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Let $f$ be given and choose $Y\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. Each $f(y)$ is countable, so the axiom of choice and infinite cardinal arithmetic giveSince the continuum is larger, choose $x$ outside this union. Then $x\notin f(y)$ for every $y\in Y$. The countable set $f(x)$ cannot contain $Y$, so choose $y\in Y\setminus f(x)$. ThusThe two points are distinct because $x\notin Y$. This proves the countable-valued free-pair criterion and the required direction of the Freiling axiom of symmetry without any measurability assumption on $f$.

Fix $a,\vec{p}\in L$ and a formula $\theta (x,\vec{p})$. The reflection theorem for definable hierarchies supplies a stage $L_{\beta }$ containing these parameters at which the finitely many subformulas of $\theta$ have the same truth values as in $L$, for arguments in $L_{\beta }$.

This reflection can be proved in the ambient ZF without presupposing Axiom schema of separation in $L$: starting at a stage containing the parameters, collect for each relevant existential subformula and every tuple in the current stage the least constructible stage in which a witness occurs, whenever a witness exists in $L$. Ambient Axiom schema of replacement bounds these ordinals. Iterate these witness-stage bounds for $\omega$ steps and take their supremum. Induction on subformulas gives the required finite reflection at the resulting union stage. Only the least stage is chosen, so no ambient axiom of choice is needed.

Since $L_{\beta }$ is transitive and contains $a$, the desired separated subset isIt is definable over $L_{\beta }$ with parameters there, so $b\in \mathrm{Def}(L_{\beta })=L_{\beta +1}\subset L$. Reflection gives $b=\{x\in a:{\theta }^L(x,\vec{p})\}$. Hence ZF proves every Axiom schema of separation instance relativized to $L$. This is a direct Separation proof in the constructible universe, not an appeal to the assertion being proved.

Let $R={\prod }_i{\mathcal{M}}_i/F$ be the reduced product by a proper filter on a set. Its underlying equivalence relation is $f\sim g$ when $\{i:f(i)=g(i)\}\in F$. Operations are interpreted coordinatewise, and a relation holds of the classes exactly when its coordinate truth set belongs to $F$.

Evaluation of a first-order term commutes with passage to the quotient, by mathematical induction on terms. Consequently the desired equivalence holds for every atomic formula, including logical equality. For a formula $\theta$ and representatives $\bar{f}$, write $S_{\theta }=\{i:{\mathcal{M}}_i\models \theta (\bar{f}(i))\}$.

For logical conjunction, $S_{\theta \wedge \psi }=S_{\theta }\cap S_{\psi }$. The filter on a set axioms giveThus the induction hypothesis transfers a conjunction in both directions.

For existential quantification, first suppose $R\models \exists y\theta (y,[\bar{f}])$. Choose a representative $g$ of a witness. Induction gives $S_{\theta (g,\bar{f})}\in F$, and this set is contained in $S_{\exists y\theta }$. Upward closure therefore gives $S_{\exists y\theta }\in F$.

Conversely, suppose $A=S_{\exists y\theta }\in F$. For each $i\in A$, choose a coordinate witness $g(i)$, and choose an arbitrary element of $M_i$ outside $A$. These simultaneous choices use the axiom of choice, as does the usual product construction. Then $A\subseteq S_{\theta (g,\bar{f})}$, so that truth set belongs to $F$. Induction gives $R\models \theta ([g],[\bar{f}])$, providing the required witness. Thereforefor every primitive positive formula. The exam's tame formulas are exactly this fragment, built using logical conjunction and existential quantification. No ultrafilter dichotomy was used.

First use the fact that a small set is absent from a complete nonprincipal ultrafilter. Indeed, if $A\subseteq \kappa$ and $|A|<\kappa$, then every $\kappa \setminus \{\alpha \}$, $\alpha \in A$, belongs to $U$ by nonprincipality. Kappa-completeness gives $\kappa \setminus A\in U$, so $A\notin U$. In particular every final segment $\{i:i>\alpha \}$ belongs to $U$.

Suppose for contradiction that the ultrapower has at most $\kappa$ elements. List representatives $f_{\alpha }:\kappa \to M$ for all its classes, indexed by $\alpha <\kappa$; repetitions are allowed. At coordinate $i<\kappa$, fewer than $\kappa$ values occur among $f_{\alpha }(i)$ with $\alpha <i$. Since $|M|\ge \kappa$, chooseThis diagonal argument for ultrapower cardinality uses the axiom of choice. For each fixed $\alpha$, the functions $g$ and $f_{\alpha }$ disagree throughout the final segment $i>\alpha$, which belongs to $U$. Their equality set therefore cannot belong to $U$, and $[g]\ne [f_{\alpha }]$.

