Axiom of Choice Implies Axiom of Dependent Choice

Theorem
The axiom of choice implies the axiom of dependent choice.

Proof
Let $\RR$ be a binary endorelation on a non-empty set $S$ such that:
 * $\forall a \in S: \exists b \in S: a \mathrel \RR b$

For an element $x \in S$, define:
 * $\map R x = \set {y \in S: x \mathrel \RR y}$

By assumption, $\map R x$ is non-empty for all $x \in S$.

Now, consider the indexed family of sets:
 * $\family {\map R x}_{x \mathop \in S}$

Using the axiom of choice, there exists a mapping $f: S \to S$ such that:
 * $\forall x \in S: \map f x \in \map R x$

That is:
 * $x \mathrel \RR \map f x$

So, for any $x \in S$, the sequence:
 * $\sequence {x_n}_{n \mathop \in \N} = \sequence {\map {f^n} x}_{n \mathop \in \N}$

where $f^n$ denotes the composition of $f$ with itself $n$ times, is a sequence such that:
 * $x_n \mathrel \RR x_{n + 1}$

for all $n \in \N$, as desired.

Also see

 * Axiom of Dependent Choice Implies Axiom of Countable Choice