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Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal $\kappa$ is greater than ${\omega }_1$: ZFC in $V_{\kappa }$ contains the full power set $\mathcal{P}(\omega )$, and the axiom of choice there gives a bijection between it and an ordinal $\gamma <\kappa$. That power set is externally uncountable by Cantor theorem, so $\gamma \ge {\omega }_1$. Infinity rules out the finite cardinals and $\omega$ before this argument is applied.

By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure $X\prec (V_{\kappa },\in )$. Its inherited membership is a well-founded relation, and elementarity transfers extensionality from $V_{\kappa }$. Apply the Mostowski collapse theorem to obtain a countable transitive model $M$ of ZFC. Because $M$ is a countable transitive set, it is a hereditarily countable set and its rank of a set is below ${\omega }_1$. In particular, $M\in V_{\kappa }$.

Now take the Skolem hull in $V_{\kappa }$ of the seedCall it $H$. The seed has external cardinality ${\aleph }_1$ and lies inside $V_{\kappa }$. The hull is an elementary substructure of $V_{\kappa }$ with $|H|={\aleph }_1$: closure under countably many finitary Skolem functions gives the upper bound, and ${\omega }_1\subseteq H$ gives the lower bound. Let $\pi :H\to N$ be its Mostowski collapse theorem map. Then $N$ is a transitive model of ZFC.

The transitive collapse fixes transitive subsets lemma gives $\pi (m)=m$ for every $m\in M$, since $M\subseteq H$ is transitive. Since also $M\in H$, the collapse sends that element to $M$ itself. Thus $M\in N$, and transitivity of $N$ yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.

Let $\theta =M\cap \mathrm{Ord}$ be the ordinal height of a model of set theory. First $\theta \ge {\omega }_1$. Indeed, if $\theta$ were countable, the internal axiom of choice would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_{\xi }^M$, $\xi <\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, sowould be a countable union of countable sets, a contradiction. This is why an uncountable transitive set model has uncountable ordinal height.

For $\xi <\theta$, absoluteness of constructible levels gives $L_{\xi }^M=L_{\xi }$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; transfinite recursion then also makes limit stages agree. Transitivity consequently givesIt remains to locate a countability witness. Given an ambient countable ordinal $\alpha$, choose an injection $f:\alpha \to \omega$. Since $V=L$, it belongs to the constructible universe. Its transitive closure, together with $f$ itself, is countable. Choose a countable elementary substructure $X\prec L_{\eta }$ of a sufficiently large limit level containing this closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_{\beta }\subseteq L_{{\omega }_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha <{\omega }_1\le \theta$. Being an injection between these fixed sets is a bounded formula in set theory, so $M$ recognizes $f$ as a countability witness. This proves countable-ordinal correctness under constructibility:The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.

The two Ramsey properties and the topologies. Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, and $[A{]}^{\omega }$ for the infinite subsets of $A$. The positive integers are used throughout. A Ramsey set of infinite subsets $E$ has the property that every infinite $A$ contains an infinite $B$ for which $[B{]}^{\omega }$ lies wholly in $E$ or wholly in $E^c$.

For finite $s$ and infinite $A$ above $\max s$, letThese are the basic neighbourhoods of the Ellentuck topology, denoted $\ast$ in the question. They form a basis: if $X$ belongs to two such sets, their stems are comparable initial segments of $X$; take the longer stem and the intersection of the two tails beyond that stem. This intersection contains the infinite remaining tail of $X$ and yields a basic set contained in both.

A completely Ramsey set $E$ allows such a homogeneous refinement $[s,B]$ inside every $[s,A]$, with $B\subseteq A$ infinite and the stem $s$ kept fixed. Taking $s=\varnothing$ shows that complete Ramseyness implies Ramseyness. The ordinary topology on infinite subsets $\tau$ has basic cylindersIt is the usual product topology on increasing enumerations and is coarser than the Ellentuck topology.

Examples distinguishing the properties. A non-Ramsey example uses the axiom of choice. Let $\kappa =2^{{\aleph }_0}$ and enumerate all infinite sets as $(A_{\alpha }{)}_{\alpha <\kappa }$. Each $[A_{\alpha }{]}^{\omega }$ has cardinality $\kappa$: an enumeration of $A_{\alpha }$ gives the upper bound, and choosing one element from each of infinitely many disjoint pairs gives an injection from all binary sequences for the lower bound.

