Re: Separation,Power and Countability.

On Jun 20, 12:54 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Jun 19, 10:25 pm, zuhair <zaljo...@xxxxxxxxx> wrote:





On Jun 20, 12:15 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Jun 19, 9:36 pm, zuhair <zaljo...@xxxxxxxxx> wrote:

On Jun 19, 11:05 pm, Keith Ramsay <kram...@xxxxxxx> wrote:

On Jun 19, 7:26 am, zuhair <zaljo...@xxxxxxxxx> wrote:
|I asked for the proof of the following:
|For any set d that is a member of Pw and is indefinable what is the
|prove of d being equinumerous to w?

Meaning, I suppose, how can one prove that each
undefinable subset A of w is equinumerous with w.

If A is finite, then A is definable as {n : n is
a natural number and (n=a0 or n=a1 or ... or
n=a_m)} for some a0,...,a_m.

If A is undefinable, then, A is infinite and each
infinite subset of w is equinumerous with w. Let
a0 be the least element of A, and inductively let
a_{k+1} be the least element of A not among a1,...,
a_k. Then a_i gives a 1-1 correspondence between
A and w.

Keith Ramsay

This depends on choice.

No, it doesn't; think about it for a moment.

it doesn't for this particular example since w is well ordered, I
know. But I was talking about the bigger picture.

No, you weren't. You asked about w. Now you have a DIFFERENT
questions.

for example take PPw , and let d be a member of PPw
that is not a member of Pw and let d be indefinable
and uncountable.
Now what is the proof that Ef(f:Pw -> d)

what is that: the question is what is the proof that
Ef(f:Pw -> d, f is injective).


Let yed. Then let f map Pw into d by f(x) = y.

Oh, that's not what you had in mind, is it? Well, it answers what you
wrote, sorry.

Anyway, without the continuum hypothesis, I don't know why all such d
are equinumerous with Pw. Maybe they are, but I don't know why one
would think that they are.

well my question is of course without the continuum hypothesis. But
let's assume the continuum hypothesis
SHOW me an an injective function from Pw to d, Remember d is an
uncountable and indefinable subset of Pw.

Zuhair

of course I am speaking in a set theory without choice.

MoeBlee- Hide quoted text -

- Show quoted text -


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