Question

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

Answer 1

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.

Answer 2

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.

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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.

Answer 3

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

Answer 4

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.

Answer 5

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 $R$.