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!

  • 1
    $\begingroup$ You do not need $\mathcal A$ to be a subset of $A,$ or a subset of $\bigcup\mathcal A.$ What you need is that every element of $\mathcal A$ is a nonempty subset of $\bigcup\mathcal A.$ $\endgroup$
    – bof
    Jul 17, 2017 at 7:18

1 Answer 1


But I don't think $\mathcal{A}$ is a subset of $\bigcup \mathcal{A}$.

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.


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