By transfinite recursion, choose two distinct, previously unused elements $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$ at each stage. Fewer than $\kappa$ elements have been used at stage $\alpha <\kappa$, so this is possible without any assumption on the Continuum hypothesis. Put $R=\{X_{\alpha }:\alpha <\kappa \}$. All the $Y_{\alpha }$ remain outside $R$, since subsequent choices also avoid previously chosen elements. Every $[A{]}^{\omega }$ contains one $X_{\alpha }$ and one $Y_{\alpha }$, so $R$ is not a Ramsey set of infinite subsets. This is the non-Ramsey set from transfinite selection construction.

Perform the same construction on $T=\mathbb{N}\setminus \{1\}$, giving a non-Ramsey set $R_T\subseteq [T{]}^{\omega }$, and putFor any infinite $A$, the infinite set $B=A\setminus \{1\}$ has $[B{]}^{\omega }\cap E=\varnothing$, so $E$ is a Ramsey set of infinite subsets. But no $[\{1\},B]$ with infinite $B\subseteq T$ is homogeneous, because $[B{]}^{\omega }$ meets both $R_T$ and its complement. Therefore $E$ is not a completely Ramsey set.

Every star-open set is completely Ramsey. We give the full fusion proof for open Ellentuck sets, without quoting an infinite Ramsey theorem. Fix a star-open set $O$ and a basic set $[s,A]$. For finite $a\subseteq A$ and an infinite tail $D\subseteq A$ above $\max a$, say $D$ accepts $a$ if $[s\cup a,D]\subseteq O$; it rejects $a$ if no infinite subset of $D$ accepts $a$. Both properties persist under passing to infinite subsets. Every infinite tail can be thinned to decide $a$: take an accepting subset if one exists, and otherwise the original tail rejects it.

First thin $A$ to decide $a=\varnothing$. If it accepts, the required homogeneous neighborhood has already been found. Otherwise keep a rejecting tail. We now fuse to an infinite set $B$ deciding every finite $a\subseteq B$, where the decision for $a$ is made by $B/a=\{b\in B:b>\max a\}$.

To do this, choose $b_1<b_2<\cdots$ successively from an infinite current tail. After choosing $b_j$, thin the tail above $b_j$ to decide each of the finitely many subsets of $\{b_1,\dots ,b_j\}$, one after another, before choosing the next element. For any finite $a\subseteq B$, at the stage when its largest element is selected, the current tail decides $a$ and contains every later selected element. Heredity therefore gives a decision on $B/a$. For $a=\varnothing$, rejection persists from the initial tail.

If $B/a$ rejects $a$, only finitely many $n\in B/a$ can have $B/n$ accepting $a\cup \{n\}$. Otherwise let $D$ be the infinite set of those $n$. Every member of $[s\cup a,D]$ has a first new element $n$, and its remaining tail lies in $B/n$. It therefore belongs to $O$ by acceptance of $a\cup \{n\}$. This would make $D$ accept $a$, contradicting rejection.

Thin $B$ a second time to $C=\{c_1<c_2<\cdots \}$ so that every finite $a\subseteq C$ is rejected. Start with the rejected empty set. At a finite stage, all subsets of the selected elements are rejected. For each of these finitely many subsets, avoid its finitely many next elements that would give an accepted extension, and choose the next $c_j$ beyond all excluded elements. Since $B$ decides every finite extension, each new extension is rejected. Induction gives the asserted property of $C$.

We claim $[s,C]\cap O=\varnothing$. If $X\in [s,C]\cap O$, openness provides a basic set $[t,D]\subseteq O$ containing $X$. Extend $t$, if necessary, to an initial segment $u$ of $X$ containing $s$. Write $u=s\cup a$ with finite $a\subseteq C$. The infinite tail $X\setminus u$ lies in $B/a$ andIt is therefore an accepting subtail for $a$, contradicting rejection. We have found either an accepting neighborhood or a neighborhood disjoint from $O$. Thus

Basic neighborhoods are clopen. Consider $[s,A]$ and a point $X$ outside it. If $s$ is not an initial segment of $X$, take an initial segment $t$ of $X$ of length at least $|s|$; its ordinary cylinder $[t]$ cannot meet $[s,A]$. Otherwise some element of $X\setminus s$ is outside $A$; take $t$ through that element, and again $[t]$ is disjoint from $[s,A]$.

