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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.

An ordinal $\kappa$ is a cardinal number precisely when no smaller ordinal has a bijection onto it. Ordinalhood and the graph predicate for a bijection are bounded formulas in set theory, so $\mathrm{Ord}(\kappa )\wedge \forall f\forall \alpha \in \kappa \neg \mathrm{Bij}(f,\alpha ,\kappa )$ is a Pi-one formula modulo ZF. This does not assert that every set can be assigned an ordinal cardinal without the axiom of choice.

If the ground model has axiom of choice, well-order the set of subnames appearing in a forcing name for an arbitrary extension set. Evaluate those activated by the generic filter. Each element of the extension set has a least preimage index; ordering by these indices well-orders the set. Only ground choice and the ZF part of the forcing theorem are needed.

A set-indexed coproduct in a category of projective objects in a category is projective. Restrict a map out of the coproduct to each summand, lift each restriction through the given epimorphism, and assemble the lifts by the coproduct universal property. Choosing an arbitrary family of lifts uses the usual axiom of choice. The empty coproduct, an initial object, is projective as well.

If $\alpha$ is a countable ordinal and $V_{\alpha }\models \mathsf{ZFC}$, then $\alpha$ must be a limit ordinal. Choose a countable cofinal sequence $({\alpha }_n)$ in $\alpha$. The internal axiom of choice gives, for every $n$, a bijection between $V_{{\alpha }_n}$ and some ordinal below $\alpha$; that ordinal is externally countable, so every $V_{{\alpha }_n}$ is countable. Hence $V_{\alpha }={\bigcup }_n{V}_{{\alpha }_n}$ is countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem, a contradiction.

Given finite lists ${\ell }_0,{\ell }_1,\dots$ in a set $X$, their union of entries has an explicit enumeration from a subset of ${\mathbb{N}}^2$: use the existing positions in each list and a Cantor pairing function. If the union is infinite, successively take the first occurrence of an entry not previously taken, in the order of its natural-number code. The well-order on natural numbers makes each step uniquely determined and yields an injection $\mathbb{N}\to X$. No axiom of choice is needed. This does not assert that an arbitrary countable family of unordered finite sets has a countable union in ZF without the axiom of choice.

A set $X$ is Dedekind-finite when no injective function $\mathbb{N}\to X$ exists. In classical Zermelo–Fraenkel set theory this is equivalent to $X$ not being in bijection with a proper subset: an injective non-surjective self-map generates distinct iterates from an element outside its image, and a countably infinite subset permits a shift fixing its complement. A finite set is Dedekind-finite. Without the axiom of choice, some infinite sets can also be Dedekind-finite.

For a family of first-order structures, take free filters on the index set as worlds, ordered by inclusion, and product sequences as parameters. Force an atom when its coordinate truth set belongs to the filter. Conjunction is local, implication and negation quantify over filter extensions, and a disjunction is forced when every extension has a further extension forcing one disjunct. Existence is witnessed by a product sequence and universality holds for all sequences at all extensions. With the axiom of choice, induction on formulas identifies forcing with membership of the classical coordinate truth set. Equivalently, every ultrafilter extension satisfies the formula in its ultraproduct. Thus the semantics validates classical logic. A local disjunction rule would fail this identification: a set and its complement can both be absent from a free filter, while their union is the whole index set.

A functor $F:\mathcal{C}\to \mathcal{D}$ is essentially surjective when every object of $\mathcal{D}$ is isomorphic to $FA$ for some $A\in \mathcal{C}$. Together with full and faithful, this characterizes an equivalence of categories under the appropriate axiom of choice convention. For large categories, choosing a quasi-inverse on all objects requires a universe or a class-choice convention.

Assuming the axiom of choice, the well-ordering theorem makes every set equipotent to an initial ordinal. The infinite initial ordinals are exactly the ${\aleph }_{\alpha }$, so every infinite cardinal is ${\aleph }_{\alpha }$ for a unique ordinal $\alpha$.

An infinite Dedekind-finite set is an infinite set with no subset that is a countably infinite set. The name Dedekind set is sometimes used for this combination of properties. Under the axiom of choice no such set exists; results about these sets must avoid obtaining arbitrary enumerations of finite subsets by unstated choices.

An infinite set is a set with infinitely many elements. Equivalently, assuming the axiom of choice, it contains a countably infinite set.

An inner model $W$ of ZF containing all ambient subsets of $\omega$ computes the same countable ordinals as the ambient universe. Every ambient countably infinite ordinal has a well-order code coded by a subset of $\omega$, hence that code belongs to $W$; finite ordinals are already shared. The Mostowski collapse theorem inside $W$ recovers the actual ordinal and its countability witness. Conversely any countability witness in $W$ remains one outside. If $W$ also contains the full power set of $\omega$ as a set, its Continuum hypothesis would give an ambient bijection from ${\omega }_1$ onto that same power set. Thus an ambient failure of this well-orderable formulation of the Continuum hypothesis is preserved in such an inner model. No internal axiom of choice is required for this argument.

Using the axiom of choice, enumerate all infinite subsets $A_{\alpha }$ of the positive integers in order type $2^{{\aleph }_0}$. By transfinite recursion, choose two previously unused infinite subsets $X_{\alpha },Y_{\alpha }$ of each $A_{\alpha }$. This is possible because each $[A_{\alpha }{]}^{\omega }$ has cardinality $2^{{\aleph }_0}$ and fewer choices have been made at stage $\alpha$. The set $\{X_{\alpha }\}$ is not a Ramsey set of infinite subsets, since every $[A_{\alpha }{]}^{\omega }$ meets both it and its complement.

An infinite game of perfect information with an open set of winning plays for I in the product topology on a discrete move set has a winning strategy in an infinite game for one player. Construct a winning-position attractor in an infinite game by transfinite recursion; ranks give I a terminating descent strategy inside it, and its complement gives II a strategy avoiding every winning prefix. The move set need not be countable, in the usual set theory with axiom of choice.