mm-2859 === Subject: Re: New to Events as sets - I don't get how we can use sets to describe... > I sure hope you can find the time to help me some more > So event is basically the end result or the outcome that did happen out of > all possible outcomes event had in its bag ? Not quite. The outcomes are what _could_ happen. The event is what you _want_ to happen. Throw two dice and add them up and you could get any integer 2, ..., 12. You win if they add to 7 or 11 so the event you are interested in is {7, 11}. > Actually, terminology confuses be a bit here. Why don't > we call a set of all possible outcomes a possibility > and an outcome that does happen an event, instead of > calling both the set and the thing that did happen an event? Well, the mathematical definition of a word may not have much to do with its english definition. The idea is that an event is a collection of outcomes. The outcomes are the possible results of individual trials. Take our dice again. We (at least imagine that) we throw two of them. One set of outcomes could be what they add up to X = {2, 3, ..., 12}. Each throw would result in one of the outcomes. An event might be that they add to an even number. E = {2, 4, 6, 8, 10, 12}. There are other possible outcomes. Y = {(1,1), 1,2), ..., (1,6),(2,1),..., (6,6)}. Now, the event that they add to an even number is H = {(1,1), (1,3),...,(1,5), (2,2),...,(2,6),...}. Which one we use depends on what we want to do next. >One event is that the two dice add to >7: S = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. > Or if I decide event A is that two dice add to any > combination of numbers? Wouldn't then A U B = G ? Depends on what G is. Let X be the outcomes above and E the event that the dice add to an even number and F the event that they add to a multiple of 3. Then E / F = {2, 3, 4, 6, 8, 9, 10, 12} which isn't X. >To say an outcome belongs to A / G says that either >the outcome belongs to A (A happens) or the >outcome belongs to G (G happens).A / G = G says that >if either A or G happens then it is certain that G >happened. In our example suppose A = {(2,5), (4,3), >(6,1)} A / S = S.We exclude A but not S. > So if everytime A is subset of S we can say > A / S = S? Yes. > Now for Venn diagrams : > Diagram for A / B is same as for a intersection of sets. Yes > The intersection of sets are those elements which belong to all intersected > sets. Yes. > Which isn't the case with A / B. So why the same Venn > diagrams? Why not? With E and F as above E / F = {6, 12}. [...] > I can't imagine any real world examples where we would say > event A - event B = event A That would tell me that no outcome in event B was also in event A. To recap the example; X = {2, 3, ..., 12} = the outcomes; E = {2, 4, 6, 8, 10, 12} = the event that the outcome is even; F = {3, 6, 9, 12} = the event that the outcome is a multiple of 3; E / F = {6, 12} = the event that the outcome is both even and a multiple of 3; E / F = {2, 3, 4, 6, 8, 9, 10, 12} = the event that the outcome is even or a multiple of 3; X - F = the event that the outcome is not a multiple of 3; S = {7} is the event that the outcome is a multiple of 7 ( _is_ 7); F - S = F no multiple of 3 is a multiple of 7; X / F = F; X / E = E. -- Paul Sperry Columbia, SC (USA) === Subject: Re: New to Events as sets - I don't get how we can use sets to describe... I'm truly sorry for taking you so much time and I hope you don't give up on me just one more time -So event is a collection of outcomes that we are interested in? If we take two dices for our example, then there are many outcomes but event is only a set of outcomes that do perform the way we want ( like two dices add up to an even number ) ? be considered a single outcome or two separate outcomes ? >One event is that the two dice add to >7: S = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. >> Or if I decide event A is that two dice add to any >> combination of numbers? Wouldn't then A U B = G ? >Depends on what G is. When we think of G as a set of all possible outcomes, exactly what outcomes are we talking about? Only the ones we are interested in, or all outcomes in general? Say we throw two dices and are interested in outcomes that add up to even numbers, would G in this case represent all possible combinations or just all even numbers? >Let X be the outcomes above You mean X = {2, 3, ..., 12}? > and E the event that the dice add to an even number >and F the event that they add to a multiple of 3. >Then E / F = {2, 3, 4, 6, 8, 9, 10, 12} which isn't X. But X does include all those numbers so wouldn't that make A and F subsets of X ? Lordy am I confused :( thank you === Subject: Re: New to Events as sets - I don't get how we can use sets to describe... <26759998.1130335672348.JavaMail.jakarta@nitrogen.mathforum.org>, > I'm truly sorry for taking you so much time and I hope you don't give up on > me just one more time > -So event is a collection of outcomes that we are > interested in? If we take two dices for our example, then > there are many outcomes but event is only a set of > outcomes that do perform the way we want ( like two > dices add up to an even number ) ? It might help for me to say where I think all this is heading. Judging from the terminology, this is leading to probability. So, the _real_ question might be If two dice are thrown, what is the probability that they add to 7 or 11? It is up to you to come up with a suitable set of outcomes and a suitable event and then find the probability of that event. All that is going on is regular