Why is the axiom of dependent choice weaker than AC?

Question

I'm looking for an intuition ( not a formal proof) on why the axiom of dependent choice is weaker than the axiom of choice. Perhaps this would also explain the "dependent choice". I'm thinking along the following lines:

The axiom of choice ensures the existence of a choice function on any family $ \cal{F}$ of non-empty sets. No structure is assumed on $\cal{F}$. In particular there doesn't have to be a relation between the members of $\cal{F}$.

On the other hand the axiom of dependent choice assumes a treelike structure on $\cal{F}$, provided by the entire relation $R$ on $X$. It ensures the existence of a branch of this tree, i.e. a sequence $(x_n)$ s.t. $\forall n\ x_nRx_{n+1}$.

So the difference is that AC assumes no structure but DC assumes a (treelike) structure.

Is my thinking correct?

Answer 1

Well, it isn't the tree-like structure that is the reason (we can make the relation to ignore it yielding the weaker axiom of countable choice). Instead, it has to do with only making countably many choices here, not the arbitrary cardinality of AC.

Axiom of Dependent Choice (DC) is actually the first nontrivial example of the dependent choice principle, introduced by Levy:

Principle of $\kappa$-dependent choice $\mathsf{DC}_\kappa$, where $\kappa$ is an initial ordinal (i.e., an aleph): If $S$ is a nonempty set, $R\subseteq S^{<\kappa}\times S$ a total binary relation (i.e. for every $s\colon\beta\to S$ with $\beta<\kappa$, there is $x\in S$ with $sRx$), then there exists a function $f\colon\kappa\to S$ such that $(f\vert_{\beta})Rf(\beta)$ for all $\beta<\kappa$.

Clearly $\mathsf{DC}_\kappa\Rightarrow\mathsf{DC}_\lambda$ when $\kappa>\lambda$. Levy proved that AC is equivalent to $(\forall\kappa)\mathsf{DC}_\kappa$ under reasonable set theory like ZF (note $\Leftarrow$ is not as trivial as you might think: we don't know every infinite cardinal is an aleph without AC-equivalents so on the face value we might not have enough cardinals in the $\forall\kappa$).