Re: Separation,Power and Countability.

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)

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.

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

MoeBlee

.

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