Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 24 Jun 2010 03:13 On 24 Jun., 02:39, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 8:44 PM, WM wrote: > > > > > > > On 23 Jun., 06:47, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >> Rather than argue that VMs proposition fails on that point, I wanted to > >> address the flaw, in order to find a more substantial objection. > > > My initials ar WM. > > > There is no flaw in the argument: Your bijection either contains all > > constructed antidiagonals. Then a list contains also its antidiagonal, > > because every list has an antidiagonal and your bijection (that is > > only a permutation of my list) contains all lines and antidiagonals. > > I.e., there is no missing antidiagonal of a "limit" list outside of > > the bijection. > > > Or there is a last diagonal of the limit list that does not belong to > > your bijection (and to my list). Then there is a countable set (the > > set that includes this last diagonal) that is not listable. > > > Think over that only a little bit. It is not difficult to understand. > > > Regrads, WM > > Thing is, I just cannot see that you've constructed a countable set that > should contain its own anti-diagonal when expressed as a list. When it does not contain it, then the antidiagonal is not listed. That is all. > All > you've done is construct a countable set of countable sets, each of > which contains the anti-diagonals of its predecessors, but none of which > by construction can be expected to contain its own anti-diagonal. That is why I did it. The set of elements of L0 and antidiagonals A0, A1, ... is countable, but cannot be listed. > > You've then claimed, but without any kind of formal argument, that if > you take this to the limit, Why should there be a limit? Why should I agrre to take this to a limit? There is no limit in n. Every one is followed by another one. > the result is a countable set that should > contain its own anti-diagonal when expressed as a list. Then Cantor's argument would fail. There are no other escapes: Either there is some list that contains its antidiagonal or there is no list that contains it. Regards, WM
From: Virgil on 24 Jun 2010 03:34 In article <41a66545-c0bb-44e3-bb41-1fdaf0c83d71(a)j4g2000yqh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 24 Jun., 01:02, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > > a) Does this list contain the anti-diagonal of > > > (..., An, ... A2, A1, A0, L0)? > > > > This is not a list of numbers. �L0 is not a number, it is a list. > > > > Therefore (..., An, ... A2, A1, A0, L0) does not have an anti-diagonal. > > You are wrong. > > The symbols above abbreviate the sequence of lists > > An > ... > A0 > L0 > > Each of them has an antidiagonal that either is not in the list (then > the set of all of them is unlistable) or is in a list. Then Cantors > argument is wrong. If one has only a list of lists, then its union can be listed and therefore has many "antidiagonals".
From: Virgil on 24 Jun 2010 03:48 In article <04127510-c0d4-4b53-b536-f87e1ac60a2c(a)j8g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 24 Jun., 02:39, Sylvia Else <syl...(a)not.here.invalid> wrote: > > On 23/06/2010 8:44 PM, WM wrote: > > > > > > > > > > > > > On 23 Jun., 06:47, Sylvia Else<syl...(a)not.here.invalid> �wrote: > > > > >> Rather than argue that VMs proposition fails on that point, I wanted to > > >> address the flaw, in order to find a more substantial objection. > > > > > My initials ar WM. > > > > > There is no flaw in the argument: Your bijection either contains all > > > constructed antidiagonals. Then a list contains also its antidiagonal, > > > because every list has an antidiagonal and your bijection (that is > > > only a permutation of my list) contains all lines and antidiagonals. > > > I.e., there is no missing antidiagonal of a "limit" list outside of > > > the bijection. > > > > > Or there is a last diagonal of the limit list that does not belong to > > > your bijection (and to my list). Then there is a countable set (the > > > set that includes this last diagonal) that is not listable. > > > > > Think over that only a little bit. It is not difficult to understand. > > > > > Regrads, WM > > > > Thing is, I just cannot see that you've constructed a countable set that > > should contain its own anti-diagonal when expressed as a list. > > When it does not contain it, then the antidiagonal is not listed. That > is all. > > > All > > you've done is construct a countable set of countable sets, each of > > which contains the anti-diagonals of its predecessors, but none of which > > by construction can be expected to contain its own anti-diagonal. > > That is why I did it. The set of elements of L0 and antidiagonals A0, > A1, ... is countable, but cannot be listed. Nonsense. {L0_0, A0, L0_1, A1, L0_2, A2, ...} is just such a list as WM clams cannot exist. > > > > You've then claimed, but without any kind of formal argument, that if > > you take this to the limit, > > Why should there be a limit? Why should I agrre to take this to a > limit? There is no limit in n. Every one is followed by another one. The elements of N have no limit, but N still exists. Any process that can be endlessly iterated need not have any "limit" but the set of all its iterations still can exist, just as N does. > > > the result is a countable set that should > > contain its own anti-diagonal when expressed as a list. > > Then Cantor's argument would fail. Which is why it never happens, at least with things like countable sets of reals or endless binary sequences. For each such there is an "antidiagonal" which is not a member of it (in fact lots and lots of antidagonals). > There are no other escapes: Either > there is some list that contains its antidiagonal or there is no list > that contains it. If the listed things are functions from N to a set of at least two elements, every one has at least one "antidiagonal". In fact lots and lots of them. At least as many antidiagonals as members.
