Choice Function Exists for Well-Orderable Union of Sets

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Theorem

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

$\forall S \in \mathbb S: S \ne \varnothing$

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

Let the union $\bigcup \mathbb S$ be well-orderable.

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

$\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$

Thus, if every member of $\mathbb S$ is a well-ordered, then we can create a choice function $f$ defined as:

$\forall S \in \mathbb S: f \left({S}\right) = \inf \left({S}\right)$

True, we may be making infinitely many choices, but we have a rule for doing so.

The Well-Ordering Theorem, which states that every set is well-orderable, implies the truth of the Axiom of Choice.

Suppose $T = \bigcup \mathbb S$ is well-orderable.

Then we can create a well-ordering $\preceq$ on $T$ so as to make $\left({T, \preceq}\right)$ a well-ordered set.

From the definition of well-ordered set, every subset of $T$ is itself well-ordered.

From Subset of Union: General Result we have that $\forall S \in \mathbb S: S \subseteq T$.

So every $S \in \mathbb S$ is well-ordered and Choice Function Exists for Set of Well-Ordered Sets applies.

$\blacksquare$

Let $\mathbb S$ be any set of sets.

That is, that every set is well-orderable.

Then, in particular, $\bigcup \mathbb S$ is well-orderable.

The result follows directly from the main theorem.

$\blacksquare$