No Identity Bijection for Omega

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

On Sep 13, 7:30 am, William Hughes <wpihug...@hotmail.com> wrote:
> 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) ...}

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

X = A intersect B = {0,2,3,...,w}

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


Russell
- 2 many 2 count

On Sep 13, 10:49 pm, logic...@comcast.net wrote:
> On Sep 13, 7:30 am, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > 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) ...}
>
> F'' = {<0,w>, <w,0>, <1,2>, (2,3), (3,4) ...}
>
> X = A intersect B = {0,2,3,...,w}
>
> G(0) = {<0,0>, <1,2>, (2,3), (3,4), ..., <w,w>}
> G(2) = {<0,0>, <1,3>, (2,2), (3,4), ..., <w,w>}
> G(3) = {<0,0>, <1,4>, (2,2), (3,3), ..., <w,w>}
> ...
> G(w) = {<0,0>, <1,z>, (2,2), (3,3), ..., <w,w>}
>

and since z does not exist, G(w) does not exist.
Your method will not work for F''


- William Hughes

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.)

Consider:

(F\{<w w> <x-1 x> <x F(x)>}) u {<x x> <x-1 F(x)>}.

I.e, consider:

(F\{<w w> <x-1 x> <x x+1>}) u {<x x> <x-1 x+1>}

> Call this new bijection G(x).

For x e w\{0}, we have G(x) = (F\{<w w> <x-1 x> <x x+1>}) u {<x x>
<x-1 x+1>}.

> Assume the elements of X can be ordered.

We don't need to assume it. There are all kinds of orderings of X.

> G(x_i+1) = G(x_i)

The domain of G is w. So you don't need extra notation 'x+i1'.

All you need is G(x), G(x+1), G(x-1) (if ~x=0), etc.

> - {<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).

Whatever I would get from sorting through your bad notation, you've
ALREADY defined G(x) for all x in w, so now defining G(x+1) or
whatever, should not conflict with the definition of G for all x,
including each x+1. So, I'm just going to assume we're working with G
as you already defined it.

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

No, w is not in the domain of G. When you defined G, you said, "Assume
x is not paired with itself in F."

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

No (as I said, I'm using your original definition of G), G(2) = G(x)
for x=2.
So G(2) = (F\{<w w> <1 2> <2 3>}) u {<2 2> <1 3>} =
{<0 1> <1 3> <2 2> <3 4> <4 5> <5 6> ...}

And w is not in the domain of G.

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

w is not in the domain of G. And 'z' is some letter that has no given
meaning by you other than, presumably to stand for G(w)(0), which is
not coherent since w is not in the domain of G.

If you want to define G so that w is in the domain of G, and with
coherent notation, then let me know. And please do use clear notation
such as '<x F(x)>' rather than ridicuously inept notation such as
'<x,b_n>' and such atrocious garbage as '{<x_i+1,x_i+1>,<m,n>}' .

MoeBlee

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

On Sep 17, 10:30 pm, logic...@comcast.net wrote:

>
> 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.

If <s t> e F, then either

s>t or s=t=w

So z does not exist.

- William Hughes

On Sep 17, 7:30 pm, logic...@comcast.net wrote:

> 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:

I mentioned meA and neB because you did and I thought you'd do
something with them toward a universal generalization. But you
didn't.

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

You're defining G as a function. You need to say what the domain is.
Also now 'n' and 'm' are free variables here when the only free
variable should be 'x'. If you are definining the function G with one
argument x, then the right side of the definition can't have any free
variables other than 'x'. Or, if you allow other free variables, then
the definition is still "relative" to those free variables as
"parameters" and you will have not completely "nailed" just what G is.

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

Okay, now that is good. (Use 'u' instead of '+' though.) You dropped
'n' and 'm', which is good. But you still need to say what the domain
of G is. That is, what values can 'x' take. Unless you say otherwise,
I will suppose that the domain of G is the domain of F = w+1.

Notice, though, that when <F(x) x> not in F, then there is no result
of "taking out". That's okay, but keep it in mind just in case.

In particular for x=w, we have:

G(w) = (F\{<w w> <w w>}) u {<w w> <w w>} = F.

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

I think that's right.

> 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.)

Now you say X in a way that suggests it's the domain of of G(x).

Also, why do you say "assuming X can be ordered"? There are all kinds
of orderings of X. We don't need to assume there is one, as it's
proven (and blazingly obvious anyway) that there is one. Which one do
you want? The standard ordering?

> 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>}

Okay.

> G({1 2}) = G2 = G1 - {<2 G1(2)> <G1(2) 2>} + {<2 2> <G1(2) G1(2)>

Now, you're putting {1 2} in the domain of G. So the domain of G was
not w+1 after all. So what exactly is the domain of G?

I'll stop there until you get it straightened out.

