Axiom of Dependent Choice Implies Axiom of Countable Choice

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

Proof
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of non-empty sets.

Define:
 * $S = \left\{{\left({x, n}\right) : n \in \N, \, x \in S_n}\right\}$

Let $\mathcal R$ be the binary endorelation on $S$ defined by:
 * $\left({x, m}\right) \ \mathcal R \ \left({y, n}\right) \iff n = m + 1$

Note that $\mathcal R$ satisfies:
 * $\forall a \in S : \exists b \in S : a \ \mathcal R \ b$

Using the axiom of dependent choice, there exists a sequence $\left\langle{y_n}\right\rangle_{n \in \N}$ in $S$ such that $y_n \ \mathcal R \ y_{n + 1}$ for all $n \in \N$.

Letting $y_n = \left({s_n, N_n}\right)$ for all $n \in \N$, it follows by the definition of $\mathcal R$ that $N_{n + 1} = N_n + 1$.

A straightforward application of mathematical induction shows that $N_n = n + N$ for some $N \in \N$.

So $s_n \in S_{n + N}$ for all $n \in \N$.

Applying: and it follows that the cartesian product of a finite number of non-empty sets is non-empty.
 * the non-emptiness of the cartesian product of two non-empty sets
 * another straightforward application of mathematical induction

So there exists a finite sequence $x_1, x_2, \ldots, x_N$ with $x_n \in S_n$ for all natural numbers $n \leq N$.

Now, define $x_n = s_{n - N}$ for all natural numbers $n > N$.

Then $x_n \in S_n$ for all $n \in \N$.

Also see

 * Axiom of Choice Implies Axiom of Dependent Choice