5

The Axiom of Infinity states that at least one inductive set exists. Inductive sets are infinite, but not all infinite sets are inductive.

Suppose that we take ZFC with the negation of the Axiom of Infinity. Is it still possible for infinite sets to exist? I feel that this might be a dicey question, since ZFC's internal notion of an infinite set relies on ordinal numbers, and (I assume) the negation of the Axiom of Infinity precludes the existence of infinite ordinals. If this question is actually ambiguous as written, I'd be happy to have that be the answer to the question.

Thanks!

templatetypedef
  • 10,206
  • I'm fairly certain that any axiom that implies the existence of an infinite set (not a dedekind-infinite set) would imply the existence of the ordinal $\omega_0$ by Replacement. But then again there are models where some infinite set do not contain a countable subset without AC. I'd wait for Asaf for an authoritative answer. –  Jul 29 '14 at 14:53
  • 1
    Since ZFC include the axiom of infinity, "ZFC with the negation of the Axiom of Infinity" is inconsistent. Being serious now, what is your definition of infinite? – William Jul 29 '14 at 14:59
  • @William that is so pedantic that it hardly really even qualifies as a joke. (ed ajf) – Thomas Andrews Jul 29 '14 at 15:01
  • 3
    @Thomas It is not, on two accounts: 1. When we say $\mathsf{ZF}$ with choice, we understand $\mathsf{ZFC}$. The question is using the standard language incorrectly, and it is not entirely clear what is meant. Most likely it is $\mathsf{ZFC}-\mathrm{Inf}+\lnot\mathrm{Inf}$, where $\mathrm{Inf}$ is the standard axiom of infinity, but this is not for sure, since 2. The axiom of infinity specifically asserts that there are inductive sets, so it is conceivable to have a $\mathsf{ZFC}'$ where we have an axiom asserting the existence of infinite sets, and specifying that none are inductive. – Andrés E. Caicedo Jul 29 '14 at 15:15
  • @Thomas In particular, it may be good to specify what meaning of "infinite" is being considered here. The actual details of the theory being asked may matter, as different equivalent definitions may fail to be equivalent now. – Andrés E. Caicedo Jul 29 '14 at 15:17
  • 1
    You still have the axiom of choice so every set can be well ordered. Given a reasonable definition of finite, the existence of a not finite set should imply the ordinal $\omega$ exists. Hence an inductive set. – William Jul 29 '14 at 15:20
  • 1
    The question has been asked here before. You need to be more precise, but probably you mean the theory where we remove the axiom of infinity, and replace it with the axiom asserting that there are no inductive sets, this theory is mutually interpretable with Peano Arithmetic, so there is not much in terms of infinite sets here. We actually have bi-interpretability if we state foundation as an axiom schema (this is a delicate technical detail, it is needed to prove the existence of transitive closures). (Cont.) – Andrés E. Caicedo Jul 29 '14 at 15:25
  • 3
    See here, here, and here for details. It was also asked on this site whether we can simply state infinity as "there is an infinite set", rather than as "there is an inductive set", so your question is perhaps whether the existence of inductive sets follows from the existence of infinite sets. The answer is yes: If there is an infinite set, its double power set is Dedekind-infinite. Now see here. – Andrés E. Caicedo Jul 29 '14 at 15:28
  • 1
    (And, as William commented above, you really want $\mathsf{ZF}$ rather than $\mathsf{ZFC}$.) – Andrés E. Caicedo Jul 29 '14 at 15:29
  • http://math.stackexchange.com/questions/763544/non-constructive-axiom-of-infinity/ seems relevant. – Asaf Karagila Jul 29 '14 at 15:31

0 Answers0