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In ZFC, every ${\aleph }_1$-Suslin set is a union of ${\aleph }_1$ many analytic sets. A branch through a representing tree uses a countable sequence of countable ordinals. That sequence is bounded below ${\omega }_1$, so restrict the labels successively below $\alpha <{\omega }_1$.

Each restricted label set is countable and can be encoded by natural numbers; its projection is therefore an analytic set. The union of these projections is the original set. This is the $\kappa ={\aleph }_0$ case of Successor-Suslin decomposition. The proof explicitly uses the uncountable cofinality of ${\omega }_1$ and countable recoding, hypotheses that need care without the axiom of choice.

The Beth numbers ${\beth }_0={\aleph }_0$, ${\beth }_{n+1}=2^{{\beth }_n}$ form a strictly increasing countable sequence by Cantor's theorem. Their supremum ${\beth }_{\omega }$ has cofinality $\omega$ because no finite subset is cofinal. It exceeds $2^{{\aleph }_0}={\beth }_1$. It is also a strong limit cardinal: if $\rho <{\beth }_{\omega }$, some $n$ has $\rho <{\beth }_n$, so $2^{\rho }\le {\beth }_{n+1}<{\beth }_{\omega }$. Thus the singular cardinal example needs no Continuum hypothesis.

There is an uncountable cardinal number $\kappa$ with ${\sigma }_{\kappa }=\kappa$. Iterate ${\kappa }_{n+1}={\sigma }_{{\kappa }_n}$ starting with ${\aleph }_0$. If there is no earlier fixed point, the increasing supremum $\kappa$ has countable cofinality, hence is a singular cardinal. The singular cardinal enumeration is cofinal in $\kappa$ below index $\kappa$, and continuity at this singular supremum gives equality.

If $\kappa$ is a regular cardinal and $\mathbb{P}$ has the $\kappa$-chain condition, then forcing with $\mathbb{P}$ preserves every cardinal number and cofinality at least $\kappa$. Indeed, a proposed surjection $f:\mu \to \lambda$ with $\mu <\lambda$ has its range contained in ${\bigcup }_{\xi <\mu }{B}_{\xi }$ from the possible-values lemma for chain-condition forcing. The regularity of $\kappa$ and infinite cardinal arithmetic make this union have cardinality below the ground-model cardinal $\lambda$, a contradiction.

The cofinality of a linear order is the least cardinality of a cofinal subset. It is preserved by order isomorphisms. The order sum of ${\omega }_1$ copies of the rational order has cofinality ${\aleph }_1$, whereas appending one final rational block changes its cofinality to ${\aleph }_0$ without changing its cardinality or the dense-linear-order axioms.

If $\alpha$ is a nonzero limit ordinal and $({\gamma }_{\xi }{)}_{\xi <\alpha }$ is strictly increasing, then $\mathrm{cf}({\sup }_{\xi <\alpha }{\gamma }_{\xi })=\mathrm{cf}(\alpha )$. A cofinal subsequence of the index order gives one inequality. Conversely, a cofinal family in the supremum determines a cofinal family of indices by choosing an index past each of its values. This also explains the cofinality of limit-indexed aleph numbers.

For an infinite cardinal number $\kappa$, use cardinal arithmetic with the axiom of choice. If $\lambda =2^{\kappa }$ had cofinality $\mu \le \kappa$, choose cofinal ordinals ${\alpha }_i<\lambda$, $i<\mu$. Their cardinalities $a_i$ satisfy ${\sum }_i{a}_i=\lambda$. The König theorem for cardinal numbers gives $\lambda <{\prod }_i{a}_i^{+}\le {\lambda }^{\mu }\le (2^{\kappa }{)}^{\kappa }=\lambda$, a contradiction.

A forcing is cofinality-preserving over a transitive model $M$ if every generic extension has the same cofinality for every ground ordinal. The values are compared as ordinals, which forcing preserves.

Under the axiom of countable choice, the first uncountable ordinal has cofinality ${\omega }_1$. Indeed, the supremum of any countable set of countable ordinals is still a countable ordinal, so no countable sequence is cofinal in ${\omega }_1$.

