# Consequences of the negation of the Axiom of Dependent Choice

It seems to me that a proper reason to include The Axiom of Choice as a foundational axiom of set theory should be based on the observation that the negation of The Axiom of Choice has absurd implications. I do not know of any implications of the negation of The Axiom of Choice that I would at once deem absurd, which motivates the search for a weaker form 'choice' whose negation should imply something absurd. A reasonable candidate would be The Axiom of Dependent Choice. The wiki page identifies an equivalent formulation of The Axiom of Dependent Choice as 'every (nonempty) pruned tree has a branch'. The negation of this statement seems absurd to me. The wiki page also points out that Baire's Theorem is equivalent to The Axiom of Dependent Choice, but the negation of Baire's Theorem is not immediately absurd to me.

My question is now: Are there other known implications of the negation of The Axiom of Dependent Choice?

Well.

As with the failure of the axiom of choice in full, I should point out that the failure of any such arbitrary choice principle is not limited to "interesting sets". The axiom of choice could be in full force when considering all the sets relevant for most mathematicians. Then, it might fail badly somewhere up in the clouds of the set theoretic universe, where it will have no consequence.

But the failure of the principle of dependent choice means not just that equivalent statements fail, but also that anything stronger than it will fail.

1. The Lowenheim-Skolem theorem for countable languages will fail (it is equivalent to $\sf DC$). Namely there is a countable language $\cal L$, and a theory $T$, such that for some uncountable model $M$ there is no countable elementary submodel.

2. The axiom of choice for families of well-ordered sets will fail. It is true, for countable families it will hold, but other than that we have no guarantees. We do know, however, that for some ordinal $\alpha$ the axiom of choice for families of size $\alpha$ will fail. Because the statement $\forall\alpha(\alpha\in\sf Ord\rightarrow AC_\alpha)$ implies $\sf DC$.

3. The Baire category theorem that you have mentioned will fail. There will be a complete metric space and some countable collection of dense open sets, whose intersection is not dense. This space will not be separable, though.

It should be pointed out that a lot of mathematics that we do can be scraped by using only countable choice, which is in fact weaker. And again, even the failure of countable choice might not be enough to say anything substantial.

Let me finish by pointing out that the Boolean Prime Ideal theorem may consistently hold without $\sf DC$ (as in Cohen's first model), and it can be used to prove many theorems useful throughout mathematics. From extending filters to ultrafilters, the compactness theorem for first-order logic, Tychonoff's theorem for Hausdorff spaces, the Hahn-Banach theorem, and many many more.

So the failure of $\sf DC$, while severe, is not enough to determine any "interesting" consequences on daily mathematics; let alone since it may fail and $\sf BPI$ will still be true.

• Thank you for this very excellent answer! I never considered the fact that The Baire Category Theorem on separable spaces only needs countable choice, and for some reason always assumed that The Boolean Prime Ideal Theorem was stronger than The Axiom of Dependent Choice. Maybe I should have asked about consequences of the negation of The Axiom of Countable Choice instead, though you hint that still one would not be able to say much. – Jonas Dahlbæk Apr 11 '15 at 8:26
• You're welcome. One point, though, Baire Category Theorem for separable complete metric spaces doesn't require any choice. – Asaf Karagila Apr 11 '15 at 8:59
• How surprising! I guess it is due to the fact that countable sets come a priori with a bijection to the naturals and therefore an order? – Jonas Dahlbæk Apr 11 '15 at 9:55
• Yes, this is because countable sets can be well-ordered by definition. Most theorems we know using choice can be proved without choice when the structures at hand can be well-ordered (or some significant part of them can be well-ordered, e.g. a dense subset). – Asaf Karagila Apr 11 '15 at 10:16
• @user254665: Huh? – Asaf Karagila Aug 7 '15 at 9:09