
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  2 many 2 count