set manipulations. Because of the use the sets are going to be put to, special names are applied. The set of outcomes is just a universal set - its subsets are called events. As far as sets go, there is nothing new - only the application is new. > be considered a single outcome or two separate > outcomes ? It doesn't make any difference. The experiment could be throwing two dice and the outcome would be the sum of their faces. Or, the experiment could be to throw one die, then throw it again. Your outcome could be (number on first throw, number on second throw) or, maybe, the outcome could just be the sum of the first and second throws. As is the case with many problems there are two steps: the set-up and the solution. The set-up is not unique and is up to you (providing you eventually end up with the correct solution). The only real requirement is that each experiment have a unique outcome associated with it. >One event is that the two dice add to >7: S = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. > Or if I decide event A is that two dice add to any >> combination of numbers? Wouldn't then A U B = G ? >Depends on what G is. > When we think of G as a set of all possible outcomes, > exactly what outcomes are we talking about? What constitutes an outcome is up to you. If G is going to be all of your possible outcomes (whatever they may be) you need to say so. > Only the ones we are interested in, or all outcomes in > general? Say we throw two dices and are interested in > outcomes that add up to even numbers, would G in this > case represent all possible combinations or just all even > numbers? Having G just be even numbers wouldn't work becouse an odd sum wouldn't have an outcome. >Let X be the outcomes above > You mean X = {2, 3, ..., 12}? Yes. > and E the event that the dice add to an even number >and F the event that they add to a multiple of 3. >Then E / F = {2, 3, 4, 6, 8, 9, 10, 12} which isn't X. > But X does include all those numbers so wouldn't that make A and F subsets of X ? Yes. If X _didn't_ include A then A wouldn't be an event. The idea is that you are going to do something - say toss a red die and a green die. You want to associate with each toss an outcome that somehow records what happened. What the outcome should be depends very much on the rest of the question. If I was interested in the sum of the values on the dice maybe that sum would be the outcome. If I was interested in whether the number on the red die was a multiple of the number on the green die, maybe I would want the outcome to be the separate numbers or, In the same instance, the outcome could be yes or no. As you will see, when you get there, some outcome sets are much preferable to others. -- Paul Sperry Columbia, SC (USA) === Subject: Re: New to Events as sets - I don't get how we can use sets to describe... Again, I don't want to be a pest, but I really hope you can help me at least once more. I finally feel I'm on to something but I can't get there without your help If again we take two dices as an example >> Or if I decide event A is that two dice add to any >> combination of numbers? Wouldn't then >>A U B = G ? >Depends on what G is. >>When we think of G as a set of all possible >>outcomes, exactly what outcomes are we >>talking about? >What constitutes an outcome is up to you. If G is >going to be all of your possible outcomes >( whatever they may be ) you need to say so. First I have to clear something up: The definition for certain even(which I call G) is at each repetition of an experiment an event will always happen . So it's not on me to decide what outcomes will G cover Wouldn't that suggest that we can only use G (for example A U G) when experiment will alway produce the result we want. So to say we want an event A where two dices to add to even( but don't want odd numbers ) then we can't use A U G if we are not prepared to allow an outcomes where two dices add to odd number, since G by definition means no matter what the outcome is it is an event G ? >X = {2, 3, ..., 12} = the outcomes; >X / F = F; >X / E = E. X is an outcome not an event so I don't see how we can use it in X / F form. Shouldn't this form only be used for events (event / event)? >The set-up is not unique and is up to you >(providing you eventually end up with the correct >solution). The only real requirement is that each > experiment have a unique outcome associated >with it. Could you please elaborate a bit. What do you mean set-up is not unique and that 'each experiment have a unique outcome '? I really appreciate your help === Subject: Re: New to Events as sets - I don't get how we can use sets to describe... <9690845.1130420704683.JavaMail.jakarta@nitrogen.mathforum.org>, [...] >X = {2, 3, ..., 12} = the outcomes; >X / F = F; Yes, I _did_ write that didn't I?. I meant X / F = X. >X / E = E. > X is an outcome not an event [...] No, X _is_ an event; it is the collection of all possible outcomes (at least). >The set-up is not unique and is up to you >(providing you eventually end up with the correct >solution). The only real >requirement is that each > experiment have a unique outcome associated >with it. > Could you please elaborate a bit. What do you mean > set-up is not unique and that 'each experiment have a unique outcome '? > I really appreciate your help OK. Let's start over. The experiment is to throw a red die and a green die. First, I need to know what it is about the experiment that interests me. Suppose I'm interested in the sum of the up faces on the two dice. Now, I want