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

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 Russell Easterly Posts: 811 Registered: 12/13/04
Re: No Identity Bijection for Omega
Posted: Sep 20, 2007 12:16 AM

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.

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.

Russell
- 2 many 2 count

Date Subject Author
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