# No Identity Bijection for Omega - Theory and Concepts

This is a discussion on No Identity Bijection for Omega - Theory and Concepts ; Let B be the identity bijection of w+1 = w U {w} . B is an ordered set of ordered pairs. B = &lt; &lt;0,0&gt;, &lt;1,1&gt;, &lt;2,2&gt;, &lt;3,3&gt;, ..., &lt;w,w&gt; &gt; For each element, B_i, let C_i be the set ...

1. ## No Identity Bijection for Omega

Let B be the identity bijection of w+1 = w U {w} .
B is an ordered set of ordered pairs.

B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >

For each element, B_i, let C_i be the set of
all B_i that appear in the bijection before B_i.
Let C_i also includes B_i. (Skeptics - this is the one to attack.)

Let D_i be the set that C_i is a bijection for.
(D_i is the set of all natural numbers that appear in C_i).

B_0 = <0,0>
B_1 = <1,1>
B_2 = <2,2>
....

B_w = <w,w>

C_0 = < <0,0> >
C_1 = < <0,0>, <1,1> >
C_2 = < <0,0>, <1,1>, <2,2> >
....
C_w = < <0,0>, <1,1>, <2,2>, ..., <w,w> >

D_0 = {0}
D_1 = {0,1}
D_2 = {0,1,2}
....
D_w = {0,1,2,...,w}

Assume D_k = w for some k.
D_k = {0,1,2,...,k} and k must be the largest natural number.
Therefore, no D_k = w.

Assume there exists a C_k that is a bijection for w.
Then there exists D_k.

Assume there exists an identity bijection, A, for w.
A must be an proper subset of B.
A must also be an initial segment of B.
If A is an initial segment of B then A = C_k for some k.

There is no identity bijection for Omega.

Russell
- 2 many 2 count

2. ## Re: No Identity Bijection for Omega

On Mon, 03 Sep 2007 19:28:20 -0700, logiclab@comcast.net
<logiclab@comcast.net> said:
> ...
> If A is an initial segment of B then A = C_k for some k.

Ooooohhhh!...*wince*...so hard to watch...

3. ## Re: No Identity Bijection for Omega

On Mon, 3 Sep 2007 logiclab@comcast.net wrote:

> Let B be the identity bijection of w+1 = w U {w} .
> B is an ordered set of ordered pairs.
>
> B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >
>

Baloney. You are over using <..>
B = { (0,0), (1,1), ... (omega_0, omega_0) }
= { (xi, xi) | xi <= omega_0 }

> For each element, B_i, let C_i be the set of

Elements are lower case. Thus for each element b_i... what's b_i?

> all B_i that appear in the bijection before B_i.

How are you ordering the elements of B?

> Let C_i also includes B_i. (Skeptics - this is the one to attack.)
>

What do you mean? Let b_i be in C_i?
Absurd, b_i can't appear before b_i.

> B_0 = <0,0>
> B_1 = <1,1>
> ...
> B_w = <w,w>
>
> C_0 = < <0,0> >

What do you mean <=? How are you ordering subsets of omega_0 + 1? Do
you intend subset? Recall you are already are using <= to mean order or
ordinals.

<snip>

> There is no identity bijection for Omega.
>

Fantastic. For your next famous theorem, prove you have no identity.

4. ## Re: No Identity Bijection for Omega

On Sep 3, 9:44 pm, William Elliot <ma...@hevanet.remove.com> wrote:
> On Mon, 3 Sep 2007 logic...@comcast.net wrote:
>
> > Let B be the identity bijection of w+1 = w U {w} .
> > B is an ordered set of ordered pairs.

>
> > B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >

>
> Baloney. You are over using <..>
> B = { (0,0), (1,1), ... (omega_0, omega_0) }
> = { (xi, xi) | xi <= omega_0 }

I defined B as an ordered set of ordered pairs.
We can do it your way if you want.

> > For each element, B_i, let C_i be the set of

>
> Elements are lower case. Thus for each element b_i... what's b_i?

The i_th element of B.

> > all B_i that appear in the bijection before B_i.

>
> How are you ordering the elements of B?

