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From: mueckenh on 11 Dec 2006 11:31 Virgil schrieb: > In article <457c6023(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > > > > To put it another way, each right branch may be considered a > > continuation of the same original infinite path, like adding a 0 to the > > right of the decimal point - it doesn't change the value. The left > > branch from every node produces a new path, creating a new value with > > the concatenation of a 1 to the string. Therefore, for every two > > branches, there is one additional path. > > From every branch point (node) there spring exactly as many paths as > from the root node, And there is no one without its personal edge. But people who cannot see that the limit of f(n) = 9n is not zero will not be able to see that every additional path splits off by an additional edge. This kind of thinking is called indirect concluding. Sme people are simply incapable of it. > and if WM, or anyone, asserts that the set of all of > them is countable, that person owes us a proof, which so far has not > been forthcoming. The set of all edges is countable. But you cannot see the conclusion resulting from that observation. > > > > > Is the diagonal longer than any line? Nope. > > Which line(s) is it not longer than? That one which contains all elements of it. Every element of the diagonal is in a line. Everey smaller one is in the same line. Again indirect concluding. > > >>> You cannot imagine the integer [pi*10^10^100]. > > > > > >> Why not, you just did! > > > > > > No. I did not. > > Well perhaps WM did not, but he provided a name for it, which is good > enough for me. Here is another name which is very important for you: "The incontradictable ZFC." That is good enough to satisfy an inconvincable set theorist. Regards, WM
From: mueckenh on 11 Dec 2006 11:42 William Hughes schrieb: > This is the argument you present. > > > > > 1) The set of lines is also the potentially infinite set of natural > > numbers. > > 2) Every element of the diagonal is in at least one line. > > 3) Every initial segment of the diagonal is (in) a line. > > No, there is one initial segment of the diagonal that is > not in a line. Then name the element of the diagonal please, which supports your claim. > > > 4) The limit (n -->oo, for the number n of elements) of the diagonal is > > oo. > > Yes and it attains this limit (i.e the limit is a maximum as > well as a supremum) > > > 5) The limit (m -->oo, for the number m of elements of a line) is less > > than infinite. > > No. The limit is oo. (The limit cannot be any finite number) And it cannot be an infinite number. There is no infinite natural number. > However, this limit is not attained (i.e. it is a > supremum but it is not a maximum.) That is not sufficient to make every element of the diagonal (with maximum) an element of a line(without maximum). Therefore you have to name that element which makes the diagonal a set assuming its maximum but leaving the set of lines not assuming the maximum. > . > > 6) Therefore not every initial segment of the diagonal can be (in) a > > line. > > 7) Contradiction between (3) an (6). > > Since, both (3) and (5) are false, there is no contradiction. (3) and (5) are not false but they are evidently true from the definition of diagonal and of natural number, respectively. Regards, WM
From: Tony Orlow on 11 Dec 2006 11:51 Dik T. Winter wrote: > In article <1165761763.908889.34550(a)80g2000cwy.googlegroups.com> Han.deBruijn(a)DTO.TUDelft.NL writes: > ... > > Let P(a) be the probability that an arbitrary natural is divisible by > > a fixed natural a. Then P(a) = 1/a . Forbidden by set theory. > > No. Not specifically forbidden by set theory. Forbidden because there are > no appropriate definitions for the words you are using (they are not used > conforming to standard definitions, so you better supply definitions). > In probability theory (as is commonly use) you have to define how you > *select* your arbitrary natural. You have not done so, so probability > theory does not have an answer. Why does that matter? This is the same thing as your stupid ball and vase trick. Why do you need to label anything, or know what you're choosing from the infinite set? If you claim there are aleph_0 of them, and I claim they are all equally likely to be chosen, by whatever means, then each has a possibility, but no probability, of being chosen. In other words, the probability is not a standard real, but infinitesimal. That's clear. Look, take your naturally numbered balls and insert them with your Acme Zeno Insertion Tool between 11:59 and noon. Then place the vase in your Acme Zeno Paint Can Shaker, and shake from noon until 12:01, aleph_0 times. Remove a ball. It's random. What's the chance ball 1 will be chosen? Tony
