Axiom of choice

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

If a nonempty set carries a relation in which every point has a successor, the axiom of dependent choice supplies an infinite sequence following that relation. Full choice implies this by selecting a successor at every point and applying the recursion theorem on the natural numbers.

A single class function chooses one member of every nonempty set. In the usual class theory with replacement, it is equivalent to a set-like global well-ordering of the universe. Ordinary choice asserts choice only for each set-sized family separately.

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.

This concerns arbitrary compact spaces, with no Hausdorff restriction. For nonempty sets $A_j$, adjoin a distinguished point ${\infty }_j$ and use the four-open-set topology $\{\varnothing ,A_j,\{{\infty }_j\},A_j\cup \{{\infty }_j\}\}$. Each resulting space is compact without choice. The product has the canonical all-infinity point. The closed coordinate conditions $x_j\in A_j$ have the finite intersection property, since finite choice is provable without the axiom of choice. If the countable product is compact, their intersection is nonempty, and any point in it supplies a choice function.

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.