1

Let's denote by ZF the Zermelo-Fraenkel axioms, and by AC the Axiom of Choice. I am sorry that the question is not very well formulated, but still:

can there be a "number theoretic statement" (by which I understand a 1st order logic sentence in the language of sets which is true iff a certain "number theoretic" fact is true), such that it follows from ZF+AC, and whose negation follows from ZF+$\neg$AC?

John Donn
  • 131

1 Answers1

2

There are no first-order sentence like that.

If $\varphi$ is such sentence, then it holds in $V$ if and only if it holds in $L$, which satisfies choice. So if $\varphi$ follows from choice, it must holds in $L$ and thus in $V$, so it cannot be disproved by $\lnot\sf AC$. And similarly, if $V$ satisfies $\lnot\sf AC$, and $\varphi$ holds there, it must hold in $L$.

In fact, you can push $\varphi$ to a relatively simple second-order sentence, $\Sigma^1_2$ specifically, by Shoenfield's absoluteness theorem.

Asaf Karagila
  • 393,674
  • I say we close it as duplicate of the first question I linked to. – Kenny Lau Sep 23 '17 at 14:20
  • What do you mean by V and L? (sorry, I have no formal education in set theory, and the informal one is quite limited). – John Donn Sep 23 '17 at 14:26
  • $V$ is the usual letter denoting "the" set theoretic universe. $L$ is Gödel's constructible universe where the axiom of choice is provably true. – Asaf Karagila Sep 23 '17 at 14:27
  • @KennyLau: For that to be "we" someone else had to vote as well... :P – Asaf Karagila Sep 23 '17 at 14:28
  • 1
    Although Shoenfield absoluteness is about $\Sigma^1_2$ sentences, it actually lets you eliminate AC from proofs of $\Pi^1_4$ statements. If ZFC proves $(\forall x)(\exists y)\phi(x,y)$, where $x,y$ range over reals and $\phi$ is $\Pi^1_2$, then in just ZF you can argue as follows. Given any $x$, look in $L[x]$, which satisfies AC, and find some $y\in L[x]$ such that $L[x]$ satisfies $\phi(x,y)$. By Shoenfield absoluteness, $\phi(x,y)$ holds (in $V$). – Andreas Blass Sep 23 '17 at 16:12
  • @Andreas: That is pretty damn cool. – Asaf Karagila Sep 23 '17 at 16:13
  • 1
    I think this argument for $\Pi^1_4$ is in a paper, "Eliminating the continuum hypothesis" by Richard Platek, which, despite the title, also has results about eliminating or reducing uses of AC. – Andreas Blass Sep 23 '17 at 16:17