From: Tony Orlow on 11 Dec 2006 12:15 David Marcus wrote: > cbrown(a)cbrownsystems.com wrote: >> mueckenh(a)rz.fh-augsburg.de wrote: >>> cbrown(a)cbrownsystems.com schrieb: >>> >>>> Because your argument is of the form: "There exists a bijection f >>>> between the naturals and the edges. >>> This is evident... >> I'm glad you agree. >> >>> ... and has nothing to do with shares. >> On the contrary; we will use f to prove that there is a bijection >> between the shares composing any given path p and N: each path p is >> composed of a countable number of shares of edges. >> >>>> There exists a bijection h between >>>> the paths and the reals. >>> Again tat is evident ... >> I'm glad you agree that h exists. >> >> ... and has nothing to do with shares. >> >> On the contrary; we will use h to prove that there is a bijection >> between shares of an edge and R: each edge is composed of an >> uncountable number of shares of edges. >> >>>> There exists a surjection T between edges and >>>> paths. Therefore, the composition h o T o f is a surjection of the >>>> naturals onto the reals; contradiction." >>>> >>>> I assume you know what "a surjection T from edges onto paths" is, yes? >>>> It doesn't map 1/2 of an edge to 1/4 of a path! It is a function that >>>> maps each edge /in the original diagram/ to a path; so that every path >>>> is in the image of the function. >>> Of course. But what I do is somewhat more sophisticated. It requires to >>> kow that 1/2 + 1/2 = 1 and some stuff that like. >> "... 1/2 + 1/2 = 1 and some stuff that like" is hardly a mathematical >> definition, sophisticated or otherwise. >> >> Either you have a surjection from edges in the original diagram to >> paths, or you don't. If you do, you have proven your claim that paths >> are both countable and uncountable. If you don't, your claim that edges >> are equinumerous with paths must be proved some other way. >> >> <snip> >> >>> Every edge is divided into shares, namely in as many shares as there >>> are paths to which this edge belongs. >> Sweet Mary! Finally, something resembling a definition! >> >> Let B(e) be the set of paths to which the edge e "belongs". >> >> Then the set of shares of an edge e is a set of elements called >> "shares", such that there exists a bijection from these shares to B(e). >> >> I suppose that if a share s is a share of edge a and s is also a share >> of edge b, then edge a = edge b; is that correct? I.e., a "share of an >> edge" is always "marked" as a share of some /specific/ edge e in the >> original diagram, yes? >> >> If that is the case, I would suggest that we define "shares of edges" >> as members of the set (paths X edges). We then define "the set of >> shares of the edge e in the original diagram" as: >> >> S(e) = {(p,e) : p in B(e)}. >> >> There is an obvious bijection from S(e) <-> B(e); so as required every >> edge is "divided" into "as many shares as there are paths" to which e >> "belongs". And if s is a share of edge a, it is not a share of edge b, >> unless a = b; so a "share of an edge" is uniquely associated with an >> edge in the original diagram. >> >> By these definitions, if edge e belongs to path p, then the share of >> edge e that belongs to path p is (p,e). Every share of an edge e (p,e) >> belongs to some path p in B(e). Every path p is "composed" of shares >> (p,e) where e belongs to p. >> >> To anticipate, let us also define "the composition of path p" as >> >> C(p) = {(p,e) : p in B(e)} >> >> So C(p) contains a unique "share of an edge" for each edge e that >> "belongs to" the path p. >> >> Do you agree with the above definitions of B, shares, S and C? >> >> Now, B(e) = {p : e "belongs to" p}. We haven't defined "belongs to" >> explicitly, but if you mean the "obvious" thing, then the set of paths >> to which edge e belongs can be bijected with the set of all paths >> (informally, by appending some unique finite length prefix associated >> with the edge e to all paths). >> >> Next, recall that we have assumed a bijection h : paths <-> R. >> Therefore, by composition, we have a bijection B(e) <-> paths <-> R for >> every edge e; and thus a bijection S(e) <-> R for every edge e; in >> other words, every edge is "divided" into an uncountable number of >> "shares". >> >> Clearly, the cardinality