say it is some large, but necessarily finite number, then it is finite, period. That vague definition of arbitrarily large but finite is most troublesome in this area, and all sorts of incorrect conclusions seem to be drawn regarding it. >I'm sorry but those ideas make no sense, and I can't agree to any of >that. >>I know. Just like you don't agree there are infinitely many finite >>integers -- even though that implies that the set of finite integers >>(which should thus be finite) has to have a maximum element, which you >>also deny :-) > > The exact same math applies to that situation, so no, I cannot agree to that > either, as it is mathematically nonsensical. You fellows seem to put a lot of > weight on there being a maximal element in any finite set of numbers, as > generalized from finite sets where a maximal element is identified, but that > generalization is unwarranted in this case. It is not a universal fact that > every finite set MUST have an identifiable maximal element. > Yes it is, with respect to some (arbitrarily assigned) ordering relation. Fine, then why can't I say, in my arbitrary ordering, that 1 is the smallest whole number and 2 is the largest, and count up through the odds and down through the evens, and have an infinite ordered set with min and max, and infinite values in the middle? > It IS a universal > fact that a tree with finite branchings and finite number of levels has finite > numbers of nodes, branches and paths. > > Yes it is; but we were talking about trees without a finite number of levels. I believe it was stated that all paths are finite, therefore there are a finite number of levels, since each path goes through all levels and is measure in terms of the number of levels long it is. If the paths are finite, then you have a finite number of levels. > Matt >>Jan > > Do you really think you can produce an infinite number of branches, producing > them in pairs, over a finite number of iterations? Show me the math on that > one, and no, I don't mean with your bijections and Cantorian argument. I mean > real math, with numbers, that actually says something. > -- Smiles, Tony === Subject: Re: Cantor and the binary tree Matt Gutting said: > Matt Gutting said: > >David Kastrup said: >David Kastrup said: >David Kastrup said: >>It assumes no such thing. It assumes that the list is indexed by >>natural numbers, and that the digits can be indexed by natural >>numbers. By the full entire set of natural numbers, both vrtically and >horizontally? the it is square. >>Last time I looked, a square has four corners, not one. Yeah, like any grid. >>Any grid has four corners? Do you even read the lunacies you are >>babbling? Look, i suppose you can imagine an infinite grid with NO corners, but when you >are talking about a list of digital numbers, you are talking about some kind of >rectangle. That shouldn't be debatable. >>Part of the definition of a square, or of a rectangle, is the existence of >>four sides, and therefore four corners. A rectangle has a top and a bottom, >>a right and a left. Otherwise, it is not a rectangle. That shouldn't be >>debatable. > > And when you decalre that the number of digits is the same as the number of > naturals, then you have a list aleph_0 wide, by aleph_1 long. > >You see one corner, and continuations to infinity horizontally and >vertically, and can't imagine an infinite square? >>A square has four sides and four corners. A quarterplane has two >>sides and one corner. So, you don't know whether there are aleph_0 or aleph_1 digits? Think of it >this way: If there are aleph_0 bits, and you are supposedly listing all the >reals, then there are aleph_1 strings, which is much larger. The grid is >greatly elongated, and diagonal traversal does not cover all strings. >>Elongated only applies to entities which have finite length. > > > Wrong. What is the ration of length vs width for a rectangle that is x wide by > 2x long, as x approaches infinity? It's still twice as long as wide. Do the > math, and stop playing bad logic games, and declaring nonexistent differences > between the finite and infinite. Most of them are nonsense. You can't just > declare them different. Use some real math occasionally. > > I am using real math. As x approaches infinity is a shorthand for as x > increases without bound. As x increases without bound, x remains finite. You > are thus noting a property of finite numbers, and generalizing the property > to something that is not finite. The infinite series 1/2^n from 1 to oo is 1. If it never reached infinity, it would be less than 1. This is hogwash in deference to Kronecker. >Must you look on the finite local level and pretend there is no >structure to this list, after having asserted that there is? Do you >still think the Earth is flat? This list is the limit as n->oo of >any list of all digital numbers of length n. Any such list is >exponentially longer than it is wide in digits. This is simple >stuff. >>You are babbling nonsense. Again. Nice declaration, but don't talk to yourself. People will think you're nuts. >Otherwise, a diagonal traversal on an elongated rectagular grid will >obviously miss many lines of the grid. >>It is nonsensical to talk about elongated rectangular for a quarter >>plane. A rectangle has four corners, not one. It is a Nx2^N, or Nx10^N, rectangle. >>No, it isn't. If N is something you can calculate with, it is finite. Not true, sorry. >If you are saying that there are infinite natural numbers with the the same >>properties (including calculation) as finite numbers, your statement entails >>rejection of the Axiom of Choice. Is this what you intend? > > Absolutely. Axiom of choice is only necessary because of the paucity of real > truth in the system. It's a kludge, a band-aid. > I am glad you cleared that up. If you are denying the Axiom of Choice, then you > are not doing standard math, and we're working at cross purposes. I never claimed to eb doing standard cardinality. I think cardinality stinks for a variety of reasons. > >>That the digits can be indexed by natural numbers is obvious since >>they obey the laws: >> For every digit with significance 10^-n, there is a following digit >> with significance 10^-{n+1}. >> Digits with different significances have different significances of >> the following digit. >> There is a digit with significane 10^-1. >> The digit with significance 10^-1 has no preceding digit. >> Any set including the digit with significance 10^-1, and for each of >> its digits also containing the following digit contains all digits of >> the number in question. >>And that is in obvious correspondance with the Peano laws for the list >>indices. So it is easy to establish a bijection between list indices >>and digits. >yes, but if it is a digital number, then there are S^L of them. If L >is the same as |N|, and S is 2 (binary, you have a grid N wide, but >2^N long. In decimal it's worse, being 10^N long. In any case, the >list is much longer than it is wide, and connot be fully traversed >diagonally. >>Nonsense. It is equally wide as long since the sequence of digit >>positions and the sequence of list positions obeys exactly the same >>laws. That is absolutely incorrect. Here is a list of all 3-digit binary numbers >000 >001 >010 >011 >100 >101 >110 >111 Are there three of them? No, there are eight. Why? Because 2^3=8. Wake up. >If you allow N bits, do you get N binary numbers? No. You get 2^N. >>You are again confusing 3 and the size of the set of naturals. >>Again. Stop your wishful thinking. No, I am giving a trivial example that shows you're wrong. >I am not aware that it shows him to be wrong, since 3 and infinity cannot >>act the same way. > > They can in many ways, but that's not what we're discussing. Digital number > systems share the property of N=S^L, for any N, S and L, finite or infinite. N > depends on S and L. > >>The list cannot be fully traversed either across, diagonally, or >>downward anyway, since it has no end in either direction. >The the proof doesn't prove anything at all, does it? What do you >think the logical argument IS in that proof? What exactly does it >demonstrate, in your mind? >>It shows that there can be a number (as a limit of a sequence) that >>for every finite place in the list will differ by at least a finite >>difference from the number at the place. So what? How does that conflate to an uncountable set of reals? >Um, you can make a complete list of any countable set. (That is, >>you can index all the members of the set by using the natural numbers.) > > So you ARE assuming an NxN grid, as I said. (N is equivalent to aleph_0) > > I'm not assuming a grid; I'm assuming an N-length list of numbers, each of > which is the limit of an infinite series, whose elements are countable. And the assumption you make in proving them to be uncountable is that all countable sets are the same. Therefore, since you find this set to be larger than N, you deem it uncountable, but that initial assumption is false. >>If you can show that there is no way of indexing *all* the members >>of the set using the natural numbers, then the set is not countable >>by definition, and therefore it is uncountable. > > That is a leap. Just because they are counted differently, doesn't mean they > are uncountable. If they are a larger set than the naturals, then that is a > valid conclusion, perhaps, but to say they can't be enumerated like the > naturals (with the very same representation, no less), is just wrong. In fact, > a list of N naturals also doesn't require N digits, but log(N), since the very > same symbolic string arithmetic applies. All the proof proves is that the list > is longer than it is wide, and so there are different infinities. That's > significant, but why stop there? There are an infinite number of infinities > between N and R, an infinite number less than N, and an infinite number > grreater than R, which can all be placed in realtive order using actual math, > once you drop your hocus-pocus approach. > Again, you can't treat N as a natural number (by, for example, taking log(N)) > unless you are denying the Axiom of Choice. This is non-standard math, which is > just fine, but it can't be extended to work in the system most mathematicians > work in most of the time. This doesn't mean that either is wrong, simply that > they are incompatible - conclusions drawn from either system can't be applied > to the other. But they both apply to set size, and if they totally disagree, they can't both be right. > >>Squares do not come into play here at all, neither do >>diagonals. Only some convenient partial visualization shows >>something akin to a diagonal, but it is not relevant to the >>proof. >It is absolutely crucial. The conclusion of the proof is that the >numbers can't be listed, because a diagonal traversal can be used >to form a number, the antidiagonal, which not on the line of >traversal, and therefore not on this list. >>The proof does not depend on the name diagonal, nor is there >>anything leading from one corner to another corner. The proof >>relies on a 1:1 correspondence/bijection of digits and list places. >>Both obey they laws of natural numbers, and so can be put into >>correspondence. That is so far off base I can't even imagine what you're >thinking. it assumes something that is simply not true, >>What? That the diagonal traversal covers all strings in the list. >It doesn't *assume* this, it *proves* it. > > (sigh) Incorrect, again. It assumes it, then proves by contradiction that it is > NOT the case that the traversal covers all items in the list, through the use > of the antidiagonal, which is not in the list traversed. That original > assumption, though, rather than being considered disproved, is clung to, and a > different conclusion s drawn, that a list cannot be generated. It's very faulty > logic. > No; the initial assumption is not that the diagonal traversal covers all > strings in the list. The initial assumption is that a countable and complete > list of real numbers (not strings) can exist. This assumption, as you note, > is disproved, and the appropriate conclusion is drawn - that such a list cannot > exist (not that it cannot be generated; the proof says nothing about > generating the list). That's perfectly reasonable and standard logic: assume > something is true, show that the assumption leads to a contradiction, deny > the assumption. I noted that the assumption is CONSIDERED disproved, but as I've said, that assumption is disproved with another assumption which is false, that all countable sets are the same size. You see, if you run into a contradiction, it may derive from places you don't expect. You need to examine ALL the assumptions/axioms that go into the statements that contradict each other, in order to really tell exactly where the flaw is. > >and jumps to conclusions about the significance of a diagonal >traversal and inversion of bits, which you think has nothing to do >with the diagonal. Maybe you NEED to smoke something. >>Maybe you should stop smoking your stuff. Ouch! >This rests on the assumption that the diagonal covers every >digital number on the list, which is clearly not the case, since >the grid is longer than it is wide. >>It does not end horizontally, and it does not end vertically. So >>you can walk on in diagonal direction (or any other direction) as >>long as you want to without exhausting either dimension. But, what does the diagonal traversal and the generated antidiagonal >signify to you, if you don't believe the diagonal has exhausted all >the numbers in the set, and yet still has some left? That is the >whole point of the proof. As far as I can see, you don't know WHAT >it's supposed to mean. >>Uh, it appears you don't know what it means. The proof shows that for >>every _finite_ place in the list, the constructed number will differ >>by at least non-zero finite value from the number at the place in the >>list. Yes, so? Is that the conclusion of the proof, the big significant discovery of >Cantor? Maybe you should ask your classmates for their notes, because you >obviously weren't paying attention that day. >>Yes, it is the big significant discovery of Cantor. That's the whole >>point of the proof. > > The significance of the discovery si simply that there are different > infinities, which IS significant. Don't conflate it with uncountability, > though. That's an unwarranted leap. > ??? You don't understand that infinities can be different, and still be countable? No, of course you don't. That's not standard....yet. > >No, it is correct thinking. How many different values can you >represent with three decimal digits? Three? A thousand? Who's >blowing smoke now? >>You are. Because three is a fixed finite number, and n can take >>on any value. For any n-length decimal number, there are 10^n possible >values. >>oo is not n-length. No payoff. So, while the rectangles get more and more elongated as n approaches oo, they >suddenly become perfectly sqaure at infinity. That's just ridiculous. >>What is this at infinity? You can only speak of something being in a given >>state at infinity if it can actually reach infinity. Which, by definition, >>is not possible. Similarly, n does not approach oo, it merely increases >>without bound (maintaining finite values all the time). > > What is the limit of N^x/x, for positive N, as x approaches oo? Is it 1? No? > Then why do you think the aspect ratio of the infinite grid is 1? It's > infinite, at infinity. > There's no such thing as at infinity, and there's no such thing as > approaching infinity. There's only arbitrarily large (but finite), and > increasing without bound (but remaining finite) - at least if you're talking > about natural numbers. Okay so the infinite series n=1->oo 1/2^n is less than 1, since n never reaches infinity? gimme a break from this tongue-tied silliness. I don't bow to Kronecker. > I didn't say anything about aspect ratio, that's your term. As I understand > aspect ratio, it applies only to those rectangular objects with finite length > and height. This is not the case for a complete list of reals. How would you know whether it applies to infinite rectangles? You don't believe there are such things, and obviously have not pur much thought into it, probably because it is thusly discouraged. > In any event, as I posted elsewhere, it's possible to prove Cantor's result > without reference to diagonals, aspect ratios, grids, rectangles, or any other > geometric analogies. Yes I saw that, but when you are talking about digit d_jj, you are describing the diagnonal, and the same argument can be made that for any real number after the first N of them, there IS no d_jj, because the row index is already larger than the column index will ever be. Your e is in that part of the list. The list exists, and is countable, but a larger infinity. There are myriad countable infinities. Now, if you really want an uncountable infinity, use an infinite set of sysmbols, as well as length in digits. > >>In any case, as David has said, neither rectangle nor square describes >>the grid/list we have, as the list doesn't fulfill the requirements of >>either geometric figure. > > Every complete list of digital numbers forms a rectangle as I have described. > You are purposely closing your eyes to this fact. > > Rectangles have left and right edges; they have top and bottom edges. Where > is the right edge to your rectangle? Where is the bottom? N digits to the right of the left edge, and R numbers below the top. >Three is an example of n. Take any n>0, and try it out. See whether >the realtionship changes as you have bigger sets. Ask yourself why >this constant equality would suddenly disappear at infinity, >>There is no suddenly. Infinity is a different ballpark. You are >>still applying your wishful thinking. Yeah, it's like me saying that, even though 2^n is always bigger than n for any >n>=0, that 2^n=n for n=oo. It's simply untrue, stupid, and ridiculous. >There is no natural number n such that n=oo. (Again, assuming the Axiom of >>Choice.) Your second equality is empty of content. > > I choose to dismiss the axiom of choice. Answer about the limit above stated > twice. > I think you're asking me to evaluate the limit, as n increases without bound, of > 2^n/n. That's simple. The limit does not exist, because the sequence {2^n/n} > diverges for increasing n. By diverge, and don't bother correcting me, you mean that the limit is infinite, not oscillating or otherwise indeterminate. 2^n/n>1 for all n in N, and increases without bound, so that when n is infinite, so is 2^n/n. > Again, if you dismiss the Axiom of Choice, I'm not sure that we can talk > further without a recognition that you are working with non-standard math. > (Standard math includes the Axiom of Choice.) Yes, I am working in a non-standard way, as I am apt to do. > Matt > >Bull, as you have just had your nose rubbed in. What kind of >mathematician contends that three digits can only represent htree >values in a digital number? Sounds like chicken-scratch math to me. >>Your problem is still that you are confusing three with cardinality >>of the naturals. They don't obey the same laws. >Pay attention. I used 3 as an example of a number of bits, to show >you concretely that three digits does NOT mean three possible >strings. If that example wasn't concrete enough for you then you're >hopeless. >>You are still thinking that 3 is governed by the same laws as infinity. I am thinking that symbolic systems are governed by the laws of symbolic >systems, and showing you that those laws apply for infinite sets as well as >finite sets, since the laws governing finiteness apply to the formulas that >govern strings. The number of strings of length L is S^L, so if that number of >strings is infinite, then either S or L is infinite, since S and L finite >implies S^L finite. It's pretty sad that you're so indoctrinated in this hare >krishna-style math that you can't think straight any more. >The laws governing finiteness apply to the formulas that govern strings of >>finite length. If they applied to strings of infinite length, we would have >>no way of distinguishing finite from infinite strings. > > The laws governing finiteness are what distinguishes between finite and > infinite values for formulas, given the finiteness or infinitude of the terms > it contains. For instance, for a^b to be finite, a and b must be finite. > >>The number of strings of length L (L a natural number), drawn from an alphabet >>of S characters (S a natural number), is indeed S^L. But if L is infinite, >>the expression S^L has no meaning as a natural number. > > It assumes an infinite value, which represents the sze of the set of strings. > Either you need infinite S or infinite L to get infinite S^L, which is my point > about the infinite set of naturals requiring either infinite strings, or an > infinite number base. No response to these last two points? Well, i have bee out of touch anyway, so who am I to complain? > >>Matt > > -- Smiles, Tony === Subject: Re: Cantor and the binary tree > It assumes an infinite value, which represents the sze of the set of > strings. > Either you need infinite S or infinite L to get infinite S^L, which is my > point > about the infinite set of naturals requiring either infinite strings, or > an > infinite number base. > No response to these last two points? Well, i have bee out of touch anyway, > so > who am I to complain? For all L in N, { n^L : n in N} has no finite upper bound, as the sequence n^L diverges towards the infinite. For all S > 1 and S in N, {S^m : m in N} has no finite upper bound, as the sequence S^m diverges towards the infinite. === Subject: Re: Cantor and the binary tree Matt Gutting said: > Matt Gutting said: > David is right in saying that the proof does not depend on a diagonal; >>it can be rephrased without even relying on any geometric analogy, and >>still holds true. > > Give it a shot. I'd love to hear that form of convoluted nonsense. I have never > heard any other real version of this proof. Enlighten me. > Okay, here goes. > Let the sequence {a_i}, indexed by the natural numbers, be a list of all > real numbers in the closed interval [0,1] (or, if you like, [0,.9999999...]). > That is, suppose we can make an ordered list of all the reals in this interval, > and suppose further that we can label each real a_1, a_2, a_3, etc. > Now each real a_i is equal to the limit, as n increases without bound, of > sum(j goes from 1 to n)[d_ij * 10^(-j)], where d_ij, the jth (let's say decimal) > digit of a_i, is a nonnegative integer less than ten. > (It is important to note that a_i is not somehow constructed by taking partial > sums; it is _defined_ as a single number that is a limit of these partial sums. > It is possible to evaluate a_i as the limit without