mm-490 === Subject: Re: Info about a particular permutation distance > I've scanned the net searching for various > permutation distance definitions, and didn't find > anything specifically related with this: > A *move* is a permutation for which only one element > is repositioned. For example, in > 1 2 3 4 5 6 7 8 > 1 2 3 8 4 5 6 7 > the element 8 is repositioned from the 8th to the > 4th place. Please note that a move is not a > transposition: in fact, a transposition involves two > moves in general. > Now, the move distance between two permutations > P1 and P2 is the minimum number of moves needed > to obtain P2 from P1. > Has this been studied anywhere? I'm looking > for algorithms (exact or approximate) to compute > moves. There seems to be abundant literature > about transpositions, but not for this other type > of simple operation. > Joaqu.92n M L.97pez Mu.96oz > Telef.97nica, Investigaci.97n y Desarrollo Se puede encontrar esto y mucho mas sobre metrics on permutations en el libro: Group Representations in Probablity and Statistics. por Persi Diaconis. He describes several metrics on S_n. The 5th one on page 112 is the one you describe. He attributes this to Cayley and mentions that Cayley proved that the distance from permutation f to permutation g in S_n is given by n - # cycles in the disjoint cycle decomposition of fg^(-1). Edwin Clark --------------------------------------------------------- W. Edwin Clark, Math Dept, University of South Florida http://www.math.usf.edu/~eclark/ --------------------------------------------------------- === Subject: Re: Info about a particular permutation distance > A *move* is a permutation for which only one element > is repositioned. For example, in > 1 2 3 4 5 6 7 8 > 1 2 3 8 4 5 6 7 > the element 8 is repositioned from the 8th to the > 4th place. Please note that a move is not a > transposition: in fact, a transposition involves two > moves in general. Sorry, but re-reading your definition of your metric your metric is not the one Diaconis attributes to Cayley. Rather it is the one he call's Ulam's distance. As he states it, it is motivated by sorting, say books on a shelf. Take a book out and insert it in another place. This seems to be what you are doing. He mentions that it is used by biologists and computer scientists and he gives references to Knuth and Gordon. If you have trouble finding Diaconis' book let me know and I will send you the references to these other sources. Best wishes, Edwin Clark > Se puede encontrar esto y mucho mas sobre metrics on permutations en > el libro: > Group Representations in Probablity and Statistics. > por Persi Diaconis. > He describes several metrics on S_n. The 5th one on page 112 is the one > you describe. He attributes this to Cayley and mentions that Cayley > proved that the distance from permutation f to permutation g in S_n is > given by n - # cycles in the disjoint cycle decomposition of fg^(-1). > Edwin Clark === Subject: Re: Info about a particular permutation distance Edwin Clark ha escrito: A *move* is a permutation for which only one element > is repositioned. For example, in 1 2 3 4 5 6 7 8 > 1 2 3 8 4 5 6 7 the element 8 is repositioned from the 8th to the > 4th place. Please note that a move is not a > transposition: in fact, a transposition involves two > moves in general. Sorry, but re-reading your definition of your metric your > metric is not the one Diaconis attributes to Cayley. > Rather it is the one he call's Ulam's distance. As he states it, it is > motivated by sorting, say books on a shelf. Take a book out and > insert it in another place. This seems to be what you are doing. > He mentions that it is used by biologists and computer scientists > and he gives references to Knuth and Gordon. If you have trouble > finding Diaconis' book let me know and I will send you the > references to these other sources. I don't have access to Diaconis' book, but I can try to find this subject in my Knuth's. Do you happen to have the exact reference into Joaqu.92n M L.97pez Mu.96oz Telef.97nica, Investigaci.97n y Desarrollo === Subject: Re: Info about a particular permutation distance <413526E1.7030508@math.usf.edu> <41353889.2020006@math.usf.edu> <41358301.60470BCC@tid.es Edwin Clark ha escrito: A *move* is a permutation for which only one element > is repositioned. For example, in 1 2 3 4 5 6 7 8 > 1 2 3 8 4 5 6 7 the element 8 is repositioned from the 8th to the > 4th place. Please note that a move is not a > transposition: in fact, a transposition involves two > moves in general. Sorry, but re-reading your definition of your metric your > metric is not the one Diaconis attributes to Cayley. > Rather it is the one he call's Ulam's distance. As he states it, it is > motivated by sorting, say books on a shelf. Take a book out and > insert it in another place. This seems to be what you are doing. > He mentions that it is used by biologists and computer scientists > and he gives references to Knuth and Gordon. If you have trouble > finding Diaconis' book let me know and I will send you the > references to these other sources. > I don't have access to Diaconis' book, but I can try to find this > subject in my Knuth's. Do you happen to have the exact reference into > Joaqu.92n M L.97pez Mu.96oz > Telef.97nica, Investigaci.97n y Desarrollo The reference simply says See Knuth (1978, 5.1.4), Gordon (1983) has suggested it for statistical tasks. I only have volumes I and II, of Knuth. This must be in Volume III. In the references at the end of the book I find only Knuth (1981) The Art of Comp. Prog. Vol II. and two listings for Gordon: Gordon, A.D., (1983) A measure of agreement between rankings. Biometrika, 66, 7-15 Gordon, L., (1983) Successive sampling in large finite populations. Ann.Statist. 11, 702-706 So your guess is as good as mine on which (maybe both) is relevant. In Diaconis' book he has the following on Ulam's distance. Here we write think of f as a permutation. So apparently in your notation this corresponds to a sequence (f(1),f(2),...,f(n)). The the Ulam distance L(f,g) between two permutations f and g is given by L(f,g) = n - length of the longest increasing subsequence in fg^(-1), that is f composed with the inverse of g. An increasing subsequence of a permutation h is a subsequence of the form: i_1 < i_2 <... <41353889.2020006@math.usf.edu> <41358301.60470BCC@tid.es> Hi Edwin, Edwin Clark ha escrito: [...] > In Diaconis' book he has the following on Ulam's distance. Here we write > think of f as a permutation. So apparently in your notation this > corresponds to a sequence (f(1),f(2),...,f(n)). The the Ulam distance > L(f,g) between two permutations f and g is given by > L(f,g) = n - length of the longest increasing subsequence in fg^(-1), that > is f composed with the inverse of g. > An increasing subsequence of a permutation h is a subsequence of the form: > i_1 < i_2 <... Lemma 2 states that the smallest number of moves (in your sense--I think) > to sort a permutation of length n is n - length of longest increasing > subsequence in the permutation. [...] > Also, searching Google on Ulam's distance I get 4 hits and some look > promising. But Ulam's metric gives even more hits! > Bottom Line: Search Google for Ulam's metric. That was the thread to pull from! After goggling a bit, I've come to the very problem I'm trying to solve, namely that of finding the longest increasing subsequence. There's plenty of stuff about this. In case you're curious, I need this algorithm to find efficient differences of STL sequences in C++, so as to be able to store them in disk as compactly as possible. Joaqu.92n M L.97pez Mu.96oz Telef.97nica, Investigaci.97n y Desarrollo === Subject: Re: Finding a basis for space spanned by vectors..... with Basis Theorem X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 10:44 PM, emailzul@starhub.net.sg (Zul) said: >Therefore how should I find a basis for the space spanned by the >following vectors: How therefore? The space spanned by your vectors is not R^4, but is only of dimension three, and the theorem that you quoted doesn't apply. Besides, it's easier to just work from the vectors that you gave; >By the Basis theorem does this mean that the following vectors has no >basis? No, it mean that your four vectors are not a basis for R ^4. Your first three vectors are a basis for the space spanned by all four. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: ?Direct Proof of irrationality of sqrt(2)? Does anyone know of a direct proof of the irrationality of the squareroot of 2? I've been trying to look for one, but I always find the typical proof by contradiction. Proofs written out are always === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? >Does anyone know of a direct proof of the irrationality of the >squareroot of 2? I've been trying to look for one, but I always find >the typical proof by contradiction. Proofs written out are always What about considering the sequence e_n = (sqrt(2)-1)^n ? Obviously, e_n is never 0, tends to 0, and there exist two sequences of integers a_n and b_n such that e_n=a_n*sqrt(2) + b_n. I let you show that these properties implies that sqrt(2) is not a rational. This proof is not mine but I don't know who found it first. === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? >Does anyone know of a direct proof of the irrationality of the >squareroot of 2? I've been trying to look for one, but I always find >the typical proof by contradiction. Proofs written out are always > What about considering the sequence e_n = (sqrt(2)-1)^n ? > Obviously, e_n is never 0, tends to 0, and there exist two sequences > of integers a_n and b_n such that e_n=a_n*sqrt(2) + b_n. I let you > show that these properties implies that sqrt(2) is not a rational. But wouldn't THAT proof be indirect? If sqrt(2) WERE rational, it would have some denominator q, so none of the e_n can be closer to 0 than 1/q, contradiction. > This proof is not mine but I don't know who found it first. