Re: Continuum hypothesis

Rupert replied:
True in the standard model.

That DOES NOT HELP.

I was trying to help herbzet, and it appears I succeeded.

Not to me it doesn't.

If you want me to help you,

There is nothing personal about this. You alleged that something
was "true(period)". YOU are the one who needs help.
Most of these things are true in some models and false in others.
YOU need help to avoid getting yourself convicted of lying about
the other models.

perhaps you should ask me politely.

I am better than most people think at computing when & toward whom
to grant politeness.

The standard model OF WHAT??

The third-order language of arithmetic.

This is a just plain stupid answer. You are in the context of 1st-
order ZFC! How is any model of "the third-order language of
arithmetic"
going to be ANY DIFFERENT from the UNIQUE standard model
of the FIRST-order language of arithmetic?!? Once you define N,
and restricting addition and multiplication to operating on N,
what all its subsets are and what all their subsets are is UNIQUELY
determined by "the standard model"!

At FIRST order, in the case of PA, it is reasonable to call
one of the models standard because it is known that there
are many different models. At 2nd-order, however, THE AXIOMS
ARE CATEGORICAL and it does NOT make sense to speak of
a standard model. There is only 1 model anyhow.

Two points:

(1) I had only mentioned a sentence. I hadn't mentioned any axioms.

That's cuz you're just ignorant like that.
The sentence cannot EXIST apart from the axioms.
The symbols in it can't be identified as MEANING THOSE symbols,
apart from the axioms.


I might have said "True in the unique model of the third-order Peano
axioms". That would have been an adequate answer, too.

No, it wouldn't, BECAUSE THE CONTEXT IS ZFC.
ZFC doesn't have "The Peano Axioms" at ANY order.
It INSTEAD has set-theoretical TRANSLATIONS of them.
And ONE model of FIRST-order ZFC has subsets and powersets of
ALL HIGHER "orders of arithmetic".

But my answer was just as good.
(2) If we allow Henkin models then they may be nonstandard.

But that is the whole point.
There are a whole bunch of different models definable, in 1st-order
ZFC,
but there is only 1 model of 2nd-order PA. You have to say something.

The continuum hypothesis is equivalent in ZFC

FIRST-order ZFC.

to a certain statement in the third-order language of arithmetic.
What's the problem?

The PROBLEM is that since YOU ARE IN FIRST-order ZFC,
YOU ARE NOT *IN* "the third-order" ANYthing! You are also
NOT IN "the language of arithmetic"! You have to TRANSLATE
and you have to clarify how you are mixing the orders!
There are COUNTABLE MODELS of 1st-order ZFC!
The UNIQUE model of 2nd-order PA *is NOT* countable!



.

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