# Equivalence of Forms of Axiom of Countable Choice

## Theorem

The following forms of the Axiom of Countable Choice are equivalent in $\mathrm{ZF}^-$:

### Form 1

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

The axiom of countable choice states that there exists a sequence:

$\left\langle{x_n}\right\rangle_{n \mathop \in \N}$

such that $x_n \in S_n$ for all $n \in \N$.

### Form 2

Let $S$ be a countable set of non-empty sets.

Then $S$ has a choice function.

## Proof

### Form 1 implies Form 2

Suppose that Axiom of Countable Choice: Form 1 holds.

Let $S$ be a countable set of non-empty sets.

Then $S$ is either finite or countably infinite.

If $S$ is finite, then it has a choice function by the Principle of Finite Choice.

Suppose instead that $S$ is countably infinite.

Then there exists a bijection $f: \N \to S$.

Thus by Form 1, there exists a mapping $g: \N \to \bigcup S$ such that:

$\forall n \in \N: g \left({n}\right) \in f \left({n}\right)$

Then $g \circ f^{-1}: S \to \bigcup S$ is a choice function for $S$.

As every countable set has a choice function, Axiom of Countable Choice: Form 2 holds.

$\Box$

### Form 2 implies Form 1

Suppose that Axiom of Countable Choice: Form 2 holds.

Let $\langle x_n \rangle_{n \mathop \in \N}$ be a sequence of non-empty sets.

By the Axiom of Replacement:

$I = \left\{{x_n: n \in \N}\right\}$

is a set.

By Axiom of Countable Choice: Form 2, $I$ has a choice function $h: I \to \bigcup I$.

Define a sequence in $\bigcup I$ by letting $y_n = h \left({x_n}\right)$.

Then by the definition of choice function:

$\forall n \in \N: y_n \in x_n$

As this holds for all such sequences, Axiom of Countable Choice: Form 1 holds.

$\blacksquare$