# 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:

- $\displaystyle S = \bigsqcup_{n \mathop \in \N} S_n$

where $\bigsqcup$ denotes disjoint union.

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 $n \in \N$.

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 $n \in \N$.

$\blacksquare$

#### Axiom of Dependent Choice

This proof depends on the Axiom of Dependent Choice.

Although not as strong as the Axiom of Choice, the Axiom of Dependent Choice is similarly independent of the Zermelo-Fraenkel axioms.

The consensus in conventional mathematics is that it is true and that it should be accepted.