This contradicts the assumed enumeration of the ultrapower. HenceThe proof uses only that each ordinal $i<\kappa$ has cardinality below $\kappa$; it does not require a separate assumption of regularity.

A strict partial order is a well-founded relation when every nonempty subset $Y\subseteq P$ has a minimal member:In ZFC this is equivalent to the absence of an infinite descending sequence $p_0{>}_P{p}_1{>}_P\cdots$. A descending sequence violates the minimal-member property; conversely, repeatedly choosing a predecessor from a subset with no minimal member constructs such a sequence using the axiom of choice.

First express inaccessibility by universally excluding witnesses to its failure. Besides $\kappa$ being an ordinal greater than $\omega$, require that for every $\alpha <\kappa$ there is no function from $\alpha$ cofinal in $\kappa$, and no injection from $\kappa$ into $\mathcal{P}(\alpha )$. The first condition gives regular cardinalhood; the second says that $\kappa$ is a strong limit cardinal. Under the axiom of choice these are exactly the conditions for a strongly inaccessible cardinal.

These are Pi-one formulas modulo ZF: the only unbounded quantifiers universally range over possible functions. Being a function of the specified domain, having bounded or cofinal range in $\kappa$, and being an injection whose values are subsets of $\alpha$ can all be written with bounded formulas in set theory, without using a power-set object as an unbounded existential witness. Consequently inaccessibility is downward absolute to a transitive inner model of ZFC.

Now suppose $\kappa$ is Mahlo in $V$. It is inaccessible in the constructible universe $L$, and every $V$-inaccessible ordinal below it is also inaccessible in $L$. Let $C\in L$ be any set that $L$ regards as a club set in $\kappa$. Closure and unboundedness are absolute properties of this fixed set of ordinals, so $C$ is also a club set in $V$. By Mahloness in $V$, it contains an ordinal that is inaccessible in $V$, hence in $L$. Every $L$-club therefore meets the $L$-inaccessible ordinals below $\kappa$. Thus Mahloness is downward absolute to the constructible universe:

Replace the strict partial order by its reflexive closure $\le$. Partially order all partial orders on $X$ extending $\le$ by inclusion. A chain has an upper bound given by its union: transitivity and antisymmetry each concern finitely many pairs, which all lie in one member of the chain. By Zorn lemma there is a maximal extension $R$.

If $a,b$ are incomparable under $R$, defineThis adds $a{R}^{\prime }b$. It is transitive: composing an old pair with a new one stays in the added set; composing two new pairs would require $bRa$, which is impossible. It is antisymmetric: an added pair and a reverse old pair would again imply $bRa$, and two reverse added pairs would do the same. Thus $R^{\prime }$ is a larger partial order, contradicting maximality. Therefore $R$ is total, and removing its diagonal gives a total order extending the original strict order.

A well-founded relation has a minimal element in every nonempty subset; with the axiom of choice this is equivalent to having no infinite strictly descending sequence. If an extension is a well-order, its least element in any nonempty subset is minimal for the original relation, so the original order must be well-founded.

However not every total extension need be a well-order. Take the empty strict order on $\mathbb{N}$, which is well-founded. Its extension defined by $x\prec y$ if $x>y$ in the usual order is total and has no least element.

Zorn lemma states that a nonempty partially ordered set in which every chain in a partial order has an upper bound has a maximal element. Here is a proof from the axiom of choice and Hartogs theorem.

Suppose there is no maximal element. Every chain $C$ has a strict upper bound: choose an upper bound $b$ and then an element strictly greater than $b$. The family of sets of strict upper bounds is a set of nonempty subsets of the underlying set $P$. The axiom of choice is used here to choose, simultaneously for every chain $C$, one strict upper bound $b(C)$.

By Hartogs theorem there is an ordinal $\kappa$ admitting no injection into $P$. Transfinite recursion defines $x_{\alpha }$ for $\alpha <\kappa$ byAt each stage the previous elements form a chain, and $x_{\alpha }$ is strictly above every one of them. Hence $\alpha \mapsto x_{\alpha }$ is an injection $\kappa \to P$, a contradiction. Therefore a maximal element exists.

For a finite position $s\in A^{<\omega }$, let $[s]$ be the cylinder set of infinite plays extending $s$. Because the factors have the discrete topology, these cylinder sets form a basis for the product topology. PutMembership in $D$ means that I has already secured the outcome, regardless of future moves. Since $G$ is an open set, a play belongs to $G$ exactly when one of its prefixes belongs to $D$.