From: WM on 24 Jun 2010 04:30 On 24 Jun., 09:34, Virgil <Vir...(a)home.esc> wrote: > > The symbols above abbreviate the sequence of lists > > > An > > ... > > A0 > > L0 > > > Each of them has an antidiagonal that either is not in the list (then > > the set of all of them is unlistable) or is in a list. Then Cantors > > argument is wrong. > > If one has only a list of lists, There is always, at every stage of the construction, one single list like Ln above, that either contains or not its antidiagonal. Thise prodedure is abbreviated by (..., An, ... A2, A1, A0, L0) but of course there cannot appear at any stage a list without first line. > then its union can be listed and > therefore has many "antidiagonals Before doing so, you first need the elements. There are two alternatives: 1) There is a list in the construction that contains its antidiagonal. Then Cantor's proof is wrong. 2) There is no list in the construction that contaiuns its antidiagonal, then the antidiagonal cannot be placed at position 0 at the previous list without generating another list and another antidiagonal. Therefore, the set of the lines of L0 and the set of all antidiagonals occuring during construction cannot be listed. It is as simple as that. It is not a big deal. Set theory shows many contradictions arising from the idea, that from "every n in N can be constructed" it is implied that "N can be constructed". This is wrong, because for every n, there is an infinite set of unconstructed elements of N. This is so for *every* n in N. Hence, N cannot be constructed, and no proof for all n in N has ever been valid. .... classical logic was abstracted from the mathematics of finite sets and their subsets .... Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. ... As Brouwer pointed out this is a fallacy, the Fall and Original sin of set theory even if no paradoxes result from it. [Weyl, Hermann (1946), "Mathematics and logic: A brief survey serving as a preface to a review of The Philosophy of Bertrand Russell", American Mathematical Monthly 53: 213.] Regards, WM
From: Virgil on 24 Jun 2010 05:04
In article <ff595e3b-8160-4655-a12a-641a304b86ff(a)y4g2000yqy.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 24 Jun., 09:34, Virgil <Vir...(a)home.esc> wrote: > > > > The symbols above abbreviate the sequence of lists > > > > > An > > > ... > > > A0 > > > L0 > > > > > Each of them has an antidiagonal that either is not in the list (then > > > the set of all of them is unlistable) or is in a list. Then Cantors > > > argument is wrong. > > > > If one has only a list of lists, > > There is always, at every stage of the construction, one single list > like Ln above, that either contains or not its antidiagonal. Show me a list of endless binary sequences or real numbers that you think contains what may reasonably be called an antidiagonal for it. I do not believe that any such list can exist. Cantor's argument satisfies me that for lists of binary infinite sequences, no antidiagonal can ever be a member of it. > > Thise prodedure is abbreviated by > (..., An, ... A2, A1, A0, L0) > but of course there cannot appear at any stage a list without first > line. But that list can be rearranged into a more standard form, for example with the An and the members of L0 listed alternatingly, and from such a rearrangement a non-member anti-diagonal can be constructed. > > > then its union can be listed and > > therefore has many "antidiagonals > > Before doing so, you first need the elements. > There are two alternatives: > 1) There is a list in the construction that contains its antidiagonal. Which Cantor's argument shows cannot happen. > Then Cantor's proof is wrong. > 2) There is no list in the construction that contaiuns its > antidiagonal, then the antidiagonal cannot be placed at position 0 at > the previous list without generating another list and another > antidiagonal. Therefore, the set of the lines of L0 and the set of all > antidiagonals occuring during construction cannot be listed. They can easily be listed, as I have several times shown. Wm has to get over the notion that the elements to be listed cannot be rearranged. > > It is as simple as that. > > It is not a big deal. Set theory shows many contradictions arising > from the idea, that from "every n in N can be constructed" it is > implied that "N can be constructed". In, for example, FOL+ZFC, there is no assumption that every n in N "can be constructed". What is assumed is that there exists a set, along with all its elements, having the properties that we want for N and its elements , which we then call N. This is wrong, because for every > n, there is an infinite set of unconstructed elements of N. But we do not "construct" any of them. Though we do construct various naming conventions for elements of N. > > This is so for *every* n in N. Hence, N cannot be constructed No one claims it can be. But it can be, and usually is, assumed without being constructed. There is no reason to despise any axiom system, however much it may seem to diverge from physical reality, unless it allows proof of a statement of the form "Both P and not P". WM, worshipping physics, does not understand that there are areas of mathematics completely outside of physics, so would hamper mathematics by forcing it not to look outside of physics. If WM could have forced this, internet commerce, for example, would be impossible. |