|
|
Re: No Identity Bijection for Omega
Posted:
Sep 13, 2007 1:24 AM
|
|
On Sep 12, 8:35 pm, William Hughes <wpihug...@hotmail.com> wrote:
> On Sep 12, 10:42 pm, logic...@comcast.net 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>.
>
> > 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>.
> > Similarly, the x in B must be paired
> > with some element of A, like <a_m,x>.
>
> > If x is paired with itself in F,
> > then <a_m,x> = <x,b_n> = <x,x>.
> > This is the case for w in the
> > bijection defined above.
> > <a_m,w> = <w,b_n> = <w,w>.
>
> > Assume x is not paired with itself in F.
>
> > 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>
>
> > Call this new bijection G(x).
> > Assume the elements of X can be ordered.
>
> > G(x_i+1) = G(x_i)
> > - {<x_i+1,n>, <m,x_i+1>}
> > + {<x_i+1,x_i+1>,<m,n>}
>
> > where m and n are whatever elements of A and B
> > are paired with x_i+1 in G(x_i).
>
> > F = { <0,1>, <1,2>, <2,3>, <3,4>, ..., <w,w> }
>
> > G(1) = { <0,2>, <1,1>, <2,3>, <3,4>, ..., <w,w> }
> > G(2) = { <0,3>, <1,1>, <2,2>, <3,4>, ..., <w,w> }
> > G(w) = { <0,z>, <1,1>, <2,2>, <3,3>, ..., <w,w> }
> z must be a natural number less than w, and greater than
> any natural number. Thus z does not exist. Thus G(w)
> does not exist.
z must be an element of some pair in F.
Let A' = B' = w+1
Let F' be a bijection of w+1 to itself
that is not the identity function.
F' = {<0,w>, <w,0>, <1,1>, <2,2>, ...}
X' = A' intersect B' = w+1
G'(0) = {<0,0>, <1,1>, <2,2>, ..., <w,w>}
G'(1) = {<0,0>, <1,1>, <2,2>, ..., <w,w>}
G'(2) = {<0,0>, <1,1>, <2,2>, ..., <w,w>}
...
G'(w) = {<0,0>, <1,1>, <2,2>, ..., <w,w>}
Are you saying G'(w) doesn't exist?
Russell
- 2 many 2 count
|
|