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? > Does anyone know of a direct proof of the irrationality of the > squareroot of 2? I've been trying to look for one, but I always find > the typical proof by contradiction. The proof by contradiction is a direct proof in this case, since the statement proved is a negation. === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? > Does anyone know of a direct proof of the irrationality of the > squareroot of 2? I've been trying to look for one, but I always find > the typical proof by contradiction. > The proof by contradiction is a direct proof in this case, since > the statement proved is a negation. Sorry I should have rephrased my question. Is there a proof of the irrationality of the squareroot which does not start by assuming the squareroot of 2 is rational (for the eventual sake of contradiction). -Bobby === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? Am 01.09.04 19:08 schrieb Bobby: >Does anyone know of a direct proof of the irrationality of the >squareroot of 2? I've been trying to look for one, but I always find >the typical proof by contradiction. > The proof by contradiction is a direct proof in this case, since >the statement proved is a negation. > Sorry I should have rephrased my question. Is there a proof of the > irrationality of the squareroot which does not start by assuming the > squareroot of 2 is rational (for the eventual sake of contradiction). > -Bobby The continued-fraction method, eventually. It is simple to show, that the cf-expansion of sqrt(2) is infinite and periodic. Rational numbers must have finite cf-expansion. Gottfried Helms === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? I remember reading a book called IRRATIONAL NUMBERS by Prof. Ivan Morton follow, proof of the irrationality of sqrt(2), but I'm afraid I don't remember the specifics. Maybe your university library has a copy of the shelves. [If you're an alma mater of the University of Toronto, I can guarntee that it is there as of the time of this posting.] === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? > I remember reading a book called IRRATIONAL NUMBERS by Prof. Ivan Morton > follow, proof of the irrationality of sqrt(2), but I'm afraid I don't > remember the specifics. > Maybe your university library has a copy of the shelves. [If you're an alma > mater of the University of Toronto, I can guarntee that it is there as of > the time of this posting.] === Subject: Re: ?Direct Proof of irrationality of sqrt(2)? > It has a more direct, if harder to > follow, proof of the irrationality of sqrt(2), but I'm afraid I don't > remember the specifics. I believe you mean a more informative proof rather than a more direct one. Here's an old posting of Keith Ramsay's that deals with this: === Subject: Re: sqrt(2) is irrational > Those who ask for a direct proof that the square root of 2 is >irrational - i.e. not rational - presumably want either a proof that >proves a stronger statement than the negation, or a proof that they >find more illuminating than the classical proof. Yes. The typical strengthening of an irrationality result is an explicit lower bound on the difference between the number (the square root of 2 in this case), and given candidates p/q for comparison with it. These are known as diophantine approximation results. Here is a quick argument: |sqrt(2)-p/q| = |2-p^2/q^2| / |sqrt(2)+p/q| = |2q^2-p^2| / q^2 |sqrt(2)+p/q| Note |2q^2-p^2|>0, as already shown, and is an integer. Hence it is at least 1. >= 1/ q^2 |sqrt(2)+p/q| Now suppose p/q<2 (true for any decent approximation to sqrt(2)!): > 1/ (2+sqrt(2)) q^2. If p/q>=2, then |sqrt(2)-p/q|> 1/(2+sqrt(2))>= 1/(2+sqrt(2)) q^2 anyway, so in any case we get |sqrt(2)-p/q| > 1/(2+sqrt(2))q^2. A somewhat better result is possible for the square root of 2. The exponent on q, however, is the best possible for any number; for any irrational x there are infinitely many rationals p/q for which |x-p/q|<1/q^2. The square root of 2 is, in this sense, one of the worst approximable numbers there is. -- Keith Ramsay Tender victuals === Subject: Radius of convergence I'm having trouble understanding a line in my Complex Variables book : (Part of a proof of a theorem): All sums are from n = 0 to infinity f is analytic on ball about a with radius R, so f(z) = sum [ a_n (z-a)^n ] for |z-a| < R Let F(z) = (z-a) sum [ (a_n / (n+1)) (z-a)^n ] Since lim (1+n)^(1/n) = 1, it follows that this power series F(z) has the same radius of convergence as f(z) Why does it follow that F(z) has same radius of convergence? I figured it has something to do with the formula for radius of convergence 1/R = limsup |b_n|^(1/n), for some series sum b_n (z-a)^n, but in this case, b_n = (a_n / (n+1)), right? Isaac === Subject: Re: Radius of convergence > I'm having trouble understanding a line in my Complex Variables book : > (Part of a proof of a theorem): All sums are from n = 0 to infinity > f is analytic on ball about a with radius R, so f(z) = sum [ a_n (z-a)^n ] > for |z-a| < R > Let F(z) = (z-a) sum [ (a_n / (n+1)) (z-a)^n ] > Since lim (1+n)^(1/n) = 1, it follows that this power series F(z) has the > same radius of convergence as f(z) > Why does it follow that F(z) has same radius of convergence? I figured it > has something to do with the formula for radius of convergence 1/R = limsup > |b_n|^(1/n), for some series > sum b_n (z-a)^n, but in this case, b_n = (a_n / (n+1)), right? So what do you think lim sup (|a_n|/(n+1))^(1/n) is, given that you know lim (1+n)^(1/n) = 1? === Subject: Re: Radius of convergence > I'm having trouble understanding a line in my Complex Variables book : > (Part of a proof of a theorem): All sums are from n = 0 to infinity > f is analytic on ball about a with radius R, so f(z) = sum [ a_n (z-a)^n ] > for |z-a| < R > Let F(z) = (z-a) sum [ (a_n / (n+1)) (z-a)^n ] > Since lim (1+n)^(1/n) = 1, it follows that this power series F(z) has the > same radius of convergence as f(z) > Why does it follow that F(z) has same radius of convergence? I figured it > has something to do with the formula for radius of convergence 1/R = limsup > |b_n|^(1/n), for some series > sum b_n (z-a)^n, but in this case, b_n = (a_n / (n+1)), right? > So what do you think lim sup (|a_n|/(n+1))^(1/n) is, given that you know > lim (1+n)^(1/n) = 1? : Is it true that if a_n, b_n are sequences, then as n ---> infinity, limit ( a_n / b_n) = limit (a_n) ---------- limit (b_n) ? Isaac === Subject: Re: Radius of convergence > But I have one question which the problem involves > : Is it true > that if a_n, b_n are sequences, then as n ---> infinity, > limit ( a_n / b_n) = > limit (a_n) > ---------- > limit (b_n) Looks like you should study calculus before complex analysis. Or at least review your old calculus text. === Subject: Re: Cum Hoc, Ergo Propter Hoc Okay; thank you. As far as the A was concerned, a wakeup of my memory shortly after I sent the original post convinced me that I was a too-enthusiastic memorizer 'way back when. I had pictured the upside-down 'A' as a slice of a circle, of all things. === Subject: Re: JSH: Assocation - Answer to the above three As far as the relevance is concerned, that boigraphical fact could help you gauge how James will react in one of these extended discussions. Yes, I took his word for it - because I met him in a group with at least two vets of the U.S. military, neither of whom called him on it. This isn't proof, but it's good enough for everyday credibility. You can ask him to back it up explicitly if you like. === Subject: Re: JSH: Assocation - Answer to the above three > As far as the relevance is concerned, that boigraphical fact could help you > gauge how James will react in one of these extended discussions. > Yes, I took his word for it - because I met him in a group with at least two > vets of the U.S. military, neither of whom called him on it. This isn't > proof, but it's good enough for everyday credibility. > You can ask him to back it up explicitly if you like. I don't have much doubt that James really is a vet. Some of the things he's said lead me to believe he was as much a screwup in uniform as he has been in sci.math, though of course he wouldn't see it that way. But I don't see that it matters whether he was in the military or not. I don't consider his behavior typical of the military folks I've known, past or present, and suspect other vets would have as little use for him as anyone else here does. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Assocation - Answer to the above three <4%GYc.53087$_h.42993@bignews3.bellsouth.net> <08bZc.2666$H23.25629@newscontent-01.sprint.ca> Discussion, linux) > As far as the relevance is concerned, that boigraphical fact could > help you gauge how James will react in one of these extended > discussions. How? Lots of folks are army veterans. Do they all react similarly in extended discussions? What is the reaction that this extra fact predicts? Since James has been playing his game for eight years, anyone that wants to predict his behavior is better served by looking at the reactions he's given. Whether he's a veteran or not (or a physics student, Southerner, Vanderbilt alum, whatever) gives very little insight into his reaction, even *assuming* that veterans tend to have particular reactions in extended discussions -- which is just poppycock. -- Jesse Hughes But nothing's being Dr. Ullrich is a particular case of something's being such that nothing is it: (Ex)~(Ey)(y = x) -- John Correy on the failings of first order logic === Subject: Re: JSH: Assocation - Answer to the above three === >Subject: Re: JSH: Assocation - Answer to the above three >Message-id: <08bZc.2666$H23.25629@newscontent-01.sprint.caAs far as the relevance is concerned, that boigraphical fact could help you >gauge how James will react in one of these extended discussions. >Yes, I took his word for it - because I met him in a group with at least two >vets of the U.S. military, neither of whom called him on it. This isn't >proof, but it's good enough for everyday credibility. >You can ask him to back it up explicitly if you like. You miss the point. Asking a liar for verification is a waste of time. Did James bring up his terrorist sympathies with his vet buddies present, or does he just do that on the Usenet? How about his advocating overthrow of the Constitution by foreign powers? Do they discuss that over a beer? -- Mensanator Ace of Clubs === Subject: Re: JSH: Assocation - Answer to the above three > As far as the relevance is concerned, that boigraphical fact could help you > gauge how James will react in one of these extended discussions. > Yes, I took his word for it - because I met him in a group with at least two > vets of the U.S. military, neither of whom called him on it. This isn't > proof, but it's good enough for everyday credibility. > You can ask him to back it up explicitly if you like. But do you have a null-test to prove it? === Subject: Re: Time, the need for the lack of it. Okay. I was going to bring up what little I know of physics at this point and note that the only fundamental quantities (at the high-school level) that don't have time as part of them are position and mass. Energy is iffy. So if you are trying to construct a mathematical-physics system without time, that's what you'd be confined to when dealing with standard classical (Newtonian/high school) mechanics. If someone else with a better knowledge of physics than me would like to step in, feel free to. === Subject: Re: Time, the need for the lack of it. > Okay. I was going to bring up what little I know of physics at this point > and note that the only fundamental quantities (at the high-school level) > that don't have time as part of them are position and mass. Energy is iffy. > So if you are trying to construct a mathematical-physics system without > time, that's what you'd be confined to when dealing with standard classical > (Newtonian/high school) mechanics. > If someone else with a better knowledge of physics than me would like to > step in, feel free to. nothing iffy about energy. energy is not a fundamental quantity and very much depends on time. energy = mass * (distance ^ 2) / (time ^ 2) this of course strengthens your point, you would be confined to position and mass. you wouldn't have access to energy. also, note that this fool is a crackpot and not seriously meaning what he says. === Subject: Re: Time, the need for the lack of it. > Okay. I was going to bring up what little I know of physics at this point > and note that the only fundamental quantities (at the high-school level) > that don't have time as part of them are position and mass. Energy is iffy. > So if you are trying to construct a mathematical-physics system without > time, that's what you'd be confined to when dealing with standard classical > (Newtonian/high school) mechanics. > If someone else with a better knowledge of physics than me would like to > step in, feel free to. At the age of 57, people are OLD, how old depends on their minds view of age. This old fart is still young. By reason of this, being still of youth, IT IS my DUTY to ask, Might I have more,Sir. Beings that use Math, as a way of life, or no more,or no less than others. All people tend to view what they do as a most nessary event, the more complicated their suronding can be made, the less others can understand, the more nessary they become. Simple, to me is the purest form of any problem. I, will always be saddened by