Do we need the axiom of choice in here?

Question

Axiom of choice: Given $\mathbb{F}$ is a set of non-empty sets. Then, there is a function $f$ with $\text{Dom}(f)=\mathbb{F}$ such that, for every $A \in \mathbb{F}$, $f(A) \in A$.

The function $f$ is called a choice function of $\mathbb{F}$.

Every well ordered set $X$ is isomorphic to an ordinal and has a first element $\{ x_0 \}$. So, can't we just say that we have a choice function which is: $f(A)=x_0$ for every $x \in X$?

Answer 1

No. You still need some form of the axiom of choice. Specifically in order to choose the isomorphism with the ordinal.

This is apparent in the Russell socks example, where you have a countably infinite set of pairs without a choice function. So you a situation where the family itself can be well-ordered, and all its members can be well-ordered, but no choice function.

---

In the case where each set is also given with its well-ordering, then the answer is yes. Picking the minimal of each set is a choice function. Compare this to the following theorem:

Theorem. The set $A$ can be well-ordered if and only if $\mathcal P(A)\setminus\{\varnothing\}$ admits a choice function.

The $(\Longrightarrow)$ direction is exactly this. Since $A$ can be well-ordered, fixing such ordering induces (uniformly) a well-ordering on every non-empty subset of $A$, and picking the minimal element is indeed a choice function.

The other direction is slightly more technical, and is similar to the general proof that the axiom of choice implies the well-ordering theorem.

Answer 2

Indeed if every $A\in\mathbb F$ is well-ordered (for example because $\bigcup\mathbb F$ is well-ordered). Then $A\mapsto\mathop{\rm min}A$ is a choice function. However, you do not say if and whence you got these well-orderings.