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In his Senior Thesis, Samuel Coskey answered the question of which axioms of $ZFC$ hold at each stage of the cumulative hierarchy. Here is the list of his results:

Axioms that always hold: Extensionality, Foundation, Union, Axiom Schema of Separation, Choice.

Axioms that hold in $V_{\alpha}$ iff $\alpha$$\gt$0: Empty Set.

Axioms that hold in $V_{\alpha}$ iff $\alpha$$\gt$$\omega$: Infinity.

Axioms that hold at limit ordinals: Power Set, Pairing.

Axioms that hold at inaccessible cardinals: Axiom Schema of Replacement.

My question is simply this:

What changes (if anything) in these results if Choice is dropped?

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  • $\begingroup$ @MartinSleziak: Thanks for the edit. $\endgroup$
    – Thomas Benjamin
    Commented Jul 21, 2016 at 14:29

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If you drop choice, then of course choice is no longer in the "always holds" category, but there's nothing to determine at what level of the cumulative hierarchy it first fails. The first failure of choice could occur just a few levels past $\omega$, or it could occur far beyond the first inaccessible cardinal. (The vagueness of "a few levels" comes from the details of how you formulate choice; formulations that are equivalent in ZF need not be equivalent when basic things like pairing are unavailable.)

As far as I can see, the only other thing that might change in the results you quoted is that definitions of inaccessibility that are equivalent in ZFC are no longer equivalent in ZF, so you need to be careful about which version of the definition you use. For more details than you probably want about various definitions of inaccessibility, see my joint paper with Ioanna Dimitriou and Benedikt Löwe, "Inaccessible caridnals without the axiom of choice" [Fundamenta Mathematicae 194 (2007) 179-189, DOI: 10.4064/fm194-2-3].

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  • $\begingroup$ @AndrenBlass: This is helpful. Could you give an example of a formulation of Choice which would fail, say, after $\omega_1$ in $ZF$? Is there a hierarchy of failures of formulations of choice as you go further up the cumulative hierarchy of $ZF$? What would those be (and would those results be in the paper you cited)? $\endgroup$
    – Thomas Benjamin
    Commented Jul 21, 2016 at 10:54
  • $\begingroup$ Also, apologies regarding the typo in the misspelling of your name in the previous comment. $\endgroup$
    – Thomas Benjamin
    Commented Jul 21, 2016 at 11:14
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    $\begingroup$ @ThomasBenjamin How about e.g. "$V_{\omega_2}$ is well-orderable"? This could fail while still having $V_{\omega_1}$ satisfy AC. This is a standard exercise in symmetric submodels: if $\mathbb{P}$ is $\vert V_\kappa\vert^+$-closed, then forcing with $\mathbb{P}$ doesn't change the first $\kappa$-many levels of the cumulative hierarchy, so any nontrivial symmetric submodel of such a forcing extension will introduce a failure of choice above $\kappa$ while preserving choice below. $\endgroup$
    – Noah Schweber
    Commented Jul 21, 2016 at 15:02
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    $\begingroup$ @PaceNielsen It seems to me that various formulations of the axiom of infinity will hold in $V_\alpha$ for any $\alpha>\omega$. The only exceptions would be versions that depend very strongly on pairing (as Francois mentioned) and ones that essentially build in some form of choice (like calling a set infinite iff it has a nonprincipal ultrafilter). "Reasonable" definitions of "infinite" should, in particular, say that $\omega$ is infinite, so there will be an infinite set in $V_\alpha$ for any $\alpha>\omega$. $\endgroup$
    – Andreas Blass
    Commented Jul 22, 2016 at 3:27
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    $\begingroup$ The fact that pairing holds in $V_\alpha$ for all limit $\alpha$ doesn't use the axiom of choice, just the definition of $V_\alpha$, basic facts about ordinals, and a tiny bit of ZF. If $x,y\in V_\alpha$ then, because $\alpha$ is a limit, $x,y\in V_\beta$ for some $\beta<\alpha$, so $\{x,y\}$ is an element of $V_{\beta+1}\subseteq V_\alpha$, and it clearly does (in $V_\alpha$) what the pairing axiom requires. $\endgroup$
    – Andreas Blass
    Commented Jul 22, 2016 at 15:34

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