evaluating any of the > partial sums.) > Now, define a sequence, or rather an ordered list, of nonnegative integers > {e_j}, such that e_j = 1 when d_jj = 0, and e_i = 0 when d_jj <> 0. It is > clear that such a sequence exists; further, just like a_i, it is not > constructed by evaluating each e_i in turn, but exists all at once, according > to the definition. It is not even necessary to determine the values of any > member e_i in the sequence. > There exists a real number e which is the limit, as n increases without bound, > of sum(j goes from 1 to n)[e_j * 10^(-j)]. > Now any real number in the interval can be represented as I represented a_i > above, or as I represented e. Further, any two numbers expressed as > limit as n increases without bound (sum (j from 1 to n)[s_j * 10^(-j)]) > and > limit as n increases without bound (sum (j from 1 to n)[t_j * 10^(-j)]) > are equal if and only if s_j = t_j for all j (that is, the two numbers are > equal if and only if all corresponding digits are identical). > But for all j, e_j <> d_jj; thus e differs from each a_j in at least one > place (the jth place); thus e is not equal to any a_j in the list, so that > e is a real number in the given interval but not in the list. But > by hypothesis, the list {a_j} included all the real numbers in this interval, > so that e *must* be in the list. This is a contradiction; therefore our > hypothesis (that I can have a list of all reals in this interval, and index > this list with natural numbers) must be false. Therefore, there are more > reals in this interval than can be indexed with the natural numbers; that is, > there are more than countably many reals in the interval. Indeed it shows that there are more numbers than digits, so there are more reals this way than naturals, it would seem, but it doesn't prove that e is not anywhere on the list. e is a number whose index is greater than N, but which is on the list. That is, e is a number for which there is no d_jj, because N elements have already been listed. The bogus assumption here is your hypothesis that if this contradiction arises, it is because the list is not enumerable, when really the reason is that it always contains more numbers than digits. This seems obvious to me, and the conclusion normally drawn seems to be based on an unfounded assumtpion that all countable infinities are the same, based on bijections with sloppiness, and this mediochre use of counting to define infinity. You know, the same proof can be done with strings representing the naturals. If you have a list of natural digital numbers, you can construct an antidiagonal that is not on the list as well, for the very same reason of N=S^L. Why does the equivalent proof not prove the naturals are uncountable? Because that would disprove the very basis for the proof, and would show that to produce N numbers, you really only need log(N) digits, but that would be a smaller infinity than N, so it doesn't fit into your scheme. Well, it's time it did. -- Smiles, Tony === Subject: Re: Cantor and the binary tree > Matt Gutting said: > Matt Gutting said: > > >David is right in saying that the proof does not depend on a >>diagonal; it can be rephrased without even relying on any >>geometric analogy, and still holds true. > > Give it a shot. I'd love to hear that form of convoluted > nonsense. I have never heard any other real version of this > proof. Enlighten me. > > Okay, here goes. > > Let the sequence {a_i}, indexed by the natural numbers, be a list > of all real numbers in the closed interval [0,1] (or, if you like, > [0,.9999999...]). That is, suppose we can make an ordered list of > all the reals in this interval, and suppose further that we can > label each real a_1, a_2, a_3, etc. > > Now each real a_i is equal to the limit, as n increases without > bound, of sum(j goes from 1 to n)[d_ij * 10^(-j)], where d_ij, the > jth (let's say decimal) digit of a_i, is a nonnegative integer less > than ten. > > (It is important to note that a_i is not somehow constructed by > taking partial sums; it is _defined_ as a single number that is a > limit of these partial sums. It is possible to evaluate a_i as the > limit without evaluating any of the partial sums.) > > Now, define a sequence, or rather an ordered list, of nonnegative > integers {e_j}, such that e_j = 1 when d_jj = 0, and e_i = 0 when > d_jj <> 0. It is clear that such a sequence exists; further, just > like a_i, it is not constructed by evaluating each e_i in turn, > but exists all at once, according to the definition. It is not even > necessary to determine the values of any member e_i in the > sequence. > > There exists a real number e which is the limit, as n increases > without bound, of sum(j goes from 1 to n)[e_j * 10^(-j)]. > > Now any real number in the interval can be represented as I > represented a_i above, or as I represented e. Further, any two > numbers expressed as > > limit as n increases without bound (sum (j from 1 to n)[s_j * > 10^(-j)]) > > and > > limit as n increases without bound (sum (j from 1 to n)[t_j * > 10^(-j)]) > > are equal if and only if s_j = t_j for all j (that is, the two > numbers are equal if and only if all corresponding digits are > identical). > > But for all j, e_j <> d_jj; thus e differs from each a_j in at > least one place (the jth place); thus e is not equal to any a_j in > the list, so that e is a real number in the given interval but not > in the list. But by hypothesis, the list {a_j} included all the > real numbers in this interval, so that e *must* be in the list. > This is a contradiction; therefore our hypothesis (that I can have > a list of all reals in this interval, and index this list with > natural numbers) must be false. Therefore, there are more reals in > this interval than can be indexed with the natural numbers; that > is, there are more than countably many reals in the interval. > > Indeed it shows that there are more numbers than digits, so there are > more reals this way than naturals, it would seem, but it doesn't > prove that e is not anywhere on the list. e is a number whose index > is greater than N, but which is on the list. That is, e is a number > for which there is no d_jj, because N elements have already been > listed. The bogus assumption here is your hypothesis that if this > contradiction arises, it is because the list is not enumerable, when > really the reason is that it always contains more numbers than > digits. This seems obvious to me, and the conclusion normally drawn > seems to be based on an unfounded assumtpion that all countable > infinities are the same, based on bijections with sloppiness, and > this mediochre use of counting to define infinity. The basic definition of a *list* of members of a set S is a mapping f: N --> S. So that, contrary to TO's delusions above, anything not indexed by a member of N CANNOT be in the list. What seems so obvious to TO involves ignoring and contradicting essential properties of what he is talking about === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Indeed it shows that there are more numbers than digits, so there > are more reals this way than naturals, it would seem, And that's what the proof is about, and nothing else. > but it doesn't prove that e is not anywhere on the list. Uh, it proves that for any list enumeratable by N, a number can be constructed that is not on it. There is not the list, but _any_ list has this property of being incomplete. > e is a number whose index is greater than N, but which is on the > list. This is nonsensical since list, for the purpose of the proof, is defined as being a mapping of N to the reals, so any definition of a list involving a concept greater than N is not talking about an entity useful for the proof. You can, of course, define list to mean something arbitrary, but the proof is about a bijection of N and R, and so any definition leaving the scope of N is useless for the purpose of the proof and shere nonsense. > That is, e is a number for which there is no d_jj, because N > elements have already been listed. The bogus assumption here is your > hypothesis that if this contradiction arises, it is because the list > is not enumerable, when really the reason is that it always contains > more numbers than digits. Your mistake is manifest in the of the word always since there is only a single non-ending list with numbers and digits both being mapped to N, and not a class of such lists. It is like saying The door is always open right now. Can you close it?. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> than any prime? Than any given prime, yes! Pick any prime and there are more than that number of primes. > I do not want to pick. I speak of any prime within the set of primes. > But this is the general reason for our disagreement again: If you pick > a fixed number 2n of the set of all even numbers, then there are > certainly more elements in the set. But I do not ask for any fixed and > picked number!!! > But if you claim, as you do, that there are only finitely many, The number of members is restricted by the values of the finite numbers. > you must > be prepared to back up your claim. My challenge to your claim is that > should any number be the number of naturals there are guaranteed to > be more by merely adding 1 to that number. > If you cannot refute that challenge, your claim fails. But if you claim, as you do, that there are only finite naturals, you must be prepared to back up your claim. My challenge to your claim is that should any number limit the values of naturals there is guaranteed to be a larger one by merely adding 1 to that value. If you cannot refute that challenge, your claim fails. === Subject: Re: Cantor and the binary tree If there is no largest prime, then the number of all primes is larger > than any prime? Than any given prime, yes! Pick any prime and there are more than that number of primes. I do not want to pick. I speak of any prime within the set of primes. But this is the general reason for our disagreement again: If you pick > a fixed number 2n of the set of all even numbers, then there are > certainly more elements in the set. But I do not ask for any fixed and > picked number!!! > But if you claim, as you do, that there are only finitely many, > The number of members is restricted by the values of the finite > numbers. Which values of finite numbers? WM admits that there is no largest prime but also claims that there is a largest even. These are mutually contradictory and cannot both be valid. > you must > be prepared to back up your claim. My challenge to your claim is that > should any number be the number of naturals there are guaranteed to > be more by merely adding 1 to that number. > If you cannot refute that challenge, your claim fails. > But if you claim, as you do, that there are only finite naturals, you > must be prepared to back up your claim. My definition of a natural being finite is that no set of naturals bounded above by that natural allow any injection into a proper subset. Based on my definition, one can easily prove that 1 is finite and (less easily) prove that if n is a finite natural then so is n+1. Thus by induction, all naturals are finite. By that same definition of finiteness (no injections to proper subsets) one can prove that the set of ALL naturals, as defined by the Peano axioms, is not finite. > My challenge to your claim is that > should any number limit the values of naturals there is guaranteed to > be a larger one by merely adding 1 to that value. What claim of mine is this supposed to be challenging? My two claims that seem to be at issue here are (1) Each natural number is finite (in the sense defined above) (2) The set of all naturals is not finite (allows injection to proper subsets). And what is the values of the naturals? > If you cannot refute that challenge, your claim fails. I do not even undersatnd the challenge as yet. But if it is what I suspect, I can refute it. === Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> numbers. > Which values of finite numbers? Those which are in the set. I repeat it as clearly as I can, but it is an unfamiliar idea for you. So ponder for a while about it and then come up with your questions, if there are some remaining. {1,2,3,...,omega} is an actually infinite set. N = {1,2,3,...} is not an actually infinite set. In N there are only finite numbers. Every natural number n is finite and is surpassed by another finite natural number n+1. Every natural n is the maximum of its finite initial sequence {1,2,3,...,n}. N is the limit of all finite initial sequences. N is potentially infinite, i.e., every maximum element n e N is surpassed by another maximum element n+1, but this is also finite. > WM admits that there is no largest prime but also claims that there is a > largest even. These are mutually contradictory and cannot both be valid. Every countable set like the set of primes is potentially infinite but not actually infinite because all of its elements are finite and so is any of its initial segments. Countable is just a definition for remaining finite by value *and by number*. > By that same definition of finiteness (no injections to proper subsets) > one can prove that the set of ALL naturals, as defined by the Peano > axioms, is not finite. It is potentially infinite, but not actually infinite, in the same way as are the vales of natural numbers. The axiom of infinity creates both, numbers and sets. Show why, nevertheless, according to your opinion, infinite sets but no infinite numbers are created although this axiom works in the same way in both cases. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> The number of members is restricted by the values of the finite >> numbers. >> Which values of finite numbers? > Those which are in the set. I repeat it as clearly as I can, but it is > an unfamiliar idea for you. So ponder for a while about it and then > come up with your questions, if there are some remaining. > {1,2,3,...,omega} is an actually infinite set. > N = {1,2,3,...} is not an actually infinite set. This is nonsensical, since the bijection omega <-> 1 1 <-> 2 2 <-> 3 3 <-> 4 ... exists. It does not make sense to label the cardinality of two sets that can have their elements placed in strict bijection with different names. > In N there are only finite numbers. For the cardinality of a set, it is irrelevant whether the members of a set are finite, infinite, pink or yellow. The cardinality of a set is completely independent from the properties of its elements: it is the privilege of cardinality not to look at the properties of elements. > Every natural number n is finite and is surpassed by another finite > natural number n+1. Which makes the cardinality infinite. > Every natural n is the maximum of its finite initial sequence > {1,2,3,...,n}. Sure. > N is the limit of all finite initial sequences. There is no limit in the strict mathematical sense, and apart from that, the limit of a sequence is not required to be an element of the sequence. 0, while being the limit of the sequence {1/n}, is not an element of it. > N is potentially infinite, i.e., every maximum element n e N is > surpassed by another maximum element n+1, but this is also finite. The word potentially is nonsensical in this context since N is static and not dependent on any variable. Either it is infinite or not. > Every countable set like the set of primes is potentially infinite > but not actually infinite because all of its elements are finite and > so is any of its initial segments. > Countable is just a definition for remaining finite by value > *and by number*. No. Countable is defined by being placeable in bijection with natural numbers. The set {aleph_0, aleph_1, aleph_2, ...} is in bijection with natural numbers, and thus is countable, even though each of its members is infinite by value. Cardinality is not worried by values at all. >> By that same definition of finiteness (no injections to proper >> subsets) one can prove that the set of ALL naturals, as defined by >> the Peano axioms, is not finite. > It is potentially infinite, but not actually infinite, in the same > way as are the vales of natural numbers. Since there is no variable involved in the definition of the set, there is no potential. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree > The number of members is restricted by the values of the finite > numbers. > Which values of finite numbers? > Those which are in the set. I repeat it as clearly as I can, but it is > an unfamiliar idea for you. So ponder for a while about it and then > come up with your questions, if there are some remaining. > {1,2,3,...,omega} is an actually infinite set. Then so is {omega, 1,2,3,...} actually infinite, as it is the same set. And as there is an obvious bijection between {omega, 1,2,3,...} and { 1, 2, 3, 4, ...}, they are of the same cardinallity. > N = {1,2,3,...} is not an actually infinite set. And as it is not actually a not-infinite set either, WM must be claiming that it is not a set at all. > In N there are only finite numbers. In [0,1] there are only finite numbers. But in both N and [0,1] there are infinitely many of them. > Every natural number n is finite and is surpassed by another finite > natural number n+1. > Every natural n is the maximum of its finite initial sequence > {1,2,3,...,n}. > N is the limit of all finite initial sequences. N is the UNION of all initial sequences, but as there are more than any finite number of such sequences, that union cannot be finite. > N is potentially infinite, WM is potentially sane. > WM admits that there is no largest prime but also claims that there is a > largest even. These are mutually contradictory and cannot both be valid. > Every countable set like the set of primes is potentially infinite but > not actually infinite because all of its elements are finite and so is > any of its initial segments. This mantra, no matter how often repeated, will not bring enlightenment to those who are stuck with it. > Countable is just a definition for remaining finite by value *and > by number*. Not so. A set is countable (but not necessarily either finite or infinite) by Cantor's definition, if it is injectable into N. Countable sets are finite only if not injectable into a subset of themselves and infinite if injectable into any subset of themselves. These are Cantor's definitions, and WM has given nno satisfactory reason not to accept them. By those definitions, the set of even naturals is infinite since x --> x+2 injects it into a proper subset of itself. > By that same definition of finiteness (no injections to proper subsets) > one can prove that the set of ALL naturals, as defined by the Peano > axioms, is not finite. > It is potentially infinite, but not actually infinite, in the same way > as are the vales of natural numbers. Then 1 and 2 and 3 and so on are only potentially a natural numbers, and all mathematics is only potential and not actual. There are no such things in the physical world as numbers, so it is all equally potential. > The axiom of infinity creates both, numbers and sets. Show why, > nevertheless, according to your opinion, infinite sets but no infinite > numbers are created although this axiom works in the same way in both > cases. Because, except in it name, that axiom does not mention or use either the quality of being finite or of being infinite. The axiom, in one form, say merely that there exists sets S such that {} is a member of S and for every object x which is a member of S, the object (x union {x}) must also be member of S. It is trivial that the intersection of all such sets is such a set, and that intersection is taken to be the set of naturals, N, though nowdays the first member, {}, is taken to be 0 instead of 1 as it usually was when I was young. The quality of a set being finite or infinite is only defined later and in terms of N as being s small as a set cold get and still be infinite. And the objects in that intersection, by reason of that definition of infiniteness of sets are finite sets. === Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> nevertheless, according to your opinion, infinite sets but no infinite > numbers are created although this axiom works in the same way in both > cases. > Because, except in it name, that axiom does not mention or use either > the quality of being finite or of being infinite. Interesting new insights. So the name is misleadimg, even deceiving? > The axiom, in one form, say merely that there exists sets S such that {} > is a member of S and for every object x which is a member of S, the > object (x union {x}) must also be member of S. > It is trivial that the intersection of all such sets is such a set, Oh, it is equally trivial that the union of all initial segments is an initial segment. >and > that intersection is taken to be the set of naturals, N, though nowdays > the first member, {}, is taken to be 0 instead of 1 as it usually was > when I was young. It is taken, including nought? But what is the intersection really? What would follow, if politicians decided that all numbers > 1000 did not belong to N? > The quality of a set being finite or infinite is only defined later It is not a matter of the axioms? Perhaps infinity does not exist at all after all? > and > in terms of N as being s small as a set cold get and still be infinite. > And the objects in that intersection, by reason of that definition of > infiniteness of sets are finite sets. No. Having pressed you to come up with such answers is an important achievement. One has to put the right questions only, I see. But you are wrong. The condition a u {a} e A guarantees the infinity of A if A is taken to be a set and also if it is taken to be the variable of numerical values n. It is very cheap to find a bijection between initial segments {1,2,3...,n} and numerical values n. You are now in the situation to defend the position; n is always finite but the union of all segments {1,2,3...,n} created or guaranteed by the axiom of infinity becomes infinite somewhere. It is simply silly. PS. I will leave this thread because it has become so long that we soon can distinguish isolated paths. I have started two new threads. You are invited to contribute. === Subject: Re: Cantor and the binary tree > The axiom of infinity creates both, numbers and sets. Show why, > nevertheless, according to your opinion, infinite sets but no infinite > numbers are created although this axiom works in the same way in both > cases. > Because, except in it name, that axiom does not mention or use either > the quality of being finite or of being infinite. > Interesting new insights. So the name is misleadimg, even deceiving? > The axiom, in one form, say merely that there exists sets S such that {} > is a member of S and for every object