EDB — 02H

[02H]Prerequisites:[01M]↺↻, [026]↺↻,[020]↺↻ . Let \(I\) be a non-empty set of indexes, let \(A_ i\) a family of non-empty sets indexed by \(i∈ I\). Recall that, by definition, the Cartesian product \(∏_{i∈ I}A_ i\) is the set of functions \(f:I→ ⋃_{i∈ I}A_ i\) such that \(f(i)∈ A_ i\) for each \(i∈ I\).

Show that the following are equivalent formulations of the axiom of choice.

• The Cartesian product of a non-empty family of non-empty sets is non-empty.

• Given a family \(A_ i\) as above, such that the sets are not-empty and pairwise disjoint, there is a subset \(B\) of \(⋃_{i∈ I}A_ i\) such that, for each \(i∈ I\), \(B∩ A_ i\) contains a single element.

• Let \(S\) be a set. Then there is a function \(g: {\mathcal P}(S) \to S\) such that \(g(A) \in A\) for each nonempty \(A \in {\mathcal P}(S)\).