Axiom of Dependent Choice Implies Axiom of Countable Choice

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

Proof
Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of non-empty sets.

Define:
 * $\ds S = \bigsqcup_{n \mathop \in \N} S_n$

where $\bigsqcup$ denotes disjoint union.

Let $\RR$ be the binary endorelation on $S$ defined by:
 * $\tuple {x, m} \mathrel \RR \tuple {y, n} \iff n = m + 1$

Note that $\RR$ satisfies:
 * $\forall a \in S : \exists b \in S : a \mathrel \RR b$

Using the axiom of dependent choice, there exists a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $S$ such that $y_n \mathrel \RR y_{n + 1}$ for all $n \in \N$.

Letting $y_n = \tuple {s_n, N_n}$ for all $n \in \N$, it follows by the definition of $\RR$ 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 $n \in \N$.

The cartesian product $S_0 \times S_1 \times \cdots \times S_{N - 1}$ is non-empty.

So there exists a finite sequence $x_0, x_1, \ldots, x_{N - 1}$ with $x_n \in S_n$ for all natural numbers $n < N$.

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

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

Also see

 * Axiom of Choice Implies Axiom of Dependent Choice