x which is a member of S, the > object (x union {x}) must also be member of S. > It is trivial that the intersection of all such sets is such a set, > Oh, it is equally trivial that the union of all initial segments is an > initial segment. It may be trivial, but it is false. Initial segments have, by definition, maximal members which define them. every n in N is in that union so the union is a subset of N. N is also a subset of that union, since every n in N is in some intiial segment, therefor in the union. Thus that union MUST equal N itself. If N, being an initial segment has a maximal member then it is not the set of all naturals, as that would violate the Peano properties. WM has worked his way into a hole again. >and > that intersection is taken to be the set of naturals, N, though nowdays > the first member, {}, is taken to be 0 instead of 1 as it usually was > when I was young. > It is taken, including nought? But what is the intersection really? > What would follow, if politicians decided that all numbers > 1000 did > not belong to N? Mathematicians would ignore those stupid politicians, just like they ignore politicians who occasionally try to legislate the value of pi. > The quality of a set being finite or infinite is only defined later > It is not a matter of the axioms? Perhaps infinity does not exist at > all after all? > and > in terms of N as being s small as a set could get and still be infinite. > And the objects in that intersection, by reason of that definition of > infiniteness of sets are finite sets. > No. Having pressed you to come up with such answers is an important > achievement. One has to put the right questions only, I see. But you > are wrong. Only in what passes for WM's mind am I wrong. > The condition a u {a} e A guarantees the infinity of A if A is taken to > be a set and also if it is taken to be the variable of numerical values > n. > It is very cheap to find a bijection between initial segments > {1,2,3...,n} and numerical values n. You are now in the situation to > defend the position; n is always finite but the union of all segments > {1,2,3...,n} created or guaranteed by the axiom of infinity becomes > infinite somewhere. It is simply silly. Then the world will be ruled by what Wm sees as silliness, since that is the way things are, however silly it may seem to WM. === Subject: Re: Cantor and the binary tree rational, which is less than q 0. There can be no such q 1 by either the well ordering or by the standard ordering. In a well ordering. nothing precedes the first element, in the denseity of the standard rational ordering there is never any next ratinal to any rational. WM: 1/1, 1/2, 1/3, 1/4, 2/3, 1/5, ... is a well ordering q 0 = 1 q 1 = 1/2 < 1 q 2 = 1/3 < 1/2 ... What is your problem, please? occuraence of an impossibility, it is irrelevant. It is based on the existence of infinitely many rationals and on the possibility to perform infinitely many transpositions. Cantor *explicitly* allowed for infinitely many transpositions. He added that a well order set remains well-ordered even after infinitely many transpositions. Ich hebe noch folgendes hervor: wenn in einer wohlgeordneten Menge M irgend zwei Elemente m und m' ihre Pl.8atze in der Rangordnung wechseln, so wird dadurch der Typus nicht ver.8andert. Daraus folgt, da¤ solche Umformungen einer wohlgeordneten Menge die Anzahl derselben unge.8andert lassen, welche sich auf eine endliche oder unendliche Folge von Transpositionen je zweier Elemente zur.9fckf.9fhren lassen. (Letter to La¤witz, 15.Feb.1884) It seems you are not fully aware of the theory you defend. N can be represented as the infinite union of its infinitely many initial segments WM: Correct. But that does not create an actually infinite set. > This holds for any n e N. Therefore N is a segment. Non sequitur. That every member of n is a segment does not mean that N is a segment any more than every member of {1,3,5,7} being an odd integer means that {1,3,5,7} is an odd integer. WM: Misguiding example. The property of being an initial segment is inherited by the union of initial segments. WM's suppositions, if true, would require ejecting the baby with the bathwater, and leaving us wiwth nothing. Better some small but true knowledge than a huge error. > That is not at all an answer to this question. Infinitely many elements > a, a', a'', ... can form a set which does not include any infinite > number, even without any number. But a set of infinitely many different > natural numbers cannot actually exist with only finite numbers. That is a statement of faith, not fact, and that faith is not backed up by any facts, but only by further statements of faith. WM: No. It is easy to see that infinity (aleph 0) time one is infinity. Stalking on the steps from n to n+1 in infinity means to add infinitely infinity*1 IFF anything about infinity can be concluded at all. > Very interesting! You know that between two different irrationals there > is always a rational. And you cannot find two irrationals without a > finite rational between them. This proves clearly that there are not > more irrationals than rationals. WM: Who believes to understand it, you meant? The axiom of infinity does not create anything, it merely postulates the existence of something. Either it postulates an infinite set and an infinite number or nothing about sets and numbers. You have no given any arguments why there is a difference. Both, values and set are similarly postulated by the axiom, if von Neumann's construction is true. According to one hypothesis, the 'universe' is of finite age, but that hypothesis is by no means proven, and even among those supporting that hypothesis there is considerable disagreement on its age. There are other hypotheses that the 'universe' had no beginning in time. WM: There are hypotheses that it begun 6000 years ago. Do we have to bother about such theories here? Nevertheless physical fact is that we do not because we have not even started to leave the solar system. But N cannot have a maximal member according to the Peano properties, so that WM's argument disproves itself. As it has not an infinite member, it has only finite members. Finite members count themselves. Therefore there cannot be infinitely many members. I repeat my unanswered question, where does it say that every set must be counted by some natural? WM: Every finite natural counts its initial set. If there are only finite naturals, then there are only elements, which count their initial sets. This implies that there is a finite set for each natural. And there is nothing surpassing an initial set. > No. There is no maximal number, although all numbers are finite. This statement contradicts the statement that every set of naturals is finite. Not the least. > If N were actually existing, then the last one could not be missing. > This assumption contradicts itself. It can be missing only because N is > not actually existing. If N were not existing then we would have to create it. I do not know how science might manage without a set of naturals, but mathematics needs one. WM: Both science and mathematics do not have and cannot create a set of all naturals. Nevertheless science works fairly well. For mathematics it seems to be a necessary doctrine to assume its existence although it is irrational. > A finite set is potentially infinite, if it has not a largest member. A finite collection of naturals without a largest member is not a set, since its membership is ambiguous. WM: It is not. It is easy to identify whether a number is a natural or not (as easy as in the infinite collection which does not have a largest member). > It is taken, including nought? But what is the intersection really? > What would follow, if politicians decided that all numbers > 1000 did > not belong to N? Mathematicians would ignore those stupid politicians, just like they ignore politicians who occasionally try to legislate the value of pi. WM: Why then do they accept that one of the most unnatural numbers now belongs to the set of naturals? > It is very cheap to find a bijection between initial segments > {1,2,3...,n} and numerical values n. You are now in the situation to > defend the position; n is always finite but the union of all segments > {1,2,3...,n} created or guaranteed by the axiom of infinity becomes > infinite somewhere. It is simply silly. destroy his position. One part remains finite, the other one suddenly becomes infinite) Then the world will be ruled by what WM sees as silliness, since that is the way things are, however silly it may seem to WM. The world and its rules have nothing to do with set theory, although Cantor attempted to describe physics, chemistry, organisms and even the political states by set theory. It was developed for that sake. But that cannot raise any doubts in your mind? I know! === Subject: Re: Cantor and the binary tree > Le q 0 be the first element of the well-ordering. Let q 1 be the first > rational, which is less than q 0. > There can be no such q 1 by either the well ordering or by the standard > ordering. In a well ordering. nothing precedes the first element, in > the > denseity of the standard rational ordering there is never any next > ratinal to any rational. > WM: > 1/1, 1/2, 1/3, 1/4, 2/3, 1/5, ... is a well ordering > q 0 = 1 > q 1 = 1/2 < 1 > q 2 = 1/3 < 1/2 > ... > What is your problem, please? WM is claiming that when one lists 1/1, 1/2, 1/3, 1/4, 2/3, 1/5, in that order that 1/2 precedes 1/1 IN THAT ORDERING? If WM means the first in that well ordering less that the first according to standard ordering, he must say so specifically. > occuraence of an > impossibility, it is irrelevant. > It is based on the existence of infinitely many rationals and on the > possibility to perform infinitely many transpositions. Cantor > *explicitly* allowed for infinitely many transpositions. He added that > a well order set remains well-ordered even after infinitely many > transpositions. > N can be represented as the infinite union of its infinitely many > initial segments > WM: Correct. But that does not create an actually infinite set. If the union is not actual, then the sets being unioned are not actual and no natural number is actual. > This holds for any n e N. Therefore N is a segment. > Non sequitur. That every member of n is a segment does not mean that N > is a segment any more than every member of {1,3,5,7} being an odd > integer means that {1,3,5,7} is an odd integer. > WM: Misguiding example. The property of being an initial segment is > inherited by the union of initial segments. Only by finite unions. In order for a set of naturals to be an initial segment it is necessary for it to have a last or maximal member, but this property need not be true for infinite unions of initial segments as the sequence of initial segments of {1/2, 2/3, 3/4, ...} demonstrates. The LUB of {1/2, 2/3, 3/4, ...}is 1 which is NOT in the union of all initial segments, so that the union of its initial segments is not an initial segment. > WM's suppositions, if true, would require ejecting the baby with the > bathwater, and leaving us wiwth nothing. > Better some small but true knowledge than a huge error. Better something that works, however poorly WM may think it works, than something that doesn't work at all. > That is not at all an answer to this question. Infinitely many elements > a, a', a'', ... can form a set which does not include any infinite > number, even without any number. But a set of infinitely many different > natural numbers cannot actually exist with only finite numbers. > That is a statement of faith, not fact, and that faith is not backed up > by any facts, but only by further statements of faith. > WM: No. It is easy to see that infinity (aleph 0) time one is infinity. > Stalking on the steps from n to n+1 in infinity means to add infinitely > infinity*1 IFF anything about infinity can be concluded at all. > Very interesting! You know that between two different irrationals there > is always a rational. And you cannot find two irrationals without a > finite rational between them. This proves clearly that there are not > more irrationals than rationals. > WM: Who believes to understand it, you meant? No! who understands it better than WM does is what I meant. > merely postulates the existence of something. > Either it postulates an infinite set and an infinite number or nothing > about sets and numbers. The axiom of infinity does not mention numbers, they only come into it after the fact. it only postulates a set with certain members. > You have no given any arguments why there is a > difference. Both, values and set are similarly postulated by the axiom, > if von Neumann's construction is true. The existence of {} is postulated elsewhere, as is the existence of {x} for any object x and (y union z) fora any sets y and z. The axiom of infinite does not need to postulate the existence of any of the members of the set whose existence it postulates, and therefore does not postulate what is postulated elsewhere. > According to one hypothesis, the 'universe' is of finite age, but that > hypothesis is by no means proven, and even among those supporting that > hypothesis there is considerable disagreement on its age. > There are other hypotheses that the 'universe' had no beginning in > time. > WM: > There are hypotheses that it begun 6000 years ago. Do we have to bother > about such theories here? Nevertheless physical fact is that we do not > because we have not even started to leave the solar system. So now WM is changing his arguments because they were wrong in the first place. It would be so much nicer if he could only make them right from the beginning and not continnually have to revise them to cover his ass. > But N cannot have a maximal member according to the Peano properties, > so that WM's argument disproves itself. > As it has not an infinite member, it has only finite members. Finite > members count themselves. Therefore there cannot be infinitely many > members. Non sequitur. Unless WM's counting is Cantor's being in on-to-one correspondence with, in which case N does count itself, as does every set. > I repeat my unanswered question, where does it say that every set must > be counted by some natural? > WM: > Every finite natural counts its initial set. If there are only finite > naturals, then there are only elements, which count their initial sets. > This implies that there is a finite set for each natural. So far okay. > And there is > nothing surpassing an initial set. Ambiguous. There is, in fact, something surpassing any initial set, infinitely many such somethings starting with the next larger initial set. WM probably intended to say something like that there is nothing in N exceeding all the members of N, which is quite rue, but for any member of N there is a larger one. This distinction appears to be too fine for WM to discern. > No. There is no maximal number, although all numbers are finite. > This statement contradicts the statement that every set of naturals is > finite. > Not the least. WM is challenging the red queen here. > If N were actually existing, then the last one could not be missing. Presumes a situation contrary to fact, that N must have a last one. The whole point of N is that it does not have any last one. For WM to demand something so contrarian is stupid. > This assumption contradicts itself. It can be missing only because N is > not actually existing. > If N were not existing then we would have to create it. I do not know > how science might manage without a set of naturals, but mathematics > needs one. > WM: Both science and mathematics do not have and cannot create a set of > all naturals. Mathematics needs one and has one. WM may chose to call it potential rather than actual, but all mathematics, being in the world of ideals rather than actuals, is equally potential. Let WM stick to his mundane actual world and cease so ineffectually trying to impede mathematicians from exploring their potential one. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Le q_0 be the first element of the well-ordering. Let q_1 be the first >> rational, which is less than q_0. > There can be no such q_1 by either the well ordering or by the standard > ordering. In a well ordering. nothing precedes the first element, in > the > denseity of the standard rational ordering there is never any next > ratinal to any rational. > WM: > 1/1, 1/2, 1/3, 1/4, 2/3, 1/5, ... is a well ordering > q_0 = 1 > q_1 = 1/2 < 1 > q_2 = 1/3 < 1/2 > ... > What is your problem, please? > occuraence of an > impossibility, it is irrelevant. > It is based on the existence of infinitely many rationals and on the > possibility to perform infinitely many transpositions. Cantor > *explicitly* allowed for infinitely many transpositions. He added that > a well order set remains well-ordered even after infinitely many > transpositions. You are confusing arbitrarily and infinitely again. > Ich hebe noch folgendes hervor: wenn in einer wohlgeordneten Menge M > irgend zwei Elemente m und m' ihre Pl.8atze in der Rangordnung wechseln, > so wird dadurch der Typus nicht ver.8andert. Daraus folgt, da¤ solche > Umformungen einer wohlgeordneten Menge die Anzahl derselben unge.8andert > lassen, welche sich auf eine endliche oder unendliche Folge von > Transpositionen je zweier Elemente zur.9fckf.9fhren lassen. (Letter to > La¤witz, 15.Feb.1884) Looks like you are in good company here. Cantor appears obviously wrong in this particular letter, judging from the limited context. I suppose he realized this later. Nobody claimed all his utterings were to be taken as gospel. > It seems you are not fully aware of the theory you defend. Which theory? > But N cannot have a maximal member according to the Peano properties, > so > that WM's argument disproves itself. > As it has not an infinite member, it has only finite members. Finite > members count themselves. Therefore there cannot be infinitely many > members. That presumes that every set can be counted by one of its elements. You never proved that pretty absurd assumption of yours. > I repeat my unanswered question, where does it say that every set must > be counted by some natural? > WM: > Every finite natural counts its initial set. If there are only finite > naturals, then there are only elements, which count their initial sets. > This implies that there is a finite set for each natural. And there is > nothing surpassing an initial set. Confusion of words again. There is no single finite subset that would not be surpassed by some other initial set. And that is the _exact_ reason why the set of naturals can't be a finite subset. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> The axiom of infinity creates both, numbers and sets. Show why, >> nevertheless, according to your opinion, infinite sets but no infinite >> numbers are created although this axiom works in the same way in both >> cases. >> Because, except in it name, that axiom does not mention or use either >> the quality of being finite or of being infinite. > Interesting new insights. So the name is misleadimg, even deceiving? >> The axiom, in one form, say merely that there exists sets S such that {} >> is a member of S and for every object x which is a member of S, the >> object (x union {x}) must also be member of S. >> It is trivial that the intersection of all such sets is such a set, > Oh, it is equally trivial that the union of all initial segments is an > initial segment. Prove it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree finite, and has a maximal number that you conveniently put in the left > hand side. > I use the fact that any set of even numbers equal or less than 2n has > cardinality n. This fact is obviously correct. > Any bounded set of naturals is finite, but that does not prove that > every set of naturals is bounded, so it does not prove that every seet > of naturals is finite. Any natural defines a bounded set. There are no naturals which have infinite values. Hence, all sets defined by naturals are finite. === Subject: Re: Cantor and the binary tree Nope, your proof also uses the fact that the set on the right hand side > is > finite, and has a maximal number that you conveniently put in the left > hand side. I use the fact that any set of even numbers equal or less than 2n has > cardinality n. This fact is obviously correct. > Any bounded set of naturals is finite, but that does not prove that > every set of naturals is bounded, so it does not prove that every set > of naturals is finite. > Any natural defines a bounded set. There are no naturals which have > infinite values. Hence, all sets defined by naturals are finite. WM is saying that any set of naturals that is bounded above by some natural is bounded above by some natural. OK! But whah WM does not say, and cannot prove since it is false, is that every set of naturals is bounded above by a natural. === Subject: Re: Cantor and the binary tree every set of naturals is bounded, so it does not prove that every set > of naturals is finite. > Any natural defines a bounded set. There are no naturals which have > infinite values. Hence, all sets defined by naturals are finite. > WM is saying that any set of naturals that is bounded above by some > natural is bounded above by some natural. OK! > But whah WM does not say, and cannot prove since it is false, is that > every set of naturals is bounded above by a natural. Of course every set of naturals is bounded by a natural. What else should be in there? A television set? We cannot determine the magnitude of it, but we know that it is finite and is maximum of its initial sequence. Every countable set like N is potentially infinite but not actually infinite because all of its elements are finite and so is any of its initial segments. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> But whah WM does not say, and cannot prove since it is false, is that >> every set of naturals is bounded above by a natural. > Of course every set of naturals is bounded by a natural. What else > should be in there? Wrong question. The set of naturals is not bounded by any of its elements (apart from the lower bound 0, of course), like the lower bound of the set of unit fractions {1/n} is not in the set. The difference between a maximum and an upper bound is standard material in high school courses. A pity that WM, who does not recollect eleventh year school material, purports to teach math. > A television set? We cannot determine the magnitude of it, but we > know that it is finite and is maximum of its initial sequence. What nonsense. The set of naturals has no single initial sequence according to your usage of the word. > Every countable set like N is potentially infinite but not actually > infinite because all of its elements are finite and so is any of its > initial segments. Please come up with a mathematically tractable definition of potentially infinite and actually infinite that is applicable to a static set. Mathematics just knows the single definition infinite that is defined by the existence of a bijection to a proper subset. If you want to use terms of your own making, you have to define them. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree > Any bounded set of naturals is finite, but that does not prove that > every set of naturals is bounded, so it does not prove that every set > of naturals is finite. Any natural defines a bounded set. There are no naturals which have > infinite values. Hence, all sets defined by naturals are finite. > WM is saying that any set of naturals that is bounded above by some > natural is bounded above by some natural. OK! > But whah WM does not say, and cannot prove since it is false, is that > every set of naturals is bounded above by a natural. > Of course every set of naturals is bounded by a natural. What else > should be in there? A television set? We cannot determine the magnitude > of it, but we know that it is finite and is maximum of its initial > sequence. We don't know any such thing. In fact, the Peano postulates specifically forbid the existence os any such thing. > Every countable set like N is potentially infinite but not actually > infinite Then WM is only potentially numerate but not actually numerate. WM's problem is that a set that is only potential is not a set at all. Sets must be well defined. Whatever it is that WM is describing is not a well defined set. === Subject: Re: Cantor and the binary tree should be in there? A television set? We cannot determine the magnitude > of it, but we know that it is finite and is maximum of its initial > sequence. > We don't know any such thing. In fact, the Peano postulates > specifically forbid the existence os any such thing. Look here: The natural number n e N is nothing else than an abbreviation of its initial segment {1,2,3,...,n} c N. N consists exclusively of elements n. Similarly N consists exclusively of subsets = initial segments (all of which include 1). There is no element of N which is not an element of such a subset. And there is not a pair of different elements n and n' of N, which satisfy the following condition: n belongs to an initial segment S which does not contain n' and n' belongs to an initial segment S' which does not contain n in short: n e S and n' !e S and n' e S' and n !e S'. As this requirement is impossible to satisfy, the segment of n includes all elements less than n. This holds for any n e N. Therefore N is a segment. This segment is potentially infinite, because there is no largest n and because it can be mapped on a proper subset. Nevertheless, it is not actually infinite. I am sure you would be intelligent enough to understand that, unless set theory had spoiled the minds of you to a large extent. > WM's problem is that a set that is only potential is not a set at all. > Sets must be well defined. Whatever it is that WM is describing is not a > well defined set. There are no well defined sets other than finite sets. === Subject: Re: Cantor and the binary tree >> Of course every set of naturals is bounded by a natural. What else >> should be in there? A television set? We cannot determine the magnitude >> of it, but we know that it is finite and is maximum of its initial >> sequence. >> We don't know any such thing. In fact, the Peano postulates >> specifically forbid the existence os any such thing. >Look here: The natural number n e N is nothing else than an >abbreviation of its initial segment {1,2,3,...,n} c N. Not in ZFC it isn't. That set is not well-founded, and only well-founded sets exist in ZFC. >N consists exclusively of elements n. Similarly N consists exclusively >of subsets = initial segments (all of which include 1). There is no >element of N which is not an element of such a subset. And there is not >a pair of different elements n and n' of N, which satisfy the following >condition: >n belongs to an initial segment S which does not contain n' >and >n' belongs to an initial segment S' which does not contain n >in short: >n e S and n' !e S and n' e S' and n !e S'. >As this requirement is impossible to satisfy, the segment of n includes >all elements less than n. This holds for any n e N. Therefore N is a >segment. Non sequitor. Martin === Subject: Re: Cantor and the binary tree <03inc1h75156gehu78dsn33ossou4q88jr@4ax.comLook here: The natural number n e N is nothing else than an >abbreviation of its initial segment {1,2,3,...,n} c N. > Not in ZFC it isn't. That set is not well-founded, and only > well-founded sets exist in ZFC. And as I have shown by more than one case tha infinite well founded sets are in contradiction with mathematics, ZFC is contradiction with mathematics. Whether or not it is in contradiction with itself is unintersting though. > Non sequitor. can be written down easily, even if it is void of substance. === Subject: Re: Cantor and the binary tree >Look here: The natural number n e N is nothing else than an >abbreviation of its initial segment {1,2,3,...,n} c N. > Not in ZFC it isn't. That set is not well-founded, and only > well-founded sets exist in ZFC. > And as I have shown by more than one case tha infinite well founded > sets are in contradiction with mathematics, ZFC is contradiction with > mathematics. What WM has shown is that he does not understand what (pure) mathematics is all about. He does not play by the rules of the game, but claims the power to dictate what those rules must be. > Whether or not it is in contradiction with itself is > unintersting though. Except to those who actually do mathematics. === Subject: Re: Cantor and the binary tree > Of course every set of naturals is bounded by a natural. What else > should be in there? A television set? We cannot determine the magnitude > of it, but we know that it is finite and is maximum of its initial > sequence. > We don't know any such thing. In fact, the Peano postulates > specifically forbid the existence of any such thing. > Look here: The natural number n e N is nothing else than an > abbreviation of its initial segment {1,2,3,...,n} c N. > N consists exclusively of elements n. N can be represented as the infinite union of its infinitely many initial segments > This holds for any n e N. Therefore N is a segment. Non sequitur. That every member of n is a segment does not mean that N is a segment any more than every member of {1,3,5,7} being an odd integer means that {1,3,5,7} is an odd integer. > WM's problem is that a set that is only potential is not a set at all. > Sets must be well defined. Whatever it is that WM is describing is not a > well defined set. > There are no well defined sets other than finite sets. Then there are no well defined sets at all, at least in any of the standard axiomatic constructions of set theory. WM's suppositions, if true, would require ejecting the baby with the bathwater, and leaving us wiwth nothing. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Of course every set of naturals is bounded by a natural. What else >> should be in there? A television set? We cannot determine the magnitude >> of it, but we know that it is finite and is maximum of its initial >> sequence. >> We don't know any such thing. In fact, the Peano postulates >> specifically forbid the existence os any such thing. > Look here: The natural number n e N is nothing else than an > abbreviation of its initial segment {1,2,3,...,n} c N. > N consists exclusively of elements n. Similarly N consists exclusively > of subsets = initial segments (all of which include 1). There is no > element of N which is not an element of such a subset. And there is not > a pair of different elements n and n' of N, which satisfy the following > condition: > n belongs to an initial segment S which does not contain n' > and > n' belongs to an initial segment S' which does not contain n > in short: > n e S and n' !e S and n' e S' and n !e S'. > As this requirement is impossible to satisfy, the segment of n includes > all elements less than n. This holds for any n e N. Therefore N is a > segment. Nope. Your Therefore, again, is a piece of hogwash and a non sequitur. There is nothing whatsoever to the left of this Therefore that would justify the conclusion. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <85zmt4chmc.fsf@lola.goethe.zz Look here: The natural number n e N is nothing else than an > abbreviation of its initial segment {1,2,3,...,n} c N. > N consists exclusively of elements n. Similarly N consists exclusively > of subsets = initial segments (all of which include 1). There is no > element of N which is not an element of such a subset. And there is not > a pair of different elements n and n' of N, which satisfy the following > condition: > n belongs to an initial segment S which does not contain n' > and > n' belongs to an initial segment S' which does not contain n > in short: > n e S and n' !e S and n' e S' and n !e S'. > As this requirement is impossible to satisfy, the segment of n includes > all elements less than n. This holds for any n e N. Therefore N is a > segment. > Nope. Your Therefore, again, is a piece of [censored] and a non > sequitur. There is nothing whatsoever to the left of this Therefore > that would justify the conclusion. The conclusion is justified by the bijection of n on its initial segment. There is no largest n, so there is no largest segment. But any n is finite, so any segment is finite. Outside of segments there are no natural numbers. *Therefore* the set of all n, called N, is a segment. PS: You will not prevent this obvoious fact by cursing and shouting. So stop foaming and leave things as they are. But you will keep on swearing like a trooper, I know. The only question is, why can't you and Hughes behave like civilized persons? === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Look here: The natural number n e N is nothing else than an >> abbreviation of its initial segment {1,2,3,...,n} c N. >> N consists exclusively of elements n. Similarly N consists exclusively >> of subsets = initial segments (all of which include 1). There is no >> element of N which is not an element of such a subset. And there is not >> a pair of different elements n and n' of N, which satisfy the following >> condition: >> n belongs to an initial segment S which does not contain n' >> and >> n' belongs to an initial segment S' which does not contain n >> in short: >> n e S and n' !e S and n' e S' and n !e S'. >> As this requirement is impossible to satisfy, the segment of n includes >> all elements less than n. This holds for any n e N. Therefore N is a >> segment. >> Nope. Your Therefore, again, is a piece of [censored] and a non >> sequitur. There is nothing whatsoever to the left of this Therefore >> that would justify the conclusion. > The conclusion is justified by the bijection of n on its initial > segment. > There is no largest n, so there is no largest segment. But any n is > finite, so any segment is finite. Outside of segments there are no > natural numbers. > *Therefore* the set of all n, called N, is a segment. Again, misuse of therefore. Outside of segments, there are no natural numbers, but that does not mean that N is a segment. It merely means that N is the union of all segments. You have not proven that the union of an _infinite_ number of segments is again a segment, and indeed you couldn't. And indeed, it is the mark of a segment in your use of the word that it contains its largest n. To quote yourself: there is no largest n, so there is no largest segment. So N can't be a segment, by your very own words. > PS: You will not prevent this obvoious fact by cursing and > shouting. I don't need to curse and shout. You contradict yourself already. You are just too stupid to realize it. > So stop foaming and leave things as they are. Good advice, you ought to heed it. > But you will keep on swearing like a trooper, I know. The only > question is, why can't you and Hughes behave like civilized persons? Calling hogwash hogwash is not uncivilized. Anyway, civilization is marked by increasing the knowledge of your culture and building upon it. You are too stupid to even fathom the current state, and are actively trying to destroy it by teaching complete nonsense to your disciples. This is the mark of an illiterate caveman by choice and malice, someone who is so jealous of living and dead persons' intellects that all he can think of is to spread incoherent nonsense about them and hope that nobody notices. You are the last person fit to talk about civility. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree infinite > Then WM is only potentially numerate but not actually numerate. Look here: The natural number n e N is nothing else than an abbreviation of its initial segment {1,2,3,...,n} c N. N consists exclusively of elements n. Similarly N consists exclusively of subsets = initial segments (all of which include 1). There is no element of N which is not an element of such a subset. And there is not a pair of different elements n and n' of N, which satisfy the following condition: n belongs to an initial segment S which does not contain n' and n' belongs to an initial segment S' which does not contain n in short: n e S and n' !e S and n' e S' and n !e S'. As this requirement is impossible to satisfy, the segment of n includes all elements less than n. This holds for any n e N. Therefore N is a segment. This segment is potentially infinite, because there is no largest n and because it can be mapped on a proper subset. Nevertheless, it is not actually infinite. > WM's problem is that a set that is only potential is not a set at all. > Sets must be well defined. Whatever it is that WM is describing is not a > well defined set. There are no well defined sets other than finite sets. === Subject: Re: Cantor and the binary tree Every countable set like N is potentially infinite but not actually > infinite > Then WM is only potentially numerate but not actually numerate. > Look here: The natural number n e N is nothing else than an > abbreviation of its initial segment {1,2,3,...,n} c N. > N consists exclusively of elements n. Similarly N consists exclusively > of subsets = initial segments (all of which include 1). There is no > element of N which is not an element of such a subset. And there is not > a pair of different elements n and n' of N, which satisfy the following > condition: > n belongs to an initial segment S which does not contain n' > and > n' belongs to an initial segment S' which does not contain n > in short: > n e S and n' !e S and n' e S' and n !e S'. > As this requirement is impossible to satisfy, the segment of n includes > all elements less than n. This holds for any n e N. Therefore N is a > segment. Not so (unless WM insists that N is a member of itself, which situation is prohibited by most set theories). N not being a member of itself means N does not have to have any property that its members have. > WM's problem is that a set that is only potential is not a set at all. > Sets must be well defined. Whatever it is that WM is describing is not a > well defined set. > There are no well defined sets other than finite sets. Then there is no such thing as a set of natural number or a set of rational numbers or a set of real numbers, and all of arithmetic must be thrown out. WM conflates the theory (pure mathematics) with the practice (science). === Subject: Re: Cantor and the binary tree > [...] This holds for any n e N. Therefore N is a segment. Consider a set of cows, C, i.e. given some c e C, c is a cow. This holds for any c e C. Therefore C is a cow. Nevertheless, it does not have horns. F. === Subject: Re: Cantor and the binary tree Consider a set of cows, C, i.e. given some c e C, c is a cow. > This holds for any c e C. Therefore C is a cow. > Nevertheless, it does not have horns. With your pseudonym you should be able to distinguish between cows and linearly ordered sets. === Subject: Re: Cantor and the binary tree [...] This holds for any n e N. Therefore N is a segment. Consider a set of cows, C, i.e. given some c e C, c is a cow. > This holds for any c e C. Therefore C is a cow. > Nevertheless, it does not have horns. > With your pseudonym you should be able to distinguish between cows and > linearly ordered sets. But apparently WM cannot distinguish between sets and their members. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Every countable set like N is potentially infinite but not actually >> infinite >> Then WM is only potentially numerate but not actually numerate. > Look here: The natural number n e N is nothing else than an > abbreviation of its initial segment {1,2,3,...,n} c N. > N consists exclusively of elements n. Similarly N consists exclusively > of subsets = initial segments (all of which include 1). There is no > element of N which is not an element of such a subset. And there is not > a pair of different elements n and n' of N, which satisfy the following > condition: > n belongs to an initial segment S which does not contain n' > and > n' belongs to an initial segment S' which does not contain n > in short: > n e S and n' !e S and n' e S' and n !e S'. > As this requirement is impossible to satisfy, the segment of n includes > all elements less than n. This holds for any n e N. Therefore N is a > segment. You just love to use words like Therefore, Thus and similar where they don't make an inkling of sense. Your left side of therefore has absolutely nothing to do with your right side of therefore. > This segment Since N does not meet your definition of segment above, you need to come up with a broader definition of segment, or quit calling N a segment. Either way, your above statements about segments don't generalize to N. > is potentially infinite, because there is no largest n and because > it can be mapped on a proper subset. Nevertheless, it is not > actually infinite. It does not make sense to talk about potentially and actually for something that is fixed and not dependent on any variable. And N is a fixed set. >> WM's problem is that a set that is only potential is not a set at >> all. Sets must be well defined. Whatever it is that WM is >> describing is not a well defined set. > There are no well defined sets other than finite sets. Wrong. It is conceivable to argue that there are no well-defined singular sets for the purposes of mathematics other than sets defined by a finite set of rules. But since the natural numbers form a set of this class, and since its finite sets of rules clearly imply that it can be bijected onto a proper subset, they are both infinite (according to the well-defined property of infinite sets) as well as well-defined. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Any bounded set of naturals is finite, but that does not prove that >> every set of naturals is bounded, so it does not prove that every >> seet of naturals is finite. > Any natural defines a bounded set. There are no naturals which have > infinite values. Hence, all sets defined by naturals are finite. You mean: hence, all bounded sets defined by naturals are finite. The set of naturals is no such set. The set of your mathematical abilities seems to be the set of bad socks when one follows your logic: it has a lot of holes in it and stinks. But why should a set have a hole because all its members have? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree How many paths does one bunch contain? Many in the upper part, less in the middle and single paths at the bottom, IF such numbers do exist. > Either the paths remain forever in the bunch without isolating > themselves. > What is that supposed to mean? Either there are no individual sequences of bits at all in the realm of bits enumerated by finite numbers. === Subject: Re: Cantor and the binary tree >> How many paths does one bunch contain? > Many in the upper part, less in the middle and single paths at the > bottom, IF such numbers do exist. How do you define the bottom (or middle) of an infinite tree? Jan === Subject: Re: Cantor and the binary tree > And there are no isolated paths in a maximal binary tree. > As you said in another post, they exist, if there is a law defining the > sequences of bits. This law is in the tree: Create every sequence of > bits. Non-responsive. WM defined a path as _isolated_ by saying that it had a branch that was contained in no other path. That would be equivalent to saying that for some finite sequence of binary bits, there can be only one extension to an infinite sequence of binary bits. In fact, any such finite sequence has as many extensions to an infinite sequence of binary bits as there are such infinite sequences. > How many paths does one bunch contain? > Many in the upper part, less in the middle and single paths at the > bottom, IF such numbers do exist. Every path in every bunch starts at the bottom (root). More paths may coincide near the root than later, but there are just as many of them. > Either the paths remain forever in the bunch without isolating > themselves. > What is that supposed to mean? > Either there are no individual sequences of bits at all in the realm of > bits enumerated by finite numbers. Or what? You explanation does not explain anything. === Subject: Re: Cantor and the binary tree Being finished and having a last digit are not equivalent. Once a rule > is esablished for determining ALL digits of a number, the process is > ended, even though the string of digits may not have an end. There is a very simple rule for the binary tree: Create each binary representation of each real number of [0,1]. And the tree obeys it. > There is > always another digit to be exchanged. The same is true for both, > antidiagonal and paths of the tree. > WRONG! Complete induction can only prove that a particular set contains > all naturals, if it does, nothing else. > The set of all numbers, for instance, which are the largest of a finite > set. > Unless WM can provide that largest natural for our perusal, his claim > that there is one is empty. Every natural number n is the largest of its sequence {1,2,3,...n}. And there is no further natural number to boost the set such that it becomes larger than any of its elements. === Subject: Re: Cantor and the binary tree You can neither imagine Cantor's antidiagonal being finished. > Being finished and having a last digit are not equivalent. Once a rule > is esablished for determining ALL digits of a number, the process is > ended, even though the string of digits may not have an end. > There is a very simple rule for the binary tree: Create each binary > representation of each real number of [0,1]. And the tree obeys it. If WM means by that that a maximal binary tree has one path for each such binary representation, and vice-versa, he is, for once, right. > There is > always another digit to be exchanged. The same is true for both, > antidiagonal and paths of the tree. WRONG! Complete induction can only prove that a particular set contains > all naturals, if it does, nothing else. The set of all numbers, for instance, which are the largest of a finite > set. > Unless WM can provide that largest natural for our perusal, his claim > that there is one is empty. > Every natural number n is the largest of its sequence {1,2,3,...n}. And > there is no further natural number to boost the set such that it > becomes larger than any of its elements. For which of {1}, {1,2}, {1,2,3},..., does WM there no further natural number? There is always a larger natural number to boost the set to larger than its prior size! I repeat: Unless WM can provide that largest natural for our perusal, his claim that there is one is empty. === Subject: Re: Cantor and the binary tree there is no further natural number to boost the set such that it > becomes larger than any of its elements. > For which of {1}, {1,2}, {1,2,3},..., does WM there no further natural > number? There is always a larger natural number to boost the set to > larger than its prior size! > I repeat: > Unless WM can provide that largest natural for our perusal, his claim > that there is one is empty. There is no largest natural number in the set N. But we know that every n e N is finite and has a finite initial segment. Why can't you grasp the simple truth that a proof of the finity of n does not necessarily need fix that n? Card({1,2,3,...,omega}) = aleph_0 Cantor says: Every number less than omega is a finite number and is surpassed byother finite nunmbers. But omega is not in the set N!!!! Card({1,2,3,...