mankind inability to form a bond with his surronds. It may be a real nice thing to have knowledge, but to what purpose this(knowledge) is used, shows how little man as a useful spicies, exist. When speech was still a form of grunts and hand signs, man could understand. As inventions appeared to allow man to converse with out presence of being, man had no way to understand one of the basic nessitys of speech, body lauange. Touch, sight, and that sence that raises the hairs on your neck for no apparent reason are most nessary to understand thoes to which we speak. Look not to reason for the answers, look not to God to answer you queries, look instead to yourself. If your answer can not be found within, then communicate your problem to others, ALL OTHERS. To do so may require your speech to be grunts and handsigns so all can realize what your problem is, but if you chose to ignor the abilitys of ALL then you have lost the use of a basic human need, to communicate. The wineo on the park bench, may if fact have the knowledge you seek,IF, you can relay your needs to him in a manner he, the wineo, can understand. I ask thoes that would use terms and words,yes, even numbers to simplfy thier speech to afford, knowledge of others,usefulness. If their was indeed, time, I would have just run out of it. === Subject: Re: Time, the need for the lack of it. === Subject: Characterizations of Borel Sets I'm about to begin my undergraduate thesis and I found a problem I would like to work on. I would like to prove or find a counter-example to the conjecture that If f is a holomorphic Borel function (in the sense that it maps Borel sets to Borel sets) and P is any polynomial, then f + P is a Borel function. Unfortunately, the characterizations of Borel sets I have seen are rather unwieldly--elements of the intersection of all the sigma-algebras generated by the topology on C. A quick look at this characterization shows me that every open and every closed set is Borel. As are singleton sets. And half-open intervals. And unions of things like these. I've looked in a bunch of books on measure theory and Rudin's Real and Complex Analysis but this is the only characterization I've found. If anyone has references to other characterizations, I would really appreciate them. Anything would be useful -- I'm trying to gain intuition on the topic. 'cid 'ooh === Subject: Re: Characterizations of Borel Sets >I'm about to begin my undergraduate thesis and I found a problem I >would like to work on. I would like to prove or find a >counter-example to the conjecture that If f is a holomorphic Borel >function (in the sense that it maps Borel sets to Borel sets) and P is >any polynomial, then f + P is a Borel function. >Unfortunately, the characterizations of Borel sets I have seen are >rather unwieldly--elements of the intersection of all the >sigma-algebras generated by the topology on C. A quick look at this >characterization shows me that every open and every closed set is >Borel. As are singleton sets. And half-open intervals. And unions >of things like these. >I've looked in a bunch of books on measure theory and Rudin's Real >and Complex Analysis but this is the only characterization I've >found. If anyone has references to other characterizations, I would >really appreciate them. Anything would be useful -- I'm trying to >gain intuition on the topic. >'cid 'ooh Have you seen the Canton-Granados-Pommerenke papers in === Subject: Re: Characterizations of Borel Sets >I'm about to begin my undergraduate thesis and I found a problem I >would like to work on. I would like to prove or find a >counter-example to the conjecture that If f is a holomorphic Borel >function (in the sense that it maps Borel sets to Borel sets) and P is >any polynomial, then f + P is a Borel function. >Unfortunately, the characterizations of Borel sets I have seen are >rather unwieldly--elements of the intersection of all the >sigma-algebras generated by the topology on C. A quick look at this >characterization shows me that every open and every closed set is >Borel. As are singleton sets. And half-open intervals. And unions >of things like these. >I've looked in a bunch of books on measure theory and Rudin's Real >and Complex Analysis but this is the only characterization I've >found. If anyone has references to other characterizations, I would >really appreciate them. Anything would be useful -- I'm trying to >gain intuition on the topic. >'cid 'ooh > Have you seen the Canton-Granados-Pommerenke papers in Yeah, that's where I got the problem. :-) I'm going to look into the list of references tonight though. 'cid 'ooh === Subject: Re: Characterizations of Borel Sets >I'm about to begin my undergraduate thesis and I found a problem I >would like to work on. I would like to prove or find a >counter-example to the conjecture that If f is a holomorphic Borel >function (in the sense that it maps Borel sets to Borel sets) and P is >any polynomial, then f + P is a Borel function. >Unfortunately, the characterizations of Borel sets I have seen are >rather unwieldly--elements of the intersection of all the >sigma-algebras generated by the topology on C. A quick look at this >characterization shows me that every open and every closed set is >Borel. As are singleton sets. And half-open intervals. And unions >of things like these. >I've looked in a bunch of books on measure theory and Rudin's Real >and Complex Analysis but this is the only characterization I've >found. If anyone has references to other characterizations, I would >really appreciate them. Anything would be useful -- I'm trying to >gain intuition on the topic. >'cid 'ooh The following may help. C is the complex plane with its usual topology. N^N is the space of irrationals (topologically equivalent to the countable infinite product of countable infinite discrete spaces). For a subset B of C, the following are equivalent: + B is Borel in C. + B is a continuous image of N^N, and so is the complement of B in C. + B is a continuous image of a complete separable metric space, and so is the complement of B in C. To find more, look for references on descriptive set theory. === Subject: Re: Characterizations of Borel Sets > Unfortunately, the characterizations of Borel sets I have seen are > rather unwieldly--elements of the intersection of all the > sigma-algebras generated by the topology on C. A quick look at this > characterization shows me that every open and every closed set is > Borel. As are singleton sets. And half-open intervals. And unions > of things like these. > I've looked in a bunch of books on measure theory and Rudin's Real > and Complex Analysis but this is the only characterization I've > found. If anyone has references to other characterizations, I would > really appreciate them. Anything would be useful -- I'm trying to > gain intuition on the topic. Almost certainly completely irrelevant - I know almost nothing about the subject - but suppose for simplicity one takes a compact space, say the torus X = R/Z. Let C(X) be the Borel space of continuous functions on X with the norm sup(|f|). Define a measure on X to be a continuous linear functions m: C(X) -> R. (This is the Bourbaki definition of measure, IIRC.) Then any Borel set is measurable with respect to any such measure. But is the converse true? If a subset S < X is measurable with respect to all such measures, does it follow that S is a Borel set? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Characterizations of Borel Sets boundary=------------060304010308060805080907 -------------------------------------------------------------- ------- >Unfortunately, the characterizations of Borel sets I have seen are >rather unwieldly--elements of the intersection of all the >sigma-algebras generated by the topology on C. A quick look at this >characterization shows me that every open and every closed set is >Borel. As are singleton sets. And half-open intervals. And unions >of things like these. >I've looked in a bunch of books on measure theory and Rudin's Real >and Complex Analysis but this is the only characterization I've >found. If anyone has references to other characterizations, I would >really appreciate them. Anything would be useful -- I'm trying to >gain intuition on the topic. Almost certainly completely irrelevant - >I know almost nothing about the subject - >but suppose for simplicity one takes a compact space, say the torus X = R/Z. >Let C(X) be the Borel space of continuous functions on X >with the norm sup(|f|). >Define a measure on X to be a continuous linear functions m: C(X) -> R. >(This is the Bourbaki definition of measure, IIRC.) >Then any Borel set is measurable with respect to any such measure. >But is the converse true? >If a subset S < X is measurable with respect to all such measures, >does it follow that S is a Borel set? No, the converse is not true. There are non-Borel sets in the unit interval [0,1] that are universally measurable. (If you assume the Axiom of Choice and other usual set theory assumptions.) === Subject: Re: Characterizations of Borel Sets > Almost certainly completely irrelevant - agreed > I know almost nothing about the subject - > but suppose for simplicity one takes a compact space, say the torus X = R/Z. > Let C(X) be the Borel space of continuous functions on X > with the norm sup(|f|). > Define a measure on X to be a continuous linear functions m: C(X) -> R. > (This is the Bourbaki definition of measure, IIRC.) > Then any Borel set is measurable with respect to any such measure. > But is the converse true? > If a subset S < X is measurable with respect to all such measures, > does it follow that S is a Borel set? No, such sets are called universally measurable and there are non-Borel universally measurable sets. So-called analytic sets, for example. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Characterizations of Borel Sets If I remember and understood something mentioned in passing by Henry Helson long ago, you can generate the Borel sets by inductively taking countable unions and intersections through the ordinals. I guess he meant the following. Let B0 a subbase for the topology on X. For any ordinal a, let B_(a+1) = the collection of all countable unions and all countable interections of sets in B_a. If a is a limit ordinal, let B_a = U{B_b: b If I remember and understood something mentioned in passing by Henry > Helson long ago, you can generate the Borel sets by inductively taking > countable unions and intersections through the ordinals. I guess he > meant the following. > Let B0 a subbase for the topology on X. > For any ordinal a, let B_(a+1) = the collection of all countable > unions and all countable interections of sets in B_a. > If a is a limit ordinal, let B_a = U{B_b: b Then U{B_a in P(P(X)): a is an ordinal} is precisely the collection > of Borel sets. > This might require second countability. Any experts care to verify? > I don't know if this will help the OP. Not quite right; B0 must be larger. If X is second countable, let B0 be the union of a subbase and the collection of complements of the sets in this subbase. (The sufficiency of this B0 requires at least countable choice.) Otherwise, let B0 be the collection of sets which are either open or closed. I think when working in the reals, you can let B0 be the set of all intervals (a, b] where a and b are rational. Note that B is an increasing sequence (over the ordinals) which is eventually constant. Thus, there exists an ordinal a such that the B_a is the collection of Borel sets. I wonder if we can explictly determine the minimum value of this a for the reals. