Let $A$ be a set, then there is a function $c :P(A)\setminus\lbrace ∅\rbrace→ A$ such that $c(X) ∈ X$ for every nonempty subset $X$ of $A$.$(*)$
The axiom of choice is stated as:
Let $\mathcal A$ be a set of nonempty sets. Then there is a function $f : \mathcal A→\bigcup \mathcal A$ such that $f(A) ∈ A$ for all $A ∈\mathcal A$.
I need to prove that the first statement implies the axiom of choice:
Here is an answer I found in this site:
3 Statements of axiom of choice are equivalent
I expand it as:
Let $\mathcal A$ be a set of non-empty sets,then $\bigcup \mathcal A$ is also a set by MK4. From $(*)$ we know that there is a function $c :P(\bigcup\mathcal A)\setminus\lbrace ∅\rbrace→ \bigcup\mathcal A$ such that $c(X) ∈ X$ for every nonempty subset $X$ of $\bigcup\mathcal A$.
And then the author says "restrict to $\mathcal A$" then we have a choice function.
But I don't think $\mathcal A$ is a subset of $\bigcup\mathcal A$.
So what does the restriction means. How may I deal with that?
Or other way to prove it would be also appreciated. Thanks!
But certainly $P \left( \bigcup \mathcal{A} \right) \setminus \{ \varnothing \}$ is a superset of $\mathcal{A}$ (check it), so restriction $c \upharpoonright \mathcal{A}$ makes sense.