The Gimel function determines all infinite cardinal powers. For a regular base, $2^{\kappa }=\gimel (\kappa )$. For a singular base, put $s={\sup }_{\rho <\kappa }{2}^{\rho }$ and $\theta =\mathrm{cf}(\kappa )$; then $2^{\kappa }=s^{\theta }$, which is $s$ if attained below $\kappa$ and $\gimel (s)$ otherwise. After these powers are known, fix an infinite exponent $\lambda$ and recurse on the base. Below $\lambda$ use $2^{\lambda }$; at successors use the Hausdorff formula for cardinal exponentiation. At a limit base greater than $\lambda$, put $a={\sup }_{\rho <\kappa }{\rho }^{\lambda }$. The answer is $a$ if $\mathrm{cf}(\kappa )>\lambda$ or the supremum is attained, and $\gimel (a)$ otherwise. In the nonattained cases, the cofinal-index argument identifies the cofinality of the supremum.

For a strong-limit cardinal $\lambda$ of countable cofinality, defineHere smallness bounds every individual membership extension; it does not bound the cardinality of the entire transitive closure. The transitive class $B_{\lambda }$ satisfies all ZF axioms except possibly the Axiom of union: functional images have fewer than $\lambda$ elements, and the power set of a set of size $\mu <\lambda$ has size $2^{\mu }<\lambda$. These facts give Axiom schema of replacement and Axiom of power set; pairs and definable subsets also remain locally small. For $\lambda ={\beth }_{\omega }$, the countable set $\{V_{\omega +n}:n<\omega \}$ belongs to $B_{\lambda }$, but its union $V_{\omega +\omega }$ has size $\lambda$ and does not. This gives an explicit relative-independence model for the Axiom of union.

Assume the Generalized continuum hypothesis. For necessity, König theorem for cardinal numbers gives ${\rho }^{\mathrm{cf}(\rho )}>\rho$ for every infinite cardinal. But $\lambda ={\kappa }^{\mu }$ implies ${\lambda }^{\mu }=\lambda$ by the currying law for cardinal exponentiation, so $\mathrm{cf}(\lambda )>\mu$. This direction uses only ZFC. Conversely, uncountable cofinality bounds the range of every countable sequence in $\lambda$. Under GCH, each smaller infinite cardinal $\nu$ satisfies ${\nu }^{{\aleph }_0}\le 2^{\nu }={\nu }^{+}\le \lambda$. There are only $\lambda$ ordinal bounds to consider, giving ${\lambda }^{{\aleph }_0}=\lambda$ and the representation with base $\lambda$ and exponent ${\aleph }_0$.

A subset $X\subseteq \lambda$ is non-reflecting in the indicated sense if it is not stationary below any $\delta <\lambda$ of uncountable cofinality. The subset itself need not be stationary. Nonreflection is inherited by subsets. With this definition every subset of ${\omega }_1$ is non-reflecting, since all smaller ordinals have countable cofinality.

A cardinal number is an initial ordinal: an ordinal not in bijection with any smaller ordinal. In $ZFC$ every set has the cardinality of a unique such ordinal.

For an infinite cardinal $\kappa$, its cofinality is the least cardinality of an unbounded subset of the ordinal $\kappa$. Equivalently,A subset is unbounded when no ordinal below $\kappa$ bounds all its elements. A cofinal sequence can be taken increasing, and $\mathrm{cf}(\kappa )$ is an infinite regular cardinal.

For an infinite cardinal regarded as an ordinal, its cofinality $\theta =\mathrm{cf}(\kappa )$ is the least cardinality of a cofinal subset of $\kappa$. Equivalently there is a family $({\alpha }_i{)}_{i<\theta }$ of ordinals below $\kappa$ whose supremum is $\kappa$, and no smaller indexed family has that property. Since an infinite initial ordinal is a limit ordinal, $\theta$ is infinite and in particular nonzero.

Use the diagonal mechanism of König theorem for cardinal numbers directly. Suppose $F:\kappa \to {}^{\theta }\kappa$ were surjective. For each $i<\theta$, the sethas cardinality at most $|{\alpha }_i|<\kappa$. Choose the least $b_i\in \kappa \setminus E_i$. The resulting function $b:\theta \to \kappa$ is not in the range: for any $\alpha <\kappa$, cofinality supplies $i$ with $\alpha <{\alpha }_i$, and then $b_i\ne F(\alpha )(i)$.

Thus no such surjection exists. Constant functions provide an injection $\kappa \to {}^{\theta }\kappa$, and an injection in the reverse direction would give a surjection by the same inverse-and-default construction as in part (i). ConsequentlyThis argument also handles $\kappa ={\aleph }_0$, when cofinal initial segments are finite; it does not incorrectly assume that smaller cardinals themselves must form a cofinal sequence in every infinite cardinal.