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Characterizations of Borel Sets > If I remember and understood something mentioned in passing by Henry > Helson long ago, you can generate the Borel sets by inductively > taking countable unions and intersections through the ordinals. I > guess he meant the following. > Let B0 a subbase for the topology on X. > For any ordinal a, let B_(a+1) = the collection of all countable > unions and all countable interections of sets in B_a. > If a is a limit ordinal, let B_a = U{B_b: b Then U{B_a in P(P(X)): a is an ordinal} is precisely the collection > of Borel sets. > This might require second countability. Any experts care to verify? > I don't know if this will help the OP. > Not quite right; B0 must be larger. If X is second countable, > let B0 be the union of a subbase and the collection of complements > of the sets in this subbase. (The sufficiency of this B0 requires > at least countable choice.) Otherwise, let B0 be the collection of > sets which are either open or closed. > I think when working in the reals, you can let B0 be the set of all > intervals (a, b] where a and b are rational. > Note that B is an increasing sequence (over the ordinals) which is > eventually constant. Thus, there exists an ordinal a such that the > B_a is the collection of Borel sets. I wonder if we can explictly > determine the minimum value of this a for the reals. The minimum a is OMEGA (the first uncountable ordinal). http://www.numdam.org/item?id=AIF_1974__24_4_47_0 === Subject: Re: Set theory question .... > Read this question somewhere and couldn't figure out how to arrive at the > answer: > At a survey, > a. 70% respondents said that they watch TV serial 'A' > b. 75% respondents said that they watch TV serial 'B' > c. 80% respondents said that they watch TV serial 'C' > d. 85% respondents said that they watch TV serial 'D' > What is the minimum percentage of respondents that watch all the four TV > serials. > The answer has been stated as 10%. > TIA > Raquel. > HINT1: > How many respondents said they DIDN'T watch 'A' ? > How many respondents said they DIDN'T watch 'B' ? > How many respondents said they DIDN'T watch 'C' ? > How many respondents said they DIDN'T watch 'D' ? > What arrangement gives you the minimum number that watch all four channels? > HINT2: > Do the same question, but with only serials A and B. (Answer is 45%). (the bonehead I am) don't know *why*. Taking cue from your HINT2, I believe the way it works is: - 30% respondents didn't watch 'A'. - 25% respondents didn't watch 'B'. Respondents that watch both 'A' and 'B' = 100 - (30+25) = 45 But I don't understand the logic behind the above method of arriving at the solution. Would you please explain it to me. I really want to get into the real logic behind this. Raquel. === Subject: Cofinal Totally Ordered Subsets of Directed Posets I was wondering how one could characterize those partially ordered directed sets that admit totally ordered cofinal subsets. Or do they all? Stefan Waner === Subject: Re: Binary degrees? > I have a question which keeps me busy for some time. Well, for math experts > like you my question is just a piece of cake. > As we all know, we use Radians, Degrees and Grad to express the size of > angles. > In computers we use hexadecimal bytes which does not 'fit' these units at > all.. > Now my question; wouldn't it be easier to use in computers a sort of Binary > Degree (BDeg) in which, let's say 2*pi = 255 or X'FF BDeg. > Would this make sin, cos etc. calculations easier using tablesearches, > Taylor expansions or whatever? I have to disagree with the other responses. I've used exactly this sort of scheme in real-time signal processing applications that used integer arithmetic for speed. > I am specially interested in this in 8-bit microcontroller applications. I > have the feeling that this would make things easier. But does it really? Depends on how expensive floating point operations are, and how often they have to be done. Designing just-precise-enough integer arithmetic involves a large investment in programming time, but can give you huge time savings if that is an issues. When I was doing this work, it was. I remember poring over code trying to shave a couple of microseconds off assembly-language loops. - Randy === Subject: diophantine equation -- now rational hi, i believe that if we have AX + BY = 1, and A and B are given irrational numbers, there are either 0 or 1 solutions (X, Y) in the rationals. How can we go about finding the 1 solution in the case where it exists. any insight would be appreciated. === Subject: Re: diophantine equation -- now rational === Subject: diophantine equation -- now rational >hi, i believe that if we have >AX + BY = 1, and A and B are given irrational numbers, >there are either 0 or 1 solutions (X, Y) in the rationals. Notice when there is a solution, abxy /= 0. Let ax + by = 1 = au + bv be two solutions. a = (1 - by)/x = 1/x - by/x a = 1/u - bv/u 1/x - by/x = 1/u - bv/u 1/x - 1/u = b(y/x - v/u) A rational equals a rational product of an irrational? No. Thus y/x - v/u = 0 1/x - 1/y = 0 x = u, y = v >How can we go about finding the 1 solution in the case where it As ax + by = 1 there are rational numbers r,s with a = r + bs Thus 1 = ax + by = rx + bsx + by = rx + b(sx + y) and 1 = rx, 0 = sx + y Hence x = 1/r, y = -sx = -s/r In other words, first produce the example. Pick any b in RQ and r,s in Q, s /= 0 for a linearly related a in RQ. ---- === Subject: Re: diophantine equation -- now rational > hi, i believe that if we have > AX + BY = 1, and A and B are given irrational numbers, there are either 0 or 1 > solutions (X, Y) in the rationals. How can we go about finding the 1 solution > in the case where it exists. any insight would be appreciated. Use LLL, PSLQ, or Ferguson-Forcade algorithm. There is info on the WWW about these. Here is one link to get you started: http://mathworld.wolfram.com/PSLQAlgorithm.html === Subject: parabola - conic section hi again, I was just reading a book on conic sections. How can one prove that if cone is sliced by a plane parallel to a lateral side of the cone, we get a parabola? ... === Subject: Re: parabola - conic section >I was just reading a book on conic sections. How can one prove that if cone is >sliced by a plane parallel to a lateral side of the cone, we get a parabola? Take the equation of a right cone x^2 + y^2 - z^2 = 0 [1] and rotate on the y axis by 45 degrees using the substitution x = (u-v)/sqrt(2) z = (u+v)/sqrt(2) to get (u^2 - 2uv + v^2)/2 + y^2 - (u^2 + 2uv + v^2)/2 = 0 so that y^2 = 2uv [2] The planes we are interested in are those where v = constant. Equation [2] is a parabola. Rob Johnson take out the trash before replying === Subject: Re: parabola - conic section > I was just reading a book on conic sections. How can one prove that if cone is > sliced by a plane parallel to a lateral side of the cone, we get a parabola? Let pi be some fixed plane. Let D be some fixed line in pi. Let F be some fixed point in pi, but not in D. A parabola may be defined as the locus of a point, in pi, which moves so that its distance from F is equal to its distance from D. F is called the focus of the parabola. D is called the directrix of the parabola. We may use a Dandelin sphere to prove that if a cone, C, is intersected by a plane, pi, which is parallel to one of the generators, G, of C, then the curve of intersection is a parabola. Let S be a sphere (a Dandelin sphere) inscribed in C and touching pi. [I will not prove the existence (or uniqueness) of such a sphere.] Let delta be the plane in which S touches C. Let D be the line of intersection of delta and pi. Let F be the point at which S touches pi. Let the vertex of the cone be the point V. Let A be the point at which G (the generator of C to which pi is parallel) intersects delta. Let L be the line in pi, and through F, which is parallel to G. Let B be the point of intersection of L and D. Clearly, angle VAB = angle ABF (since VA [the line G] is parallel to BF [the line L]). Let us call this angle t. Clearly, all the generators of C meet delta at angle t. Let P be any point on the intersection of C and pi. Let the generator of C which passes through P intersect delta at Q. Let the foot of the perpendicular from P to delta be the point R. Let the foot of the perpendicular from P to D be the point C. Clearly, the plane PCR is parallel to the plane FBA and so the angle PCR is equal to t. Hence the triangles PCR and PQR are congruent (they are both right-angled with a common side PR and equal corresponding angles [since angle PQR = t = PCR]). So PQ=PC. Also, PF=PQ (they are both tangents from P to S, and hence of equal length). Hence, PF=PC. So, the distance from P to F is the same as the distance from P to D. Thus, P lies on a parabola with focus F and directrix D. Q.E.D. ~~~~ For info on Dandelin, see: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Dandelin.html http://scienceworld.wolfram.com/biography/Dandelin.html For a slightly whimsical take on all this, see: http://www.clowder.net/hop/Dandelin/Dandelin.html by Hop David -- Clive Tooth http://www.clivetooth.dk === Subject: Re: parabola - conic section > I was just reading a book on conic sections. > How can one prove that if cone is sliced by a > plane parallel to a lateral side of the cone, > we get a parabola? It depends on how much you want to assume. In a simple case, a cone in 3-space has equation z^2 = x^2+y^2. Its upper nappe (one cone with vertex down) has equation z = sqrt(x^2+y^2). In 3-space, one plane parallel to a lateral side has equation z = y + c, where c > 0. To find an equation of the interesction eliminate in the system: z = sqrt(x^2+y^2) z = y + c. We get sqrt(x^2+y^2) = y+c. Squaring both sides gives us: x^2+y^2 = (y+c)^2 x^2+y^2 = y^2+2cy+c^2 x^2 = 2cy + c^2 y = (1/2c)(x^2-c^2) This last equation may be identifiable to you as an equation of a parabola. === Subject: Re: JSH: Math proofs <87vfeze6cf.fsf@phiwumbda.org> Discussion, linux) >Seems to me that we're not really avoiding assumptions in natural >deduction so much as moving to a sequent-style calculus. I'm not sure >that this does anything for us. > Well, yes, sorry. > A ->formal<- proof, defined as a finite sequence etc., will > necessarily begin with either a tautology or an axiom. I don't understand this claim either. There are formal systems of natural deduction. In these formal systems, a proof my begin with an assumption. I suspect that when you learned logic, you just didn't do natural deduction. If one wants to study logic, then natural deduction is an annoying formalization to use --- all those damn subproofs with their assumptions make for special cases in proving things. But natural deduction is a name for a class of related *formal* logics in which assumptions and assumption-discharging rules play a role. The sequent calculi are much handier for the logician and this could be what you learned. In that case, the transformation T,X |- Y -------- T |- X -> Y (perhaps written T,X => Y |- T => X -> Y ) is an explicit rule of inference. [...] > The proof at issue originally here, the proofs contained in Wiles's > paper, are not formal proofs. And most of them do not begin with a > true statement. Theorems 0.2 and 0.3, for example, begin with > assumptions, since they are proofs of A->B via the usual method method > of assuming A and deducing B, without ever invoking the Deduction > Metatheorem (which is unnecessary outside of formal logic, for the > most part). The proof of Theorem 5.2, that all semistable elliptic > curves over Q are modular, begins with Suppose that E is a semistable > elliptic curve over Q, which is certainly not a true statement. Indeed. -- Destiny is a funny thing. Once I thought I was destined to become Emperor of Greenland, sole monarch over its 52,000 inhabitants. Then