to make up tally sheet to record the results of my experiment. The first column will have labels Try 1, Try 2, .... Each row will contain some check boxes. How do I label the check box columns? What I need to make sure of is that there is one and only one check box for every possible result of a Try. I also need to make sure that if a box is checked, it tells me something about the sum of the two dice. The column labels, all together, make up my set of outcomes. There are at least two ways to label the check box columns: 2 3 4 5 6 7 8 9 10 11 12 Try 1 Try 2 Try 3 . . . or (1,1) (1,2) ... (1,6), (2,1), .... (6,6). Try 1 . . . In the second case, there are 36 check box columns and for example (3,2) means 3 on the red die and 2 on the green die. In the first case if a box labeled, say, 6 is checked it means the dice added to 6. In the second case if a box labeled, say, (5, 4) is checked I can still get the information I want: 5 + 4 = 9. Suppose that I want to know how often the dice add to 5. In the first case I would look at the column labeled 5; the event is {5}. In the second case I would look at the columns labeled (1,4), (2,3), (3,2) and (4,1). Now the event is {(1,4), (2,3), (3,2), (4,1)}. Either way, If I did 100 tries, I could, in the first case, count how many times column 5 was checked or, in the second case, count how many times one of the event columns (1,4), (2,3), (3,2) or (4,1) was checked. Both methods would tell me how often the dice added to 5. Suppose instead, I am interested in how often the number on the red die is smaller than the number of the green die. I could use the second case above and the event {(1,2), (1,3),...,(5,6)} (15 outcomes in all) and count how many times an event column was checked. Or I could do it this way: B S Tie Try 1 . . . B means the red die is bigger than the green one, S means red is smaller. The event is {S} and I count how many times the {S} column is checked. If I had left off Tie I would not have had an outcome set for the experiment: (5,5) would not have had an outcome. I cannot consider (5,5) to be both B and Tie. Only one outcome to a trial. Yes No Try 1 . . . Now the event is {Yes} and I count how many times the Yes column is checked. Yes No Tuesday Dog Try 1 . . . Would be OK - the last two columns just never get checked. Finally, set theory plays only a tiny role in all this. Eventually you will find out how to find the frequency of events which are combined using /, / and -. -- Paul Sperry Columbia, SC (USA) === Subject: Re: evaluate sum =?ISO-8859-1?Q?=85?= kx^2k ( k from 0 to infinity) > Hi > trying to evaluate Sum (kx^2k) (k from 0 to infinity, x isn't specified > to be < 1) > It diverges ??? Am I right??? You've seen the relevance of x isn't specified to be < 1, so you'll need to do some specifying yourself: for such-and-such values of x the series diverges, for such-and-such values it converges. When it does converge, you'll need some trick to find the sum. It's a bit like the geometric series Sum(x^2k), so can you modify any trick you know for summing that? Ken Pledger. === Subject: Re: Really need help-don't understand it <5595074.1130260455012.JavaMail.jakarta@nitrogen.mathforum.org>, >> a) familly members sit together , >The two three seat loges hold family J and the four >seat loge holds family S? > yes >> a) I would think that formula would be same as c) >> since familly S is in a room for four and familly J is >>in two rooms for >> three ! But the correct result is 60. >>How is that case different from >> c) ? >That depends on what (a) means. Interpreted as I did >above, the answer _is_ >the same as (c). Maybe it >would help to give us the question in its original >language (if it's Dutch, I'm out of luck). > It's not dutch, but still I doubt you know my language since it's slavic You are right about that. Somehow I got the impression you were posting from Belgium. -- Paul Sperry Columbia, SC (USA) === Subject: functions ETAuAhUAx5lK+/UvOhPeTuOyeVd64nKrgl8CFQCAqkutfNNwkxFP719IuUn1nHOc/A== If I draw a random graph: specify the x-y intercepts,visually estimate the derivate at several points, and note concavity how would I convert this to function form? === Subject: Re: functions Wed, 26 Oct 2005 01:26:42 -0400 from dd ss : > If I draw a random graph: specify the x-y intercepts,visually > estimate the derivate at several points, and note concavity how would I > convert this to function form? Without more information, you can't. Different functions can look very similar. Here's a (trivial) example: f(x) = x+1 g(x) = (x^2-999999x-1000000)/(x-1000000) Those functions are identical everywhere, except at one point. -- Stan Brown, Oak Road Systems, pkins County, New York, USA http://OakRoadSystems.com/ That was a stupid lie, easy to expose, not worthy of you. George Sanders as Addison Dewitt in /All About Eve/ (1950) === Subject: Re: functions > If I draw a random graph: specify the x-y intercepts,visually > estimate the derivate at several points, and note concavity how would I > convert this to function form? There will be infinitely many functions satisfying the kinds of restrictions you propose. === Subject: Question with golden mean Let P:= (1+ sqrt(5))/2 and f(x)= 1+ P*cos(x)+A*cos(2x)+B*cos(3x}) with real parameters A and B. It is known that f(x) >= 0 for all x in [0,pi]. Find A and B. === Subject: A= 2 x 2 matrix = ? C is the set of complex numbers, a,b are given in C , b =/= 0, D is a subset of C and f : D ---> C a differentiable function. Find all 2 x 2 complex matrices A with the spectrum in