Well-Orderable Set has Choice Function

Theorem

Let $S$ be a well-orderable set.

Then $S$ has a choice function.

Proof

Let $\preccurlyeq$ be a well-ordering on $S$.

Let $T \subseteq S$ be an arbitrary non-empty subset of $S$.

As $S$ is a well-ordered set, $T$ has a unique smallest element by $\preccurlyeq$.

Thus, we may define the choice function $C: \powerset S \setminus \set \O \to S$ as:

$\forall T \in \powerset S \setminus \set \O: \map C T$ is the smallest element of $T$ under $\preccurlyeq$

This is the choice function we require.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.9$