I thought I was destined to build a Polynesian longship in my garage. I was wrong then, but I've got it now. -- The Tick === Subject: Re: JSH: Math proofs >Seems to me that we're not really avoiding assumptions in natural >deduction so much as moving to a sequent-style calculus. I'm not sure >that this does anything for us. > Well, yes, sorry. It's still valid, provided that you turn P -> Q into (~P v Q). But it does take some scribbling to work through. [Incidentally this transform is the basis behind W. Dale Hall's note that If F, then T and If F, then F are both valid.] === Subject: Re: Maths & finance exercise about forward contrat > I found in a book the exercise and correction below. > I don't see why I'm wrong. Do you? > Eric > **Exercise: > Suppose that A(0)=100 and A(1)=105 dollars, the present price of pound So you have a 5% interest rate > sterling is S(0)=1.6 dollars, and the forward price is F=1.50 dollars And an appreciation rate of less than 0% on sterling (but more than -10%). So set up two risk-free arbitrage opportunities (as you did below), but you can pretty much see that investing money can't yield more than 5% or less than -10%. So it's up to you to go and see what equation was wrong and why. Another recommendation is to put labels (units) on everything. Then you cancel units -- and that will tell you where your mistake is. Because you'll have and equation with something like (weeks)^2= ft. Doesn't make any sense at all. Jon Miller > to a pound with delivery date 1. > How much should a sterling bond cost today if it promises to pay 100 > at time 1? > Hint: the forwad contract is based on an asset involving negative > carrying costs (the interest earnd by investing in sterling bonds > **correction: > suppose that a sterling bond promising to pay 100 at time 1 for 1.5$ > to a pound is selling for x pounds at time 0. To find x consider the > following startegy. > At time 0: > Borrow 1.6 x dollars and change the sum into x pounds. > Purchase a sterling bond for x pounds. > Take a short forward position to sell 100 for $1.5 to a pound with > delivery date 1. > Then, at time 1: > Cash the bond, collecting 100 > Close the short forward position by selling 100 for $1.5 > Repay the cash loan with interest, that is, 1.68x dollars in total > The balance of this transaction is 150 - 1.68x dollars, which must be > equal to zero or else an arbitrage opportunity would arise. It follows > that sterling bond promising to pay 100 at time 1must sell for > x=150/1.68=89.29 pounds at time 0. > **What I have done: > suppose that a sterling bond promising to pay 100 at time 1 for 1.5$ > to a pound is selling for x pounds at time 0. To find x consider the > following startegy. > At time 0: > Borrow 100 dollars and change the sum into 100/1.6 pounds. > Purchase a x/(100/1.6)=1.6x/100 sterling bond for 100/1.6 pounds. > Take a short forward position to sell (100*1.6x/100 ) for $1.5 to a > pound with delivery date 1. > Then, at time 1: > Cash the bond, collecting (100*1.6x/100)=1.6x > Close the short forward position by selling 1.6x for $1.5 > Repay the cash loan with interest, that is, 105 dollars in total > The balance of this transaction is 1.6x*1.5 - 105 dollars, which must > be equal to zero or else an arbitrage opportunity would arise. It > follows that sterling bond promising to pay 100 at time 1must sell > for x=105/(1.6*1.5)=43.75 pounds at time 0. === Subject: Isomorphism between (Z_2^k)^* and Z_2 x Z_2^(k-2) An isomorphism from Z_2 x Z_2^(k-2) onto (Z_2^k)^* for k >= 3, is (s,n) -> (-1)^s * 5^n Lemma. In (Z_2^k)^*, the order of 5 = o(5) = 2^(k-2) This assures the map is well defined. Details left to the sagacious reader. ---- === Subject: Re: The real numbers, and general comments > It is because R is essentially a geometic concept. The inspiration to > construct all of R came solely from geometric intuitions of > continuity, not from analysis. > No, R can be thought of as a geometric concept. It need not be thought > of that way. There is a single concept, but multiple models. Each > model that correctly represents the topological structure of R is > acceptable. The real number line is one such model. Dedekind cuts are > another. Show me the set of all Dedekind cuts, or all Cauchy sequences. > When we say that R is uncountable (since uncountable sets are > impossible) whaat we are really saying is that a line (modeled by R) > is not made up of points, because it can not be covered with any set > of points (Cantor's argument). > If uncountable sets were impossible, we wouldn't say R is uncountable. > The rest of your above sentence makes no sense. I am saying that the continuous (the line) can not and should not be modeled by the discrete (points). R is not a set of numbers but a concept. > But, still, the zeroes of any function can be found, and certainly are > numbers. This provides the method by which we can extend Q. And, as I > stated in 'Uncountable sets in CZF', the union of all possible such > field extensions of Q, is the computable numbers. So we can rightly > say that the computables are all the NUMBERS that exist. > No we can't. There are provably numbers that are not computable. You > don't have to like them, but they do exist. In particular, there are > only countably many computable numbers, so there must be real numbers > that are not computable since there are uncountably many reals. No, you can't diagonalise on the computables, S, because S itself is not computable. Hence you can't construct a number that isn't computable. I see no reason to consider a thing that can't be constructed to be a number. Also, this 'procedure' can't show that there are uncountably many 'numbers'. > You are right about the abstraction. Whether it is ugly or not is a > matter of interpretation. If you find it ugly, perhaps you should work > on something else. The only justification for pure mathematics is is intellectual beauty. > For example, projective geometry is one of the most beautiful > discoveries in math. It is so obviously the correct theory for > Euclidean geometry, that one can not deny that Apollonius and > Archimedes would have adopted it, had they known. Yet, the workers of > the past century have made even this ugly; pick up a modern book of > projective geometry and all you will see is dense, abstract formulae. > Projective geometry is no more correct than any other. In particular, > projective geometry *is not* Euclidean geometry. They are different. Euclidean geometry naturally embeds in projective geometry. I perceive metric and projective techniques is simply different methods of proving results in geometry, not as different fields. > I have also observed that this trend does not seem to have affected > and thus properly think about the value of their work. Modern math has > made this impossible; I know that they would reply that this is > because mathematics is more intellectually sophisticated. But, as Alan > Sokal so memorably demonstrated, certain fields of the humanities are > intellectually vacuous despite (or because of?) their esoteric jargon. > Could this not be partially true of mathematics today? > Math is a tool of the sciences. That does not mean it is a science in > the same sense that physics/chemistry/etc are. No, math is not a science but is closer allied to philosophy. I wouldn't doubt that modern philosophy suffers from the same defect. Andrew Usher === Subject: Re: The real numbers, and general comments >It is because R is essentially a geometic concept. The inspiration to >construct all of R came solely from geometric intuitions of >continuity, not from analysis. >No, R can be thought of as a geometric concept. It need not be thought >of that way. There is a single concept, but multiple models. Each >model that correctly represents the topological structure of R is >acceptable. The real number line is one such model. Dedekind cuts are >another. > Show me the set of all Dedekind cuts, or all Cauchy sequences. I can show you how to construct any particular Dedekind cut, but I don't have time to type out uncountably many countably infinite sets. If you mean show you the set-builder notation for them, look here: http://en.wikipedia.org/wiki/Dedekind_cut >When we say that R is uncountable (since uncountable sets are >impossible) whaat we are really saying is that a line (modeled by R) >is not made up of points, because it can not be covered with any set >of points (Cantor's argument). >If uncountable sets were impossible, we wouldn't say R is uncountable. >The rest of your above sentence makes no sense. > I am saying that the continuous (the line) can not and should not be > modeled by the discrete (points). R is not a set of numbers but a > concept. That approach makes calculus/analysis somewhat tricky to work with. Perhaps if you elaborate a little I'll be able to see how you plan to do the standard work with R when it is not viewed as numbers. Or would you rather have us work strictly with Dedekind cuts, etc? >But, still, the zeroes of any function can be found, and certainly are >numbers. This provides the method by which we can extend Q. And, as I >stated in 'Uncountable sets in CZF', the union of all possible such >field extensions of Q, is the computable numbers. So we can rightly >say that the computables are all the NUMBERS that exist. >No we can't. There are provably numbers that are not computable. You >don't have to like them, but they do exist. In particular, there are >only countably many computable numbers, so there must be real numbers >that are not computable since there are uncountably many reals. > No, you can't diagonalise on the computables, S, because S itself is > not computable. Hence you can't construct a number that isn't > computable. I see no reason to consider a thing that can't be > constructed to be a number. Also, this 'procedure' can't show that > there are uncountably many 'numbers'. S is enumerable, which is sufficient to demonstrate a number that is enumerable but not computable. >You are right about the abstraction. Whether it is ugly or not is a >matter of interpretation. If you find it ugly, perhaps you should work >on something else. > The only justification for pure mathematics is is intellectual beauty. My justification for pure mathematics is that it's fun. Whether it has intellectual beauty is a matter of taste. I've seen some proofs that I thought were wonderfully elegant, and others that are horribly inelegant. Both types of proof work, however, which is the most important aspect. >For example, projective geometry is one of the most beautiful >discoveries in math. It is so obviously the correct theory for >Euclidean geometry, that one can not deny that Apollonius and >Archimedes would have adopted it, had they known. Yet, the workers of >the past century have made even this ugly; pick up a modern book of >projective geometry and all you will see is dense, abstract formulae. >Projective geometry is no more correct than any other. In particular, >projective geometry *is not* Euclidean geometry. They are different. > Euclidean geometry naturally embeds in projective geometry. I perceive > metric and projective techniques is simply different methods of > proving results in geometry, not as different fields. Projective geometry has the following axioms for the projective plane: 1. Given any two distinct points, there is exactly one line incident with both of them. 2. Given any two distinct lines, there is exactly one point incident with both of them. 3. There are four points such that no line is incident with more than two of them. 2. is a direct contradiction of the parallel postulate in Euclidean plane geometry. How do you figure they are the same? >I have also observed that this trend does not seem to have affected >and thus properly think about the value of their work. Modern math has >made this impossible; I know that they would reply that this is >because mathematics is more intellectually sophisticated. But, as Alan >Sokal so memorably demonstrated, certain fields of the humanities are >intellectually vacuous despite (or because of?) their esoteric jargon. >Could this not be partially true of mathematics today? >Math is a tool of the sciences. That does not mean it is a science in >the same sense that physics/chemistry/etc are. > No, math is not a science but is closer allied to philosophy. I > wouldn't doubt that modern philosophy suffers from the same defect. Having studied philosophy, I can't agree with this statement, either. It does offer insight into your approach to math, however. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments > It is because R is essentially a geometic concept. The inspiration to > construct all of R came solely from geometric intuitions of > continuity, not from analysis. > No, R can be thought of as a geometric concept. It need not be thought > of that way. There is a single concept, but multiple models. Each > model that correctly represents the topological structure of R is > acceptable. The real number line is one such model. Dedekind cuts are > another. > Show me the set of all Dedekind cuts, or all Cauchy sequences. I'm slightly puzzled as to the exact mathematical usage of the word show here, so could you give an example of a set that you can show, and the rules that we can follow that mean you're happy that we have indeed shown something. === Subject: Re: Help needed for solving this differential equation === >Subject: Help needed for solving this differential equation [(y-1)^2 * (n^2-1)+ n^2*n^x](dy/dx)^2 - 2*x*(y-1)(dy/dx) - x^2 = 0 [(y-1)^2 (n^2-1) + n^2 n^x] y - 2x(y-1)y' - x^2 = 0 > I would be inclined to first solve for dy/dx, using the quadratic > formula. Of course, that gives an awfully complicated result. > [(y-1)^2 * (n^2-1) + n^2 * n^x] * ( y' )^2 - 2 * x * (y-1) * y' - x^2 = 0. > if you did substitutions, u = (y-1) and u' = y', then > [u^2 * (n^2-1) + n^2 * n^x] * (u')^2 - 2xu u' - x^2 = 0. let v = [u^2 (n^2 - 1) + n^2 n^x] u' = (2xu +- sqr(4x^2 + 4vx^2)) / 2v = x(u +- sqr(1 + v) / v = yicks === Subject: Re: Uncountable sets in CZF? R^V[G] not in R^V, that is R, the set of all real numbers. >Let the set F of forcing conditions be the set of all functions p whose >domain is a finite subset of N and whose range is a subset of {0,1}, i.e. >F = {p : p is a function, dom(p) is a finite subset of N, range(p) is a >subset of {0,1}}. Define q <= p (q is stronger than p) if q extends p, >i.e. p is a subset of q, i.e. dom(p) is a subset of dom(q) and p(n) = q(n) >for all n in dom(p). It follows that p and q are compatible iff >p(n) = q(n) for all n in the intersection of dom(p) and dom(q). >Let G be a generic set in F (G is not necessarily an element of V). >Let n be an element of N, and let D = {p in F : n in dom(p)}. If q is >a forcing condition, and n is an element of dom(q), then q is stronger >than q and q is an element of D. >If q is a forcing condition and n is not an element of dom(q), let >r = q u {(n,0)}, then r is a function, dom(r) is a finite subset of N >and range(r) is a subset of {0,1}, and so r is a forcing condition, >r is stronger than q, and n is an element of dom(r), and so r is an >element of D. >It follows that for all forcing conditions q, there exists a forcing >condition r stronger than q such that r is an element of D. So D is >dense. Since G is a generic set, then G intersects D. Let p be a common >element of G and D, then n is an element of dom(p), and so n is an element >of dom(UG) (where, for a set A, UA is the union of A). >Since n is an arbitrary element of N, then dom(UG) = N. >Since range(p) is a subset of {0,1} for all p in G, then range(UG) is a >subset of {0,1}. >Suppose UG is not a function, then there exists n in N such that (n,0) and >(n,1) are both elements of UG. It follows that there exist p in G and >q in G such that p(n) = 0 and q(n) = 1. Since n is an element of the >intersection of dom(p) and dom(q), and p(n) and q(n) are not equal, then >p and q are incompatible, contradicting the genericity of G. Since the >assumption that UG is not a function leads to a contradiction, then UG is >a function UG : N -> {0,1} in V[G]. >Let A = {n in N : (UG)(n) = 1}, then A is a subset of N in V[G]. >Let B be a subset of N in V, and define D = {p in F : exists n in dom(p) >((n in B and p(n) = 0) or (n not in B and p(n) = 1))}. If q is a forcing >condition and q is an element of D, then q is stronger than q and q is an >element of D. >Suppose q is a forcing condition and q is not an element of D, then >for all n in dom(q) ((n in B and q(n) = 1) or (n not in B and q(n) = 0)). >Since dom(q) is finite, then N-dom(q) is nonempty. Let m be an element >of N-dom(q), and define r = q u {(m,0)} if m is an element of B, and >r = q u {(m,1)} if m is not an element of B. Then r is a function, >dom(r) is a finite subset of N, and range(r) is a subset of {0,1}, >so that r is a forcing condition, r is stronger than q, and ((m in B and >r(m) = 0) or (m not in B and r(m) = 1)). It follows that r is an element >of D. >So D is a dense set. Since G is generic, then G intersects D, and so >there exist p in G and n in dom(p) such that ((n in B and p(n) = 0) or >(n not in B and p(n) = 1)). If n is an element of B, then p(n) = 0, so >(UG)(n) = 0, so n is not an element of A. If n is not an element of B, >then p(n) = 1, so (UG)(n) = 1, so n is an element of B. It follows that >n is an element of either A or B, but not both, and so A and B are not >equal. That is, for any subset B of N in V, there is a natural number n >such that n is an element of the symmetric difference of A and B, and so >A and B are not equal. It follows that A is not a subset of N in V, i.e. >A is an element of V[G], but not an element of V. >Let x = sum_{n in N} 2^{-(UG)(n)}, then x is a real number in [0,2] in >V[G], but x is not an element of V, i.e. x is a new real number. >That should have been x = sum_{n in N} 2^{-n (UG)(n)}. That should have been x = sum_{n in N} (UG)(n) 2^{-n}. >Note, by the way, that G is not an element of V. David ----- === Subject: Re: Uncountable sets in CZF? >Unless... we have just proved that there exist no models of ZFC in which >N is the 'real' N and R is the 'real' R, a.l.f.o. > That might be the case. >Sure. >But: using the very same argument, it would not be possible for a model >of ZFC to have an N that is countable-a.l.f.o, and an R-in-V that is >uncountable-a.l.f.o. And i don't think that is true (you actually >I must be missing something, but what? > I was giving it some thought, and I think that it may be that, while the > existence of a generic set or a generic ultrafilter is consistent relative > to ZFC, that does not mean that such a generic set or generic ultrafilter > must exist. And if not, Skolem-Loewenheim is wrong, and there is not bijection between N and R? >Personally, i don't find it disturbing in itself that there are models >of ZFC in which the 'counterparts of real sets' are not the same as the >'real' Platonic ones. I got over that, long time ago. ;-) > On the other hand, the above suggests that the totality of the real > numbers exceeds in cardinality any set in the model under investigation. Which would suggest than ZFC is not capable of constructing uncountable sets. Of course that makes sense to me, given my beliefs about uncountability. Andrew Usher === Subject: Re: Uncountable sets in CZF? >Unless... we have just proved that there exist no models of ZFC in which >N is the 'real' N and R is the 'real' R, a.l.f.o. >That might be the case. >Sure. >But: using the very same argument, it would not be possible for a model >of ZFC to have an N that is countable-a.l.f.o, and an R-in-V that is >uncountable-a.l.f.o. And i don't think that is true (you actually >I must be missing something, but what? >I was giving it some thought, and I think that it may be that, while the >existence of a generic set or a generic ultrafilter is consistent relative >to ZFC, that does not mean that such a generic set or generic ultrafilter >must exist. > And if not, Skolem-Loewenheim is wrong, and there is not bijection > between N and R? If not, then it seems to me that David's statement, made earlier: > (...) given any model V of ZFC, and any two infinite sets A and B in > V, there exists a generic extension V[G] of V in which A and B are > bijective. needs at least some slight modification (in terms of 'consistent to assume', or so, rather than actual existence.) (Otherwise it seems to me that we have just proven ZFC inconsistent. Now i wouldn't be worried proving ZFC inconsistent, but not in these few steps. :-) ) -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? > This is essentially the same as my beliefs. If a bijection can be > constructed between N in R[V] in some model V', then given that the > size of N, and of R[V], has not been changed, that would seem to say > that R[V] is countable, only unprovably so in V. Since we know that > the real R is not countable, that would seem to say that ZFC is > incomplete, or wrong. > But that is already well known for a long time. > There is no way to characterize 'the real model of ZFC' (whatever that > means) with new axioms or so. It was once known as the Skolem-Loewenheim > paradox. I see, ZFC is incomplete. Is this related to Godel's theorem? > But i think the issue i am touching is more serious. > If in some extension, there does exist a bijection between the sets, > and the sets have not received new elements, this necessarily means the > sets are 'really' bijective when looked upon from the outside world. Yes, it certainly seems that set should be considered 'really' bijective if they biject in any model. A logical analogy would be that Wiles's proof of FLT is acceptable within number theory, even though it uses fields outside of number theory. Andrew Usher === Subject: Re: Uncountable sets in CZF? >This is essentially the same as my beliefs. If a bijection can be >constructed between N in R[V] in some model V', then given that the >size of N, and of R[V], has not been changed, that would seem to say >that R[V] is countable, only unprovably so in V. Since we know that >the real R is not countable, that would seem to say that ZFC is >incomplete, or wrong. >But that is already well known for a long time. >There is no way to characterize 'the real model of ZFC' (whatever that >means) with new axioms or so. It was once known as the Skolem-Loewenheim >paradox. > I see, ZFC is incomplete. Is this related to Godel's theorem? The Skolem-Loewenheim theorem/paradox is older (early 1920s). -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? R^V[G] not in R^V, that is R, the set of all real numbers. >Let the set F of forcing conditions be the set of all functions p whose >domain is a finite subset of N and whose range is a subset of {0,1}, i.e. >F = {p : p is a function, dom(p) is a finite subset of N, range(p) is a >subset of {0,1}}. Define q <= p (q is stronger than p) if q extends p, >i.e. p is a subset of q, i.e. dom(p) is a subset of dom(q) and p(n) = q(n) >for all n in dom(p). It follows that p and q are compatible iff >p(n) = q(n) for all n in the intersection of dom(p) and dom(q). >Let G be a generic set in F (G is not necessarily an element of V). >Let n be an element of N, and let D = {p in F : n in dom(p)}. If q is >a forcing condition, and n is an element of dom(q), then q is stronger >than q and q is an element of D. >If q is a forcing condition and n is not an element of dom(q), let >r = q u {(n,0)}, then r is