D, such that / a b f(A) = 0 a/ . Note: Case a=1, b=1966,f(x)= sin(x) was proposed at Putnam Competition-1996. === Subject: Re: Poly.degree 3 having all roots rational > If the equation x^3-Ax^2+ BX- 1=0 > has only rational roots, then A < 1 or A >1 ? > [Key Words: > possible references= Werner Mnich= ??, > J.W.S.Cassels, W.Sierpinski, G.Sansone,...] >>No. For A=B=1 the equation has only one real root: x=1 which is >>rational. >When the OP says all the roots are rational, don't we understand that >all are real? Evidently this depends on the value of we. ************************ === Subject: Re: Discreet Question >[...] >Don't hit post message more than once. Certainly not three times in >40 second intervals. But they were not all the same message! There were at least three different messages in the four posts - note the spelling of discrete in the subject and the first line of the text. (Actually I'm a little disappointed there were not four different versions - coulda been a discrete question, given the two spellings and the two places the word appears, how many different posts are possible...) ************************ === Subject: Multinomial distribution Consider the Multinomial distribution for the case of three possible outcomes with probabilities (p1, p2, p3). If we observe (y1,y2,y3) numbers of outcomes in the three categories write down the log-likelihood for these data. The answer I found is: l=Sum(p1log(y1)+p2log(y2)+p3log(y3) Is it correct? === Subject: Re: Multinomial distribution The problem then states that: In order to maximise this log-likelihood under the constraint p1+p2+p3=1 write p3 in terms of the other probabilities and rewrite the log-likelihood in terms of p1 and p2. That's easy if am thinking correctly... p3=1-p1-p2 and log(P(Y1=y1,Y2=y2,Y3=y3))=y1log(p1)+y2log(p2)+y3log(1-p1-p2) This is correct right? The difficult part is: Obtain the derivatives of the log-likelihood with respect to p1 and p2. Setting these to zero, express the resulting equation in terms of p1/y1, p2/y2, p3/y3 Am really stack in this part. I can understand that the second part is talking about MLE but I can't go there if I don't have the derivative. Please help me! === Subject: Re: Multinomial distribution >The problem then states that: >In order to maximise this log-likelihood under the >constraint p1+p2+p3=1 write p3 in terms of the other >probabilities and rewrite the log-likelihood in terms of >p1 and p2. >That's easy if am thinking correctly... >p3=1-p1-p2 and >log(P(Y1=y1,Y2=y2,Y3=y3))=y1log(p1)+y2log(p2)+y3log(1-p1->p2) >This is correct right? yes, it's correct >The difficult part is: >Obtain the derivatives of the log-likelihood with >respect to p1 and p2. Setting these to zero, express the >resulting equation in terms of p1/y1, p2/y2, p3/y3 >Am really stack in this part. I can understand that the >second part is talking about MLE but I can't go there if >I don't have the derivative. Please help me! d/d(p1) (y1log(p1)+y2log(p2)+y3log(1-p1-p2)) = y1/p1 - y3/(1-p1-p2) d/d(p2) (y1log(p1)+y2log(p2)+y3log(1-p1-p2)) = y2/p2 - y3/(1-p1-p2) Setting y1/p1 - y3/(1-p1-p2) = 0 and y2/p2 - y3/(1-p1-p2) = 0 and solving for p1 and p2, you'll get p1=y1/(y1+y2+y3), p2=y2/(y1+y2+y3), p3=y3/(y1+y2+y3) as a maximum likelihood estimation for p1, p2 and p3. Best wishes Torsten. === Subject: Re: Multinomial distribution >Consider the Multinomial distribution for the case of >three possible outcomes with probabilities (p1, p2, p3). >If we observe (y1,y2,y3) numbers of outcomes in the >three categories write down the log-likelihood for these >data. >The answer I found is: >l=Sum(p1log(y1)+p2log(y2)+p3log(y3) >Is it correct? Hi Maria, for the multinomial distribution, P(Y1=y1,Y2=y2,Y3=y3) = (y1+y2+y3)!/(y1! y2! y3!)* p1^y1 * p2^y2 * p3^y3 So if you are allowed to ignore the combinatorial term (y1+y2+y3)!/(y1! y2! y3!) (which does not depend on p1, p2 or p3), you get log (P(Y1=y1,Y2=y2,Y3=y3)) = y1*log(p1) + y2*log(p2) + y3*log(p3), so your answer is correct. Best wishes Torsten. === Subject: Re: Multinomial distribution >Consider the Multinomial distribution for the case of >three possible outcomes with probabilities (p1, p2, p3). >If we observe (y1,y2,y3) numbers of outcomes in the >three categories write down the log-likelihood for these >data. >The answer I found is: >l=Sum(p1log(y1)+p2log(y2)+p3log(y3) >Is it correct? >Hi Maria, >for the multinomial distribution, >P(Y1=y1,Y2=y2,Y3=y3) = (y1+y2+y3)!/(y1! y2! y3!)* >p1^y1 * p2^y2 * p3^y3 >So if you are allowed to ignore the combinatorial term >(y1+y2+y3)!/(y1! y2! y3!) (which does not depend on >p1, p2 or p3), you get >log (P(Y1=y1,Y2=y2,Y3=y3)) = >y1*log(p1) + y2*log(p2) + y3*log(p3), >so your answer is correct. sorry, I misread your answer: you interchanged the pi's and the yi's. Best wishes Torsten. === Subject: Re: Munkres, Topology > On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean > in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. [...] I thought I would use this problem to blow the dust of my long unused topology. The problem also appears as an exercise early on in Kelly. I am wondering about the phrase A and B form a separation of X-Y. I assumed that meant A and B were separated in X-Y. It turns out, if A and B are separated in X, I can work it - otherwise not (yet). It is a matter of closures. Do you have a proof for A and B separated in X-Y? All I want here is Yes or No (or maybe Maybe). -- Paul Sperry Columbia, SC (USA) === Subject: Re: Munkres, Topology <261020051735300815%plsperry@sc.rr.com> === Subject: Re: Munkres, Topology > On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > I am wondering about the phrase A and B form a separation of X-Y. > I assumed that meant A and B were separated in X-Y. It turns out, if > A and B are separated in X, I can work it - otherwise not (yet). It > is a matter of closures. > Do you have a proof for A and B separated in X-Y? > All I want here is Yes or No (or maybe Maybe). No. I assumed A,B separated subset X. However note when A,B separated subsets X, then A / Y, B / Y separated subsets subspace Y Conversely when A,B (subsets X) separate Y, A / Y, B / Y are nonnul separated subsets X A disconnected iff some separated nonnul U,V with A = U/V U,V consider subset X. That's the nice thing about separated sets, an effective independence from subspaces. -- Anyway, just saw your later post. Would you like to review my proof? ---- === Subject: Re: Munkres, Topology === > Subject: Re: Munkres, Topology > On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > I am wondering about the phrase A and B form a separation of X-Y. > I assumed that meant A and B were separated in X-Y. It turns out, if > A and B are separated in X, I can work it - otherwise not (yet). It > is a matter of closures. > Do you have a proof for A and B separated in X-Y? > All I want here is Yes or No (or maybe Maybe). > No. I assumed A,B separated subset X. > However note when A,B separated subsets X, then > A / Y, B / Y separated subsets subspace Y > Conversely when A,B (subsets X) separate Y, > A / Y, B / Y are nonnul separated subsets X > A disconnected iff some separated nonnul U,V with A = U/V > U,V consider subset X. > That's the nice thing about separated sets, > an effective independence from subspaces. Got it: For subsets A, B of Y < = X, A and B are separated in Y iff they are separated in X. I just needed someone to rub my nose in it. > -- > Anyway, just saw your later post. > Would you like to review my proof? I think I have it now - see my subsequent reply to . -- Paul Sperry Columbia, SC (USA) === Subject: Re: Munkres, Topology alt.math.undergrad: >> On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean >> in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > [...] > I thought I would use this problem to blow the dust of my long unused > topology. The problem also appears as an exercise early on in Kell[e]y. My own is sadly long unused, and I'm supposedly a topologist by trade! > I am wondering about the phrase A and B form a separation of X-Y. I > assumed that meant A and B were separated in X-Y. It turns out, if A > and B are separated in X, I can work it - otherwise not (yet). It is a > matter of closures. > Do you have a proof for A and B separated in X-Y? All I want here is > Yes or No (or maybe Maybe). When I posted originally, I had no detailed proof at all, just an approach that looked a lot more promising than Simon's. I've just taken a quick look at the details, and they seem to work out okay. It was a pretty quick look, though, so call it 'probably'. === Subject: Re: Munkres, Topology > alt.math.undergrad: >> On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean >> in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > [...] > I thought I would use this problem to blow the dust of my long unused > topology. The problem also appears as an exercise early on in Kell[e]y. > My own is sadly long unused, and I'm supposedly a topologist > by trade! > I am wondering about the phrase A and B form a separation of X-Y. I > assumed that meant A and B were separated in X-Y. It turns out, if A > and B are separated in X, I can work it - otherwise not (yet). It is a > matter of closures. > Do you have a proof for A and B separated in X-Y? All I want here is > Yes or No (or maybe Maybe). > When I posted originally, I had no detailed proof at all, > just an approach that looked a lot more promising than > Simon's. I've just taken a quick look at the details, and > they seem to work out okay. It was a pretty quick look, > though, so call it 'probably'. > Kelley (spelled it right this time) says X-Y = A / B where A and B are separated. Let Y' = X - Y. Cl( ) means closure in X. We have Y' = A / B with (presumeably) [Cl(A) / Y'] / B = [Cl(B) / Y'] / A = empty. Suppose we have A / Y = T / V with [Cl(T) / (A / Y)] / V = [Cl(V) / (A / Y)] / T = empty. A little bit of distributivity gets Cl(T) / Y / V = Cl(V) / Y / T = empty. [Also Cl(T) / A / V = Cl(V) / A / T = empty for what that's worth.] Y = (T / Y) / (V / Y) and [Cl(T / Y) / Y] / (V / Y) <= [Cl(T) / (Y / A)]/ V = empty. Similarly Cl(V / Y) / Y / (T / Y) = empty. Since Y is connected, one of T / Y and V / Y is empty. Wlog, suppose T / Y = empty and V / Y = Y. So, T <= A and Y <= V. I've now worked my way down to your suggestion. I guess I'm now supposed to disconnect X somehow. The obvious choice is T and B / V. X = A / (B / Y) = T / (B / V). Cl(T) / (B / V) = (Cl(T) / B) / (Cl(T) / V). Those are the the wrong closures. For example T <= A so Cl(T) <= Cl(A) but we (or at least _I_) don't know Cl(A) / B is empty nor do I know Cl(T) / V is empty. There is the same problem with Cl(B / V) / T. I don't see any way around the fact that the only closures I know anything about are in X-Y or A / Y and I need to know about closures in X. My topological intuition isn't what it once was. I no longer believe I can do it even if A and B are separated in X although that helps some. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Munkres, Topology alt.math.undergrad: >> alt.math.undergrad: >> On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean >> in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. [...] > I am wondering about the phrase A and B form a separation of X-Y. I > assumed that meant A and B were separated in X-Y. It turns out, if A > and B are separated in X, I can work it - otherwise not (yet). It is a > matter of closures. > Do you have a proof for A and B separated in X-Y? All I want here is > Yes or No (or maybe Maybe). >> When I posted originally, I had no detailed proof at all, >> just an approach that looked a lot more promising than >> Simon's. I've just taken a quick look at the details, and >> they seem to work out okay. It was a pretty quick look, >> though, so call it 'probably'. > Kelley (spelled it right this time) says X-Y = A / B where A and B > are separated. > Let Y' = X - Y. Cl( ) means closure in X. > We have Y' = A / B with (presumeably) > [Cl(A) / Y'] / B = [Cl(B) / Y'] / A = empty. > Suppose we have A / Y = T / V with > [Cl(T) / (A / Y)] / V = [Cl(V) / (A / Y)] / T = empty. A little > bit of distributivity gets Cl(T) / Y / V = Cl(V) / Y / T = empty. > [Also Cl(T) / A / V = Cl(V) / A / T = empty for what that's worth.] > Y = (T / Y) / (V / Y) and > [Cl(T / Y) / Y] / (V / Y) <= [Cl(T) / (Y / A)]/ V = empty. > Similarly Cl(V / Y) / Y / (T / Y) = empty. Since Y is connected, > one of T / Y and V / Y is empty. Wlog, suppose T / Y = empty > and V / Y = Y. > So, T <= A and Y <= V. > I've now worked my way down to your suggestion. I guess I'm now > supposed to disconnect X somehow. The obvious choice is T and B / V. > X = A / (B / Y) = T / (B / V). > Cl(T) / (B / V) = (Cl(T) / B) / (Cl(T) / V). Those are the the > wrong closures. For example T <= A so Cl(T) <= Cl(A) but we (or at > least _I_) don't know Cl(A) / B is empty You know that the closure of A in XY is disjoint from B. But the closure of A in XY is Cl(A) / XY, so Cl(A) / XY / B = ø. But XY / B = B, so Cl(A) / B = ø. > nor do I know Cl(T) / V is empty. {V, T} is a separation of A / Y, so V is disjoint from the closure of T in A / Y: V / (A / Y) / Cl(T) = ø. V <= A, so V / B = ø. Thus, ø = [V / (A / Y) / Cl(T)] / (V / B) = V / (A / Y / B) / Cl(T) = V / Cl(T). [...] === Subject: Re: Munkres, Topology > alt.math.undergrad: >> alt.math.undergrad: >> On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean >> in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > [...] > I am wondering about the phrase A and B form a separation of X-Y. I > assumed that meant A and B were separated in X-Y. It turns out, if A > and B are separated in X, I can work it - otherwise not (yet). It is a > matter of closures. > Do you have a proof for A and B separated in X-Y? All I want here is > Yes or No (or maybe Maybe). >> When I posted originally, I had no detailed proof at all, >> just an approach that looked a lot more promising than >> Simon's. I've just taken a quick look at the details, and >> they seem to work out okay. It was a pretty quick look, >> though, so call it 'probably'. > Kelley (spelled it right this time) says X-Y = A / B where A and B > are separated. > Let Y' = X - Y. Cl( ) means closure in X. > We have Y' = A / B with (presumeably) > [Cl(A) / Y'] / B = [Cl(B) / Y'] / A = empty. > Suppose we have A / Y = T / V with > [Cl(T) / (A / Y)] / V = [Cl(V) / (A / Y)] / T = empty. A little > bit of distributivity gets Cl(T) / Y / V = Cl(V) / Y / T = empty. > [Also Cl(T) / A / V = Cl(V) / A / T = empty for what that's worth.] > Y = (T / Y) / (V / Y) and > [Cl(T / Y) / Y] / (V / Y) <= [Cl(T) / (Y / A)]/ V = empty. > Similarly Cl(V / Y) / Y / (T / Y) = empty. Since Y is connected, > one of T / Y and V / Y is empty. Wlog, suppose T / Y = empty > and V / Y = Y. > So, T <= A and Y <= V. > I've now worked my way down to your suggestion. I guess I'm now > supposed to disconnect X somehow. The obvious choice is T and B / V. > X = A / (B / Y) = T / (B / V). > Cl(T) / (B / V) = (Cl(T) / B) / (Cl(T) / V). Those are the the > wrong closures. For example T <= A so Cl(T) <= Cl(A) but we (or at > least _I_) don't know Cl(A) / B is empty > You know that the closure of A in XY is disjoint from B. > But the closure of A in XY is Cl(A) / XY, so > Cl(A) / XY / B = ø. But XY / B = B, so Cl(A) / B = ø. OK - I forgot B <= X - Y. > nor do I know Cl(T) / V is empty. > {V, T} is a separation of A / Y, so V is disjoint from the > closure of T in A / Y: V / (A / Y) / Cl(T) = ø. V <= A, Can't be. Above we have T <= A so Y / A = T / V <= A but A and Y are disjoint. However, William has set me straight: For U, W <= R <= S, U and W are separated in R iff they are separated in S. I should have read Kelley more carefully. A and B are separated in X - Y and hence in X. T and V are separated in A / Y and hence in X. So, Cl(T) / (B / V) = (Cl(T) / B) / (Cl(T) / V) <= (Cl(A) / B) / (Cl(T) / V) = empty. Cl(B / V) / T = (Cl(B) / T) / (Cl(V) / T) <= (Cl(B) / A) / (Cl(V) / T) = empty. > so V / B = ø. Thus, > ø = [V / (A / Y) / Cl(T)] / (V / B) = > V / (A / Y / B) / Cl(T) = V / Cl(T). > [...] > -- Paul Sperry Columbia, SC (USA) === Subject: Re: Munkres, Topology alt.math.undergrad: >> alt.math.undergrad: >> alt.math.undergrad: >> On Tue, 25 Oct 2005 00:56:09 EDT, Simon Dean >> in alt.math.undergrad: > Hi. Need help on the following: > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > I am wondering about the phrase A and B form a separation of X-Y. I > assumed that meant A and B were separated in X-Y. It turns out, if A > and B are separated in X, I can work it - otherwise not (yet). It is a > matter of closures. [...] > Let