Countable Set has Choice Function
Jump to navigation
Jump to search
Theorem
Let $S$ be a countable set.
Let $\mathbb S = \powerset S \setminus \set \O$ be the power set of $S$ excluding the empty set $\O$.
Then there exists a choice function $C$ for $S$:
- $\forall x \in \mathbb S: \map C x \in x$
Proof
From Countable Set is Well-Orderable, we have that $S$ is a well-orderable set.
The result follows from Well-Orderable Set has Choice Function.
$\blacksquare$
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