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Topic: No Identity Bijection for Omega
Replies: 116   Last Post: Sep 22, 2007 5:58 PM

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 Jesse F. Hughes Posts: 9,776 Registered: 12/6/04
Re: No Identity Bijection for Omega
Posted: Sep 19, 2007 4:20 PM

logiclab@comcast.net writes:

> On Sep 18, 3:11 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> logic...@comcast.net writes:
>> > 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.
>
>> What happens is that the sequence of functions does not converge.
>
> How does a bijection converge?

This is a literally meaningless question.

Maybe you think that if we have a sequence of bijections, then the
sequence converges (and furthermore converges to a bijection). This
is clearly false.

> I descibe a family of bijections.
> 0 is paired with some natural number
> in every one of these bijections.
> I just want to know which natural number
> 0 is paired with when every other natural
> number has been paired with itself.

This question doesn't make any particular sense. I suppose what you
mean is this: You've defined a sequence of functions G(0), G(1),
... and you want to discuss the limit G(w). Namely, you want to ask:
What is G(w)(0)? But you have not shown that the limit G(w) exists
and it is indeed trivial to show that it does not exist.

Maybe we should be explicit about what we mean when we say that a
sequence G(0), G(1), ... converges to a function H. Let G(i): A -> B
be a function for each i in N. We say that the sequence G(0),
G(1),... converges to H if for each n in N, there is a j in N such
that for all k > j:

G(k)(n) = G(j)(n).

Now it is trivial to show that, in your particular example, the
sequence does not converge: For any j in N,

G(j)(0) != G(j+1)(0).

So all these questions of what happens in the limit are silly. Your
sequence of functions diverges.

>> What also happens is that you pretend it *does* converge and use the
>> obvious contradiction to introduce a fictional number z.

>
> I only rearrange the pairs in F.
> z is a member of some pair in F.

Nonsense. There is no number z satisfying your silly delusions.

--
Mo memorized the dictionary
But just can't seem to find a job
Or anyone who wants to marry "Memorizin' Mo",
Someone who memorized the dictionary. Shel Silverstein

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