Build the winning-position attractor in an infinite game by transfinite recursion. Start with $W_0=D$, use unions at limit ordinals, and setThis sequence is increasing: the defining operation is order-preserving, and $W_0\subseteq W_1$. It stabilizes at some ordinal. Indeed, if it strictly increased at every successor stage below the successor cardinal number of $|A^{<\omega }|$, choosing one newly added position per stage would inject that larger cardinal into $A^{<\omega }$. Denote the stable set by $W$. This constructs the needed fixed point directly, without an appeal to an unstated fixed-point theorem.

Every position in $W$ has a least entry ordinal, its rank. Rank zero means membership in $D$. A positive rank is a successor $\alpha +1$: at an I-position some extension belongs to $W_{\alpha }$, while at a II-position every extension does. If the empty word belongs to $W$, I always chooses an extension of smaller rank until $D$ is reached. II's choices also decrease rank before that happens. There is no infinite strictly decreasing sequence of ordinals, so some finite prefix lies in $D$. This is a winning strategy in an infinite game for I.

If the empty word is outside $W$, the fixed point equation says that every extension of an I-position outside $W$ is still outside $W$, and that a II-position outside $W$ has at least one extension outside $W$. II chooses such an extension. Every resulting prefix avoids $D$, hence the whole play avoids $G$ by openness. This is a winning strategy in an infinite game for II.

The choices can be made simultaneously by fixing a well-order of $A$, as allowed by the usual axiom of choice. The argument does not assume that $A$ is countable. Every such open game is determined:

Write $\mathfrak{c}=2^{{\aleph }_0}$. Its cofinality is ${\aleph }_1$, so choose a strictly increasing sequence of infinite cardinal numbers $({\kappa }_{\xi }{)}_{\xi <{\omega }_1}$ below $\mathfrak{c}$ with supremum $\mathfrak{c}$. Such a sequence is obtained by refining a cofinal sequence and choosing larger cardinals recursively; $\mathfrak{c}$ is a singular cardinal, hence a limit cardinal.

By the axiom of choice, fix a bijection $b:\mathfrak{c}\to \mathbb{R}$. SetEvery $A_{\xi }$ is infinite, $|A_{\xi }|={\kappa }_{\xi }<\mathfrak{c}$, and distinct indices give distinct cardinalities. The cofinality of the sequence ensuresThus $|\mathcal{A}|={\aleph }_1$. Pairwise different sizes means different sizes for distinct members; without that qualification the parenthetical condition would contradict the case $A=B$.

Use the Lévy hierarchy, with bounded formulas in set theory as the quantifier-free level for this purpose. Let $\mathrm{Ord}(\kappa )$ be the bounded ordinal predicate, and let $\mathrm{Bij}(f,\alpha ,\kappa )$ be the bounded predicate saying that $f$ is a graph of a bijection from $\alpha$ onto $\kappa$. Then cardinalhood is expressed byAll quantifiers in $\mathrm{Ord}$ and $\mathrm{Bij}$ are bounded by the displayed sets and the graph. Ordered-pair membership, graph functionality, injectivity and surjectivity have bounded expansions using the usual set coding of ordered pairs. The only unbounded quantifier shown is universal over $f$. The formula is therefore ${\Pi }_1$ and in particular a Pi-one formula modulo ZF.

In ZF, a cardinal number is an ordinal not in bijection with any smaller ordinal, so this formula has exactly the required meaning. Thus cardinalhood is Pi-one definable; the argument does not use the axiom of choice.

An equivalence of categories consists of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ together with invertible natural transformations $GF\cong 1_{\mathcal{C}}$ and $FG\cong 1_{\mathcal{D}}$. A strict isomorphism of categories instead requires a functor with an inverse whose composites are literally identity functors.

First suppose $F$ belongs to an equivalence of categories. The isomorphisms $FGD\cong D$ give essential surjectivity. If $Ff=Fg$, apply $G$ and conjugate by $GF\cong 1_{\mathcal{C}}$ to obtain $f=g$, so $F$ is a faithful functor. Similarly $G$ is a faithful functor. Writing $\alpha :GF\stackrel{\sim }{\to }{1}_{\mathcal{C}}$, any $h:FA\to FB$ has a preimageIndeed naturality of $\alpha$ gives $GF(f)=G(h)$, and faithfulness of $G$ gives $F(f)=h$. Thus $F$ is a full and faithful functor.

