In article <04127510-c0d4-4b53-b536-f87e1ac60a2c@j8g2000yqd.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 24 Jun., 02:39, Sylvia Else <syl...@not.here.invalid> wrote: > > On 23/06/2010 8:44 PM, WM wrote: > > > > > > > > > > > > > On 23 Jun., 06:47, Sylvia Else<syl...@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.
