Re: Continuum hypothesis

On Sep 3, 1:39 am, herbzet <herb...@xxxxxxxxx> wrote:
herbzet wrote:
Um, is the range of the second-order quantifiers not usually
the set of all _ordered_ subsets of the naturals?

No.


I mean:

At first-order we have eg Fabc is true if <a,b,c> is an
element of the set of ordered triples that constitute F.

Please!
At first-order, we MAY not even be DOING *set*-theory!
We may have a much weaker axiom-set!
It is normal for the MODEL of the 1st-order theory to be built
using 1st-order ZFC (in which case these ordered triples, and
sets of them, may exist), but just because F is usually going to
be INTERPRETED (in this context) to a relation does NOT mean
that F *is* a relation! First-order theories do not NEED models!
They can go on perfectly well existing without them!


So at second order AF(Fabc) is true if every 3-place
predicate in the domain contains <a,b,c> as an element.

Set theory axioms are even LESS likely to be present or relevant
at 2nd-order, or any higher order. The whole point of 1st-order ZFC
is
that it allows you to approximate the higher orders while remaining
technically
first-order. Once you decide to bite the bullet and actually ascend
to the higher
order, there is simply no more need for set-theoretical notation of
any kind.

So the domain is a set that has as elements every set of
ordered triples.

So the domain is the set of (at least) all sets of ordered n-tuples.

Or am I totally screwed up here?

You are mixing paradigms.
You need to be either in set theory or in higher-order logic,
and the twain do NOT meet.

.

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