If you start your construction by defining G, then say exactly what
the domain of G is. Then define other functions G1, G2, whatever, but
don't change what G is after you've already defined it.

MoeBlee.

On Sep 18, 10:00 am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Sep 17, 7:30 pm, logic...@comcast.net wrote:
>
> > 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:
>
> I mentioned meA and neB because you did and I thought you'd do
> something with them toward a universal generalization. But you
> didn't.

As George points out, I need to use the
free variables m and n:

X = A/\B
xeX

Ex(<m x> e F) -> meA
Ex(<x n> e F) -> neB

Since X can be ordered, let x1 be the first
element of X.

Define a bijection, G1:

G1(x1) = F - {<x1 n> <m x1>} u {<x1 x1> <m n>}

Now define another bijection, G2.

G2(x2) = G1 - {<x2 n> <m x2>} u {<x2 x2> <m n>}

Repeat this process for every element of X.

Several people have pointed out this process
"never ends" and/or "doesn't converge".

With some tips from you and others, I have
a simpler proof that makes recursive calls to F.

A = w+1
B = A-{0}
X = A/\B= B

F = {<0 1> <1 2> <2 3> ... <w w>}

I want to prove that if F is
a bijection between A and B,
there exists a bijection, S,
between A-B and B-A.

Take the first (and only)
element of A-B = 0.

If F(0) is not an element of X,
then F(0) must be an element
of B-A proving there exists a
bijection between A-B and B-A.

Assume F(0) is an element of X.
Then <F(0) F(F0))> must be an element of F.

Define a new bijection, S1.

S1 = F - {<0 F(0)> <F(0) F(F(0))>}
u {<0 F(F(0))> <F(0) F(0)>}

Consider S1(0) = F(F(0)).

If S1(0) is an element of X
then <F(F(0)) F(F(F(0))> is an element of F.

Define a new bijection, S2.

S2 = F - {<0 F(0)> <F(0) F(F(0))>}
u {<0 F(F(0))> <F(0) F(0)>}

Continue this process until
Sz(0) is an element of B-A.

This process must terminate
if F is a bijection.

It is clear that

Si(0) = F(F(...F(0)...))

where i is the number of times F(0)
has been iterated.

F(0) can be iterated at most |X| times.


Russell
- 2 many 2 count

On Sep 20, 12:16 am, logic...@comcast.net wrote:
> On Sep 18, 10:00 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
>
>
> > On Sep 17, 7:30 pm, logic...@comcast.net wrote:
>
> > > 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:
>
> > I mentioned meA and neB because you did and I thought you'd do
> > something with them toward a universal generalization. But you
> > didn't.
>
> As George points out, I need to use the
> free variables m and n:
>
> X = A/\B
> xeX
>
> Ex(<m x> e F) -> meA
> Ex(<x n> e F) -> neB
>
> Since X can be ordered, let x1 be the first
> element of X.
>
> Define a bijection, G1:
>
> G1(x1) = F - {<x1 n> <m x1>} u {<x1 x1> <m n>}
>
> Now define another bijection, G2.
>
> G2(x2) = G1 - {<x2 n> <m x2>} u {<x2 x2> <m n>}
>
> Repeat this process for every element of X.
>
> Several people have pointed out this process
> "never ends" and/or "doesn't converge".
>
> With some tips from you and others, I have
> a simpler proof that makes recursive calls to F.
>
> A = w+1
> B = A-{0}
> X = A/\B= B
>
> F = {<0 1> <1 2> <2 3> ... <w w>}
>
> I want to prove that if F is
> a bijection between A and B,
> there exists a bijection, S,
> between A-B and B-A.
>
> Take the first (and only)
> element of A-B = 0.
>
> If F(0) is not an element of X,
> then F(0) must be an element
> of B-A proving there exists a
> bijection between A-B and B-A.
>
> Assume F(0) is an element of X.
> Then <F(0) F(F0))> must be an element of F.
>
> Define a new bijection, S1.
>
> S1 = F - {<0 F(0)> <F(0) F(F(0))>}
> u {<0 F(F(0))> <F(0) F(0)>}
>
> Consider S1(0) = F(F(0)).
>
> If S1(0) is an element of X
> then <F(F(0)) F(F(F(0))> is an element of F.
>
> Define a new bijection, S2.
>
> S2 = F - {<0 F(0)> <F(0) F(F(0))>}
> u {<0 F(F(0))> <F(0) F(0)>}
>
> Continue this process until
> Sz(0) is an element of B-A.
>

Since B-A is the empty set this process will
never terminate.

> This process must terminate
> if F is a bijection.
>
> It is clear that
>
> Si(0) = F(F(...F(0)...))
>
> where i is the number of times F(0)
> has been iterated.
>
> F(0) can be iterated at most |X| times.
>


So if |X| is infinite, then F(0) can be iterated
at most an infinite number of times.

- William Hughes