Axiom of Choice Implies Axiom of Dependent Choice

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

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

For an element $x \in S$, define:
 * $R \left({x}\right) = \left\{{y \in S : x \ \mathcal R \ y}\right\}$

By assumption, $R \left({x}\right)$ is non-empty for all $x \in S$.

Now, consider the family of sets $\left\langle{R \left({x}\right)}\right\rangle_{x \in S}$.

Using the axiom of choice, there exists a mapping $f : S \to S$ such that $f \left({x}\right) \in R \left({x}\right)$ for all $x \in S$.

That is, $x \ \mathcal R \ f \left({x}\right)$.

So, for any $x \in S$, the sequence $\left\langle{x_n}\right\rangle_{n \in \N} = \left\langle{f^n \left({x}\right)}\right\rangle_{n \in \N}$, where $f^n$ denotes the composition of $f$ with itself $n$ times, is a sequence such that:
 * $x_n \ \mathcal R \ x_{n+1}$

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

Also see

 * Axiom of Dependent Choice Implies Axiom of Countable Choice