For an infinite cardinal number $\kappa$, its cofinality isviewing $\kappa$ as an initial ordinal. Equivalently it is the smallest cardinal number which is the domain size of a cofinal map into $\kappa$.

Put $\lambda =2^{\kappa }$ and $\mu =\mathrm{cf}(\lambda )$. Suppose $\mu \le \kappa$, and choose cofinal ordinals ${\alpha }_i<\lambda$ for $i<\mu$. Put $a_i=|{\alpha }_i|<\lambda$. Their union is $\lambda$, so infinite cardinal arithmetic givesApply the König theorem for cardinal numbers to $a_i<a_i^{+}$. Each successor cardinal $a_i^{+}$ is at most $\lambda$, whenceHere the last equalities use the currying law for cardinal exponentiation and square of an infinite cardinal. This contradiction proves the cofinality of a power-set cardinal bound

Independence here is a relative consistency assertion: from a ground first-order model satisfying ZFC, construct a first-order model of each remaining theory in which the omitted axiom fails. The ground axiom of choice is a tool for constructing these countermodels, not an extra axiom imposed on the theories being tested. The constructible universe supplies the usual relative-consistency implication from ZF to ZFC, so this ground choice assumption requires no additional consistency strength. These constructions can be formalized inside an arbitrary ground model; they do not require the stronger assumption that a transitive model exists.

Union. Put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe ground cardinal $\lambda$ is a strong limit cardinal of countable cofinality. This transitive class contains the empty set and $\omega$. Pairing preserves the bound, as does taking a subset, so separation holds for every formula relativized to this class. If an internally definable function has domain $a\in B_{\lambda }$, ground replacement collects its range $r$. Since $|r|\le |a|<\lambda$ and every descendant of an element of $r$ already has fewer than $\lambda$ members, $r\in B_{\lambda }$. Thus the full functional Axiom schema of replacement holds. For $x\in B_{\lambda }$, every actual subset of $x$ belongs to the class andConsequently its entire power set belongs to the class. Extensionality and foundation are inherited from the ground membership relation; $\omega$ witnesses infinity. However $a=\{{\beth }_n:n<\omega \}$ belongs to $B_{\lambda }$, while its only possible union is $\lambda$, which does not. All the other ZF axioms hold, and union fails. Bounding the entire transitive closure by $<\lambda$ instead would give the wrong model: functional ranges at this singular bound need not be hereditarily small.

Power set. Use the transitive class $H_{{\omega }_1}$ of hereditarily countable sets. Pairing, union and separation preserve hereditary countability. A functional image of a countable domain is countable, and its members have countable transitive closures; their countable union is countable in the ground ZFC model. This proves replacement. The remaining structural axioms and infinity are inherited. Every actual subset of $\omega$ belongs to $H_{{\omega }_1}$, but a putative internal power set of $\omega$ would contain all these subsets and hence be uncountable by Cantor's theorem. Power set fails in $H_{{\omega }_1}$, with every other ZF axiom intact.

Replacement. Use $V_{\omega +\omega }$ in the cumulative hierarchy. At a limit ordinal cutoff, pairing, union, power set and every instance of separation remain below the cutoff; foundation, extensionality and infinity hold as well. With parameter $V_{\omega }$, define $f(n)=V_{\omega +n}$ by $n$ finite iterations of power set. The finite iteration witnesses and each value are inside $V_{\omega +\omega }$, so this is an internally definable total function on $\omega$. Its range has rank of a set $\omega +\omega$ and is absent from the model. This instance of replacement fails.

Extensionality. Build well-founded coded sets over one urelement $a$, retaining a distinct ordinary empty set $e$. Membership of an atom is always false, and membership of a set code is its coded extension. Thus $a\ne e$ but they have identical empty extensions. Pairing, union, separation and replacement are performed on set codes; pure $\omega$ witnesses infinity, and the well-founded construction gives foundation. There is a small point concerning the unmodified, untyped Axiom of power set: the atom is vacuously a subset of every object. Therefore the internal power-set object for $x$ contains all ordinary set-subsets of $x$ and the atom. It has exactly the required extension. All remaining ZF axioms hold, but extensionality fails. In the usual typed atom theory, the power-set axiom instead quantifies only over set-subsets.