a function, dom(r) is a finite subset of N >and range(r) is a subset of {0,1}, and so r is a forcing condition, >r is stronger than q, and n is an element of dom(r), and so r is an >element of D. >It follows that for all forcing conditions q, there exists a forcing >condition r stronger than q such that r is an element of D. So D is >dense. Since G is a generic set, then G intersects D. Let p be a common >element of G and D, then n is an element of dom(p), and so n is an element >of dom(UG) (where, for a set A, UA is the union of A). >Since n is an arbitrary element of N, then dom(UG) = N. >Since range(p) is a subset of {0,1} for all p in G, then range(UG) is a >subset of {0,1}. >Suppose UG is not a function, then there exists n in N such that (n,0) and >(n,1) are both elements of UG. It follows that there exist p in G and >q in G such that p(n) = 0 and q(n) = 1. Since n is an element of the >intersection of dom(p) and dom(q), and p(n) and q(n) are not equal, then >p and q are incompatible, contradicting the genericity of G. Since the >assumption that UG is not a function leads to a contradiction, then UG is >a function UG : N -> {0,1} in V[G]. >Let A = {n in N : (UG)(n) = 1}, then A is a subset of N in V[G]. >Let B be a subset of N in V, and define D = {p in F : exists n in dom(p) >((n in B and p(n) = 0) or (n not in B and p(n) = 1))}. If q is a forcing >condition and q is an element of D, then q is stronger than q and q is an >element of D. >Suppose q is a forcing condition and q is not an element of D, then >for all n in dom(q) ((n in B and q(n) = 1) or (n not in B and q(n) = 0)). >Since dom(q) is finite, then N-dom(q) is nonempty. Let m be an element >of N-dom(q), and define r = q u {(m,0)} if m is an element of B, and >r = q u {(m,1)} if m is not an element of B. Then r is a function, >dom(r) is a finite subset of N, and range(r) is a subset of {0,1}, >so that r is a forcing condition, r is stronger than q, and ((m in B and >r(m) = 0) or (m not in B and r(m) = 1)). It follows that r is an element >of D. >So D is a dense set. Since G is generic, then G intersects D, and so >there exist p in G and n in dom(p) such that ((n in B and p(n) = 0) or >(n not in B and p(n) = 1)). If n is an element of B, then p(n) = 0, so >(UG)(n) = 0, so n is not an element of A. If n is not an element of B, >then p(n) = 1, so (UG)(n) = 1, so n is an element of B. It follows that >n is an element of either A or B, but not both, and so A and B are not >equal. That is, for any subset B of N in V, there is a natural number n >such that n is an element of the symmetric difference of A and B, and so >A and B are not equal. It follows that A is not a subset of N in V, i.e. >A is an element of V[G], but not an element of V. >Let x = sum_{n in N} 2^{-(UG)(n)}, then x is a real number in [0,2] in >V[G], but x is not an element of V, i.e. x is a new real number. That should have been x = sum_{n in N} 2^{-n (UG)(n)}. Note, by the way, that G is not an element of V. David ----- === Subject: Re: Uncountable sets in CZF? <2pg9j4FklatcU2@uni-berlin.de A fact which could be considered disturbing is that for any > transfinite cardinal not cofinal with omega, there exists a model of > ZFC in which the cardinality of R is the given cardinal. > Sorry to bother you once more, but... can you provide a short argument > for this, as well? >Moreover, what's the status of there exists a transfinite cardinal not >cofinal with omega ? In ZFC, all successor cardinals are regular, so none of them are cofinal with omega. In ZF, I don't know, but I suspect that there might be models in which the statement is false. David ----- === Subject: Re: Uncountable sets in CZF? >A fact which could be considered disturbing is that for any >transfinite cardinal not cofinal with omega, there exists a model of >ZFC in which the cardinality of R is the given cardinal. >Sorry to bother you once more, but... can you provide a short argument >for this, as well? >Moreover, what's the status of there exists a transfinite cardinal not >cofinal with omega ? > In ZFC, all successor cardinals are regular, so none of them are cofinal > with omega. Oh, ok. Apparantly my set theory has become a bit rusty. What's your definition of 'cofinal with omega'? -- Herman Jurjus === Subject: Re: Uncountable sets in CZF? <2pg9j4FklatcU2@uni-berlin.de> <2ples6Fmdsl5U2@uni-berlin.de A fact which could be considered disturbing is that for any >transfinite cardinal not cofinal with omega, there exists a model of >ZFC in which the cardinality of R is the given cardinal. >Sorry to bother you once more, but... can you provide a short argument >for this, as well? >Moreover, what's the status of there exists a transfinite cardinal not >cofinal with omega ? > In ZFC, all successor cardinals are regular, so none of them are cofinal > with omega. >Oh, ok. Apparantly my set theory has become a bit rusty. >What's your definition of 'cofinal with omega'? I may not have the terminology exactly right. By omega, I mean the smallest limit ordinal, so that omega is the order type of N. When I describe an ordinal alpha as being cofinal with omega, then I mean that there is an ascending function f : omega -> alpha such that sup_{n < omega} f(n) = alpha. In the case of a cardinal in ZFC, I mean that the cardinal kappa is cofinal with omega iff it is cofinal with omega as an ordinal, i.e. iff there is an ascending function f mapping omega to ordinals of cardinality less than kappa, such that sup_{n < omega} f(n) = kappa. The cofinality of the limit ordinal alpha is the smallest ordinal sigma such that there is an ascending function f : sigma -> alpha such that sup_{tau < sigma} f(tau) = alpha. The cofinality is necessarily a regular transfinite cardinal (i.e. the cofinality of sigma is sigma itself). The cofinality of an aleph is the cofinality of its initial ordinal. In ZFC, if A has cardinality lambda, then the cofinality of the power set of A necessarily exceeds lambda. David ----- === Subject: Re: Uncountable sets in CZF? were sets as opposed to a theory where numbers are primary objects. ZF and ZFC are therefore what you call plain set theories. In ZF and ZFC, numbers are defined AS SETS. I wonder where you got the erroneous idea that numbers were primary objects in ZF and ZFC??? A natural number is defined to be either the empty set or a successor ordinal whose elements are all either empty or successor ordinals, a set n is a natural number if n is empty or [n is a successor ordinal and for all m in n (m is empty or m is a successor ordinal)]. Equivalently, a natural number is an element of the smallest limit ordinal. The smallest limit ordinal omega is a model of Peano's Axioms with 0 being given by the empty set and the successor function being given by S(x) = x u {x}, where for sets A and B, A u B denotes the union of A and B. Addition on N (the set of natural numbers, i.e. omega) is defined by recursion by m + 0 = m, m + S(n) = S(m + n). Multiplication on N is defined by recursion by m.0 = 0, m.S(n) = m.n + m. Define the relation == on N x N by (a,b) == (c,d) iff a + d = b + c, then == is an equivalence relation on N x N. Define Z = (N x N)/==, and define an integer to be an element of Z, i.e. an integer is an equivalence class in N x N, and therefore a subset of N x N. Addition is defined on Z by [(a,b)] + [(c,d)] = [(a + c,b + d)]. Multiplication is defined on Z by [(a,b)].[(c,d)] = [(a.c + b.d,a.d + b.c)]. Both addition and multiplication on Z are well-defined. N is embedded in Z by mapping m to [(m,0)]. Define the relation == on Z x (Z-{0}), where 0 in Z is identified with 0 in N by the embedding, by (a,b) == (c,d) iff a.d = b.c, then == is an equivalence relation on Z x (Z-{0}). Define Q = (Z x (Z-{0}))/==, and define a rational number to be an element of Q, i.e. a rational number is an equivalence class in Z x (Z-{0}), and therefore a subset of Z x (Z-{0}). Addition on Q is defined by [(a,b)] + [(c,d)] = [(a.d + b.c,b.d)]. Multiplication on Q is defined by [(a,b)].[(c,d)] = [(a.c,b.d)]. Both addition and multiplication on Q are well-defined. Z is embedded in Q by mapping x to [(x,1)], where 1 in Z is identified with 1 in N by the embedding of N in Z. So, we have definitions of natural numbers, integers and rational numbers completely in terms of sets. There are two possible definitions of real numbers. A sequence in Q is defined as a function x : N -> Q. A Cauchy sequence in Q is a sequence x in Q such that for all rational d > 0, there exists k in N such that if m and n are natural numbers such that m > k and n > k, then |x(m) - x(n)| < d, i.e. for all d in Q (d > 0 => exists k in N for all m in N for all n in N [(m > k and n > k) => |x(m) - x(n)| < d]). Define the relation == on Cauchy sequences in Q by x == y if for all rational d > 0, there exists k in N such that for all natural numbers n > k, |x(n) - y(n)| < d, i.e. for all d in Q (d > 0 => exists k in N for all n in N (n > k => |x(n) - y(n)| < d)). Let CauchyQ be the set of Cauchy sequences in Q, then == is an equivalence on CauchyQ. Define R = CauchyQ/==, and define a real number to be an element of R, so that a real number is a set of Cauchy sequences in Q. Define addition and multiplication on CauchyQ by (x + y)(n) = x(n) + y(n) for all n in N, (x.y)(n) = x(n).y(n) for all n in N. Define addition and multiplication on R by [x] + [y] = [x + y], [x].[y] = [x.y]. Addition and multiplication are well-defined on R. For r in Q, define the sequence x_r in Q by x_r(n) = r for all n in N, then x_r is a Cauchy sequence in Q. Q is embedded in R by mapping r to [x_r] for all r in Q. An alternative definition of real number is the Dedekind definition: a real number is a set x of rational numbers such that (1) x is not empty, (2) x is not equal to Q, (3) if r is an element of x and s is a rational number such that s < r, then s is an element of x, (4) if r is an element of x, then there exists a rational number s such that r < s and s is an element of x. So a real number x is a nonempty proper subset of Q such that for all r in x for all s in Q (s < r => s in x), for all r in x exists s in Q (r < s and s in x). Let R be the set of all real numbers. Addition is defined on R by x + y = {r + s : r in x, s in y}. then x + y is an element of R. 0 in R is defined equal to {r in Q : r < 0}. Order is defined on R by x <= y (x is less than or equal to y) iff x is a subset of y, so that x < y iff x is a proper subset of y. For x in R, -x is defined by -x = {r in Q : exists s in Q (r < -s and s not in x)}, then -x is an element of R. Multiplication on R is defined by x.y = 0 if x = 0 or y = 0, x.y = {r in Q : r <= 0} u {r.s : r in x, s in y, r > 0, s > 0} if x > 0, y > 0, x.y = -(x.(-y)) if x > 0, y < 0, x.y = -((-x).y) if x < 0, y > 0, x.y = (-x).(-y) if x < 0, y < 0, then x.y is an element of R. Q is embedded in R by mapping r to {s in Q : s < r}. So a natural number is an ordinal, an integer is an equivalence class of ordered pairs of natural numbers, a rational number is an equivalence class of ordered pairs of integers, and real number is an equivalence class of Cauchy sequences of rational numbers (Cauchy definition) or a set of rational numbers (Dedekind definition). C can be defined by C = R x R. Addition on C is given by (a,b) + (c,d) = (a + c,b + d). Multiplication on C is given by (a,b).