Re: Separation,Power and Countability.
- From: MoeBlee <jazzmobe@xxxxxxxxxxx>
- Date: Wed, 20 Jun 2007 10:37:32 -0700
On Jun 20, 9:24 am, zuhair <zaljo...@xxxxxxxxx> wrote:
On Jun 20, 12:54 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
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).
Yes, you need to say "f is injective" if that is what you mean.
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.
First, if d is not defined as a particular set, then, it is not clear
to me what we can show any PARTICULAR function that has d as a
superset of the range. Second, even WITH the generalized continuum
hypothesis (I meant 'generalized' in my first mention also), I don't
know that there is a proof that Pw dominates d. (With the generalized
continuum hypothesis, d is either equinumerous with Pw or with PPw.)
Third, just to note, now you've been explicit that f needs only to be
an injection (not necesarily a bijection).
More basically, I don't know what you think is at stake in showing
that d is dominated by Pw.
of course I am speaking in a set theory without choice.
Then you're not allowing the generalized continuum hypothesis.
MoeBlee
.
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