Thus the complement of $[s,A]$ is already open in the ordinary topology on infinite subsets, and hence in the finer Ellentuck topology. A basic neighbourhood is open by definition, so

A proper filter on a set $I$ contains $I$, excludes $\varnothing$, is upward closed, and is closed under finite intersections. Order its proper extensions by inclusion. The union of a nonempty chain is again a proper filter: finitely many members lie together in one chain member. Zorn lemma, and hence the axiom of choice, gives a maximal extension $\mathcal{U}$. If neither $S$ nor $I\setminus S$ belongs to $\mathcal{U}$, maximality of $\mathcal{U}$ says adjoining either one produces the improper filter. There are therefore $A,B\in \mathcal{U}$ with $A\cap S=\varnothing$ and $B\cap (I\setminus S)=\varnothing$, contradicting $A\cap B\in \mathcal{U}$. Thus $\mathcal{U}$ is an ultrafilter.

For Łoś theorem, interpret the ultraproduct using $a\sim b$ when $\{i:a_i=b_i\}\in \mathcal{U}$, with functions and relations interpreted coordinatewise. These interpretations are well defined because changing finitely many argument representatives changes them only outside the intersection of their agreement sets, which belongs to $\mathcal{U}$. For any first-order formula and sequences of parameters,Prove this by structural induction. Atomic formulas follow from the interpretation of terms. Conjunction uses finite intersections; negation uses the ultrafilter dichotomy between a set and its complement. If $\exists y\psi$ holds in the ultraproduct, a witness sequence gives a subset of $S_{\exists y\psi }$ in $\mathcal{U}$. Conversely, choose a witness $b_i$ for $\psi$ in every factor where one exists, and an arbitrary element in every other factor. This is the second explicit use of the axiom of choice; all factors have nonempty carriers. The inductive hypothesis applied to $\psi (\mathbf{a},\mathbf{b})$ supplies the ultraproduct witness. The other connectives and universal quantification follow from these cases.

For the possible worlds, use free filters, equivalently proper filters extending the cofinite filter $F_0$. Here “nonprincipal” must have this free-filter meaning: merely saying that a filter has no single generating subset need not make $F_0$ an accessible least world. The designated world is $F_0$, and all worlds have carrier $D={\prod }_i{M}_i$. Atoms, including equality, are forced byEquality may identify distinct sequences at a world; it is a persistent congruence, and at an ultrafilter world its quotient is the ordinary ultraproduct.

The required equivalence with all ultraproducts uses dense-extension filter semantics. Its recursive clauses areEvery extension in these clauses is a free proper filter. The dense clause for disjunction is essential; for propositional formulas, local disjunction would instead give a Kripke model for intuitionistic propositional logic and would invalidate the claimed equivalence.

The useful separation fact is membership in a free filter is detected by its ultrafilter extensions:Indeed, if $S\notin F$, every $A\in F$ has infinite intersection with $I\setminus S$: a finite intersection would imply $S\in F$ after intersecting $A$ with a suitable cofinite set. Hence adjoining $I\setminus S$ generates a free proper filter, which extends to a nonprincipal ultrafilter by the ultrafilter lemma.

Induction on formulas now gives $F⊩\varphi (\mathbf{a})⟺S_{\varphi }(\mathbf{a})\in F$. Negation and implication follow by adjoining the positive set on which their classical truth conditions fail. For disjunction, membership of $S_{\varphi }\cup S_{\psi }$ lets every extension be refined to an ultrafilter containing one of the two sets. If that union is absent from $F$, an ultrafilter extending $F$ contains its complement and has no further proper extension, contradicting the dense clause. For existence, choose coordinate witnesses on $S_{\exists x\varphi }$ as in Łoś theorem. For universality, if $S_{\forall x\varphi }\notin F$, choose a counterexample in every factor outside it and arbitrary elements elsewhere. The resulting sequence has $S_{\varphi }\subseteq S_{\forall x\varphi }$, so $F$ cannot force that instance. The converse follows from upward closure and persistence.

At the designated world, therefore,This semantics validates classical first-order logic, including law of excluded middle. For example, every free filter has an ultrafilter refinement deciding an atomic formula, so its excluded middle is forced even when neither disjunct is forced at the original world. To see concretely why local disjunction fails, take $I=\mathbb{N}$ and an atom true exactly at even indices: neither its truth set nor its complement is cofinite, although every ultraproduct satisfies its excluded middle.

Let $M$ be the given transitive model and suppose that its ordinal height of a model of set theory $\mathrm{Ord}\cap M$ were a countable set. For every $x\in M$, the internal Axiom of choice gives a bijection from $x$ to an ordinal of $M$; transitivity makes this an actual bijection, and the ordinal is externally countable. Thus every element of $M$ is externally countable.

