Re: Separation,Power and Countability.

On Jun 20, 12:37 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
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.)

d cannot be equinumerous with PPw. because d is a subset of Pw. then d
cannot be supernumerous to Pw.
PPw is supernumerous to Pw, then d cannot be equinumerous with PPw.

Third, just to note, now you've been explicit that f needs only to be
an injection (not necesarily a bijection).

if there is injection from Pw to d, then this means that there is a
bijection from Pw to d.
Why?
Because since d is a subset of Pw, then f:d->Pw
f(x)=x is an injection.
Accordingly from Cantor-Bernsteing-Schroeder theorem we have that:
there exist a bijection between Pw and d.

More basically, I don't know what you think is at stake in showing
that d is dominated by Pw.

How can d be dominated by Pw. According to the continuum hypothesis, |
Pw|=Aleph-1
which is the set of all countable ordinals.
Since d is not countable, then d cannot be strictly dominated by Pw,
since by then it would be countable which is a paradox.

d is a subset of Pw, then d cannot be supernumerous to Pw.
So d is not subnumerous to Pw , nor it is supernumerous to Pw .So d
should be equinumerous to Pw.

But I asked for anyone in this forum to SHOW me an injection from Pw
to d. were d is uncountable and indefinable subset of Pw.

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

Then you're not allowing the generalized continuum hypothesis.

Ok no problem lets assume choice.



MoeBlee


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Relevant Pages

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  • Re: Separation,Power and Countability.
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  • Re: Separation,Power and Countability.
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