Re: what makes it true?

mensanator@xxxxxxxxxxx wrote:
>
> Timothy Little wrote:
>> For CH, it turned out that we can assert the existence of such an
>> entity in ZFC without introducing any new inconsistency.
>
> I couldn't, at first, figure out what an "entity" means in regard to
> CH. Would it be a set of numbers whose cardinality is greater than
> the natural numbers but less than the reals?

A set of any sort, but a set of numbers quickly follows. If a set X
of suitable size existed, then an injection would exist from it to the
reals (because |X| < |R|), and we can take the image of that
injection -- which contains numbers.


>> It would be weirder, because one would expect that if there is a
>> consistent statement asserting the existence of a natural number
>> satisfying some property, we should be able to find one (in
>> principle) by just counting upward until we hit it.
>
> This is partly what's confusing me. It seems to me that this means
> it is decidable but maybe that's not what decidable means.

I didn't explicitly specify that I was referring to two different
systems: ZFC (in which GC may be undecidable), and ZFC + not-GC (in
which it is decidable).

In the latter system a counterexample exists, but the only way to
produce it is by opaquely invoking the not-GC axiom. Much like we
opaquely invoke the Axiom of Infinity to assert the existence of N,
even though we can never finish writing all its elements.


>> However, such a thing would not be without precedent: Goodstein's
>> theorem is undecidable in Peano arithmetic. The negation asserts
>> the existence of a natural number having a certain property. That
>> negation is consistent with the rest of PA.
>
> But you could (in principle) count upwards until you find one with
> that property.

Given what we intuitively mean by a natural number, we should be able
to. However, we know (from ZFC) that if our natural numbers are
modelled by finite ordinals, that will never happen. So the axioms of
PA allow 'natural numbers' that are *not* modelled by finite ordinals.
Our counterexample will be one of those.


>> However, we "know" (from ZFC) that if you plug in any natural
>> number, it will fail to have the property. So weird things do
>> happen!
>
> I'm not familiar with Goodstein's theorem. Is that a case where
> undecidable in PA ends up being true because ZFC proves that no
> number can be a counter example?

In ZFC, we can show that the nonzero terms in the sequence are bounded
above by a strictly decreasing sequence of ordinals, and we know that
every such sequence of ordinals is finite.

In PA, the concept of an ordinal cannot even be expressed. The fact
that the proposition is undecidable in PA can be interpreted as
meaning that the axioms of PA do not tightly enough specify what we
mean by a natural number. We can form models that obey the axioms of
PA, but behave differently.

Goedel's 1st incompleteness theorem can be interpreted as saying that
we cannot ever specify exactly what we mean by a natural number.


- Tim
.

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