For each $\alpha \in \mathrm{Ord}\cap M$, the internal rank $V_{\alpha }^M$ belongs to $M$ and is therefore countable. Every $x\in M$ lies in one of these ranks, so $M$ is a countable union of countable sets and is itself countable, contrary to the hypothesis. By uncountable transitive set model has uncountable ordinal height, $M$ contains uncountably many ordinals.

Suppose that the countable ordinal $\alpha$ satisfied $V_{\alpha }\models \mathsf{ZFC}$. The axioms force $\alpha$ to be a limit ordinal above $\omega$, so choose an externally countable cofinal function $n\mapsto {\alpha }_n$ into $\alpha$, with ${\alpha }_n+1<\alpha$. The internal Axiom of choice gives a bijection in $V_{\alpha }$ between each $V_{{\alpha }_n}$ and some ordinal below $\alpha$. Every such ordinal is externally a countable set, hence every $V_{{\alpha }_n}$ is externally countable. The countable union of countable sets is countable, sowould be countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem. This contradiction is the result Countable rank-initial segment cannot model ZFC, and therefore

Zorn lemma states that if every chain in a nonempty partially ordered set has an upper bound in that set, then the set has a maximal element. It is equivalent, over the usual axioms of set theory, to the axiom of choice.

By part ii every subset of the Baire space of sequences is $2^{{\aleph }_0}$-Suslin. If $2^{{\aleph }_0}\le {\aleph }_2$, part i would make every such set ${\aleph }_2$-Suslin. Therefore

Now suppose $2^{{\aleph }_0}>{\aleph }_2$. The axiom of choice gives a set $B\subseteq {\omega }^{\omega }$ of cardinality exactly ${\aleph }_2$. If $B$ were ${\aleph }_1$-Suslin, the Aleph-one-Suslin decomposition into analytic sets would write it as a union of ${\aleph }_1$ analytic sets. If all those analytic sets were countable, their union would have cardinality at most ${\aleph }_1$, so one of them is uncountable. The perfect set property for analytic sets then makes that member, and hence $B$, have cardinality $2^{{\aleph }_0}$, contradictingThus $B$ is not ${\aleph }_1$-Suslin, and

Suppose ZFC proved that every ${\aleph }_1$-Suslin set is determined. The assumed consistency of ZFC and the relative consistency of the Continuum hypothesis would then give a model ofIn that model $2^{{\aleph }_0}={\aleph }_1$. By Every set of reals is continuum-Suslin, every subset of ${\omega }^{\omega }$ is therefore ${\aleph }_1$-Suslin and hence determined. This is the axiom of determinacy.

But the axiom of choice produces an undetermined set of reals, so ZFC and the axiom of determinacy are incompatible. The displayed theory cannot have a model, contradicting the relative consistency of ZFC plus the continuum hypothesis. Hence ZFC cannot prove that all ${\aleph }_1$-Suslin sets are determined.

For cardinals, $\kappa <\lambda$ means that there is an injection $\kappa \to \lambda$ but no bijection. Under the axiom of choice, every set is well-orderable, so every infinite cardinal is the initial ordinal ${\aleph }_{\alpha }$ for a unique ordinal $\alpha$. Cantor theorem givesand minimality of the successor cardinal givesThe strict inequality ${\aleph }_{\alpha +1}<2^{{\aleph }_{\alpha }}$ printed in the conversion is not provable and can be false under the generalized continuum hypothesis; the intended symbol must be $\le$.

Assume the axiom of choice. By the well-ordering theorem, every set is bijective with an ordinal, and hence with an initial ordinal. Any two ordinals are comparable, so for any sets $K,L$ either $K$ injects into $L$ or $L$ injects into $K$. Transitivity is immediate, and antisymmetry on cardinalities follows from the Cantor-Schröder-Bernstein theorem. Thus cardinal comparison is a total order.

Conversely, assume the cardinal comparability principle. Given any set $X$, let $h(X)$ be its Hartogs ordinal. By Hartogs theorem, there is no injection $h(X)\to X$. Comparability therefore supplies an injection $X\to h(X)$. Pulling the ordinal order back to $X$ gives a well-order of $X$. Hence every set can be well-ordered, and the well-ordering theorem implies the axiom of choice.

Under the axiom of choice, identify every infinite cardinal $\kappa$ with its initial ordinal. We prove ${\kappa }^2=\kappa$ by transfinite induction on infinite cardinals. The countable case follows from the usual diagonal enumeration of $\mathbb{N}\times \mathbb{N}$.