(c,d) = (a.c - b.d,a.d + b.c). R is embedded in C by mapping x to (x,0). The elements of C are called the complex numbers. There are other definitions of complex numbers (the most satisfying probably being C = R[x]/(x^2 + 1)). >So you have that in the generic extension that there are more elements >in R, There *may* be more elements in R. It depends on the extension. >but no more elements in N. R is the complete set of real >numbers in V, but you have that R in V is a proper subset of R in the >extension V[G]. Then you claim there is a bijection between N^V[G] >and R^[V]. In V[G], for an *appropriate* extension. Never a bijection in V. >N^V[G] has no more elements that N^V, yet you have it >bijectively mapping to R^V, In V[G]. >which you claim has less elements than >R^V[G] as a proper subset or R^V. Conversely, R^V is the complete set >of real numbers, each real number is defined in R^V, and N^V[G] is not >different than N^V in that for each n E N^V there exists n E N^V[G] >and vice versa. >N^V and N^V[G] are the same set in that they differ in no elements, >they are each no different than N, yet you have that N^V[G] >bijectively maps to R^V but not to R^V[G], and claim that R^V[G] is a >proper superset of R^V. R^V would be a proper subset of R^{V[G]}, not vice versa. >Please name or describe a real element of >R^V[G] not in R^V, that is R, the set of all real numbers. Let the set F of forcing conditions be the set of all functions p whose domain is a finite subset of N and whose range is a subset of {0,1}, i.e. F = {p : p is a function, dom(p) is a finite subset of N, range(p) is a subset of {0,1}}. Define q <= p (q is stronger than p) if q extends p, i.e. p is a subset of q, i.e. dom(p) is a subset of dom(q) and p(n) = q(n) for all n in dom(p). It follows that p and q are compatible iff p(n) = q(n) for all n in the intersection of dom(p) and dom(q). Let G be a generic set in F (G is not necessarily an element of V). Let n be an element of N, and let D = {p in F : n in dom(p)}. If q is a forcing condition, and n is an element of dom(q), then q is stronger than q and q is an element of D. If q is a forcing condition and n is not an element of dom(q), let r = q u {(n,0)}, then r is a function, dom(r) is a finite subset of N and range(r) is a subset of {0,1}, and so r is a forcing condition, r is stronger than q, and n is an element of dom(r), and so r is an element of D. It follows that for all forcing conditions q, there exists a forcing condition r stronger than q such that r is an element of D. So D is dense. Since G is a generic set, then G intersects D. Let p be a common element of G and D, then n is an element of dom(p), and so n is an element of dom(UG) (where, for a set A, UA is the union of A). Since n is an arbitrary element of N, then dom(UG) = N. Since range(p) is a subset of {0,1} for all p in G, then range(UG) is a subset of {0,1}. Suppose UG is not a function, then there exists n in N such that (n,0) and (n,1) are both elements of UG. It follows that there exist p in G and q in G such that p(n) = 0 and q(n) = 1. Since n is an element of the intersection of dom(p) and dom(q), and p(n) and q(n) are not equal, then p and q are incompatible, contradicting the genericity of G. Since the assumption that UG is not a function leads to a contradiction, then UG is a function UG : N -> {0,1} in V[G]. Let A = {n in N : (UG)(n) = 1}, then A is a subset of N in V[G]. Let B be a subset of N in V, and define D = {p in F : exists n in dom(p) ((n in B and p(n) = 0) or (n not in B and p(n) = 1))}. If q is a forcing condition and q is an element of D, then q is stronger than q and q is an element of D. Suppose q is a forcing condition and q is not an element of D, then for all n in dom(q) ((n in B and q(n) = 1) or (n not in B and q(n) = 0)). Since dom(q) is finite, then N-dom(q) is nonempty. Let m be an element of N-dom(q), and define r = q u {(m,0)} if m is an element of B, and r = q u {(m,1)} if m is not an element of B. Then r is a function, dom(r) is a finite subset of N, and range(r) is a subset of {0,1}, so that r is a forcing condition, r is stronger than q, and ((m in B and r(m) = 0) or (m not in B and r(m) = 1)). It follows that r is an element of D. So D is a dense set. Since G is generic, then G intersects D, and so there exist p in G and n in dom(p) such that ((n in B and p(n) = 0) or (n not in B and p(n) = 1)). If n is an element of B, then p(n) = 0, so (UG)(n) = 0, so n is not an element of A. If n is not an element of B, then p(n) = 1, so (UG)(n) = 1, so n is an element of B. It follows that n is an element of either A or B, but not both, and so A and B are not equal. That is, for any subset B of N in V, there is a natural number n such that n is an element of the symmetric difference of A and B, and so A and B are not equal. It follows that A is not a subset of N in V, i.e. A is an element of V[G], but not an element of V. Let x = sum_{n in N} 2^{-(UG)(n)}, then x is a real number in [0,2] in V[G], but x is not an element of V, i.e. x is a new real number. >That's why I interpreted what you said as affirming the existence of a >bijection between N and R, because N^V[G] has no element not in N and >bijectively maps to R, the set of all real numbers. Only if you do not take enough care to precisely specify your meaning. >EF is basically n/oo defined for n E N. That is not a proper definition of EF. Define oo. If oo is meant to be infinity, then your definition is no definition at all. >There is much discussion of >it on sci.math. >The average value of all hypercomplex numbers is zero. Define hypercomplex number if you don't mean quaternion, and explain what you mean by average value in this context. >I try to not use models in theory, Then how do you know what you are calling a set and what you are not calling a set? The model is the collection of all sets. >because I think a theory should be >just a theory, If you have just a theory, then you don't have any concrete examples of your objects. >and to not use classes in set theory, because a set >theory should have only sets. Which ZF and ZFC do. >Applying a metatheory or extended model or denying that anything is a >set in set theory only leads to obfuscation and not a real addressing >of the resolution of basically set existence issues within the set >theory itself. You need axioms to specify what you are calling a set. What are your axioms? The rest is cut out until Ross supplies convincing evidence that he knows what he is talking about. David ----- === Subject: Re: Explaining the foundations of math === Subject: Re: Explaining the foundations of math >I have similar concerns, but consider the experience in CA, where >I live. Our CA schools were #1 in the nation in the 50s and early >60s. Then some conservative didn't live having to finance education Yes, I took a couple of classes while in California. >Anyway, Prop 13 was passed, I forget the details, but property taxes >couldn't be raised to pay for schools, and things have gone downhill >ever since. Now CA's schools are falling apart, literally, Bill >Moyers did a very depressing piece on this, and our students now rank >49th or 50th in the nation. Yicks! I heard is was going down hill, but that far? >Don't parents care about their children any more? They're too busy having to work their butts off. >Has the cult of give me money? taken over completely every >other possible value, so that people are trying to live >meaningful lives by accumulating money and consumer junk? Yes, Kapitalismus Uber Alles. >Its a sad sad tale. Indeed, 'Safe for democracy' is code word for capitalist take over. ---- === Subject: Re: Explaining the foundations of math === Subject: Re: Explaining the foundations of math >Agreed. I find it interesting that in the 50's the US was near >the top of the curve in quality of education. Now that we've >started worrying about social issues in the schools, we're >plummeting fast. > Sigh. Soon we'll see US plummeting scientifically. >Soon? I figured the programming jobs were fleeing to India for >this very reason. I wouldn't be surprised to see the other sciences >similarly affected? 'Tis the zenith of best news, for the earnings of the CEO gods have increased even faster by outsourcing jobs of over paid US workers. As for outsourcing research, US double-think citizens will have no difficultly being made to understand that buying prescription drugs from bad foreign countries like Canada is dangerous, but using drugs developed and tested in nice unsanitary third world countries is safe. > But less I stray too far off topic, what are they doing in grade > school, are they still teaching numbers, ie arithmetic? Or are > youngsters spared such demeaning demoralization by being taught > to use hand calculators instead? >I know that graphing calculators are required in middle school/high >school in South Carolina for algebra classes. I have yet to hear >anyone explain *why* a graphing calculator should be required. It's the fad, technology is our salvation. >For that matter, I have yet to understand why they're allowed on the >SAT now. Is it not obvious? So they can get better scores. > These days community colleges are promoting and emphasizing use of > graphic calculators as algebra. Does this forebode of yet more > mathematical incompetence? >If a student doesn't actually understand what a graphing calculator >does, how can that student be expected to make sense of of the >calculus methods for graphing? Indeed. When tutoring I often demanded a student not turn on that &#(*#_) contraption until they understood the problem, the math and what's needed to solve the problem. With that in hand, or in mind, the calculor, if at all useful by now, could be used effectively. Otherwise it'd be just GIGO, garbage in garbage out, or as some students prefer, to flounder around using the graphic calculor as a guessing machine. ---- === Subject: Re: Explaining the foundations of math >But less I stray too far off topic, what are they doing in grade >school, are they still teaching numbers, ie arithmetic? Or are >youngsters spared such demeaning demoralization by being taught >to use hand calculators instead? >I know that graphing calculators are required in middle school/high >school in South Carolina for algebra classes. I have yet to hear >anyone explain *why* a graphing calculator should be required. > It's the fad, technology is our salvation. Until we have pocket AIs, it will not be our salvation. Once we get pocket AIs, it could be our destruction. >For that matter, I have yet to understand why they're allowed on the >SAT now. > Is it not obvious? So they can get better scores. Funny, I thought the point of the SAT is to see what the student knows. How silly of me. > These days community colleges are promoting and emphasizing use of > graphic calculators as algebra. Does this forebode of yet more > mathematical incompetence? >If a student doesn't actually understand what a graphing calculator >does, how can that student be expected to make sense of of the >calculus methods for graphing? > Indeed. When tutoring I often demanded a student not turn on that &#(*#_) > contraption until they understood the problem, the math and what's needed > to solve the problem. With that in hand, or in mind, the calculor, if at > all useful by now, could be used effectively. Otherwise it'd be just > GIGO, garbage in garbage out, or as some students prefer, to flounder > around using the graphic calculor as a guessing machine. I've found that this approach to be effective as well. The students who are understanding the material are usually the ones that do not grab a calculator as step one. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Explaining the foundations of math >I know that graphing calculators are required in middle school/high >school in South Carolina for algebra classes. I have yet to hear >anyone explain *why* a graphing calculator should be required. >For that matter, I have yet to understand why they're allowed on the >SAT now. > Is it not obvious? So they can get better scores. > Funny, I thought the point of the SAT is to see what the student knows. > How silly of me.