The standard ordering for the naturals.

> > Let C_i also includes B_i. (Skeptics - this is the one to attack.)

>
> What do you mean? Let b_i be in C_i?
> Absurd, b_i can't appear before b_i.

I define c_k as the set of all ordered pairs, <i,j>, in B such that
i<=k.

> > B_0 = <0,0>
> > B_1 = <1,1>
> > ...
> > B_w = <w,w>

>
> > C_0 = < <0,0> >

>
> What do you mean <=? How are you ordering subsets of omega_0 + 1? Do
> you intend subset? Recall you are already are using <= to mean order or
> ordinals.

I am using <> for ordered sets.
c_0 = <<0,0>>

>
> <snip>
>
> > There is no identity bijection for Omega.

>
> Fantastic. For your next famous theorem, prove you have no identity.

I don't go away that easily.

Russell
- The universe is one dimensional

5. ## Re: No Identity Bijection for Omega

On Mon, 3 Sep 2007 logiclab@comcast.net wrote:
> On Sep 3, 9:44 pm, William Elliot <ma...@hevanet.remove.com> wrote:
> > On Mon, 3 Sep 2007 logic...@comcast.net wrote:
> >
> > > Let B be the identity bijection of w+1 = w U {w} .
> > > B is an ordered set of ordered pairs.

> >
> > B = { (0,0), (1,1), ... (omega_0, omega_0) }
> > = { (xi, xi) | xi <= omega_0 }

>

I use the Greek letter 'xi' in accordance to the convention of
using Greek letters for ordinals,

> > > For each element, B_i, let C_i be the set of

> >
> > Elements are lower case. Thus for each element b_i... what's b_i?

>
> The i_th element of B.
>

What is the i_th element of R?

> > > all B_i that appear in the bijection before B_i.

> >
> > How are you ordering the elements of B?

>
> The standard ordering for the naturals.
>

As the elements of B aren't integers, how are you ordering them?

> > > Let C_i also includes B_i. (Skeptics - this is the one to attack.)

> >
> > What do you mean? Let b_i be in C_i?
> > Absurd, b_i can't appear before b_i.

>
> I define c_k as the set of all ordered pairs, <i,j>, in B such that
> i<=k.
>

C_k = { (i,j) in B | i <= k } = { (i,i) | i <= k }

Ok, clearer than before and of course
C_omega0 = { (xi,xi) | xi <= omega_0 } = B

> > > B_0 = <0,0>
> > > B_1 = <1,1>
> > > ...
> > > B_w = <w,w>

> >

Now finally you get around to it and define
B_xi = (xi, xi)
for all xi <= omega_0.

> > > C_0 = < <0,0> >

> >
> > What do you mean <=? How are you ordering subsets of omega_0 + 1? Do
> > you intend subset? Recall you are already are using <= to mean order or
> > ordinals.

>
> I am using <> for ordered sets.

Ok, what order does <> imply?

> c_0 = <<0,0>>
>

Do I understand this?
C_0 is the ordered set containing only <0,0>
which is the ordered set for 0?

> > > There is no identity bijection for Omega.

> > Fantastic. For your next famous theorem, prove you have no identity.

> I don't go away that easily.
>

Ok, give a less shabby presentation and I'll review it.

6. ## Re: No Identity Bijection for Omega

On Mon, 03 Sep 2007 19:28:20 -0700, logiclab@comcast.net wrote:

>Let B be the identity bijection of w+1 = w U {w} .
>B is an ordered set of ordered pairs.
>
>B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >
>
>For each element, B_i, let C_i be the set of
>all B_i that appear in the bijection before B_i.
>Let C_i also includes B_i. (Skeptics - this is the one to attack.)
>
>Let D_i be the set that C_i is a bijection for.
>(D_i is the set of all natural numbers that appear in C_i).
>
>B_0 = <0,0>
>B_1 = <1,1>
>B_2 = <2,2>
>...
>
>B_w = <w,w>
>
>
>C_0 = < <0,0> >
>C_1 = < <0,0>, <1,1> >
>C_2 = < <0,0>, <1,1>, <2,2> >
>...
>C_w = < <0,0>, <1,1>, <2,2>, ..., <w,w> >
>
>
>D_0 = {0}
>D_1 = {0,1}
>D_2 = {0,1,2}
>...
>D_w = {0,1,2,...,w}
>
>
>Assume D_k = w for some k.
>D_k = {0,1,2,...,k} and k must be the largest natural number.
>Therefore, no D_k = w.
>
>Assume there exists a C_k that is a bijection for w.
>Then there exists D_k.
>
>Assume there exists an identity bijection, A, for w.
>A must be an proper subset of B.