Foundation. In the ground universe let $\pi$ exchange $\varnothing$ and $\{\varnothing \}$ and fix all other objects. On the same domain define the Rieger-Bernays permutation modelSince $\pi$ is a bijection, two objects with the same $E$-members are equal, so extensionality holds. Any ground collection $B$ of desired members is represented by the object ${\pi }^{-1}(B)$. In particular, an $E$-pair is ${\pi }^{-1}(\{a,b\})$ and the $E$-union of $a$ isAn $E$-subset of $a$ is an object $z$ with $\pi (z)\subseteq \pi (a)$, so its $E$-power-set object isTranslate every formula by replacing membership with $E$; ground separation and replacement collect the required extensions and the same representation supplies their internal witnesses. The $E$-empty object is $e={\pi }^{-1}(\varnothing )$, and its $E$-successor operation is $s(x)={\pi }^{-1}(\pi (x)\cup \{x\})$. Ground well-founded recursion and replacement collect the iterates of $s$ from $e$; representing this collection gives an $E$-inductive object. Thus infinity holds. But for $q=\varnothing$, $\pi (q)=\{q\}$, whence $qEq$ and $q$ has precisely itself as its $E$-member. Foundation fails.

Infinity. The hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Each internal domain is externally finite, so every internally definable functional image is finite and belongs to $V_{\omega }$, proving replacement. No finite object can contain the empty set and all its successive Finite von Neumann ordinals. Infinity fails, and all other ZF axioms hold.

Choice with atoms. For the negative model use the Basic Fraenkel permutation model. Let $A$ be an infinite set of urelements and let every permutation of $A$ act recursively on sets. Keep precisely the hereditarily finitely supported objects: an object has a finite support $F\subseteq A$ if every permutation fixing $F$ pointwise fixes the object, and every object occurring in its membership ancestry must also have such a support. The class of these objects contains the pure universe and $A$ itself.

Supports for a pair are united; union preserves a support; a formula defining a subset uses the finite union of the supports of its domain and parameters. An internally definable functional range has the same support, since its values are uniquely specified, and its elements are already hereditarily supported. This proves the full separation and replacement schemas. All hereditarily supported set-subsets of a fixed set form its symmetric internal power set; this collection is invariant under the pointwise stabilizer of a support of the original set. For the untyped version add all the atoms, as above. Pure $\omega$, ground well-foundedness and set-extensionality give the remaining axioms.

The family $[A{]}^2$ of unordered pairs of atoms is hereditarily supported. If it had a choice function $f$, let $F$ be a finite support of $f$. Choose distinct $a,b\in A\setminus F$. The transposition exchanging $a,b$ fixes $F$ and $\{a,b\}$, but exchanges its two possible chosen elements. Invariance of $f$ would imply $f(\{a,b\})$ is fixed by that transposition, a contradiction. Conversely, the full well-founded universe over the atoms, before restricting to finite supports, satisfies choice when the ground universe does. Both choice and its negation are relatively consistent with the stated weakening of extensionality.

Forcing constructs new first-order models while controlling which ground-model properties survive. Fix a ground model $M$ and a partially ordered set $P\in M$, writing $q\le p$ when $q$ is the stronger condition. A generic filter is upward closed, directed towards stronger conditions, and meets every ground-model dense subset of a forcing order. If $M$ is externally a countable transitive model, enumerate its dense subsets and successively choose stronger conditions in them. The resulting filter is $M$-generic. External countability is crucial: this is not a claim that the ground model regards its own universe as countable.

A forcing name is built recursively from pairs $(\sigma ,p)$ of earlier names and conditions. For a generic filter $G$, its evaluation of a forcing name isThe canonical forcing name $\check{x}=\{(\check{y},1):y\in x\}$ evaluates to $x$, so $M\subseteq M[G]$. Recursive evaluation also proves transitivity. Name ranks bound the ranks of their evaluations, so an ordinal in the extension is below some ground ordinal and is already a ground ordinal. Thus forcing preserves ordinals.

The central link between syntax and generic interpretation is the forcing theorem. The relation $p⊩\phi ({\tau }_1,\dots ,{\tau }_n)$ is definable in the ground model, and the forcing truth lemma saysFor an atomic membership assertion, densely many refinements must witness equality with a member-name whose associated condition is compatible with the refinement. Equality is obtained by recursive mutual inclusion of names. For negation, $p⊩\neg \phi$ means no stronger condition forces $\phi$; for an existential quantification, witnesses must be obtainable densely below $p$. Induction on name rank establishes the atomic clauses, and induction on formulas establishes definability and the truth lemma. The generic filter meets the relevant dense sets at the existential step. An arbitrary semantic assertion about one chosen generic filter is insufficient: the definable relation must work uniformly for all generic interpretations.

