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Apply the stated cardinal comparability principle to $x$ and its Hartogs ordinal $h(x)$. The Hartogs theorem rules out an injection $h(x)\to x$, so comparability supplies an injectionOrder $x$ by $u\prec v$ exactly when $j(u)\in j(v)$. This is the restriction of an ordinal well-order and therefore well-orders $x$. Hence every set can be well-ordered. The well-ordering theorem is equivalent to the axiom of choice: for a family of nonempty sets, well-order its union and choose the least member of each set. The comparison statement therefore implies choice.

For a transitive set $X$, use $L_0(X)=X$, $L_{\alpha +1}(X)=\mathrm{Def}(L_{\alpha }(X))$ and unions at nonzero limit ordinals. The successor operation is the definable power set. Each level is transitive and the levels increase; moreover $X\in L_1(X)$. Their union is $L(X)$, an inner model of ZF. The initial-set convention differs from a hierarchy using a predicate, or one starting at $X\cup \{X\}$. Axiom of choice need not hold for an arbitrary base without an internally available well-order.

For a transitive set $X$, the relative constructible universe $L(X)$ is the union of the relative constructible hierarchy, starting at $L_0(X)=X$, applying the definable power set at successors and taking unions at limit ordinals. It is the smallest inner model of ZF containing $X$ and all its members. For an arbitrary set $A$, take the base $X$ to be the transitive closure of $\{A\}$, so that $A\in X$. Axiom of choice need not hold for an arbitrary base; the construction uses initial parameters rather than a predicate relativization.

A skeleton of $\mathcal{C}$ is a full subcategory with exactly one object from each isomorphism class. The axiom of choice supplies skeletons of small categories. Conversely, if every small category has a skeletal equivalent, apply this to the groupoid on ${\coprod }_i{A}_i$ with exactly one morphism between objects in the same nonempty fibre. A quasi-inverse from the skeletal category selects one object per fibre, giving a choice function. This converse uses an equivalence of categories with specified quasi-inverse functors, not just the existence of a full essentially surjective functor in one direction.

Assuming the axiom of choice, every infinite cardinal satisfiesOne proof takes a least counterexample $\kappa$, identifies cardinals with initial ordinals, and well-orders $\kappa \times \kappa$ first by $\max \{\alpha ,\beta \}$. Every proper initial segment then has cardinality below $\kappa$ by minimality, so the whole order has cardinality at most $\kappa$, contradicting the choice of $\kappa$.

For infinite cardinals $\kappa$ and $\lambda$, the axiom of choice and the square of an infinite cardinal giveThe maximum is a lower bound for both operations, while both are bounded above by respectively $\mu +\mu$ and ${\mu }^2$, where $\mu =\max \{\kappa ,\lambda \}$.

Let $M$ be a transitive set model of ZFC. If $\mathrm{Ord}\cap M$ were countable, then every $x\in M$ would be countable: the internal axiom of choice supplies a bijection from $x$ to an ordinal of $M$, which is externally countable. For every $\alpha \in \mathrm{Ord}\cap M$, the internal rank $V_{\alpha }^M$ is an element of $M$ and hence countable. The Axiom schema of replacement inside $M$ gives $M={\bigcup }_{\alpha \in \mathrm{Ord}\cap M}{V}_{\alpha }^M$, a countable union of countable sets. Thus every uncountable transitive set model has uncountably many ordinals.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.

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

ZFC is Zermelo–Fraenkel set theory together with the axiom of choice.