Re: CH Question

From: Charlie-Boo (chvol_at_aol.com)
Date: 10/23/04 

Date: 23 Oct 2004 01:57:41 -0700

mtx014@linux.services.coventry.ac.uk (Robert Low) wrote
> Charlie-Boo <chvol@aol.com> wrote:
> >generally "functions of P(N)" such as P(N)u{0}. Then I wondered if
> >anyone knew of any other way of defining (c.f. proving) a set A that
> >satisfies N<A and posted the general question of A<B where B is not a
> >function of P(A). ?tay?
>
> So, is the example I gave an example of what you asked for?
>
> A = N
> B = continuous functions from R to R.
>
> Or am I currently thrashing about on deck with a hook in
> the corner of my mouth?

(Or perhaps I am the captain with a hook for a hand - explaining why
my writing is cryptic?)

It's not a really good example for my purposes. There is a (simple)
1-1 correspondence between subsets of N and real numbers between 0 and
1. So the set of real numbers is the set of subsets of N i.e. P(N)
and a map from the set of real numbers is a map from P(N), which is a
function of P(N). It doesn't give me a way to construct an A such
that N<A and we don't know immediately that A >= 2^N.

(What interpretations of 2^N do we know of (subset of N, function over
N, real number, infinite string)? Is one more general than the rest
in that they are all special cases of it while it is not a special
case of any of the others?)

BTW The original question is not a strict mathematical decision. I am
not asking if such sets formally exist. That would require a more
precise definition of "function of P(N)". The point is, how do we
define a set A such that N<A? Methods used that produce only sets A
>= 2^N do not satisfy the second equation A<2^N.

C-B

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