of C(p) is the same as the cardinality of some >> subset of the edges in the original diagram; and so card(C(p)) <= >> card(edges). Equally clearly, the cardinality of C(p) is at least >> countable ("every path is infinite"). Therefore, by the bijection f, it >> follows that there is a bijection C(p) <-> N; in other words, every >> path is composed of a countable number of shares of edges. >> >> <snip> >> >>> But it is no necessary to divide an edge at all, because ... >> Let's stick to one argument at a time, shall we? Or have you given up >> on trying to prove your "rational relation" argument? >> >> You now have the following functions whose existence we have >> established: >> >> (1) A bijection f between the naturals N and the edges. >> >> (2) A bijection h between the paths and the reals, R. >> >> (3) A function B : edges /in the original diagram/ -> sets of paths, >> where B(e) = { p : edge e "belongs to" path p}, and card(B(e)) = >> card(paths) = card(R). >> >> (4) The set of all "shares" is a subset of (paths X edges); i.e., "s is >> a share" implies s is of the form (p,e) for some path p and edge e in >> the original diagram, and where e belongs to p. >> >> (5) A function S: edges /in the original diagram/ -> sets of shares, >> where S(e) = {(p,e) : p in B(e)}; and card(S(e)) = card(R) for all >> edges e. >> >> (6) A function C: paths -> sets of share, where C(p) = {(p,e) : p in >> B(e)}; and card(C(p)) = card(N). >> >> (7) A function g : (paths X edges X N) -> R, with the property that lim >> n->oo sum(over all edges) g(p,e,n) = 2. >> >> In English, these 7 statements correspond to: >> >> (1) Edges are countable. >> (2) Paths can be bijected with the reals. >> (3) B(e) is the (uncountable) set of all paths to which edge e belongs. >> (4) Shares (of edges in the original diagram) are denoted (p,e) for >> some path p and some edge e where e belongs to p. >> (5) S(e) is the (uncountable) set of all shares of edge e. >> (6) C(p) is the (countable) set of all shares which belong to path p. >> (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 >> >> Now, how do we use the above functions to produce a surjection T : >> edges /in the original diagram/ -> paths? > > That's a good question! > >> You have previously mentioned "full edges" as distinct from "edges in >> the original diagram". How do they fit into the picture? What is the >> role of the function g in your argument? > May I? All values in the tree are created by an edge adding a 1 to the right of a previous string. If 1/3 is in the tree, then there is an edge where the value is exactly 1/3. If there is no such edge, then the value is not in the tree. You are treating the paths as a set of reals in the unit interval. That is fine, if infinite strings are allowed, but then the number of edges is no longer countable. If they ARE allowed, then there exist edges infinitely far down the tree, specified by such paths. If no such edges exist, then neither do any such values in this set of reals, and so you have a set of binary rationals, as countable as the edges. There is no edge marking 1/3, because there are no infinite strings where all bit positions are finite. The standard set-theoretic reasoning stands in sharp contrast to simple logic. Each node produces two edges, one continuing the original value down the path with a 0, and the other creating a new value and path by appending a 1. Thus, for every two new edges, one new path is produced. For every additional bit, each existing string has two possibilities. How can set theory claim to be more basic than that?
From: Tony Orlow on 11 Dec 2006 12:16
David Marcus wrote: > Virgil wrote: >> In article <457c60d0(a)news2.lightlink.com>, >> Tony Orlow <tony(a)lightlink.com> wrote: >>> Mike Kelly wrote: >>>> Tony Orlow wrote: >>>>> Set theory "disagrees". Interpret that as you wish. >>>> Measure theory and probability theory do too, more to the point. >>>> >>> Great. Contradictions within mathematics. >> What contradictions "within" anything? >> >> It is just that mathematics says that given certain definitions as in >> probability theory there is no probability distribution on elements of a >> countable set in which each member has the same probability as every >> other member. > > I would think common sense says this too. I've picked random numbers > from finite sets, but I haven't a clue how to go about picking a random > positive integer. > Use your Acme Zeno Paint Can Shaker from noon until 12:01, and remove a ball. What are the chances it will be ball 1? |