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What other axioms in set theory are stronger than AC ? I mean what are those axioms that will imply AC ?

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up vote 13 down vote accepted

The axiom "every set is constructible" (denoted V = L), and the axiom "every set is definable from an ordinal parameter" (denoted often as V= HOD, and sometimes as V= OD) each implies AC, and each is provably stronger than AC.

More specifically, it is well-known that:

(a) (Within ZF), V = L implies V = HOD, and V = HOD implies AC.

(b) Neither of the above implications is reversible.

Historical Note: The axiom V = L was introduced by Gödel in his seminal work on the consistency of AC and GCH with ZF. The axiom V= HOD was first publicly introduced in a joint paper of Myhill and Scott, but their paper acknowledges that the axiom was independently considered by a number of people, including Gödel, Post, and the joint work of Vopěnka and Balcar.

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Do you know a good discussion of V = HOD? I've seen it mentioned here at MO a couple of times, and I'm curious to learn more. – arsmath Jan 1 '13 at 18:19
@arsmath: Kunen's chapter on ordinal definability in his textbook on set theory is a very good place to learn about V=HOD; Jech's text also has a discussion of it but with far less detail. – Ali Enayat Jan 1 '13 at 18:44
@arsmath: You might also look at the original paper of Myhill and Scott. It's in the proceedings of the 1967 UCLA set theory conference, published by the AMS in the series "Proceedings of Symposia in Pure Mathematics" (volume 13, part 1). – Andreas Blass Jan 1 '13 at 19:47
Since $V=L$ is mentioned in several answers, I should point out that more generally, the statement that $V$ is a fine-structural model (sometimes written as $V=K$, where $K$ stands for some flavor of the "core model") suffices, and is much less restrictive. For example, it is compatible with $V=L$ but also with $V=L[\mu]$, the smallest inner model for a measurable cardinal, or $V=M_1^\sharp$, etc. In all cases, if $V$ is a fine-structural model, it is self-wellorderable, so $V=HOD$ and AC holds. – Andrés E. Caicedo Jan 1 '13 at 23:33
@ALi: Would you explain what is "V=L" more precisely or name some references (for amateurs of course !) – user30338 Jan 7 '13 at 9:24
  1. The axiom For every infinite set $X$ if $Y$ is such that $|X|\lt|Y|$ and $|Y|\leq|\mathcal P(X)|$, then $|Y|=|\mathcal P(X)|$. which is also known as the Generalized Continuum Hypothesis.

  2. In turn this axiom is equivalent (as it turns out) to For every ordinal $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. While the equivalent may not be true in ZF, it is certainly true in ZFC. Both formulations imply the axiom of choice.

    (This axiom follows from $V=L$ mentioned in the other answers, but does not imply it. It does not follow from $V=HOD$.)

  3. There exists an ordinal $\alpha$ such that there is no set $X$ for which there is a chain of cardinals between $X$ and $\mathcal P(X)$ of order type $\alpha$. (i.e. the "distance" between $X$ and $\mathcal P(X)$ is never longer than $\alpha$).

    Equally this can be required about $X$ and $X^2$. However assuming choice $|X|=|X^2|$ for infinite sets, so this turns out to be equivalent; whereas we can use class forcing to push power sets far enough so there are such sequences of any length.

Generally speaking, let $\varphi$ be a proposition which is consistent with ZFC but not provable from ZFC, then AC+$\varphi$ is an axiom stronger than the axiom of choice.

In the fashion above, we can think about $\varphi$ being "the axiom of choice holds, and $\lozenge_\kappa$ is true for all $\kappa<\aleph_\omega$." or something like that.

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For every infinite set $X$ ... – Andrés E. Caicedo Jan 1 '13 at 20:04
Andres, yes, thank you. (Also $|X|\lt|Y|$ was incorrectly written as $\leq$.) – Asaf Karagila Jan 1 '13 at 20:05

Many large cardinal hypotheses imply AC, when stated in a certain natural way.

For example, the assertion over ZF that there are unboundedly many inaccessible cardinals, defined as uncountable regular strong limit cardinals, implies the axiom of choice.

Indeed, even having a proper class of strong limit cardinals implies the axiom of choice.

What's more, the axiom of choice is equivalent over ZF to the assertion "there are unboundedly many strong limit cardinals".

The forward direction is simply the usual proof in ZFC that there is a closed unbounded class of strong limit cardinals. For the reverse implication, we define that $\kappa$ is a strong limit if and only if it is an initial ordinal for which $\beta\lt\kappa\implies P(\beta)\lt\kappa$. Note that this implies in particular that the power set $P(\beta)$ is well-orderable for $\beta\lt\kappa$, since it injects into $\kappa$, which is well-orderable. Thus, if there are a proper class of such $\kappa$, then every well-orderable set will have a well-orderable power set. It is a non-obvious fact (and in fact a quite slippery fact, which beginners often get wrong when first trying to prove it) that if every well-orderable set has a well-orderable power set, then AC holds.

To see that there are several natural but inequivalent formulations of inaccessibility and of strong limit cardinals in ZF, see the treatment in A. Blass, I. Dimitriou, B. Loewe, Inaccessible cardinals without the axiom of choice.

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Well a proper class of strong limit cardinals is provable in ZFC. – Asaf Karagila Jan 1 '13 at 21:42
Asaf, yes, that is the reverse direction of the equivalence. – Joel David Hamkins Jan 1 '13 at 21:54
David, unfortunately, that's not quite the case, since the various forms of inaccessibility are no longer equivalent without AC, and in particular, this form of inaccessibility is not equivalent to UA in ZF alone (although it is in ZFC). – Joel David Hamkins Jan 1 '13 at 23:42
The only resson this works is that the statement that powersets are well-orderable is just a clause in the hypothesis. Inaccessibility is not relevant. – Monroe Eskew Jan 1 '13 at 23:54
Yes, I think there are analogues for UA of the issues in the Blass, Dimitriou, Loewe paper to which I link. That might make a nice topic to work out carefully... – Joel David Hamkins Jan 2 '13 at 0:28

The Axiom of Constructibility implies the Axiom of Choice, as well as many other results independent of ZFC, such as the continuum hypothesis.

