Axiom of choice

Every family of nonempty sets has a choice function. In ZF this is equivalent to the well-ordering theorem.

The axiom of countable choice allows one to choose an element from each member of a countable family of nonempty sets. It is a restricted form of the axiom of choice. In the usual proof that a countable union of countable sets is countable, it suffices to choose one enumerating injection for each set; if those enumerations are already supplied, that proof requires no further choice.

A choice function on a family $\mathcal{A}$ of nonempty sets is a function $f$ with $f(A)\in A$ for every $A\in \mathcal{A}$.

For sets $X$ and $Y$, ${\mathsf{AC}}_X(Y)$ says that every $X$-indexed family $(A_x{)}_{x\in X}$ of nonempty subsets of $Y$ has a choice function $c:X\to Y$ satisfying $c(x)\in A_x$.

Every set admits a well-order. In ZF this statement is equivalent to the axiom of choice.

Given a choice function on all nonempty subsets of $X$, recursively choose the next point from the complement of all earlier choices. Hartogs theorem forces the recursion to exhaust $X$ before it defines an injection from $h(X)$ into $X$, producing a bijection from an ordinal to $X$.

For any sets $X,Y$, either $X$ injects into $Y$ or $Y$ injects into $X$. Applying this to $X$ and its Hartogs ordinal proves the well-ordering theorem, so over ZF cardinal comparability is equivalent to choice.