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I've heard this joke thrown around:

The Axiom of Choice is obviously true, the well-ordering principle is obviously false, and who can guess about Zorn's lemma?

Of course, all three statements are equivalent. Are there non trivial reformulations of any of other axioms for ZFC?

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  • $\begingroup$ Btw. if you drop the power set axiom, these statements are no longer equivalent - something we have to keep in mind when working with models of weak fragments of $\operatorname{ZFC}$. $\endgroup$ – Stefan Mesken Nov 23 '16 at 16:00
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    $\begingroup$ Foundation admits a nice reformulation, as the schema of $\in$-induction. $\endgroup$ – Andrés E. Caicedo Nov 23 '16 at 16:01
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An interesting reformulation of ZF uses levels (the levels of the Von Neumann hierarchy, actually). We have of variables two types: sets ($x$,$y$,...) and levels ($V$, $V'$,...). Can be proved that ZF is equivalent to Extensionality, Separation, Infinity, Replacement and:

Accumulation: $$\forall V'\forall x[x\in V'\iff\exists V\in V'(x\in V\vee x\subset V)].$$

Restriction: $$\forall x\exists V(x\subset V).$$

(exercise 3.11.9 of Set Theory by Frank Drake)

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Sure, as many as you like. For example:

  • Relative to the other axioms of $\operatorname{ZFC}$, the axiom scheme of replacement and the axiom scheme of collection are equivalent. However, this is no longer true, when dropping the power set axiom.
  • $\operatorname{ZFC}$ is equivalent to $\operatorname{ZFC}^0$, where the latter doesn't allow parameters in its axiom schemes. See here.
  • There are, relative to the other axioms of $\operatorname{ZFC}$, many equivalent formulations of the axiom of infinity, e.g. "$V_{\omega}$ exists".
  • The usual formulation of $\operatorname{ZFC}$ is redundant. Therefore you can leave out any of the redundant axioms, say $\phi$. Then, relative to the other axioms of $\operatorname{ZFC}$, $\phi$ is equivalent to $\psi$ for any $\psi$ with $\operatorname{ZFC} \vdash \psi$.
  • ...
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    $\begingroup$ When you say "[t]he usual formulation of ZFC is redundant", is this proven or are you saying if that is true, we can leave out the redundant axioms? $\endgroup$ – Nitin Nov 23 '16 at 16:11
  • $\begingroup$ @Nitin I'm saying that there are axioms included, that already follow from others. In particular, separation and pairing are entirely redundant see here. $\endgroup$ – Stefan Mesken Nov 23 '16 at 16:15
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    $\begingroup$ Plus, any presentation of replacement I am aware of includes redundant axioms. $\endgroup$ – Andrés E. Caicedo Nov 24 '16 at 14:46
  • $\begingroup$ @AndrésE.Caicedo That got me thinking: Is there a $\subseteq$-minimal subset of $\operatorname{ZFC}$ equivalent to $\operatorname{ZFC}$ that is not recursively enumerable? If so, we should use that instead! $\endgroup$ – Stefan Mesken Nov 24 '16 at 18:10
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Something that I have read of, but have not studied: Take the versions of the Comprehension and Separation schemas in Kunen [1]. Kurt Godel showed that in ZF minus Comprehension, the Comprehension schema can be replaced by 8 sentences that he presented (or,equivalenty, by one cumbersome sentence.) I dk if Infinity matters here but I doubt it.

Reference: [1].Kunen, K. Set Theory: An Introduction to Independence Proofs.

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  • $\begingroup$ That would imply that ZF is finitely axiomatizable, which is not true. $\endgroup$ – Asaf Karagila Nov 24 '16 at 12:20
  • $\begingroup$ @AsafKaragila Why? The system still has replacement, I assume. If anything, comprehension can be eliminated. $\endgroup$ – Andrés E. Caicedo Nov 24 '16 at 14:01
  • $\begingroup$ Do you have a reference for this? Gödel showed how we can use a finite list of functions to code the constructible universe rather than having to formalize definability, but I do not recall him addressing comprehension specifically. (He worked in a system different from ZF anyway.) $\endgroup$ – Andrés E. Caicedo Nov 24 '16 at 14:04
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    $\begingroup$ @Andrés: I see, but in that case separtaion is just a theorem and can be removed altogether (or replaced by something like $\exists x(x=x)$ or any other valid sentence). $\endgroup$ – Asaf Karagila Nov 24 '16 at 14:06

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