This explains preservation of ZFC. Extensionality and foundation follow from transitivity; canonical names supply infinity. Pair and union names give pairing and union. For separation, restrict a name's members to conditions forcing the defining formula. For replacement, the definable forcing relation lets ground axiom schema of collection bound a set of names for all possible witnesses to the unique outputs; evaluating their collected name gives the range. For power set, let $\tau$ name the given set and let $D$ be the set of its member-names. Every subset of ${\tau }_G$ has an equivalent name drawn from $\mathcal{P}(D\times P)$, by the truth lemma, so a ground set of candidate names collects all its subsets. Finally choose a ground well-order of the relevant names; its evaluated image surjects onto the desired set, and choosing the first name for each value well-orders that set. This is why ordinary forcing over a choice model preserves choice.

For a concrete example, Cohen forcing consists of finite partial functions $\omega \to 2$, ordered by extension. The generic union $c=\bigcup G$ is total because deciding each coordinate is dense. For every ground real $x$, the set of conditions disagreeing with $x$ somewhere is dense, so $c\ne x$. This construction changes the universe by adding a genuinely new real without changing its ordinals.

The size and compatibility of conditions determine what else changes. With the countable chain condition for forcing, an ordinal-valued name has only countably many possible values at each coordinate: take a maximal antichain deciding that value. Therefore the possible range of a name for a map with infinite ground domain of cardinality $\mu$ lies in a ground set of size at most $\mu \cdot {\aleph }_0=\mu$. This prevents cardinal collapse and changes of uncountable regular cofinalities. Forcing by finite partial functions $\lambda \times \omega \to 2$ is ccc: the delta-system lemma thins any uncountable family of finite domains to a delta-system, and a further thinning makes their values agree on its finite root. Two remaining conditions are compatible. Distinct coordinates add distinct reals.

For example, start with GCH and take $\lambda ={\aleph }_2$. This forcing has size $\lambda$ and adds $\lambda$ distinct reals. A nice name for a real uses countably many antichains, and each antichain is countable. The ground arithmetic ${\lambda }^{{\aleph }_0}=\lambda$ bounds the number of such names by $\lambda$. Hencegiving a relative-consistency example of the failure of CH. The ground choice of GCH is available through the constructible universe; no claim that every starting model has this arithmetic is needed.

By contrast, countably closed forcing adds no new countable sequences of ground ordinals. Given a name, recursively strengthen a condition to decide each coordinate, then take a lower bound of the descending sequence. That condition decides the entire sequence. Closure and chain conditions thus control different features; neither should be silently substituted for the other.

Further applications include changing cofinalities, adding branches through trees, and constructing prescribed continuum patterns. If the aim is to violate choice, an ordinary generic extension of a choice model will not do: one restricts to a suitable symmetric collection of names, checking the axioms again. These examples also mark a limitation: forcing proves relative consistency, not absolute consistency of the background theory. The countable-transitive-model picture gives a concrete semantic construction; the definable forcing relation and Boolean-valued formulation supply the corresponding formal consistency arguments without asserting that a transitive ground model can be proved to exist.

For an infinite cardinal $\rho$ put ${\beth }_0(\rho )=\rho$ and ${\beth }_{n+1}(\rho )=2^{{\beth }_n(\rho )}$. The Erdős-Rado theorem for finite arities isThe partition relation requires a homogeneous subset of order type ${\rho }^{+}$ for every coloring of $(n+1)$-element subsets by at most $\rho$ colors. In particular it gives uncountable homogeneous sets. The case $n=0$ is the infinite pigeonhole principle: a union of $\rho$ sets each of size at most $\rho$ cannot have size ${\rho }^{+}$.

We give the induction step by a closed elementary-submodel construction of an end-homogeneous sequence. Let $\mu$ be infinite, $\theta =(2^{\mu }{)}^{+}$, and $c:[\theta {]}^{r+1}\to \rho$ with $\rho \le \mu$. Choose a sufficiently large regular $\chi$ and an elementary submodel $M\prec H_{\chi }$ with $c,\theta \in M$, all ordinals below ${\mu }^{+}$ in $M$, $|M|=2^{\mu }$, and closure under externally given sequences of length at most $\mu$. To construct it, take ${\mu }^{+}$ successive elementary Skolem-hull closures, at each step including all such sequences from the preceding stage. Cardinality stays $2^{\mu }$ because $(2^{\mu }{)}^{\mu }=2^{\mu }$. In the final union every sequence of length at most $\mu$ has all its entries in some earlier stage, by regularity of ${\mu }^{+}$, and so was included at the next stage.

