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Re: No Identity Bijection for Omega
Posted:
Sep 13, 2007 10:30 AM
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On Sep 13, 1:24 am, logic...@comcast.net wrote:
> 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.
z does not exist. Therefore z is not 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?
No, G'(w) is the identity bijection on w+1.
G'(w) exists. However, G'(w) does not contain
z, so G'(w) is not relevant.
Your putative G(w) does contain z. z does not exist.
Therefore G(w) does not exist.
Your method works for some bijections and does
not work for some bijections. Try starting with the following
bijection from w+1 to (w+1)\1.
F'' = {<0,w>, <w,0>, <1,2>, (2,3), (3,4) ...}
- William Hughes
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