Choice Function for Set does not imply Choice Function for Union of Set/Mistake

Source Work

 * Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
 * $\S 4$ Well ordering and choice:
 * Exercise $4.3$
 * Exercise $4.3$

Mistake

 * Show that if there exists a choice function for $S$ then there exists a choice function for $\bigcup S$.

Analysis
The statement is not provable from the axioms of Zermelo-Fraenkel Set Theory because the statement implies the Axiom of Choice but Axiom of Choice is Independent of ZF.

To show this, let $A$ be an arbitrary set.

Then, by the Axiom of Pairing, there exists a set
 * $S = \set A$

with $A$ as its only element.

There exists a choice function:
 * $f: \powerset S \setminus \set \O \to S$

for $S$ that maps the single element:
 * $\set A \in \powerset S \setminus \set \O$

in its domain to $A$.

The union of $S$ is
 * $\bigcup S = A$

If the given statement is true, then there exists a choice function for every set $A$.