Y' = X - Y. Cl( ) means closure in X. > We have Y' = A / B with (presumeably) > [Cl(A) / Y'] / B = [Cl(B) / Y'] / A = empty. > Suppose we have A / Y = T / V with > [Cl(T) / (A / Y)] / V = [Cl(V) / (A / Y)] / T = empty. A little > bit of distributivity gets Cl(T) / Y / V = Cl(V) / Y / T = empty. > [Also Cl(T) / A / V = Cl(V) / A / T = empty for what that's worth.] > Y = (T / Y) / (V / Y) and > [Cl(T / Y) / Y] / (V / Y) <= [Cl(T) / (Y / A)]/ V = empty. > Similarly Cl(V / Y) / Y / (T / Y) = empty. Since Y is connected, > one of T / Y and V / Y is empty. Wlog, suppose T / Y = empty > and V / Y = Y. > So, T <= A and Y <= V. > I've now worked my way down to your suggestion. I guess I'm now > supposed to disconnect X somehow. The obvious choice is T and B / V. > X = A / (B / Y) = T / (B / V). > Cl(T) / (B / V) = (Cl(T) / B) / (Cl(T) / V). Those are the the > wrong closures. For example T <= A so Cl(T) <= Cl(A) but we (or at > least _I_) don't know Cl(A) / B is empty >> You know that the closure of A in XY is disjoint from B. >> But the closure of A in XY is Cl(A) / XY, so >> Cl(A) / XY / B = ø. But XY / B = B, so Cl(A) / B = ø. > OK - I forgot B <= X - Y. > nor do I know Cl(T) / V is empty. >> {V, T} is a separation of A / Y, so V is disjoint from the >> closure of T in A / Y: V / (A / Y) / Cl(T) = ø. V <= A, > Can't be. Sorry, that was a typo for V <= A / Y, which still gives you V / B = ø. [...] === Subject: Re: Munkres, Topology that if A and B form a separation of X-Y, then YUA is > connected. > If we assume that YUA is not connected, then we can write > YUA as YUA = TUV for separated sets T and V. Then X = > (TUV)UB. Now, I can show that(TUV) and B are nonempty > disjoint sets. > Can someone show how to prove that neither set contains a > limit point of the other? Egads. Are you discussing topology or metric space theory? > I think that you're on the wrong track. Try a different approach. Here's my approach snipped from Will's notes: connected S,K, separated A,B, SK = A/B ==> A/K connected. Otherwise: some separated nonnul U,V with A/K = U/V ... ---- === Subject: Re: Munkres, Topology On Wed, 26 Oct 2005 01:59:05 -0700, William Elliot > Let Y be a subset of X; let X and Y be connected. Show > that if A and B form a separation of X-Y, then YUA is > connected. > If we assume that YUA is not connected, then we can write > YUA as YUA = TUV for separated sets T and V. Then X = > (TUV)UB. Now, I can show that(TUV) and B are nonempty > disjoint sets. > Can someone show how to prove that neither set contains a > limit point of the other? > Egads. Are you discussing topology or metric space theory? Obviously not the latter: there are no metrics in sight, not anything from which to produce them. The notion of limit point of set is general topological. (That's a very strange opposition anyway, since metric space theory is a part of topology.) [...] === Subject: Re: cholesky decomposition > hi, > i have been reading about the 'cholesky algorithm' to perform a cholesky > decomposition, here at :- http://en.wikipedia.org/wiki/Choleskey_decomposition#The_Cholesky_algorithm > but under the section titled 'cholesky algorithm' and under where it says > 'At step i, the matrix .... ' then it gives a matrix for A^(i) - i am > confused as to the meaning of the symbols ai,i , bi and B^(i) mean . it > doesnt explain as far as i can see. > Choleski decomposition is really really easy to understand, that page > makes it harder to understand than it really is. absolutely agree with you there :) the way i do choleskys is to carry out an LDL^T decomposition first, then check for positive definiteness, then form the cholesky matrix from L and the square root of D. but i was looking at that algorithm simply out of curiosity > Start with the matrix L. It's lower triangular, so the transpose L^T > is upper triangular. > Do the matrix multiplication M=L * L^T. The resulting matrix M will be > symmetric and positive definite if there is no zero on the diagonal of > L. > So write down the details of this multiplication. You only have to > compute the elements on and below the diagonal of M because M is > symmetric so M(1,2) is the same as M(2,1), M(1,3) = M(3,1) etc. (here > in my notation M(1,2) means element of row column 2 of M, etc.) > As an example, for a 4x4 matrix the computations are: > M(1,1)=L(1,1)^2 > M(2,1)=L(2,1) * L(1,1) > M(2,2)=L(2,1)^2 + L(2,2)^2 > M(3,1)=L(3,1) * L(1,1) > M(3,2)=L(3,1) * L(2,1) + L(3,2) * L(2,2) > M(3,3)=L(3,1)^2 + L(3,2)^2 + L(3,3)^2 > M(4,1)=L(4,1) * L(1,1) > M(4,2)=L(4,1) * L(2,1) + L(4,2) * L(2,2) > M(4,3)=L(4,1) * L(3,1) + L(4,2) * L(3,2) + L(4,3) * L(3,3) > M(4,4)=L(4,1)^2 + L(4,2)^2 + L(4,3)^2 + L(4,4)^2 > There is no mystery here, this is regular matrix multiplication of L by > L^T. > This is when we know L and want to find M. > But in Cholesky decomposition, it's the opposite, we know M and want to > find L. No problem, just do the above calculations but in reverse. > Isolate the values L(1,1), L(2,1) etc... on the left, like this: > L(1,1)= sqrt( M(1,1) ) > L(2,1)=M(2,1) / L(1,1) > L(2,2)=sqrt( M(2,2)-L(2,1)^2 ) > L(3,1)=M(3,1) / L(1,1) > L(3,2)=( M(3,2) - L(3,1) * L(2,1) )/ L(2,2) > L(3,3)=sqrt( M(3,3) - ( L(3,1)^2 + L(3,2)^2 ) ) > L(4,1)=M(4,1) / L(1,1) > L(4,2)=( M(4,2) - L(4,1) * L(2,1) ) / L(2,2) > L(4,3)=( M(4,3) - ( L(4,1) * L(3,1) + L(4,2) * L(3,2) ) / L(3,3) > L(4,4)=sqrt( M(4,4) - ( L(4,1)^2 + L(4,2)^2 + L(4,3)^2 ) ) > This is the Cholesky algorithm which is described in that wikipedia > page under the heading The Cholesky-Banachiewicz and