}) cannot be infinite, although there is no largest n. Because we know that all naturals are finite. One can even use this fact to state: A countable set, i.e. a set the elements of which can be bijected with N, is *not* actually infinite. In particular N is not actally infinite (though it is potentially infinite). Otherwise there was the following contradiction: By the axiom of infinity guarantees, after von Neumann, the existence of all the natural numbers {}, {{}}, {{},{{}}}, ... The axiom of infinity guarantees infinite sets. Both values and cardinality are created by this axiom. Why is the cardinality infinite but not the values? === Subject: Re: Cantor and the binary tree Every natural number n is the largest of its sequence {1,2,3,...n}. And > there is no further natural number to boost the set such that it > becomes larger than any of its elements. > For which of {1}, {1,2}, {1,2,3},..., does WM there no further natural > number? There is always a larger natural number to boost the set to > larger than its prior size! > I repeat: > Unless WM can provide that largest natural for our perusal, his claim > that there is one is empty. > There is no largest natural number in the set N. In that case, the mapping n --> n+1 is an injection from N to a proper subset of N, and, at least according to the Cantor definition, N is an infinite set. By what definition of finiteness versus infiniteness does WM claim that N is finite? > But we know that every > n e N is finite and has a finite initial segment. Why can't you grasp > the simple truth that a proof of the finity of n does not necessarily > need fix that n? Because it is not a simple truth but a simple falsehod. At least if the Cantor definition of a set being infinite is used. If Wm has some other criterion for separating finite sets from infinite ones, he should reveal it now. > Card({1,2,3,...,omega}) = aleph_0 > Cantor says: Every number less than omega is a finite number and is > surpassed byother finite nunmbers. But omega is not in the set N!!!! > Card({1,2,3,...}) cannot be infinite, although there is no largest n. > Because we know that all naturals are finite. One can even use this > fact to state: A countable set, i.e. a set the elements of which can be > bijected with N, is *not* actually infinite. In particular N is not > actally infinite (though it is potentially infinite). > Otherwise there was the following contradiction: > By the axiom of infinity guarantees, after von Neumann, the existence > of all the natural numbers {}, {{}}, {{},{{}}}, ... > The axiom of infinity guarantees infinite sets. > Both values and cardinality are created by this axiom. Why is the > cardinality infinite but not the values? The same holds for the set of rational numbers and the set of real numbers, the umbers are all finite but the sets are not. If, for example, the rational fractions were only finite in number then there would have to be two fractions between which no others could exist, meaning for some positive fractions a/b and c/d, the fraction (a+b)/(c+d) is not a fraction. Otherwise we could again create an injection into a proper subset. === Subject: Re: Cantor and the binary tree Because we know that all naturals are finite. One can even use this > fact to state: A countable set, i.e. a set the elements of which can be > bijected with N, is *not* actually infinite. In particular N is not > actally infinite (though it is potentially infinite). > Otherwise there was the following contradiction: > By the axiom of infinity guarantees, after von Neumann, the existence > of all the natural numbers {}, {{}}, {{},{{}}}, ... > The axiom of infinity guarantees infinite sets. > Both values and cardinality are created by this axiom. Why is the > cardinality infinite but not the values? > The same holds for the set of rational numbers and the set of real > numbers, the numbers are all finite but the sets are not. That is not at all an answer to this question. Infinitely many elements a, a', a'', ... can form a set which does not include any infinite number, even without any number. But a set of infinitely many different natural numbers cannot actually exist with only finite numbers. > If, for example, the rational fractions were only finite in number then > there would have to be two fractions between which no others could > exist, meaning for some positive fractions a/b and c/d, the fraction > (a+b)/(c+d) is not a fraction. Very interesting! You know that between two different irrationals there is always a rational. And you cannot find two irrationals without a finite rational between them. This proves clearly that there are not more irrationals than rationals. But you assert that there are always more irrationals than rationals. You see, you can argue with numbers which can never be fixed. Try to transfer this ability to the largest natural of a finite set, and you can understand a bit of what infinity is. > Otherwise we could again create an > injection into a proper subset. The proper and only logic answer to my question is: If the axiom of infinity creates an actually infinite set then it creates an actually infinite number simultaneously. Because both are created in exactly the same way. === Subject: Re: Cantor and the binary tree > Card({1,2,3,...}) cannot be infinite, although there is no largest n. > Because we know that all naturals are finite. One can even use this > fact to state: A countable set, i.e. a set the elements of which can be > bijected with N, is *not* actually infinite. In particular N is not > actally infinite (though it is potentially infinite). Otherwise there was the following contradiction: > By the axiom of infinity guarantees, after von Neumann, the existence > of all the natural numbers {}, {{}}, {{},{{}}}, ... > The axiom of infinity guarantees infinite sets. > Both values and cardinality are created by this axiom. Why is the > cardinality infinite but not the values? > The same holds for the set of rational numbers and the set of real > numbers, the numbers are all finite but the sets are not. > That is not at all an answer to this question. Infinitely many elements > a, a', a'', ... can form a set which does not include any infinite > number, even without any number. But a set of infinitely many different > natural numbers cannot actually exist with only finite numbers. That is a statement of faith, not fact, and that faith is not backed up by any facts, but only by further statements of faith. WM has not produced anything to show that any of the axiom systems mathematicians use as a basis of set theory conflicts with anything outside WM's unproved statements of faith. WM has not produced anything more persuasive thab pleas to accept his faith on faith as justification for his faith. WM should be posting his Credo's in religious newsgroups. > If, for example, the rational fractions were only finite in number then > there would have to be two fractions between which no others could > exist, meaning for some positive fractions a/b and c/d, the fraction > (a+b)/(c+d) is not a fraction. > Very interesting! You know that between two different irrationals there > is always a rational. And you cannot find two irrationals without a > finite rational between them. This proves clearly that there are not > more irrationals than rationals. Not to anyone who understands the real number system. > Otherwise we could again create an > injection into a proper subset. > The proper and only logic answer to my question is: If the axiom of > infinity creates an actually infinite set then it creates an actually > infinite number simultaneously. Because both are created in exactly the > same way. The axiom of infinity does not create anything, it merely postulates the existence of something. If WM does not like it, he is fre to make up a set theroy of his own, but there is no guarantee anyone but WM will ever bother to learn it and an almost certain guarantee that no one but WM will ever bother to use it. The set theories currently extant are all much more satisfactory (to mathematicians at least) than anything WM has so far proposed. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Card({1,2,3,...,omega}) = aleph_0 > Cantor says: Every number less than omega is a finite number and is > surpassed byother finite nunmbers. But omega is not in the set N!!!! > Card({1,2,3,...}) cannot be infinite, although there is no largest > n. Because we know that all naturals are finite. Word games. We know that _every_ natural is finite, but we also know that the _set_ of all naturals is infinite. > Otherwise there was the following contradiction: > By the axiom of infinity guarantees, after von Neumann, the existence > of all the natural numbers {}, {{}}, {{},{{}}}, ... > The axiom of infinity guarantees infinite sets. > Both values and cardinality are created by this axiom. Why is the > cardinality infinite but not the values? Because each value stands for its own, while the cardinality stands for _all_ values. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Unless WM can provide that largest natural for our perusal, his >> claim that there is one is empty. > Every natural number n is the largest of its sequence > {1,2,3,...n}. And there is no further natural number to boost the > set such that it becomes larger than any of its elements. There is no need to boost the set such that it becomes larger, since N is not defined of being of the form {1,2,3,...n} in the first place. Since it is not defined as such a set, you can't blame anybody that it indeed does not happen to be such a set. In particular when you are trying to propose to people at the same time that it should be a limited set. You are more often in contradiction with M.9fckenheim than with Cantor. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <85d5q4jii1.fsf@lola.goethe.zz> Unless WM can provide that largest natural for our perusal, his >> claim that there is one is empty. > Every natural number n is the largest of its sequence > {1,2,3,...n}. And there is no further natural number to boost the > set such that it becomes larger than any of its elements. > There is no need to boost the set such that it becomes larger, > since N is not defined of being of the form {1,2,3,...n} in the first > place. It is defined as containing exclusively natural numbers. For each of them this form is correct. Each of them shows that there is need of more natural numbers in order to obtain an infinite set. That I mean by boosting. Natural numbers were developed by counting, i.e., each being the maximum of its sequence. They did not loose this property when set theory was introduced. The set of all natural numbers did not mean anything else than it literally means. Only now it has lost any connection with mathematics in order to facilitate a model of actual infinity. > Since it is not defined as such a set, you can't blame anybody > that it indeed does not happen to be such a set. In particular when > you are trying to propose to people at the same time that it should be > a limited set. The natural numbers do not take infinite values. The natural numbers count themselves as they are simultaneously ordinal and cardinal numbers. Hence, the set of all natural numbers is not infinite. === Subject: Re: Cantor and the binary tree > Unless WM can provide that largest natural for our perusal, his >> claim that there is one is empty. Every natural number n is the largest of its sequence > {1,2,3,...n}. And there is no further natural number to boost the > set such that it becomes larger than any of its elements. > There is no need to boost the set such that it becomes larger, > since N is not defined of being of the form {1,2,3,...n} in the first > place. > It is defined as containing exclusively natural numbers. For each of > them this form is correct. Each of them shows that there is need of > more natural numbers in order to obtain an infinite set. That I mean by > boosting. As there are infinitely many of these initial sets of naturals, and it is only the union of all of them that produces the entire set of naturals, the finiteness of any one, or even all, of them is irrelevant. > Natural numbers were developed by counting, i.e., each being the > maximum of its sequence. Each such set is bounded by having a maximal member. If WM claims that the set of ALL naturals is also bounded by having a maximal member, he is obligated to produce it. > They did not loose this property when set > theory was introduced. The set of all natural numbers did not mean > anything else than it literally means. Only now it has lost any > connection with mathematics in order to facilitate a model of actual > infinity. The set of all naturals may have lost connection to the physical world, but in mathematics, that tends to be an advantage rather than a handicap. But it has not lost any connection to mathematics, and is, indeed, at the foundation of analysis, an important segment of mathematics. > Since it is not defined as such a set, you can't blame anybody > that it indeed does not happen to be such a set. In particular when > you are trying to propose to people at the same time that it should be > a limited set. > The natural numbers do not take infinite values. The set of naturals injects, by the successor operation, onto a proper subset of itself. Mathematics deems such sets to be infinite. > The natural numbers > count themselves as they are simultaneously ordinal and cardinal > numbers. Hence, the set of all natural numbers is not infinite. Well, the set of all naturals is not finite either, at least by any mathematically workable definition of finiteness of sets. What definition of a set being finite is WM going by? Certainly not mine. === Subject: Re: Cantor and the binary tree <85d5q4jii1.fsf@lola.goethe.zz> is only the union of all of them that produces the entire set of > naturals, the finiteness of any one, or even all, of them is irrelevant. There are only finitely many finite numbers. That is obvious. The set is potentially infinite but always remains finite, i.e., it never takes an actually infinite number with an actually infinite value. > Natural numbers were developed by counting, i.e., each being the > maximum of its sequence. > Each such set is bounded by having a maximal member. If WM claims that > the set of ALL naturals is also bounded by having a maximal member, he > is obligated to produce it. Every member is finite and counts the members in its initial segment. There is no chance to escape finity. > They did not loose this property when set > theory was introduced. The set of all natural numbers did not mean > anything else than it literally means. Only now it has lost any > connection with mathematics in order to facilitate a model of actual > infinity. > The set of all naturals may have lost connection to the physical world, > but in mathematics, that tends to be an advantage rather than a handicap. > But it has not lost any connection to mathematics, and is, indeed, at > the foundation of analysis, an important segment of mathematics. The set of all naturals, of course. This set is potentially infinite. But actual infinity is absolutely irrelevant to the foundation of any science. > Since it is not defined as such a set, you can't blame anybody > that it indeed does not happen to be such a set. In particular when > you are trying to propose to people at the same time that it should be > a limited set. > The natural numbers do not take infinite values. > The set of naturals injects, by the successor operation, onto a proper > subset of itself. Mathematics deems such sets to be infinite. By intermingling potential and actual infinity. This property is a clear definition for potential infiniy. Injection s only possible for finite elements. Those elements define finite initial segments. > The natural numbers > count themselves as they are simultaneously ordinal and cardinal > numbers. Hence, the set of all natural numbers is not infinite. > Well, the set of all naturals is not finite either, at least by any > mathematically workable definition of finiteness of sets. > What definition of a set being finite is WM going by? Certainly not > mine. That is simple and, moreover, it is correct. Every set is finite, i.e., it may be potentially infinite (satisfying your criterion of infinity) but certainly it is never actually infinite. === Subject: Re: Cantor and the binary tree > As there are infinitely many of these initial sets of naturals, and it > is only the union of all of them that produces the entire set of > naturals, the finiteness of any one, or even all, of them is irrelevant. > There are only finitely many finite numbers. There are infinitely many of them between 0 and 1. > Each such set is bounded by having a maximal member. If WM claims that > the set of ALL naturals is also bounded by having a maximal member, he > is obligated to produce it. > Every member is finite and counts the members in its initial segment. > There is no chance to escape finity. Not as long as one keeps stopping, but one need not ever stop. But the axiom of infinity does it all in one swell foop. And the result is the set of ALL naturals with each as actual as all others. > They did not loose this property when set > theory was introduced. The set of all natural numbers did not mean > anything else than it literally means. Only now it has lost any > connection with mathematics in order to facilitate a model of actual > infinity. > The set of all naturals may have lost connection to the physical world, > but in mathematics, that tends to be an advantage rather than a handicap. > But it has not lost any connection to mathematics, and is, indeed, at > the foundation of analysis, an important segment of mathematics. > The set of all naturals, of course. This set is potentially infinite. > But actual infinity is absolutely irrelevant to the foundation of any > science. And science is almost totally irrelevant to the precise form of any mathematics. Though scientists often find it useful to borrow from mathematics. Since it is not defined as such a set, you can't blame anybody > that it indeed does not happen to be such a set. In particular when > you are trying to propose to people at the same time that it should be > a limited set. The natural numbers do not take infinite values. > The set of naturals injects, by the successor operation, onto a proper > subset of itself. Mathematics deems such sets to be infinite. > By intermingling potential and actual infinity. All mathematics is entirely and equally mere potentiality. There is nothing actual, in WM's sense, to any of it. So trying to say that one bit is more potential thatn another is fruitless. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Unless WM can provide that largest natural for our perusal, his > claim that there is one is empty. >> Every natural number n is the largest of its sequence >> {1,2,3,...n}. And there is no further natural number to boost the >> set such that it becomes larger than any of its elements. >> There is no need to boost the set such that it becomes larger, >> since N is not defined of being of the form {1,2,3,...n} in the first >> place. > It is defined as containing exclusively natural numbers. No, it isn't. It is defined by the five Peano axioms, and natural numbers are defined by virtue of being members of that set. But that does not mean that there must exist some method of throwing together a finite or any number of natural numbers and arrive at the set. If such a procedure exists, fine, if it doesn't, also fine. The Peano axioms don't care either which way. > For each of them this form is correct. Each of them shows that there > is need of more natural numbers in order to obtain an infinite > set. That I mean by boosting. Sure, and no finite set will ever have enough natural numbers. Construction by boosting from finite sets will not work, and nobody says it will. > Natural numbers were developed by counting, i.e., each being the > maximum of its sequence. That's history. > They did not loose this property when set theory was introduced. They can be used for that without being exhausted. But that is all. Their definition is the Peano axioms, not some history of counting. And that is because a history of counting is not tractable or verifiable, whereas five small rules are. > The set of all natural numbers did not mean anything else than it > literally means. Only now it has lost any connection with > mathematics in order to facilitate a model of actual infinity. Oh, it has not lost any connection with mathematics, quite contrary. It is not completely coverable by actual counting (since you can never finish that way), but that's why it has been defined by axioms instead. >> Since it is not defined as such a set, you can't blame anybody >> that it indeed does not happen to be such a set. In particular >> when you are trying to propose to people at the same time that it >> should be a limited set. > The natural numbers do not take infinite values. The natural numbers > count themselves as they are simultaneously ordinal and cardinal > numbers. Hence, the set of all natural numbers is not infinite. Nonstandard use of Hence since the cardinality of the naturals is not natural. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree Complete induction will do so. Even Cantor was convinced that complete > > induction is sufficient for all countable sets, and with him every > > mathematician at his times. > He may be convinced, in that case he was wrong. I rather think that you > misrepresent what Cantor has stated, because you do not understand the > distinction. Yes: I understand that erste Zahlenkasse means countable. Cantor understood that. > 2n > Card({2, 4, ...., 2n}) > This is true for *each* natural n. And that is what complete induction > is. It does not state anything about N as the right hand side of the > formula. You have a serious comprehension problem. Stating something > is valid for *all* n is the same as stating that something is valid for > *each* n. In Dutch (and I think in German) the same applies. It does > not state anything of the collection of *each* n, or of *all* n, it > only states something about the individual members. I know that each n does define a finite sequence. Fom that one can conclude that all n are in a finite sequence, in German and n Dutch. Your distinction is nothing than a helpless attempt to save set-theory as has those which nhave been aplied in the past several times. Think of L.9awenstein-Skolem: Every consistent first oder theory has a countable model. iIf set theoy is consistent, then it is countable, hence inconsistent. Therefore set theory canot be consistent. All your attempts are in vain. > > So how you can get to your conclusion > > for the set of all even numbers using induction? > > because there is no even number at which the reasoning stops or gets > > incorrect. > Yes, so what? This does not imply anything about the set of even numbers, > except that for each member of that set the set constructed by taking all > even numbers less than or equal to that number Why should I talk about any fixed number??? I speak of any number of the set, which is a finite numbers. And there are no further numbers. For finite numbers we can prove that they define a finite sequence. Why do you always consider larger numbers? Where do you get them from??? My proof covers each and every finite number, simultaneously, instantanneously, just like Cantor's antidiagonal. There is nothing beyond the numbers reached by complete induction. Of course, for every n there is an n+1, but that one is also covered (if Cantor can do so by his proof). === Subject: Re: Cantor and the binary tree >> > Complete induction will do so. Even Cantor was convinced that complete >> > induction is sufficient for all countable sets, and with him every >> > mathematician at his times. >> He may be convinced, in that case he was wrong. I rather think that you >> misrepresent what Cantor has stated, because you do not understand the >> distinction. Yes: >I understand that erste Zahlenkasse means countable. Cantor understood >that. >> 2n > Card({2, 4, ...., 2n}) >> This is true for *each* natural n. And that is what complete induction >> is. It does not state anything about N as the right hand side of the >> formula. You have a serious comprehension problem. Stating something >> is valid for *all* n is the same as stating that something is valid for >> *each* n. In Dutch (and I think in German) the same applies. It does >> not state anything of the collection of *each* n, or of *all* n, it >> only states something about the individual members. >I know that each n does define a finite sequence. Fom that one can >conclude that all n are in a finite sequence, in German and n Dutch. >Your distinction is nothing than a helpless attempt to save set-theory >as has those which nhave been aplied in the past several times. Think >of L.9awenstein-Skolem: Every consistent first oder theory has a >countable model. iIf set theoy is consistent, then it is countable, >hence inconsistent. Therefore set theory canot be consistent. All your >attempts are in vain. Once again, you have been misled by your haste to declare a contradiction. The existance of a countable model for ZFC does not create any contradictions. >> > So how you can get to your conclusion >> > for the set of all even numbers using induction? >> > > because there is no even number at which the reasoning stops or gets >> > incorrect. >> Yes, so what? This does not imply anything about the set of even numbers, >> except that for each member of that set the set constructed by taking all >> even numbers less than or equal to that number >Why should I talk about any fixed number??? I speak of any number of >the set, which is a finite numbers. And there are no further numbers. >For finite numbers we can prove that they define a finite sequence. Why >do you always consider larger numbers? Where do you get them from??? My >proof covers each and every finite number, simultaneously, >instantanneously, just like Cantor's antidiagonal. Your proof doesn't cover any numbers. The domain of your proof are initial segments of even numbers. *Anything* that is not one of those is not covered. The reason everyone keeps talking about the next one is because there are sets of even natural numbers that are not Initial segmenys. One way to prrove this is to show that the set of all even natural numbers is not an initial segment. To prove this, take any initial segment. It is has a maximum element 2n for some n in N. but 2n + 2 is an even number larger than 2m. Therefore that initial segment is not equal to the set of all even naturals. Since, that was an arbitrary initial segment, none of the initial segments is the set of all even natural numbers. Therefore, there are sets of even natural numbers your proof does not cover. Martin === Subject: Re: Cantor and the binary tree contradiction. The existance of a countable model for ZFC does not > create any contradictions. We know by the proof of L.9awenheim-Skolem: Every consistent theory has a countable model. If set theory would have a countable model, it would be inconsistent. If set theory would not have a countable model it would be inconsistent. But set theory is consistent, by definition. How could it be otherwise? Even if it turned out that the real numbers contained only one element or even are an empty set, this would not create any contradictions, in your opinion. > Your proof doesn't cover any numbers. The domain of your proof are > initial segments of even numbers. *Anything* that is not one of those > is not covered. The reason everyone keeps talking about the next one > is because there are sets of even natural numbers that are not Initial > segments. Every number is the maximum of an initial segment. There is no number without initial segment. Hence the number of elements up to any number is finite (i.e., N is potentially infinite, but not actually infinite). One way to prrove this is to show that the set of all even > natural numbers is not an initial segment. To prove this, take any > initial segment. It is has a maximum element 2n for some n in N. but > 2n + 2 is an even number larger than 2m. Why don't you ask whether this number has an initial segment too? I can assurer you that it has. > Therefore that initial > segment is not equal to the set of all even naturals. That one is not, of course, but as there is no number without such an initial segment, the next one you may quote, has an initial segment too. In order to have a set which is not the limit of initial segments, you need at least one natural number without such a segment. But you cannot find one. Therefore all your arguing is in vain. === Subject: Re: Cantor and the binary tree >> Once again, you have been misled by your haste to declare a >> contradiction. The existance of a countable model for ZFC does not >> create any contradictions. >We know by the proof of L.9awenheim-Skolem: Every consistent theory has >a countable model. If set theory would have a countable model, it would >be inconsistent. How so? >If set theory would not have a countable model it >would be inconsistent. But set theory is consistent, by definition. Actually, the consistancy of the commonly used set theories is unknown. The best we can say is that we haven't found any yet. > How >could it be otherwise? Even if it turned out that the real numbers >contained only one element or even are an empty set, this would not >create any contradictions, in your opinion. Those would create contradictions. There are proofs that R is infinite. Being able to prove that R is finite, would be a contradiction. Luckily for set theorists, no (valid) proofs of that have been produced. >> Your proof doesn't cover any numbers. The domain of your proof are >> initial segments of even numbers. *Anything* that is not one of those >> is not covered. The reason everyone keeps talking about the next one >> is because there are sets of even natural numbers that are not Initial >> segments. >Every number is the maximum of an initial segment. There is no number >without initial segment. Hence the number of elements up to any number >is finite (i.e., N is potentially infinite, but not actually infinite). No ones arguing against this. > One way to prrove this is to show that the set of all even >> natural numbers is not an initial segment. To prove this, take any >> initial segment. It is has a maximum element 2n for some n in N. but >> 2n + 2 is an even number larger than 2m. >Why don't you ask whether this number has an initial segment too? I can >assurer you that it has. So what if it does? >> Therefore that initial >> segment is not equal to the set of all even naturals. >That one is not, of course, but as there is no number without such an >initial segment, the next one you may quote, has an initial segment >too. In order to have a set which is not the limit of initial segments, >you need at least one natural number without such a segment. But you >cannot find one. Therefore all your arguing is in vain. No you don't need one without an initial segment. You just need the absence of a largest one. Martin === Subject: Re: Cantor and the binary tree > > Complete induction will do so. Even Cantor was convinced that complete > > induction is sufficient for all countable sets, and with him every > > mathematician at his times. > > He may be convinced, in that case he was wrong. I rather think that you > misrepresent what Cantor has stated, because you do not understand the > distinction. Yes: > I understand that erste Zahlenkasse means countable. Cantor understood > that. I think I now understand what the meaning was: you can implement a successor function on the elements of a countable set, so induction was sufficient to proof something about all elements of the set. This does *not* mean that it proofs something about the set itself. > 2n > Card({2, 4, ...., 2n}) > This is true for *each* natural n. And that is what complete induction > is. It does not state anything about N as the right hand side of the > formula. You have a serious comprehension problem. Stating something > is valid for *all* n is the same as stating that something is valid for > *each* n. In Dutch (and I think in German) the same applies. It does > not state anything of the collection of *each* n, or of *all* n, it > only states something about the individual members. > I know that each n does define a finite sequence. Fom that one can > conclude that all n are in a finite sequence, in German and n Dutch. Yes, all n are in a finite sequence, but it is not necessarily true that all n are in the same finite sequence. That is a jump of logic. > Your distinction is nothing than a helpless attempt to save set-theory > as has those which nhave been aplied in the past several times. Think > of L=F6wenstein-Skolem: Every consistent first oder theory has a > countable model. iIf set theoy is consistent, then it is countable, > hence inconsistent. Therefore set theory canot be consistent. All your > attempts are in vain. Yes, you are a Platonist, that was clear. > > So how you can get to your conclusion > > for the set of all even numbers using induction? > > > > because there is no even number at which the reasoning stops or gets > > incorrect. > > Yes, so what? This does not imply anything about the set of even numbers, > except that for each member of that set the set constructed by taking all > even numbers less than or equal to that number > Why should I talk about any fixed number??? I speak of any number of > the set, which is a finite numbers. And there are no further numbers. > For finite numbers we can prove that they define a finite sequence. Right. > Why > do you always consider larger numbers? Where do you get them from??? By adding 2. > My > proof covers each and every finite number, simultaneously, > instantanneously, just like Cantor's antidiagonal. There is nothing > beyond the numbers reached by complete induction. Yup, so for each element of N defines a bounded set with some specific cardinality. That is what your proof shows. It shows *nothing* about the set N itself. > Of course, for every n there is an n+1, but that one is also covered > (if Cantor can do so by his proof). Cantor needs not for his antidiagonal. The antidiagonal is not an iterative process. When you give a list of real numbers, with a formula I can calculate the n-th digit of the antidiagonal without access to any preceding digit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree all n are in the same finite sequence. That is a jump of logic. You are wrong, unless you can name two different natural numbers, m and n, which are in different initial sequences {1,2,3,...,m} and [1,2,3,...,n}, such that nor n is in the first neither m is in the second. As you cannot, my proof does not include a jump of logic. > Yes, you are a Platonist, that was clear. No. I am not at all a Platonist. I do not even believe that more than 10^100 objects can exist simultaneously. > Cantor needs not for his antidiagonal. The antidiagonal is not an > iterative process. When you give a list of real numbers, with a > formula I can calculate the n-th digit of the antidiagonal without > access to any preceding digit. And I can name the initial sequence for any natural number you want, even if you add 2. That is not an iterative process either. Every natural has its initial sequence. === Subject: Re: Cantor and the binary tree > Yes, all n are in a finite sequence, but it is not necessarily true that > all n are in the same finite sequence. That is a jump of logic. > You are wrong, unless you can name two different natural numbers, m and > n, which are in different initial sequences {1,2,3,...,m} and > [1,2,3,...,n}, such that nor n is in the first neither m is in the > second. As you cannot, my proof does not include a jump of logic. Nope, but for *every* n I can name an m that is not in it's initial segment. That is sufficient to show that not all are in the same initial segment. > Yes, you are a Platonist, that was clear. > No. I am not at all a Platonist. I do not even believe that more than > 10^100 objects can exist simultaneously. What is exist? Does sqrt(5) exist? > Cantor needs not for his antidiagonal. The antidiagonal is not an > iterative process. When you give a list of real numbers, with a > formula I can calculate the n-th digit of the antidiagonal without > access to any preceding digit. > And I can name the initial sequence for any natural number you want, > even if you add 2. That is not an iterative process either. Every > natural has its initial sequence. Right. But there is no initial sequence that contains all naturals. In the same way you can state that you can find the initial start of the antidiagonal for every number n, but you can not find the initial start of the antidiagonal for *all* n at once (because there is none), and that is not needed. On the other hand, when you claim that N is finite, it must have an initial start. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree Yes, all n are in a finite sequence, but it is not necessarily true that > > all n are in the same finite sequence. That is a jump of logic. > > You are wrong, unless you can name two different natural numbers, m and > > n, which are in different initial sequences {1,2,3,...,m} and > > [1,2,3,...,n}, such that nor n is in the first neither m is in the > > second. As you cannot, my proof does not include a jump of logic. > Nope, but for *every* n I can name an m that is not in it's initial segment. > That is sufficient to show that not all are in the same initial segment. For that one you name, I can name an initial segment. And this segment includes all numbers yet considered. That is sufficient to show that all are in the same initial segment. > > Yes, you are a Platonist, that was clear. > > No. I am not at all a Platonist. I do not even believe that more than > > 10^100 objects can exist simultaneously. > What is exist? Does sqrt(5) exist? No. Not as a number. It is an idea, a notion, a wish. > > Cantor needs not for his antidiagonal. The antidiagonal is not an > > iterative process. When you give a list of real numbers, with a > > formula I can calculate the n-th digit of the antidiagonal without > > access to any preceding digit. > > And I can name the initial sequence for any natural number you want, > > even if you add 2. That is not an iterative process either. Every > > natural has its initial sequence. > Right. But there is no initial sequence that contains all naturals. Of course it is. At least all naturals which are there. Any natural which is there can be identified by bijection with its initial segment. An antidiagonal existing and being different from any number of he list must exist for all n at once. That *is* required if the proof shall be of any use. But I knew that the argument was false. === Subject: Re: Cantor and the binary tree > > > > Yes, all n are in a finite sequence, but it is not necessarily true > > that all n are in the same finite sequence. That is a jump of logic. > > > > You are wrong, unless you can name two different natural numbers, m and > > n, which are in different initial sequences {1,2,3,...,m} and > > [1,2,3,...,n}, such that nor n is in the first neither m is in the > > second. As you cannot, my proof does not include a jump of logic. > > Nope, but for *every* n I can name an m that is not in it's initial > segment. That is sufficient to show that not all are in the same > initial segment. > For that one you name, I can name an initial segment. And this segment > includes all numbers yet considered. That is sufficient to show that > all are in the same initial segment. Nope. It only shows that all numbers considered are in the same initial segment. There are a few numbers still missing. > > No. I am not at all a Platonist. I do not even believe that more than > > 10^100 objects can exist simultaneously. > > What is exist? Does sqrt(5) exist? > No. Not as a number. It is an idea, a notion, a wish. I do not understand what you mean here. When I am working in Q(sqrt(5)), I will certainly regard sqrt(5) as a number. But I think that by your reasoning, sqrt(-1) is also not a number. > Right. But there is no initial sequence that contains all naturals. > Of course it is. At least all naturals which are there. Any natural > which is there can be identified by bijection with its initial > segment. A long time ago you have stated that you would continue the discussion with the intention that there are not finitely many natural numbers. But whatever. Consider the infinitely many finite natural numbers as ideas, notions, wishes. Because that is what they are, just notions. > An antidiagonal existing and being different from any number of he list > must exist for all n at once. That *is* required if the proof shall be > of any use. But I knew that the argument was false. Again, my question. When does something exist in your opinion? Can you give a clear definition of that concept? Apparently mathematical existence conflicts with your ideas about existence. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree > For that one you name, I can name an initial segment. And this segment > includes all numbers yet considered. That is sufficient to show that > all are in the same initial segment. Nope. It only shows that all numbers considered are in the same initial segment. There are a few numbers still missing. WM: There are as many segments still missing too. > What is exist? Does sqrt(5) exist? > No. Not as a number. It is an idea, a notion, a wish. I do not understand what you mean here. When I am working in Q(sqrt(5)), I will certainly regard sqrt(5) as a number. But I think that by your reasoning, sqrt(-1) is also not a number. WM: A real number must be capable of being put in order relation (<) with any other real number. That is possible for 5 but not for sqrt(5). Because if you cut of all digits except the first 10^100, then you get a rational number which cannot and never be distinguished by the same one where the last digit is exchanged by 5, say. There fore already these rational numbers cannot be put in <-relation with each other. They are not numbers. Even less sqrt(5) is a number. sqrt(-1) is a complex entity which consists of numbers, 0 and 1 in this case, like a vector or a matrix consist of numbers. It is usual to call these entities complex numbers like vector are called vectors. Why not? > Right. But there is no initial sequence that contains all naturals. > Of course it is. At least all naturals which are there. Any natural > which is there can be identified by bijection with its initial > segment. A long time ago you have stated that you would continue the discussion with the intention that there are not finitely many natural numbers. WM: You must distinguish, and that isy fairl difficult, because sometimes we argue from the standpoint of set theory, sometimes not My standpoint is: The values of natural numbers are always finite, but there is no upper bound. There cannot exist more than 10^100 elements of set in the whole universe. Therefore, there cannot exist more than 10^100 natural numbers simultaneously. Not regarding these physical limits, I say: If the number of naturals is unlimited, then there is no upper bound for this cardinality. (I use cardinality here as a convenient abbreviation for a measure of the number of elements, not in the direct meaning of set theory). But it is always finite, though no largest cardinal number can be fixed. It is a potential infinity. When I agree to infinity (which in fact does never and nowhere exist) then I mean this potential infinity. When disproving Cantors theory, then I start from actual infinity in order to show a contradiction. For instance I use actually infinitely many transpositions to order the set of actually infinitely many rational numbers. Or I construct a binary tree with actually infinitely many levels. Again, my question. When does something exist in your opinion? Can you give a clear definition of that concept? Apparently mathematical existence conflicts with your ideas about existence. WM: Mathematical existence is nothing but being well defined. By Peano's axioms all natural numbers do exist. I distinguish between definition and existence. A natural number exists as that notion which the elements of a fundamental set have in common. (I use the definition to construct those sets. The definition is prior to existence.) 