# Choice Function Exists for Set of Well-Ordered Sets

## Theorem

Let $\mathbb S$ be a set of sets such that:

- $\forall S \in \mathbb S: S \ne \O$

that is, none of the sets in $\mathbb S$ may be empty.

Let every element of $\mathbb S$ be well-ordered.

Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ satisfying:

- $\forall S \in \mathbb S: \exists x \in S: \map f S = x$

## Proof

Every member of $\mathbb S$ is a well-ordered set.

Thus, for $S \in \mathbb S$, there is a minimal element $s$ for $S$ (with respect to the ordering of $S$).

By Well-Ordering Minimal Elements are Unique, $s$ is unique.

Therefore, we can define $f$ by:

- $\forall S \in \mathbb S: \map f S = s$

$\blacksquare$

## Warning

Note that this only applies if we are given a well-ordering for each $S \in \mathbb S$.

More formally, this means: if there is a mapping that maps $S \in \mathbb S$ to a well-ordering of $S$.

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If we just know that each $S \in \mathbb S$ is well-orderable, we generally do need AoC to get a choice function (to apply the proof above, we have to pick a well-order for each $S\in \mathbb S$, which requires AoC. This is related to the fact that generally we need AoC to show that, for example, the countable union of countable sets is countable.)