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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 17, 2007 10:30 PM

On Sep 17, 2:04 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Sep 12, 9:00 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
>
>
>
>

> > On Sep 12, 7:42 pm, logic...@comcast.net wrote:
>
> > > On Sep 12, 10:39 am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > Let <m,n> be an element of F.
> > > Since m is an element of A and
> > > n is an element of B, we can write <a_m,b_n>.

>
> > I can spend more of my mental time and energy trying to figure out
> > what you intend by such notation, and other posters indulge you by
> > gleaning what you mean despite your confused notation. But clarity is
> > not served by that.

>
> > You don't know how to express yourself in clear mathematical notation
> > and prose, because you don't read any mathematics.

>
> > Here's what we have so far:
>
> > 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}

>
> > And, as george pointed out, X = B.
>
> > Now, moving on, what you said above is as simple as:
>
> > Suppose m in A, n in B, and <m n> in F.
>
> > [I'm interrupted now. I have to leave the computer. I might not post
> > until next week, when perhaps I'll resume.]

>
> Continuing:
>
> We have:
>
> 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} = B
>
> meA
> neB
> <m n> e F
>
> [Single indent quotes are RM's now, not mine.]
>
> On Sep 12, 7:42 pm, logic...@comcast.net wrote:
>

> > If x is an element of both A and B then
> > there must exist an element of F
> > such that the x in A is paired with
> > some element of B, like <x,b_n>.

>
> If x in A/\B, then x in B, so
>
> F(x) in B and <x F(x)> in F.
>

> > Similarly, the x in B must be paired
> > with some element of A, like <a_m,x>.

>
> If x in B, then there is an m in A such that x=F(m) (i.e., <m x> in
> F).
>

> > If x is paired with itself in F,
>
> Then x=w.
>

> > then <a_m,x> = <x,b_n> = <x,x>.
>
> <w F(w)> = <w w> = <x x>
>
> is what you're saying reduces to.
>

> > This is the case for w in the
> > bijection defined above.
> > <a_m,w> = <w,b_n> = <w,w>.

>
> Again, just as I said, what you're saying reduces to:
>
> <w F(w)> = <w w> = <x x>
>

> > Assume x is not paired with itself in F.
>
> So now x is any member of w.
>

> > I can create a new bijection, G,
> > where x is paired with itself by
> > replacing the elements <a_m,x>
> > and <x,b_n> in F with <x,x> and <a_m,b_n>

>
> As far as I can tell (from your ersatz notation), what you're saying
> is:
>
> Let x be a particular member of w\{0}. (x can't be 0 since for no m do
> we have <m 0> e F.)

I am sorry I haven't explained this well enough.

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:

G(x) = F - {<x n> <m x>} + {<x x> <m n>}

or in different notation:
G(x) = F - {<x F(x)> <F(x) x>} + {<x x> <F(x) F(x)>}

If F is a bijection, G(x) is also a bijection.

I want to define a family of bijections such that
every element of X less than or equal to x
is paired with itself in Gx.
(I am assuming X can be ordered.)

Using the example above:

G(1) = G1 = F - {<1 2> <0 1>} + {<1 1> <0 2>}
G1 = {<0 2> <1 1> <2 3> <3 4> ... <w w>}

G({1 2}) = G2 = G1 - {<2 G1(2)> <G1(2) 2>} + {<2 2> <G1(2) G1(2)>
G2 = G1 - {<2 3> <0 2>} + {<2 2> <0 3>}
G2 = {<0 3> <1 1> <2 2> <3 4> ... <w w>}

G({1 2 3}) = G3 = G2 - {<3 4> <0 3>} + {<3 3> <0 4>}
G3 = {<0 4> <1 1> <2 2> <3 3> ... <w w>}

Consider what happens if I continue this process
for all the elements of X:

G(X) = {<0 z> <1 1> <2 2> <3 3> ... <w w>}

where zeB and <y z> e F for some y<z.

Any help with notation would be greatly appreciated.

Russell
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