Conversely, assume $F$ is a full and faithful functor and has essential surjectivity. Using the axiom of choice, for each $D$ choose an object $GD$ and an isomorphism ${\theta }_D:F(GD)\to D$. Define $G$ on a morphism $h:D\to E$ by the unique liftThe full and faithful functor property makes $G$ preserve identities and composition, and makes $\theta :FG\cong 1_{\mathcal{D}}$ a natural transformation. For $A\in \mathcal{C}$, lift ${\theta }_{FA}:FGFA\to FA$ uniquely to ${\alpha }_A:GFA\to A$. Lifting its inverse shows that ${\alpha }_A$ is invertible. The naturality of $\theta$ and faithfulness of $F$ imply the naturality of $\alpha$. This provesFor large categories, this choice argument is understood in a fixed universe, or with the corresponding class-choice convention; ordinary set-sized axiom of choice suffices for small categories.

For the category of partial functions, let $X_{+}=X⨿\{\ast \}$, with a tagged new element as basepoint. Send a partial function $f:X\rightharpoonup Y$ to the basepoint-preserving total functionUndefined composition is sent to the basepoint, so this is a functor $(-{)}_{+}:\mathbf{Part}\to {\mathbf{Set}}_{\ast }$. Restriction away from the basepoint recovers each partial function uniquely; hence it is a full and faithful functor. Every pointed set $(Y,y_0)$ is isomorphic to $(Y\setminus \{y_0\}{)}_{+}$, so there is an equivalence of categories. This particular equivalence can also be constructed explicitly, without choice, by deleting and adjoining the basepoint.

These actual categories are equivalent but not isomorphic. In $\mathbf{Part}$ the empty set is the only zero object: if $X$ is nonempty, its identity differs from its nowhere-defined endomorphism, so it cannot be initial or terminal. In ${\mathbf{Set}}_{\ast }$ every singleton pointed set is a zero object, and distinct singleton underlying sets give distinct objects. An isomorphism of categories is a bijection on objects preserving zero objects; it cannot take one such object onto several. This uses the categories of all actual sets, as in the paper, rather than chosen skeletal categories of representatives.

A skeletal category has no distinct isomorphic objects. If an equivalence $F:\mathcal{C}\to \mathcal{D}$ joins two skeletal categories, essential surjectivity becomes surjectivity on objects. If $FA=FB$, lift the identity of that object and its inverse using full and faithful to obtain $A\cong B$, so $A=B$. Thus $F$ is bijective on objects and on every hom-set. Its inverse on objects and morphisms is a strictly inverse functor, proving that it is an isomorphism of categories.

Under the axiom of choice, choose one object from each isomorphism class of a small category. The full subcategory on those objects is a skeleton of a category, and its inclusion is a full and faithful functor with essential surjectivity, hence an equivalence of categories.

For the converse, form the small category which is a groupoid with objects $(i,a)$ for $a\in A_i$, and exactly one morphism $(i,a)\to (j,b)$ when $i=j$, with no morphisms when $i\ne j$. Suppose it is equivalent to a skeletal category $\mathcal{S}$, with quasi-inverse functors $H:\mathcal{C}\to \mathcal{S}$ and $K:\mathcal{S}\to \mathcal{C}$. For each $i$, the objects $H(i,a)$ for $a\in A_i$ are isomorphic, hence all equal to a uniquely determined $s_i$. The isomorphism $KH(i,a)\cong (i,a)$ ensures that $K(s_i)=(i,a_i)$ for some $a_i\in A_i$. The rule $i\mapsto a_i$ is a choice function. In particular, no representative in $A_i$ had to be chosen to define $s_i$, since it is unique. Consequently

Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits $\kappa =\omega$, for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention $\kappa >\omega$.

First $|V_{\gamma }|<\kappa$ for every $\gamma <\kappa$. At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently $V_{\kappa }\models \mathsf{ZFC}$. In particular, for Axiom schema of replacement, a domain $a\in V_{\kappa }$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the rank of a set over its elements is below $\kappa$ by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below $\kappa$; uncountability supplies axiom of infinity.

Enumerate the first-order formulas as ${\phi }_0,{\phi }_1,\dots$. For each finite collection, reflection inside the set structure $V_{\kappa }$ gives a closed unbounded subset of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_{\gamma }$, and the supremum of their least witness ranks remains below $\kappa$ by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because $\kappa$ is regular and uncountable. At each resulting $\alpha$,The infinite cardinal numbers below $\kappa$ also form a closed unbounded subset: they are unbounded because ${\lambda }^{+}\le 2^{\lambda }<\kappa$ for infinite $\lambda <\kappa$, and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal $\kappa$ has size $\kappa$. HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of $\kappa$.