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The CH implies a well ordering of R, trivially. It says the reals can be put in 1-1 correspondence with an ordinal, namely $omega_1$. The GCH or any other global assertion or any other global assertion about cardinal exponentiation will imply full AC for the same reason. – Monroe Eskew Jan 1 '13 at 18:15
I meant that the axiom of constructibility implies the continuum hypothesis, as well as the generalized continuum hypothesis. – arsmath Jan 1 '13 at 18:17
CH does not imply that $\mathbb R$ can be well-ordered. That would be The Aleph Hypothesis for $\aleph_0$, whereas there are models in which $\mathbb R$ cannot be well-ordered and every set of reals is either countable or of size continuum (e.g. Solovay's model). – Asaf Karagila Jan 1 '13 at 19:35
As can be seen from the comments of mbsq and Asaf Karagila, the terminology "continuum hypothesis" becomes ambiguous in the absence of the axiom of choice. It is often formulated as $2^{\aleph_0}=\aleph_1$ (which mbsq calls CH and Asaf calls the aleph hypothesis for $\aleph_0$), and it is often formulated as "every uncountable set of reals admits a bijection to $\mathbb R$ (which is what Asaf means by CH). Analogous ambiguities afflict GCH. – Andreas Blass Jan 1 '13 at 19:42
As an original reference (not a textbook), see: Sierpiński, Wacław L'hypothèse généralisée du continu et l'axiome du choix. Fundamenta Mathematicae 34, (1947). 1–5 – Margaret Friedland Jan 1 '13 at 21:04

The axiom of global choice. Technically this isn't really an axiom: global choice (GC) states that there is a formula $\phi(x, y)$ such that the relation $$ A\le_\phi B:= V\models \phi(A, B)$$ is a well-ordering of the universe $V$. This can't be stated as a single formula, so in some sense it's a meta-axiom. It clearly implies choice, and is implied by $V=L$: we already have a partition of the universe into $L_0$, $L_1$, . . . , $L_\alpha$, . . ., and we can get from here to a full (class-)well-ordering of the universe by fixing at the outset some well-ordering of formulas in the language of set theory (since at each stage in the construction of $L$ we are only taking definable powersets; this is why this argument doesn't work in just $ZFC$).

A couple comments on why I think global choice is interesting (even though it's not expressible in the language of set theory):

  • Although the relative consistency of Choice was proven rather early, it was via $L$, which satisfies $GC$ as well; the result that, assuming the consistency of $ZFC$, there is a model of $ZFC$ with no definable well-ordering of the universe came much later [NOTE: this is based on foggy memory, and I don't recall exactly when this result happened; can someone remind me?], and generally telling whether a model of $ZFC$ satisfies $GC$ is very hard.

  • Basically by reversing the argument that it is true in $L$, one can make an informal argument that GC implies that the universe is small (contains only "buildable" things). So there's a somewhat intuitive argument for $AC+\neg GC$: the universe should be "big enough" that for each family of sets, we have a choice function, but the universe should also be big enough that any specific definable well-order "misses something."

  • $GC$ is useful in other set theories. First of all, as noted above, even stating $GC$ needs a class theory like Morse-Kelley or NBG (expansions of $ZFC$ to also talk about classes). We can also ask about the status of $GC$ in really odd set theories like New Foundations (EDIT: As Ali points out below, in NF/NFU Choice and Global Choice are essentially one and the same - but the point still stands that global choice could still be interesting in set theories different than $ZFC$.)

  • $GC$ makes sense even outside of set theories! It doesn't make sense to ask whether a given ring $R$ satisfies the axiom of choice, since elements of $R$ aren't (at least on the face of things) sets, but it does make sense to ask whether there is a formula in the language of rings which well-orders $R$.

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I've been slightly vague about my "definable" well-orderings: I think everything I said is true whether we take "definable" to mean with or without parameters. But I welcome corrections on this point! – Noah Schweber Jan 2 '13 at 4:55
Noah: (1) Despite appearances, GC as you define it (without parameters) is well-known to be equivalent to V=OD, thanks to the reflection theorem [and if you allow parameters, then this weaker version of GC is equivalent to "V is ordinal definable from a parameter], and (2) Jensen's original proof of consistency of NFU shows the consistency of NFU with global choice (in the NFU context, since there is a universal set, global choice simply stands for the statement "every set can be well-ordered"). – Ali Enayat Jan 2 '13 at 7:10
Re: (1), that's neat, I didn't know that. Re: (2), of course - welp, that's what I get for answering while sleep-deprived. – Noah Schweber Jan 2 '13 at 7:37
I think of GC not as making an assertion about definable classes, even with parameters, but rather as one of the principle axioms of GBC, stated as the second-order assertion that some class is a well-ordering of the universe V. This can be forced over any ZFC model without adding sets, by forcing with the class of set-sized well-orderings, ordered by end-extension. This is how we know GBC is conservative over ZFC. If you insist that the order is definable, even with parameters, you wouldn't get the conservativity result. – Joel David Hamkins Jan 2 '13 at 14:57
That is, what you've got (with or without parameters) is a strengthening of what is usually known as the global choice axiom. – Joel David Hamkins Jan 2 '13 at 15:19

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