Both the affirmation of the AC and its denial are consistent in set theory, but is there in any way that one can say it must be the case that one of them is actually true? I understand what it means for AC and its denial to be consistent; but is there a sense in which truth is more than merely avoiding contradictions. Perhaps it is the case that the denial of AC is the "truth of the matter" but we won't run into any contradictions if we assume AC is true and do mathematics using Zorn's lemma (which is equivalent) etc.

This question is prompted as I'm reading "Metaphysics: An Anthology" edited by Kim, Korman, and Sosa. In it, there is an article by Plantinga "Modalities: Basic Concepts and Distinctions", it is stated that "The axiom of choice and the continuum hypothesis are either necessarily true or necessarily false...."

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    Notably AC implies LEM, while (if memory serves) ~AC does not
    – Canyon
    Dec 30, 2017 at 1:07
  • To clarify..... In this instance, I'm concerned about the law of excluded middle as it pertains to the AC. So if 'p' is the statement "The Axiom of Choice is true" then I mean is it the case that (p ∨ ~p) is true? Must at least p or ~p be true?
    – Clclstdnt
    Dec 30, 2017 at 23:28
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    If you're a mathematical Platonist, then AC has a definite truth value. ‎Gödel was a Platonist in spite of being the one who discovered undecidability. If one is a formalist then no mathematical theorem is true, it's merely a logical consequence of the axioms. Pick your philosophy.
    – user4894
    Dec 30, 2017 at 23:44
  • I really like this answer. The others are good, but this gets to the heart of the question. Thank you.... it essentially boils down to one's person philosophy on the matter.
    – Clclstdnt
    Dec 31, 2017 at 0:03
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    There are a lot of distinct ideas in logic that, unfortunately, common language all squashes down into a single term "truth". In my opinion, most of the depth to questions like this are just about disentangling the different possible meanings, and are rather straightforward once you know what specifically you mean to ask!
    – user6559
    Dec 31, 2017 at 4:45

2 Answers 2


It depends what you mean by "set theory". ZF is not the only way to formalise set theory (though it is, of course, the most widely accepted). Cf. this excerpt from the Wikipedia entry on the AoC, in the context of constructive mathematics:

In constructive set theory, however, Diaconescu's theorem shows that the axiom of choice implies the law of excluded middle (unlike in Martin-Löf type theory, where it does not). Thus the axiom of choice is not generally available in constructive set theory. A cause for this difference is that the axiom of choice in type theory does not have the extensionality properties that the axiom of choice in constructive set theory does.

Some results in constructive set theory use the axiom of countable choice or the axiom of dependent choice, which do not imply the law of the excluded middle in constructive set theory. Although the axiom of countable choice in particular is commonly used in constructive mathematics, its use has also been questioned.

You might also be interested in reading the "Variants" section of the nLab page on the AoC, and this quote from the nLab page on 'topos':

The internal logic of toposes is intuitionistic higher order logic. This means that, while the law of excluded middle and the axiom of choice may fail, apart from that, every logical statement not depending on these does hold internal to every topos.

So in addition to @celtschk's comment that the answer "depends on what you think set theory describes", it also depends on the specific set theory in which one is working.


It depends on what you think set theory describes. If you think set theory describes some sort of independently existing structure (sets that exist independent of set theory), then of course for that independently existing structure either the axiom of choice holds, or it doesn't.

However a different view is that sets are defined by set theory in the very same sense as group elements are defined by group theory. In group theory, the statement "there exists a group element of which no positive power is the identity" (let's call it the "infinite order hypothesis") is independent from the group axioms. Yet nobody would consider it a problem; it just means there are models of group theory (that is, groups) where the "infinite order hypothesis" is true (such as the additive group of integers), and there are others where the "infinite order hypothesis" is false (such as additive group of the integers modulo 3).

However note that even if you assume there are some "true sets" independent of ZF set theory which set theory is meant to describe, then the independence of AC of the ZF set theory axioms does not violate the law of the excluded middle. It just means they are not sufficient to describe all aspects of those "true sets"; in particular, they do not allow to decide whether all sets admit a choice function.

Note that also the axiom of infinity is independent of the other axioms of set theory; it is consistent to assume that there are only hereditary finite sets (that is, finite sets whose elements are also finite, as are the elements of their elements, and so on). If you think of set theory as describing some "true sets", then you can also ask whether the axiom of infinity holds for then (i.e. whether there actually exist infinite sets). If, however, you view set theory as just describing a class of models, then set theory with infinity is just describing a specific subclass of models (namely those which actually have infinite sets), and then adding the axiom of choice again describes a specific subclass of those models (namely the class of models where all sets have a choice function), not unlike how the theory of abelian groups describes a subclass of the models (groups) that group theory describes (namely the subclass of groups whose group operation is commutative).

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