Consequences of the negation of the Axiom of Dependent Choice

Question

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?

Answer 1

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.

Answer 2

If I remember correctly, DC is also equivalent to Zorn's lemma for finite chains, with the two basically just being contrapositions of each other.

So negating it would imply that there is a partially ordered set such that every chain is finite and bounded, but at the same time no element is maximal.