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Topic: No Identity Bijection for Omega
Replies: 116   Last Post: Sep 22, 2007 5:58 PM

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: No Identity Bijection for Omega
Posted: Sep 4, 2007 5:33 AM

On Mon, 03 Sep 2007 19:28:20 -0700, logiclab@comcast.net wrote:

>Let B be the identity bijection of w+1 = w U {w} .
>B is an ordered set of ordered pairs.
>
>B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >
>
>For each element, B_i, let C_i be the set of
>all B_i that appear in the bijection before B_i.
>Let C_i also includes B_i. (Skeptics - this is the one to attack.)
>
>Let D_i be the set that C_i is a bijection for.
>(D_i is the set of all natural numbers that appear in C_i).
>
>B_0 = <0,0>
>B_1 = <1,1>
>B_2 = <2,2>
>...
>
>B_w = <w,w>
>
>
>C_0 = < <0,0> >
>C_1 = < <0,0>, <1,1> >
>C_2 = < <0,0>, <1,1>, <2,2> >
>...
>C_w = < <0,0>, <1,1>, <2,2>, ..., <w,w> >
>
>
>D_0 = {0}
>D_1 = {0,1}
>D_2 = {0,1,2}
>...
>D_w = {0,1,2,...,w}
>
>
>Assume D_k = w for some k.
>D_k = {0,1,2,...,k} and k must be the largest natural number.
>Therefore, no D_k = w.
>
>Assume there exists a C_k that is a bijection for w.
>Then there exists D_k.
>
>Assume there exists an identity bijection, A, for w.
>A must be an proper subset of B.

Yes.

>A must also be an initial segment of B.

Yes.

>If A is an initial segment of B then A = C_k for some k.

No. Try to give us a _proof_ that any initial segment
must be C_k for some k. Keep in mind that there's
a simple counterexample, so the proof might be tricky.

>There is no identity bijection for Omega.
>
>
>Russell
>- 2 many 2 count

************************

David C. Ullrich

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