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

 Messages: [ Previous | Next ]
 Russell Easterly Posts: 811 Registered: 12/13/04
Re: No Identity Bijection for Omega
Posted: Sep 11, 2007 10:59 PM

On Sep 11, 11:04 am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Sep 10, 7:37 pm, logic...@comcast.net wrote:
>

> > On Sep 10, 6:11 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > I realize this is a long thread,
> > so I will try to bring you up to date.

>
> Last I posted to you, I showed several mistakes in your argument. Are
> you going to repeat those mistakes (some of which are variations on
> mistakes you've been making for over two years) or is your your latest
> argument cleaned up from those mistakes?
>

> > Let F be a bijection between A and B.
>
> > A = w+1 = {0,1,2,3,...,w}
> > B = A-{0} = (1,2,3,4,...,w}

>
> > F = { <0,1>, <1,2>, <2,3>, <3,4>, ..., <w,w> }
> > Let X = A intersect B = {1,2,3,4,...,w}

>
> Okay. However, and this is only a minor point, I'd give this in the
> order in which the objects are defined:
>
> A = w+1 = {0 1 2 ... w}
> B = A\{0} = (1 2 3 ...w}
> S = the successor function on w
> F = Su{<w w>} = {<0 1> <1 2> ... <w w>}
> X = A/\B = {1 2 3 ... w}
>

> > /now define a family of bijections, G():
>
> > For each x in X replace {<x,b_n>, <a_m,x>}
> > with {<x,x>, <a_m,b_n>}

>
> That makes no sense to me. (1) Neither 'a' not 'b' have been defined.
> (2) the members of X are not of the form of an unordered pair of
> ordered pairs.
>
> MoeBlee

a_m is a member of setA.
b_n is a member of set B.
The ordered pairs are members of F.
x is a member of X and must also be
a member of both A and B.

Let x = w.
Replace <w,w> and <w,w> in F
with <w,w> and <w,w>.
This is a new bijection
(assuming F is a bijection).

G(w) = F = { <0,1>, <1,2>, <2,3>, <3,4>, ..., <w,w> }

Choose another element of x and repeat
this process using G(w) instead of F.

Let x=1
Replace <1,2> and <0,1> with
<0,2> and <1,1> in G(w)

G(w,1) = { <0,2>, <1,1>, <2,3>, <3,4>, ..., <w,w> }

Repeat this process until every element of X
is paired with itself.

Russell
- 2 many 2 count

Date Subject Author
9/3/07 Russell Easterly
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