For the induction step, well-order the Cartesian product $\kappa \times \kappa$ by first comparingand then using lexicographic order among pairs with the same maximum. The predecessors of $(\alpha ,\beta )$ lie in $(\eta +1)\times (\eta +1)$, where $\eta =\max \{\alpha ,\beta \}<\kappa$. Put $\mu =|\eta +1|<\kappa$. If $\mu$ is infinite, the induction hypothesis makes this predecessor set have cardinal at most ${\mu }^2=\mu <\kappa$; if $\mu$ is finite, the same conclusion is immediate.

Let $\theta$ be the order type of this well-order. Every proper initial segment of $\theta$ therefore has cardinal less than $\kappa$. Hence $\theta <{\kappa }^{+}$, since otherwise its initial segment of order type $\kappa$ would have cardinal $\kappa$. Thus $|\kappa \times \kappa |=|\theta |\le \kappa$. The map $\alpha \mapsto (\alpha ,0)$ gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields

Letbe the injection witnessing $\kappa \lambda \le \kappa +\lambda$, and fix $(k_{\ast },l_{\ast })\in K\times L$. Extend the inverse of $i$ to a surjectionby settingThis construction uses only the fixed pair, not the axiom of choice.

Restrict $p$ to the $K$-summand and take its second coordinate:If $r$ is onto, it is the required surjective function from $K$ to $L$.

Otherwise choose one $l_0\in L\setminus r[K]$. For each $k\in K$, surjectivity of $p$ gives a preimage of $(k,l_0)$. No such preimage lies in the $K$-summand, by the choice of $l_0$, so it lies in the $L$-summand. Unless $(k,l_0)=(k_{\ast },l_{\ast })$, this preimage is exactly $i(k,l_0)$ and is unique: all points outside the range of $i$ map only to the default pair. Distinct $k$ give distinct preimages because $i$ is injective.

If $l_0\ne l_{\ast }$, these unique preimages therefore define an injective function $K\to L$. The only exceptional case is $l_0=l_{\ast }$, where they initially define an injectionIf $j$ is onto, extend it arbitrarily at $k_{\ast }$ to obtain a surjection $K\to L$. If it is not onto, choose one $l^{\prime }\in L\setminus j[K\setminus \{k_{\ast }\}]$ and set $j(k_{\ast })=l^{\prime }$; this gives an injection $K\to L$. These cases prove the product-sum comparison lemma entirely in ZF.

Part (b) proves one implication. Conversely, assume in ZF thatfor every infinite cardinal $\kappa$. Let $X$ be any set and let $\gamma =h(X)$ be its Hartogs ordinal. Finite $X$ is already well-orderable, so suppose $X$ is infinite and form the disjoint unionBy hypothesis, $A\times A$ is bijective with $A$. Since $\gamma \times X$ injects into $A\times A$, we obtain

Apply part (c), with $K=\gamma$ and $L=X$. It gives either an injection $\gamma \to X$ or a surjection $s:\gamma \to X$. The first alternative contradicts Hartogs theorem. In the second, every fiber $s^{-1}(\{x\})$ is a nonempty set of ordinals and therefore has a least member. The mapis an injection of $X$ into the ordinal $\gamma$ and pulls its well-order back to $X$. Thus every set is well-orderable, so the well-ordering theorem gives the axiom of choice. This proves the Tarski cardinal-square theorem.

Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. The axiom of choice says that every family of nonempty sets has a choice function. The well-ordering theorem says that every set admits a well-order.

Choice gives the Hausdorff maximal principle by repeatedly choosing an element extending a chain; the union at limit stages is again a chain. A maximal chain has an upper bound, and that upper bound is maximal, proving Zorn. Conversely, apply Zorn to partial choice functions ordered by extension. A maximal partial choice function must have the full family as domain, proving choice.

Choice also well-orders a set by recursively choosing from the unchosen remainder; Hartogs theorem forces this recursion to exhaust the set before it reaches the Hartogs ordinal. Conversely, from a well-order on the union of a family, choose the least element of each member. Thus all three principles are equivalent.

By choice-function well-ordering construction, use transfinite recursion to choosewhenever the set on the right is nonempty. The chosen elements are distinct, so if this construction continued through the Hartogs theorem ordinal $h(x)$, the map $\beta \mapsto x_{\beta }$ would inject $h(x)$ into $x$, a contradiction. It therefore stops at some ordinal $\alpha <h(x)$. At the stopping stage every element of $x$ has been selected, and henceis a bijection $\alpha \to x$.

Assuming the axiom of choice, every set $x$ has such a choice function on its nonempty subsets. Transporting the membership order on $\alpha$ across the bijection well-orders $x$. Thus the axiom of choice implies the well-ordering theorem.