Yes.

>A must also be an initial segment of B.

Yes.

>If A is an initial segment of B then A = C_k for some k.

No. Try to give us a _proof_ that any initial segment
must be C_k for some k. Keep in mind that there's
a simple counterexample, so the proof might be tricky.

>There is no identity bijection for Omega.
>
>
>Russell
>- 2 many 2 count

************************

David C. Ullrich

7. ## Re: No Identity Bijection for Omega

logiclab@comcast.net wrote:
> On Sep 3, 9:44 pm, William Elliot <ma...@hevanet.remove.com> wrote:
>> On Mon, 3 Sep 2007 logic...@comcast.net wrote:
>>
>>> Let B be the identity bijection of w+1 = w U {w} .
>>> B is an ordered set of ordered pairs.
>>> B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >

>> Baloney. You are over using <..>
>> B = { (0,0), (1,1), ... (omega_0, omega_0) }
>> = { (xi, xi) | xi <= omega_0 }

>
> I defined B as an ordered set of ordered pairs.
> We can do it your way if you want.
>
>>> For each element, B_i, let C_i be the set of

>> Elements are lower case. Thus for each element b_i... what's b_i?

>
> The i_th element of B.
>
>>> all B_i that appear in the bijection before B_i.

>> How are you ordering the elements of B?

>
> The standard ordering for the naturals.

The elements of B are not natural numbers. They are not even all ordered
pairs of natural numbers - think about (omega_0, omega_0) - so even if
you order them by first element, you need to say a bit more to define a
total order for B.

Patricia

8. ## Re: No Identity Bijection for Omega

OK...I'll chime in, too.

On Sep 3, 10:28 pm, logic...@comcast.net wrote:
>
> There is no identity bijection for Omega.

Why do you bother with this bijection stuff? Just cut out all the
ordered pair confusion and prove that w doesn't exist. I think that's
what your proof is really trying to say. Something like:

Assume k = w for some k.
k = k and k must be the largest natural number.
Therefore, no k = w.

Assume there exists a k that is w.
Then there exists k.
...
There is no Omega.

I think that kind of proof is something we could all work with.

Mitch

9. ## Re: No Identity Bijection for Omega

On Sep 3, 10:28 pm, logic...@comcast.net wrote:
> Let B be the identity bijection of w+1 = w U {w} .
> B is an ordered set of ordered pairs.
>
> B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >

Then, OBVIOUSLY, the identity bijection for w is just THE
SET YOU GET WHEN YOU REMOVE <w,w> FROM B.

More to the point, it is an easy theorem of ZFC that EVERY set
has an identity bijection. EVERY one.

10. ## Re: No Identity Bijection for Omega

On Sep 3, 10:28 pm, logic...@comcast.net wrote:
> Let B be the identity bijection of w+1 = w U {w} .
> B is an ordered set of ordered pairs.

There is no such thing as an ordered set.
You can IMPOSE orderings on the set from the OUTside
but the set INherently does NOT have any particular order
of its members.

>
> B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >
>
> For each element, B_i, let C_i be the set of
> all B_i that appear in the bijection before B_i.

We ALL KNOW what "before" means. It IS in the dictionary.
So don't trip yourself up here.

> Let C_i also includes B_i. (Skeptics - this is the one to attack.)

Not "skeptics": just SPEAKERS OF ENGLISH.
NOTHING is EVER *before* ITSELF! And that really should be
"let C_i be the set of all B_*J* that appear in the bijection before
B_i".
NOt that it ever matters where anything "appears". Everybody can just
agree
that 3 comes before 4 EVEN if you write it after.