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

 Messages: [ Previous | Next ]
 William Hughes Posts: 4,277 Registered: 12/6/04
Re: No Identity Bijection for Omega
Posted: Sep 20, 2007 12:51 AM

On Sep 20, 12:16 am, logic...@comcast.net wrote:
> On Sep 18, 10:00 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
>
>

> > On Sep 17, 7:30 pm, logic...@comcast.net wrote:
>
> > > We have:
>
> > > A = w+1 = {0 1 2 ... w}
> > > B = A\{0} = (1 2 3 ... w}
> > > F = {<0 1> <1 2> ... <w w>}
> > > X = A/\B = {1 2 3 ... w} = B

>
> > > meA
> > > neB
> > > <m x> e F
> > > <x n> e F

>
> > > Define a new bijection as follows:
>
> > I mentioned meA and neB because you did and I thought you'd do
> > something with them toward a universal generalization. But you
> > didn't.

>
> As George points out, I need to use the
> free variables m and n:
>
> X = A/\B
> xeX
>
> Ex(<m x> e F) -> meA
> Ex(<x n> e F) -> neB
>
> Since X can be ordered, let x1 be the first
> element of X.
>
> Define a bijection, G1:
>
> G1(x1) = F - {<x1 n> <m x1>} u {<x1 x1> <m n>}
>
> Now define another bijection, G2.
>
> G2(x2) = G1 - {<x2 n> <m x2>} u {<x2 x2> <m n>}
>
> Repeat this process for every element of X.
>
> Several people have pointed out this process
> "never ends" and/or "doesn't converge".
>
> With some tips from you and others, I have
> a simpler proof that makes recursive calls to F.
>
> A = w+1
> B = A-{0}
> X = A/\B= B
>
> F = {<0 1> <1 2> <2 3> ... <w w>}
>
> I want to prove that if F is
> a bijection between A and B,
> there exists a bijection, S,
> between A-B and B-A.
>
> Take the first (and only)
> element of A-B = 0.
>
> If F(0) is not an element of X,
> then F(0) must be an element
> of B-A proving there exists a
> bijection between A-B and B-A.
>
> Assume F(0) is an element of X.
> Then <F(0) F(F0))> must be an element of F.
>
> Define a new bijection, S1.
>
> S1 = F - {<0 F(0)> <F(0) F(F(0))>}
> u {<0 F(F(0))> <F(0) F(0)>}
>
> Consider S1(0) = F(F(0)).
>
> If S1(0) is an element of X
> then <F(F(0)) F(F(F(0))> is an element of F.
>
> Define a new bijection, S2.
>
> S2 = F - {<0 F(0)> <F(0) F(F(0))>}
> u {<0 F(F(0))> <F(0) F(0)>}
>
> Continue this process until
> Sz(0) is an element of B-A.
>

Since B-A is the empty set this process will
never terminate.

> This process must terminate
> if F is a bijection.
>
> It is clear that
>
> Si(0) = F(F(...F(0)...))
>
> where i is the number of times F(0)
> has been iterated.
>
> F(0) can be iterated at most |X| times.
>

So if |X| is infinite, then F(0) can be iterated
at most an infinite number of times.

- William Hughes

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