Cholesky-Crout > algorithms. > It's the version that computes row by row. If you follow this > algorithm row by row on paper you'll be able to visualize easily how it > works. > Here is a speudo-code of the algorithm, only 10 lines without the > comments and parentheses: > for row=1 to nrow > for i=1 to row-1 > // Here compute the sum > // ie. sum=L(4,1) * L(3,1) + L(4,2) * L(3,2) > sum=0 > for j=1 to i-1 > sum=sum+L(row,j)*L(i,j) > // here save the sum > // ie: L(4,3)=( M(4,3) - sum ) / L(3,3) > L(row,i)=(M(row,i) - sum)/L(i,i) > // Here compute the sum for the element on the diagonal > // ie. sum=L(4,1)^2 + L(4,2)^2 + L(4,3)^2 > sum=0 > for j=1 to row-1 > sum=sum+L(row,j)^2 > // and save it under the square root: > // ie: L(4,4)=sqrt( M(4,4) - sum ) > L(row,row)=sqrt(M(row,row) - sum); > Of course it only works if the pivot element L(i,i) is not zero because > of the division, and that's why Cholesky is only good for symmetric > positive definite matrix because they never have a zero on the diagonal > of L. === Subject: Re: cholesky decomposition decomposition, here at :- > http://en.wikipedia.org/wiki/Choleskey_decomposition#The_Cholesky_algorithm > but under the section titled 'cholesky algorithm' and under where it > says > 'At step i, the matrix .... ' then it gives a matrix for A^(i) - i am > confused as to the meaning of the symbols ai,i , bi and B^(i) mean . > it > doesnt explain as far as i can see. > Choleski decomposition is really really easy to understand, that page > makes it harder to understand than it really is. > absolutely agree with you there :) the way i do choleskys is to carry out > an LDL^T decomposition first, then check for positive definiteness, then > form the cholesky matrix from L and the square root of D. > but i was looking at that algorithm simply out of curiosity I looked at it too and in the wikipedia page it refers to Golub and Van loan Matrixc Computations so I looked it up, and here is their explanation: in section 4.2.5 Outer product Cholesky it says you can decompose a nxn matrix A like this: A= [ a v^T ] [ v B ] = [ c 0 ] [ 1 0 ] [ c v^T /c ] [ v/c Id_(n-1)] [ 0 B-(v*v^T)/a ] [ 0 Id_(n-1)] where c = sqrt(a) (and a>0 because A is positive definite), v is a column vector of dimension n-1 and v^T is the corresponding row column, and v*v^T is the outer product, so that B-(v*v^T)/a is a sub-matrix of dimension (n-1)x(n-1). So you can write this as A=L_1 * A_1 * (L_1)^T. Then you apply recursively this decomposition to the sub-matrix B-(v*v^T)/a n-1 more times and in the end the matrix A_n in the middle is reduced to the identity and you have the product A=L_1*L2*...L_n * Id * L_n^T * .... L_1 ^T, which can be written as A=G*G^T where G=L_1*...*L_n, hence the decomposition. but in a more confusing way. === Subject: Re: Discrete Question >I have a quick Discreet Question on strong induction > the question is Prove by stron induction that postage of 24 cents or > more can be achieved by using only 5-cent and 7-cent stamps. > I am totally baffled by this question I've worked on it for a bit and > can't seem to come to any sort of conclusion at all > anyone out there able to help? amount = (number 5-cent stamps, number 7-cent stamps) 24 = (2, 2) congruent to 4 modulo 5 25 = (5, 0) congruent to 0 modulo 5 26 = (1, 3) congruent to 1 modulo 5 27 = (4, 1) congruent to 2 modulo 5 28 = (0, 4) congruent to 3 modulo 5 (Since every number is congruent to 0,1,2,3,4 mod 5, then by adding the correct number of 5-cents stamps to one of the 5 solns above we can land on every integer >=24) Using strong induction with the follwoing modifications..... Let D be a subset of the nonnegative integers Z with the properties that (1) the integer 24 is in D and (2) any time that the interval [24, n] is contained in D, one can show that n+1 is also in D. Under these conditions, D is all the integers >= 24 (1) As shown above, 24 is in D. (2) The interval [24, 28] is in D, and if [24, n], n>=29 is in D then by finding the highest number in the interval congruent to n+1 mod 5 and adding one 5-cent stamp, n+1 is also in D. Therfore D is the integers >=24 Joe === Subject: Re: Discrete Question >>I have a quick Discreet Question on strong induction >> the question is Prove by stron induction that postage of 24 cents or >> more can be achieved by using only 5-cent and 7-cent stamps. >> I am totally baffled by this question I've worked on it for a bit and >> can't seem to come to any sort of conclusion at all >> anyone out there able to help? > amount = (number 5-cent stamps, number 7-cent stamps) > 24 = (2, 2) congruent to 4 modulo 5 > 25 = (5, 0) congruent to 0 modulo 5 > 26 = (1, 3) congruent to 1 modulo 5 > 27 = (4, 1) congruent to 2 modulo 5 > 28 = (0, 4) congruent to 3 modulo 5 > (Since every number is congruent to 0,1,2,3,4 mod 5, > then by adding the correct number of 5-cents stamps to one of the > 5 solns above we can land on every integer >=24) > Using strong induction with the following modifications..... > Let D be a subset of the nonnegative integers Z with the properties that > (1) the integer 24 is in D and > (2) any time that the interval [24, n] is contained in D, one can show > that n+1 is also in D. > Under these conditions, D is all the integers >= 24 > (1) As shown above, 24 is in D. > (2) The interval [24, 28] is in D, > and if [24, n], n>=29 is in D then by finding typo above - should of course be n>=28 > the highest number in the interval congruent to n+1 mod 5 > and adding one 5-cent stamp, n+1 is also in D. > Therfore D is the integers >=24 > Joe