Fix the base convention $L_0(X)=X$. For a transitive set $X$, the relative constructible hierarchy isHere the definable power set isThe satisfaction for a set structure in this definition is a definable set-theoretic relation, obtained by finite syntax coding and recursion on a first-order formula. The transfinite recursion therefore defines a class in ZF.

Each level is a transitive set. For transitive $A$, every $a\in A$ is a subset of $A$ definable using the parameter $a$, so $A\subseteq \mathrm{Def}(A)$; also $A\in \mathrm{Def}(A)$ using the always-true first-order formula. Thus the levels increase and $X\in L_1(X)$. Starting instead with $X\cup \{X\}$ is another common indexing convention, but it is not the convention used here. The relative universe need not satisfy axiom of choice: arbitrary $X$ need not have an internally available well-order.

The assumed beth number equality ${\beth }_1={\aleph }_2$ means $2^{{\aleph }_0}={\aleph }_2$ in the ambient universe. Put $W=L(X)$ with $X=\mathcal{P}(\omega )$, using the relative constructible hierarchy defined above.

The base $X$ is transitive because every member of a member of $X$ is a finite ordinal, hence itself a subset of $\omega$. Moreover $X\subseteq W$ and $X\in W$, and $W$ is an inner model of ZF. The latter standard fact follows from the hierarchy and reflection: bounded-rank witnesses give pairing and union; reflection gives separation and replacement; ambient replacement bounds the stages of all relatively constructible subsets of any fixed set, giving power set. It does not require axiom of choice inside $W$. Since no new subsets of $\omega$ can appear in an inner class,

Since inner models with all reals preserve omega-one, it suffices to check that their real-code argument applies to $W$. Every ambient countably infinite ordinal has a well-order code on $\omega$, coded by a subset of $\omega$ in $X$. Finite ordinals and $\omega$ are already shared. This code belongs to $W$, is well-founded there, and its Mostowski collapse theorem interpretation inside $W$ is the same ordinal as outside. Hence the ordinal is countable in $W$. Conversely, any countability witness in $W$ remains a witness in the ambient universe. Thus ${\omega }_1^W={\omega }_1$.

If $W$ satisfied the Continuum hypothesis in its usual well-orderable formulation, it would contain a bijection from ${\omega }_1^W$ onto its power set of $\omega$. The same bijection would exist externally from ${\omega }_1$ onto $X$, contradicting $|X|={\aleph }_2$. ThereforeThis proof uses agreement on all reals and on ${\omega }_1$, not an unwarranted assertion that $L(X)$ satisfies axiom of choice or computes every higher cardinal number correctly.

Let $\mathrm{Inj}(<\omega ,X)$ denote the finite repetition-free sequences, including the empty sequence. The map $x\mapsto (x)$ is injective, so this set is infinite when $X$ is infinite. Suppose, for a contradiction, that it contains a countably infinite subset. Fix its given injective enumeration $s_0,s_1,\dots$; this is part of the supposition, not a choice from a family of sets.

Enumerate all pairs $(n,j)$ with $j<\mathrm{length}(s_n)$ by their natural-number pairing codes, recording the corresponding entries $s_n(j)$. If the union of entries were infinite, recursively selecting the first new entry would give an injection $\mathbb{N}\to X$, contradicting Dedekind-finiteness. Therefore the union is a finite set $F$, say of size $m$. A repetition-free sequence over $F$ has length at most $m$, and there are exactlysuch sequences, with the empty product equal to one. All $s_n$ would belong to this finite set, contradicting their distinctness. Finite counting and the least-code construction use no axiom of choice. This proves finite repetition-free sequences preserve Dedekind-finiteness and the required infinitude:

Work in ZF with the law of excluded middle, without the axiom of choice. The paper's terminology is an infinite Dedekind-finite set. The useful common principle is countable union of explicitly ordered finite lists without choice: given an actual sequence of finite lists, enumerate their entries by list number and position using a Cantor pairing function. If the union of entries is infinite, repeatedly take the entry with the least code not already selected. This produces an injection from $\mathbb{N}$ without choosing any enumerations of unordered sets.

Consequently, if a family of objects has an explicitly ordered finite list of labels for each object, and only finitely many objects can be made from any given finite pool of labels, then a countably infinite list of distinct objects forces a countably infinite subset of the label set. For a Dedekind-finite label set this is impossible. If the label set embeds into the object family as well, infinitude is preserved. ThusThe distinction between ordered lists supplied by the data and arbitrary finite subsets is essential: a blanket countable-union theorem for unordered finite sets would introduce a choice principle.