2 = The common property of {II, son & moon, father & mother, ...} In this way one can prove the existence of 3, 4, .... But it becomes tedious if applicable at all for larger numbers. Therefore I say: A number does exist if it can be represented in an fixed g-adic system. This representation *is* that number. In the decimal system you see here on your screen the numbers 2, 4711, 10^1000^10000. But not all numbers which can be defined, do exist. The number floor(pi*10^10^100) does not exist. And it will never exist. The 80-digit number: 1234567890123456789012345678901234567890123456789012345678901234567890123456 7890 does exist here. It will never be possible to have all possible 80-digit numbers existing simultaneously, because there are less than But each of these numbers can be brought to existence within a minute. So they have some kind of potential existence, contrary to floor(pi*10^10^100) which does not exist. If you wish, please answer in one of the new threads, because this one has become too long already. === Subject: Re: Cantor and the binary tree WM was being wrong again at great length and with little sense. === Subject: Re: Cantor and the binary tree ... > What is exist? Does sqrt(5) exist? > > No. Not as a number. It is an idea, a notion, a wish. > I do not understand what you mean here. When I am working in Q(sqrt(5)), > I will certainly regard sqrt(5) as a number. But I think that by your > reasoning, sqrt(-1) is also not a number. > WM: A real number must be capable of being put in order relation (<) > with any other real number. Why? When I am working in Q(sqrt(5)) I am not interested in order relations. > That is possible for 5 but not for sqrt(5). > Because if you cut of all digits except the first 10^100, then you get > a rational number which cannot and never be distinguished by the same > one where the last digit is exchanged by 5, say. There fore already > these rational numbers cannot be put in <-relation with each other. > They are not numbers. Even less sqrt(5) is a number. You are too focussed on decimal arithmetic. If I write numbers in the base sqrt(5), I can certainly compare sqrt(5) with all other real numbers. > sqrt(-1) is a complex entity which consists of numbers, 0 and 1 in this > case, like a vector or a matrix consist of numbers. It is usual to call > these entities complex numbers like vector are called vectors. Why not? And in Q(sqrt(5)) I work with numbers that also take the form of vectors. > A long time ago you have stated that you would continue the discussion > with the intention that there are not finitely many natural numbers. > WM: You must distinguish, and that isy fairl difficult, because > sometimes we argue from the standpoint of set theory, sometimes not Yes, indeed, sometimes you argue from one point of view, sometimes from the other point of view, and you change position without warning. If you want to show that set theory is inconsistent, you have to start with the viewpoint that there are infinitely many natural numbers. Otherwise you only show that set theory is inconsistent with your (personal) view that there are only finitely many naturals. > My standpoint is: The values of natural numbers are always finite, but > there is no upper bound. There cannot exist more than 10^100 elements > of set in the whole universe. Therefore, there cannot exist more than > 10^100 natural numbers simultaneously. Again, *what is your definition of exist*? > When disproving Cantors theory, then I start from actual infinity in > order to show a contradiction. For instance I use actually infinitely > many transpositions to order the set of actually infinitely many > rational numbers. Or I construct a binary tree with actually infinitely > many levels. But what has been stated again and again, such a process of infinitely many operations that depend on each other will never terminate. > Again, my question. When does something exist in your opinion? Can > you give a clear definition of that concept? Apparently mathematical > existence conflicts with your ideas about existence. > WM: Mathematical existence is nothing but being well defined. By > Peano's axioms all natural numbers do exist. Yes, so pi exists in the mathematical sense, because it is well-defined. > I distinguish between definition and existence. > A natural number exists as that notion which the elements of a > fundamental set have in common. (I use the definition to construct > those sets. The definition is prior to existence.) What are fundamental sets? And what (in this case) is the definition? But I understand that when you say exist you are not meaning mathematical existence, but something semi-illusionary. > If you wish, please answer in one of the new threads, because this one > has become too long already. No. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree WM: A real number must be capable of being put in order relation (<) > > with any other real number. > Why? When I am working in Q(sqrt(5)) I am not interested in order relations. Because a number is something else than just a notion or an idea. Real numbers are by axioms to be put in order. It is commonly assumed that this really an be executed. But this assumption s wrong. > You are too focussed on decimal arithmetic. If I write numbers in the > base sqrt(5), I can certainly compare sqrt(5) with all other real numbers. But then you have no relation to numbers in base 1 and te problem is the same. > > sqrt(-1) is a complex entity which consists of numbers, 0 and 1 in this > > case, like a vector or a matrix consist of numbers. It is usual to call > > these entities complex numbers like vector are called vectors. Why not? > And in Q(sqrt(5)) I work with numbers that also take the form of vectors. I do not try to hinder you. I also work with vectors. > But what has been stated again and again, such a process of infinitely > many operations that depend on each other will never terminate. You are in error. Once the well ordering is given the operations do not depend on each other, but they are all determined from the beginning, exactly defined by the given prescription. Two different computers executing that program would never deviate from each other. Therefore you cannot escape to agree that this example is *logically* equivalent to Cantor's diagonal. > Yes, so pi exists in the mathematical sense, because it is well-defined. Concerning real numbers, definition and existence do not coincide. That was assumed in the past only because mathematicians were not aware of the principle limitations. === Subject: Re: Cantor and the binary tree > > WM: A real number must be capable of being put in order relation (<) > > with any other real number. > > Why? When I am working in Q(sqrt(5)) I am not interested in order > relations. > Because a number is something else than just a notion or an idea. Real > numbers are by axioms to be put in order. It is commonly assumed that > this really an be executed. But this assumption s wrong. When I am working in Q(sqrt(5)) I am not interested in order relations, and I wish to regard sqrt(5) as a number in that case. > But what has been stated again and again, such a process of infinitely > many operations that depend on each other will never terminate. > You are in error. Once the well ordering is given the operations do not > depend on each other, but they are all determined from the beginning, > exactly defined by the given prescription. Two different computers > executing that program would never deviate from each other. Therefore > you cannot escape to agree that this example is *logically* equivalent > to Cantor's diagonal. There is a *huge* logical difference. If you (theoretically) assigned infinitely many computers to the task of calculating the antidiagonal, the operation will terminate. When you do the same with your process of exchanges the operation will not terminate. BTW, you can describe your process slightly different. Take the first number of the well-ordered find one that is smaller again. Etc. > Yes, so pi exists in the mathematical sense, because it is well-defined. > Concerning real numbers, definition and existence do not coincide. > That was assumed in the past only because mathematicians were not aware > of the principle limitations. Well, if pi does not exist, what is it? Is d/dx sin(x) only approximately cos(x)? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree But what has been stated again and again, such a process of infinitely > > many operations that depend on each other will never terminate. > > You are in error. Once the well ordering is given the operations do not > > depend on each other, but they are all determined from the beginning, > > exactly defined by the given prescription. Two different computers > > executing that program would never deviate from each other. Therefore > > you cannot escape to agree that this example is *logically* equivalent > > to Cantor's diagonal. > There is a *huge* logical difference. If you (theoretically) assigned > infinitely many computers to the task of calculating the antidiagonal, > the operation will terminate. So you would also assign a computer to the last line, theoretically? Otherwise, if it did not exist, for example, none of your computers could reach it and the process would not terminate. > When you do the same with your process > of exchanges the operation will not terminate. BTW, you can describe > your process slightly different. Take the first number of the well-ordered > find one that is smaller again. Etc. You can also describe the creation of the antidiagonal in this way. find the first one below. Etc. I think, my steps are larger, so I will be ready faster. That is the only difference. > > Yes, so pi exists in the mathematical sense, because it is well-defined. > > Concerning real numbers, definition and existence do not coincide. > > That was assumed in the past only because mathematicians were not aware > > of the principle limitations. > Well, if pi does not exist, what is it? Is d/dx sin(x) only approximately > cos(x)? Those are ideas which only for special values of x take on the character of numbers. The equations connecting these ideas are certainly as true as circumference of circle is its diameter * pi or sqrt(2) * sqrt(2) = 2. === Subject: Re: Cantor and the binary tree > > But what has been stated again and again, such a process of infinitely > > many operations that depend on each other will never terminate. > > > > You are in error. Once the well ordering is given the operations do not > > depend on each other, but they are all determined from the beginning, > > exactly defined by the given prescription. Two different computers > > executing that program would never deviate from each other. Therefore > > you cannot escape to agree that this example is *logically* equivalent > > to Cantor's diagonal. > > There is a *huge* logical difference. If you (theoretically) assigned > infinitely many computers to the task of calculating the antidiagonal, > the operation will terminate. > So you would also assign a computer to the last line, theoretically? What last line? > Otherwise, if it did not exist, for example, none of your computers > could reach it and the process would not terminate. Eh? Each of them would be able in finite time to determine the n-th digit. Even in a very short time. > You can also describe the creation of the antidiagonal in this way. > find the first one below. Etc. I think, my steps are larger, so I will > be ready faster. That is the only difference. You can describe *any* number in such an iterative process. But not all iterative processes are equivalent with non-iterative processes. That is the difference. > Well, if pi does not exist, what is it? Is d/dx sin(x) only approximately > cos(x)? > Those are ideas which only for special values of x take on the > character of numbers. The equations connecting these ideas are > certainly as true as circumference of circle is its diameter * pi or > sqrt(2) * sqrt(2) = 2. Pray explain. You have lost me again. What are the characters of numbers. And I mean numbers in your sense (they do not conform to numbers in the mathematical sense). Beating at terminology? Or what? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree So you would also assign a computer to the last line, theoretically? > What last line? > > Otherwise, if it did not exist, for example, none of your computers > > could reach it and the process would not terminate. > Eh? Each of them would be able in finite time to determine the n-th > digit. Even in a very short time. But there would always remain a large part undone, even an infinite part, that means: nearly all of the work to be done. > > You can also describe the creation of the antidiagonal in this way. > > find the first one below. Etc. I think, my steps are larger, so I will > > be ready faster. That is the only difference. > You can describe *any* number in such an iterative process. But not all > iterative processes are equivalent with non-iterative processes. That is > the difference. *Any* tranposition to be performed is determined from the beginning (given a certain initial well-ordering of the rationals of (0,1)). You can say, for *any* transposition, when it will have to occur and what will be the result. It is not an iterative process. It is equivalent to Cantor's a_nn replaced by b_n. It is impossible, however, to show the completed result, namely the ordered set of rationals. It is equally impossible to show the completed antidiagonal. The only difference is, that the latter is not so obvious. Terefore some people believe in finished infinity, as yet. > > Those are ideas which only for special values of x take on the > > character of numbers. The equations connecting these ideas are > > certainly as true as circumference of circle is its diameter * pi or > > sqrt(2) * sqrt(2) = 2. > Pray explain. You have lost me again. What are the characters of > numbers. And I mean numbers in your sense (they do not conform to > numbers in the mathematical sense). Beating at terminology? Or what? A number is an idea which can be put in oder (<) with any other number. sqrt(2) and the same idea, with digit number 10^100 exchanged by 2, cannot and never be put in this order. === Subject: Re: Cantor and the binary tree >> > You can also describe the creation of the antidiagonal in this way. >> > find the first one below. Etc. I think, my steps are larger, so I will >> > be ready faster. That is the only difference. >> You can describe *any* number in such an iterative process. But not all >> iterative processes are equivalent with non-iterative processes. That is >> the difference. >*Any* tranposition to be performed is determined from the beginning >(given a certain initial well-ordering of the rationals of (0,1)). You >can say, for *any* transposition, when it will have to occur and what >will be the result. It is not an iterative process. It is equivalent to >Cantor's a_nn replaced by b_n. >It is impossible, however, to show the completed result, namely the >ordered set of rationals. The only way to even make sense of the completed results is to use a limit. While I have some ideas for canditates, I don't know of any natural topology on sequences of orderings. And in the ideas I do have, your sequences diverges (I.e. there is no completed result) or the completed result is not what you claim it is. > It is equally impossible to show the >completed antidiagonal. It's not necessary to do so. All that is needed is proof of it's existence. >> > Those are ideas which only for special values of x take on the >> > character of numbers. The equations connecting these ideas are >> > certainly as true as circumference of circle is its diameter * pi or >> > sqrt(2) * sqrt(2) = 2. >> Pray explain. You have lost me again. What are the characters of >> numbers. And I mean numbers in your sense (they do not conform to >> numbers in the mathematical sense). Beating at terminology? Or what? >A number is an idea which can be put in oder (<) with any other number. >sqrt(2) and the same idea, with digit number 10^100 exchanged by 2, >cannot and never be put in this order. It has a unique position in the natural order on R. Whether or not we prove a particular descripition of the number to be less than or greater than it is another story. BTW, even the natural numbers have this problem. Martin === Subject: Re: Cantor and the binary tree > > So you would also assign a computer to the last line, theoretically? > What last line? > > Otherwise, if it did not exist, for example, none of your computers > > could reach it and the process would not terminate. > Eh? Each of them would be able in finite time to determine the n-th > digit. Even in a very short time. > But there would always remain a large part undone, even an infinite > part, that means: nearly all of the work to be done. With one computer per line and one digit to compute per computer, WM is positing very slow computers. > > You can also describe the creation of the antidiagonal in this way. > > find the first one below. Etc. I think, my steps are larger, so I will > > be ready faster. That is the only difference. > You can describe *any* number in such an iterative process. But not all > iterative processes are equivalent with non-iterative processes. That is > the difference. > *Any* tranposition to be performed is determined from the beginning > (given a certain initial well-ordering of the rationals of (0,1)). You > can say, for *any* transposition, when it will have to occur and what > will be the result. It is not an iterative process. It is equivalent to > Cantor's a_nn replaced by b_n. > It is impossible, however, to show the completed result, namely the > ordered set of rationals. It is impossible to show by mere iteration that the limit of a convergent sequence is the limit, but that does not mean the limit does not exist. > > Those are ideas which only for special values of x take on the > > character of numbers. The equations connecting these ideas are > > certainly as true as circumference of circle is its diameter * pi or > > sqrt(2) * sqrt(2) = 2. > Pray explain. You have lost me again. What are the characters of > numbers. And I mean numbers in your sense (they do not conform to > numbers in the mathematical sense). Beating at terminology? Or what? > A number is an idea which can be put in oder (<) with any other number. > sqrt(2) and the same idea, with digit number 10^100 exchanged by 2, > cannot and never be put in this order. It is a shame that WM cannot put his ideas in order. === Subject: Re: Cantor and the binary tree > > So you would also assign a computer to the last line, theoretically? > > What last line? > > > Otherwise, if it did not exist, for example, none of your computers > > could reach it and the process would not terminate. > > Eh? Each of them would be able in finite time to determine the n-th > digit. Even in a very short time. > But there would always remain a large part undone, even an infinite > part, that means: nearly all of the work to be done. Have you read the part you snipped? > > You can also describe the creation of the antidiagonal in this way. > > find the first one below. Etc. I think, my steps are larger, so I will > > be ready faster. That is the only difference. > > You can describe *any* number in such an iterative process. But not all > iterative processes are equivalent with non-iterative processes. That is > the difference. > *Any* tranposition to be performed is determined from the beginning > (given a certain initial well-ordering of the rationals of (0,1)). You > can say, for *any* transposition, when it will have to occur and what > will be the result. It is not an iterative process. It is equivalent to > Cantor's a_nn replaced by b_n. It is not. But you are too dense to notice. > It is impossible, however, to show the completed result, namely the > ordered set of rationals. There is no completed result. > It is equally impossible to show the > completed antidiagonal. The only difference is, that the latter is not > so obvious. Terefore some people believe in finished infinity, as yet. There is no completed antidiagonal, that is you can not give all decimal digits. On the other hand it is sufficient to show the existence of a real that is different from all the reals on your list. > > Those are ideas which only for special values of x take on the > > character of numbers. The equations connecting these ideas are > > certainly as true as circumference of circle is its diameter * pi or > > sqrt(2) * sqrt(2) = 2. > > Pray explain. You have lost me again. What are the characters of > numbers. And I mean numbers in your sense (they do not conform to > numbers in the mathematical sense). Beating at terminology? Or what? > A number is an idea which can be put in oder (<) with any other number. > sqrt(2) and the same idea, with digit number 10^100 exchanged by 2, > cannot and never be put in this order. You are indeed using your own personal definition of the concept of number. In general in mathematics you start with a set of elements with the operations + and * defined. When the operations satisfy some basic properties (the set under + forms a group, * is distributive over +) you can call the elements numbers. (Some mathematicians want a bit more structure like associativity, or sometimes even commutativity, of *, but not all want that.) When it is clear about what structure you are talking you can use the term numbers as synonym for elements. Otherwise it is best to qualify the numbers. Note that until now order relations play no role. And indeed, the 5-adic numbers can not be consistently ordered, amongst others, because the square root of -1 is a 5-adic number. The only place where, in the definitions, ordering plays a role is when you want to introduce transcendental numbers. For algebraic numbers there is no need. And indeed, it is a cornerstone of Galois theory that you can not algebraically distringuish the roots of an irreducible polynomial. