Re: Separation,Power and Countability.

On Jun 20, 5:34 pm, zuhair <zaljo...@xxxxxxxxx> wrote:
On Jun 20, 6:37 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

On Jun 20, 4:34 pm, zuhair <zaljo...@xxxxxxxxx> wrote:

Just an intuition. Nothing else.

It's not even clear what the question MEANS given that d is not
definable. How can you define a function with regard to a set that is
not itself defined?

Yes, that's the beauty of the subject.

Sorry, but inspiration of beauty dosn't strike me too much from
vagueness. I adore the beauty of clarity and the transformation of
intuitive notions into rigorous mathematics. No transformation from
intuition into mathematics, then no mathematical cigar, as it were.
What is beautiful to me is the ingeniousness of finding how to express
the intuitions AS mathematics and the ingeniousness of finding the
logical paths of proof and construction.

Review Keith Ramsy's proof of the injective function from
w to d , were d is an indefinable subset of w. It's a nice example of
such an injective function from a definable set to indefinable set.

It's 'Ramsay'.

I know that proof; it's a standard proof that any infinite subset of w
is denumerable. Undefinability has nothing to do with it except for
the fact that if a subset of w is undefinable then the subset is
infinite. In other words, the proof is about infinite subsets of w;
whether or not they're definable is irrelevent.

With choice, maybe some approach along these lines?:

Let 'f"x' stand for the image of f from x.

Let d be well ordered.

Let Pw be inexed by an ordinal K.

So, instead of an injection from Pw into d, find an injection from K
into D?

Maybe a function f on K by transfinite recursion?

f(0) = least member of d.
f(n+) = least member of f"(n+)
f(L) = the least membe of ???, for limit ordinals L.

So, how to fill in '???' to complete the definition of f?

Somehow, maybe find some relation between limit ordinals L and the
subsets of w that are indexed by those limit ordinals? Hmm, not enough
information. I don't know...

MoeBlee










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