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

Source Work

2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.):

Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:

$\S 4$ Well ordering and choice:

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 $A$.

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: Exercise $4.3$