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > > You are in error. Once the well ordering is given the operations > > do not depend on each other, but they are all determined from the > > beginning, exactly defined by the given prescription. Two > > different computers executing that program would never deviate > > from each other. Therefore you cannot escape to agree that this > > example is *logically* equivalent to Cantor's diagonal. > There is a *huge* logical difference. If you (theoretically) > assigned infinitely many computers to the task of calculating the > antidiagonal, the operation will terminate. Proof for that? This almost sounds like the infinitely should always be enough kind of wishy-washy argument we hear from our Discantors. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> > WM: A real number must be capable of being put in order relation >> > (<) with any other real number. >> Why? When I am working in Q(sqrt(5)) I am not interested in order relations. > Because a number is something else than just a notion or an > idea. Real numbers are by axioms to be put in order. It is commonly > assumed that this really an be executed. But this assumption s > wrong. >> You are too focussed on decimal arithmetic. If I write numbers in >> the base sqrt(5), I can certainly compare sqrt(5) with all other >> real numbers. > But then you have no relation to numbers in base 1 and te problem is > the same. Numbers in base 1. Wow. And you are not ashamed of purporting to lecture about mathematics. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree > No. I am not at all a Platonist. I do not even believe that more than > 10^100 objects can exist simultaneously. What is the diameter of a universe that is so restricted, and how can WM prove that the diameter of the universe is no larger? Or is 10^100 merely the limting number of shadows on the cave wall that we cannot see out of? === Subject: Re: Cantor and the binary tree 10^100 objects can exist simultaneously. > What is the diameter of a universe that is so restricted, and how can WM > prove that the diameter of the universe is no larger? Because the age of the universe is about 10^10 years and the light velocity is limited. > Or is 10^100 merely the limting number of shadows on the cave wall that > we cannot see out of? There is no cave. There is no out. === Subject: Re: Cantor and the binary tree > No. I am not at all a Platonist. I do not even believe that more than > 10^100 objects can exist simultaneously. > What is the diameter of a universe that is so restricted, and how can WM > prove that the diameter of the universe is no larger? > Because the age of the universe is about 10^10 years and the light > velocity is limited. According to one hypothesis, the 'universe' is of finite age, but that hypothesis is by no means proven, and even among those supporting that hypothesis there is considerable disagreement on its age. There are other hypotheses that the 'universe' had no beginning in time. > Or is 10^100 merely the limting number of shadows on the cave wall that > we cannot see out of? > There is no cave. There is no out. And how can WM be so certain that phenomena accurately portray noumenon? Or is that just another of his declarations of faith. === Subject: Re: Cantor and the binary tree Discussion, linux) > > Your distinction is nothing than a helpless attempt to save set-theory > > as has those which nhave been aplied in the past several times. Think > > of L=F6wenstein-Skolem: Every consistent first oder theory has a > > countable model. iIf set theoy is consistent, then it is countable, > > hence inconsistent. Therefore set theory canot be consistent. All your > > attempts are in vain. > Yes, you are a Platonist, that was clear. Platonist? That word must mean something different (and more demeaning) than I thought. -- Jesse F. Hughes When you try to kiss a girl, it's hard not to get spit on the girl. -- Quincy P. Hughes, age 3 (almost 4) === Subject: Re: Cantor and the binary tree > > Your distinction is nothing than a helpless attempt to save set-theory > > as has those which nhave been aplied in the past several times. Think > > of L=F6wenstein-Skolem: Every consistent first oder theory has a > > countable model. iIf set theoy is consistent, then it is countable, > > hence inconsistent. Therefore set theory canot be consistent. All your > > attempts are in vain. > > Yes, you are a Platonist, that was clear. > Platonist? That word must mean something different (and more > demeaning) than I thought. Every valid theory is first order... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> > Your distinction is nothing than a helpless attempt to save >> > set-theory as has those which nhave been aplied in the past >> > several times. Think of L=F6wenstein-Skolem: Every consistent >> > first oder theory has a countable model. iIf set theoy is >> > consistent, then it is countable, hence inconsistent. Therefore >> > set theory canot be consistent. All your attempts are in vain. >> Yes, you are a Platonist, that was clear. > Platonist? That word must mean something different (and more > demeaning) than I thought. Well, take a look at the conclusion of Euthydemon: don't judge the merits of an art itself by the mass of incapable idiots that pretend to be versed in it. Given that M.9fckenheim pretends to be teaching math, I find the whole dialog of Euthydemon (quite recommendable reading) fitting to a theta. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree >> > Your distinction is nothing than a helpless attempt to save >> > set-theory as has those which nhave been aplied in the past >> > several times. Think of L=F6wenstein-Skolem: Every consistent >> > first oder theory has a countable model. iIf set theoy is >> > consistent, then it is countable, hence inconsistent. Therefore >> > set theory canot be consistent. All your attempts are in vain. >> Yes, you are a Platonist, that was clear. > Platonist? That word must mean something different (and more > demeaning) than I thought. > Well, take a look at the conclusion of Euthydemon: don't judge the > merits of an art itself by the mass of incapable idiots that pretend > to be versed in it. > Given that M.9fckenheim pretends to be teaching math, I find the whole > dialog of Euthydemon (quite recommendable reading) fitting to a theta. Is there a good online source in English? === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > > Your distinction is nothing than a helpless attempt to save > > set-theory as has those which nhave been aplied in the past > > several times. Think of L=F6wenstein-Skolem: Every consistent > > first oder theory has a countable model. iIf set theoy is > > consistent, then it is countable, hence inconsistent. Therefore > > set theory canot be consistent. All your attempts are in vain. >> Yes, you are a Platonist, that was clear. >> Platonist? That word must mean something different (and more >> demeaning) than I thought. >> Well, take a look at the conclusion of Euthydemon: don't judge the >> merits of an art itself by the mass of incapable idiots that pretend >> to be versed in it. >> Given that M.9fckenheim pretends to be teaching math, I find the whole >> dialog of Euthydemon (quite recommendable reading) fitting to a theta. > Is there a good online source in English? I found . It does a worse job in my opinion of bringing across the hilarity of the dialogue than the impressive Schleiermacher translation (which actually is often used as a reference point for other translations), but of course Schleiermacher's translation is into German. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <878y0qzcny.fsf@phiwumbda.org> <85ekail7yq.fsf@lola.goethe.zz> Discussion, linux) > > Your distinction is nothing than a helpless attempt to save > > set-theory as has those which nhave been aplied in the past > > several times. Think of L=F6wenstein-Skolem: Every consistent > > first oder theory has a countable model. iIf set theoy is > > consistent, then it is countable, hence inconsistent. Therefore > > set theory canot be consistent. All your attempts are in vain. > Yes, you are a Platonist, that was clear. >> Platonist? That word must mean something different (and more >> demeaning) than I thought. > Well, take a look at the conclusion of Euthydemon: don't judge the > merits of an art itself by the mass of incapable idiots that pretend > to be versed in it. Yes, but what was clear from that paragraph was not that Mueckenheim is a Platonist. I don't know anything in Platonism that would justify such an analysis of the famous Lowenstein-Skolem [sic] theorem. Indeed, all adult Platonists accepts the correctness of the L-S theorem and that set theory is likely consistent. If Dik thinks the paragraph conveys the information that Mueckenheim is a Platonist, then I must assume that Dik thinks Platonist is synonymous with moron. Now, I know that some folks think that all Platonists are morons, but not even the most rabid anti-Platonist claims synonymy[1]. > Given that M.9fckenheim pretends to be teaching math, I find the whole > dialog of Euthydemon (quite recommendable reading) fitting to a theta. Yeah, well, it's *your* damn country. I'm just glad that my homeland doesn't have such widely visible idiots[2]. Footnotes: [1] It is a good word, synonymy. I wish I could use it more often. [2] Before making the obvious joke, just ask yourself: is it really plausible that I *accidentally* used such an obvious straight line? Really, I hate to add this footnote, but I also hate to have twenty-seven people (not Kastrup, who never can think of anything funny to say) try to get the keenest, most brilliant zinger out of this line. -- Jesse F. Hughes Dead men can't talk. Especially when they've been cremated. === Subject: Re: Cantor and the binary tree ... > > Your distinction is nothing than a helpless attempt to save > > set-theory as has those which nhave been aplied in the past > > several times. Think of L=3DF6wenstein-Skolem: Every consistent > > first oder theory has a countable model. iIf set theoy is > > consistent, then it is countable, hence inconsistent. Therefore > > set theory canot be consistent. All your attempts are in vain. > > Yes, you are a Platonist, that was clear. >> >> Platonist? That word must mean something different (and more >> demeaning) than I thought. > > Well, take a look at the conclusion of Euthydemon: don't judge the > merits of an art itself by the mass of incapable idiots that pretend > to be versed in it. > Yes, but what was clear from that paragraph was not that Mueckenheim > is a Platonist. I don't know anything in Platonism that would justify > such an analysis of the famous Lowenstein-Skolem [sic] theorem. See: : Model Existence theorem is steadily provoking the so-called Skolem's paradox. Indeed, in ZF we can prove existence of uncountable sets. Still, according to Model Existence theorem, if ZF is consistent, then there is a countable model of ZF. I.e. ZF proves existence of uncountable sets, yet it has a countable model! Is this possible? Does it mean that ZF is inconsistent? Platonists could say even more: any consistent axiomatic set theory has countable models, hence, no axiom system can represent the intended set theory (i.e. the Platonist world of sets) adequately. See also further in that page about Platonist superstitions. (Yes, it is of some type of Platonist, obviously.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor and the binary tree <87u0jexsyd.fsf@phiwumbda.org> Discussion, linux) > See: : > Model Existence theorem is steadily provoking the so-called Skolem's paradox. > Indeed, in ZF we can prove existence of uncountable sets. Still, according to > Model Existence theorem, if ZF is consistent, then there is a countable model > of ZF. I.e. ZF proves existence of uncountable sets, yet it has a countable > model! Is this possible? Does it mean that ZF is inconsistent? Platonists > could say even more: any consistent axiomatic set theory has countable models, > hence, no axiom system can represent the intended set theory (i.e. the > Platonist world of sets) adequately. > See also further in that page about Platonist superstitions. (Yes, it is > of some type of Platonist, obviously.) Obviously. Platonism doesn't justify such pathetic reasoning. The type of Platonist you mean is the stupid Platonist. Your author (Podnieks) asks questions about inconsistency without even attempting to sketch how L-S => ZF is inconsistent. He uses terms like represent ... adequately without clarification. Does he just mean that no first order of set theory is categorical? How is that say[ing] even more than ZF is inconsistent? David Kastrup's comments are more apt than I realized. I'd hate for people to think that Platonism (or philosophy of mathematics) is filled with such bad reasoning and argument. (Note: I am not defending Platonism as such, but it is certainly not as simple-minded as this. -- Who knows, maybe that may be the only way to settle this crap. It's not like it'd be that hard for me to go back and get a math degree. I can penetrate the math social group and then finish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup's comments are more apt than I realized. I'd hate for > people to think that Platonism (or philosophy of mathematics) is > filled with such bad reasoning and argument. (Note: I am not > defending Platonism as such, but it is certainly not as > simple-minded as this. Euthydemos and Dionysodoros, two youths that return to Athens to teach philosophy after they have taught fencing with little success the previous year, are pretty much in the same ballpark as a M.9fckenheim that teaches mathematical philosophy. Mathematics and natural sciences are hard sciences as you always have to bow to verifying your results. That keeps its practitioners, where they do more than scratch the surface, honest (of course excepting the jokers that are not bothered about inconsistency in their arguments). Philosophy is softer, and thus its various branches are littered with pretenders that are incapable of the mental discipline of maintaining rigid thought models even in the absence of immediate outer forces. Mainstream movements of vague thinking are also invading other branches like literary sciences, conveniently overlooking the fact that maintaining coherent models worthy of the name science and progress of knowledge requires _more_ mental capacities and discipline in soft rather than hard sciences. It also overlooks small details that encompassing theories such as intermediality (sp?) require decades of studies of centuries of literature, arts, history and culture to arrive at any worthwhile results. But instead of becoming an end-of-life occupation for the best and most extensive studiers of the vast semantic web of our culture, it is becoming a mainstream idiocy taught to students that boils down to picking arbitrary works and noting superficial similarities that can be either accidental or based on connections so entangled in time that any student would be incapable of tracking them down to the locus of their origin, anyway. There is a lot of mock science of that order going on in the soft sciences, and it is a shame. A deficient logician does not make a good philosopher. Progress is not made by babbling idiots, regardless of whether one is looking at hard or soft sciences. But the damage caused by babbling idiots might be larger on the latter. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree >Mathematics and natural sciences are hard sciences as you always >have to bow to verifying your results. That keeps its practitioners, >where they do more than scratch the surface, honest (of course >excepting the jokers that are not bothered about inconsistency in >their arguments and the fools who can neither see for themselves, nor be led to see, the inconsistency in their arguments Lee Rudolph === Subject: Re: Cantor and the binary tree <87u0jexsyd.fsf@phiwumbda.org> <87d5q1y55n.fsf@phiwumbda.org> <85k6k9in5j.fsf@lola.goethe.zz> Discussion, linux) [snip very nice analysis of the difficulties of philosophy and other soft endeavors] > A deficient logician does not make a good philosopher. So anyways, [...] I've decided that I'm a philosopher now, and, a philosopher of mathematical logic. Ross A. Finlayson He apparently decided this *precisely* because he was incapable of doing logic. Hey, we don't make philosophers of science *do* science, so why should a philosopher of mathematical logic be capable of logic? > Progress is not made by babbling idiots, regardless of whether one > is looking at hard or soft sciences. But the damage caused by > babbling idiots might be larger on the latter. Well, Ross is just a joke, of course, and causes no damage at all. But you're right that there are some bad thinkers out there that more credibly pose as philosophers and they do no good for the reputation of philosophy. -- Who knows, maybe that may be the only way to settle this crap. It's not like it'd be that hard for me to go back and get a math degree. I can penetrate the math social group and then finish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > [snip very nice analysis of the difficulties of philosophy and other > soft endeavors] >> A deficient logician does not make a good philosopher. > So anyways, [...] I've decided that I'm a philosopher now, and, a > philosopher of mathematical logic. > Ross A. Finlayson > He apparently decided this *precisely* because he was incapable of > doing logic. Hey, we don't make philosophers of science *do* > science, so why should a philosopher of mathematical logic be > capable of logic? Well, philosophy is the art of tying the pieces together. A philosopher of mathematical logic incapable of doing logic has _neither_ the pieces _nor_ the string. That's pretty much the worst fix you can be in as a metascientist. >> Progress is not made by babbling idiots, regardless of whether one >> is looking at hard or soft sciences. But the damage caused by >> babbling idiots might be larger on the latter. > Well, Ross is just a joke, of course, and causes no damage at all. > But you're right that there are some bad thinkers out there that > more credibly pose as philosophers and they do no good for the > reputation of philosophy. Actually, there are complete mainstream _branches_ in the posing and cargo cult business. It is not an isolated phenomenon there. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw [...] > Yeah, well, it's *your* damn country. I'm just glad that my > homeland doesn't have such widely visible idiots[2]. > [2] Before making the obvious joke, just ask yourself: is it really > plausible that I *accidentally* used such an obvious straight line? > Really, I hate to add this footnote, but I also hate to have > twenty-seven people (not Kastrup, who never can think of anything > funny to say) try to get the keenest, most brilliant zinger out of > this line. Is the joke supposed to be that I was born in Walnut Creek, Cal., to a mother hailing from The Hoosier State (talk about an idiotic state motto...)? Or that I am so widely visible (Google throws out my full name about 3000 times more often than that of M.9fckenheim)? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <878y0qzcny.fsf@phiwumbda.org> <85ekail7yq.fsf@lola.goethe.zz> <87u0jexsyd.fsf@phiwumbda.org> <851x6il4my.fsf@lola.goethe.zz> Discussion, linux) > [...] >> Yeah, well, it's *your* damn country. I'm just glad that my >> homeland doesn't have such widely visible idiots[2]. >> [2] Before making the obvious joke, just ask yourself: is it really >> plausible that I *accidentally* used such an obvious straight line? >> Really, I hate to add this footnote, but I also hate to have >> twenty-seven people (not Kastrup, who never can think of anything >> funny to say) try to get the keenest, most brilliant zinger out of >> this line. > Is the joke supposed to be that I was born in Walnut Creek, Cal., to a > mother hailing from The Hoosier State (talk about an idiotic state > motto...)? Um, er, yeah... That is obviously the joke I intended. Now you spoiled it. -- Jesse F. Hughes Basically there are two angry groups. I am a harsh force of one. Against me is a society of mathematicians. So far it's been a draw. -- JSH gives another display of keen insight. === Subject: Re: Cantor and the binary tree > Is the joke supposed to be that I was born in Walnut Creek, Cal., to a > mother hailing from The Hoosier State (talk about an idiotic state > motto...)? The Indiana State Motto is The Crossroads of America. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Is the joke supposed to be that I was born in Walnut Creek, Cal., to a >> mother hailing from The Hoosier State (talk about an idiotic state >> motto...)? > The Indiana State Motto is The Crossroads of America. > Ok, so it is the nickname . -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor and the binary tree <878y0qzcny.fsf@phiwumbda.org> <85ekail7yq.fsf@lola.goethe.zz> <87u0jexsyd.fsf@phiwumbda.org> [1] It is a good word, synonymy. I wish I could use it more often. The 'o' spoils it, though - syzygy is vastly better, not that I get to use it often. Brian Chandler http://imaginatorium.org The quality of entertainment on sci.math has dropped a lot recently... === Subject: Re: Cantor and the binary tree !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi$t ^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> > Complete induction will do so. Even Cantor was convinced that complete >> > induction is sufficient for all countable sets, and with him every >> > mathematician at his times. >> He may be convinced, in that case he was wrong. I rather think that you >> misrepresent what Cantor has stated, because you do not understand the >> distinction. Yes: > I understand that erste Zahlenkasse means countable. Cantor understood > that. >> 2n > Card({2, 4, ...., 2n}) >> This is true for *each* natural n. And that is what complete induction >> is. It does not state anything about N as the right hand side of the >> formula. You have a serious comprehension problem. Stating something >> is valid for *all* n is the same as stating that something is valid for >> *each* n. In Dutch (and I think in German) the same applies. It does >> not state anything of the collection of *each* n, or of *all* n, it >> only states something about the individual members. > I know that each n does define a finite sequence. Fom that one can > conclude that all n are in a finite sequence, in German and n Dutch. And here is your mistake: for all n, you can indeed find an associated finite sequence in which n is contained. But you can't find a single finite sequence in which all n would be contained. Quantifier dyslexia, your standard ailment. You should stop practicing mathematics, since this condition is a fatal impediment to it.