Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$, i.e., $f(a) \in a$ for all $a \in A$?
proof of the axiom of choice for countable sets in ZF
Andrew S.
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