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7

Let me address the question as stated first. Why does $\sf CH$ has no determinate provability from any of the axioms we throw at it? This is false. As remarked in the comments. Plenty of axioms prove $\sf CH$ or disprove it. Things like $V=L$ or $\lozenge$ imply $\sf CH$ whereas things like $\sf PFA$ and similar forcing axioms imply its negation (these in ...


7

Yes, this is inconsistent, and your argument is a way of proving this. In fact, $\mathsf{AD}^*$ is an overkill, and we can prove that there is an undetermined game with $A=\omega_1$. To see this, note that either there is an undetermined game on integers (and we are done), or else $\mathsf{AD}$ holds, so $\omega_1$ does not inject into $\mathbb R$. (The ...


4

Assuming consistency, there are infinitely many undecidable statements. The set of decidable statements is recursively enumerable, but not recursive. Which means the set of undecidable statements is not recursively enumerable - that is, any computer program which enumerates only undecidable statements will fail to enumerate all undecidable statements. So ...


4

As indicated in the other answer, there are countably many undecidable statements (the largest possible number), and this is an immediate consequence of the incompleteness theorem. The version of incompleteness that Andrews is using is that if $T$ is a consistent, recursive set of axioms that interprets a modicum of arithmetic, then the set of $T$-decidable ...


4

The first thing I'd do in your situation is to prove (if it's not already done in the source from which you quote) that the ranges of normal functions $\omega_1\to\omega_1$ are exactly the closed unbounded subsets of $\omega_1$. So the stationary sets are exactly the sets that have nonempty intersection with every closed unbounded set. Second, you should ...


2

For the purpose of applying set theory as a foundational subject, one must be able to express typical mathematical reasoning, such as Let $r$ be a real number... Let $n$ be an integer... Let $a$ be a sequence of real numbers... Let $f$ be a real-valued function of the reals... Let $S$ be a subset of the reals... And in general, if you have a kind of ...


1

I know that both NF and NFU are supposed to be quite weak. NFU is weaker than PA and it has the same consistency strength as MacLane set theory (Z with bounded quantifiers). MacLane set theory is significantly stronger than PA. It has the same consistency strength as Russellian unramified typed set theory (TST). In an answer to another question, I tried ...



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