Put $\beta =\sup (M\cap \theta )$. Regularity of $\theta$ gives $\beta <\theta$, and $\beta \notin M$ since otherwise its successor would contradict the definition of the supremum. Recursively, for $\alpha <{\mu }^{+}$ choose $x_{\alpha }\in M\cap \theta$ above all earlier $x_{\xi }$ and satisfyingThere are at most $\mu$ constraints. Their parameters and their colors belong to $M$, and closure puts their complete code in $M$. The point $\beta$ witnesses their simultaneous satisfiability in $H_{\chi }$; elementarity supplies a witness in $M$. This gives an increasing end-homogeneous sequence of length ${\mu }^{+}$: on it, the color of an $(r+1)$-tuple depends only on the first $r$ entries.

For the induction from arity $n$ to $n+1$, take $\mu ={\beth }_{n-1}(\rho )$ and $r=n$. Color the $n$-tuples of the resulting sequence by their common color with a later point, equivalently by their color with $\beta$. Its index order has type ${\mu }^{+}={\beth }_{n-1}(\rho {)}^{+}$, so the induction hypothesis supplies a homogeneous index set of type ${\rho }^{+}$. The original $(n+1)$-tuples on those indices have that same color. This completes the proof without substituting the theorem's name for its induction.

A useful ordinal partition-bound function isIt is total: choose an infinite cardinal $\rho \ge |\alpha |$ with ${\rho }^{+}>\alpha$; the pairs case supplies $(2^{\rho }{)}^{+}$ with the required homogeneous order type for all these color numbers. It is nondecreasing and satisfies $f(\alpha )\ge \alpha$. Start iterating $f$ and take a supremum $\kappa$ at a limit stage. If the increasing iterates do not attain that supremum, then for every $\alpha <\kappa$ some iterate $a_{\xi }$ satisfies $\alpha \le a_{\xi }<\kappa$, andIf the iterates have already stabilized, their supremum is a fixed point directly. Thus the relevant nontrivial case is an uncountable cardinal $\kappa$ closed below itself under $f$.

Such closure implies that $\kappa$ is a strong limit. Suppose $\lambda <\kappa$ is infinite and $\kappa \le 2^{\lambda }$. Take distinct binary strings of length $\lambda$, indexed in order by $\kappa$, and color a pair by its first differing coordinate. No three strings are homogeneous: three bits cannot be pairwise different at the same coordinate. Closure would give $f(\lambda +1)<\kappa$, and its restriction would have a homogeneous subset of order type $\lambda +1$ in $\lambda$ colors, a contradiction. Hence $2^{\lambda }<\kappa$; finite exponents are harmless for an uncountable $\kappa$.

Use the usual tree property: every set-theoretic tree of height $\kappa$ with nonempty levels of cardinality below $\kappa$ has a cofinal branch. For an uncountable cardinal satisfying this all-set-theoretic trees convention, $\kappa$ must be regular. If it were singular with cofinality $\tau <\kappa$, attach to one root $\tau$ chains whose lengths increase cofinally to $\kappa$. Every level has at most $\tau$ nodes, the height is $\kappa$, and no branch is cofinal. This contradiction proves regularity. Combined with the preceding closure argument it proves that the cardinal under discussion is strongly inaccessible.

For any $c:[\kappa {]}^2\to \nu$, $0<\nu <\kappa$, make a separated set-theoretic tree of functions. Its level $\alpha$ consists of the functions $t:\alpha \to \nu$ realized as $t(\xi )=c(\{\xi ,\beta \})$ for some $\beta \ge \alpha$, with $\beta <\kappa$. Order these functions by restriction. Levels are nonempty and have size at most ${\nu }^{|\alpha |}<\kappa$, using the strong-limit property. The tree property gives a branch $b:\kappa \to \nu$. Recursively select $x_{\eta }$ for $\eta <\kappa$, larger than all earlier selections, whose coloring function agrees with $b$ through a level above those earlier selections; regularity allows that level to remain below $\kappa$, and its node has a realizing point. Then $c(\{x_{\xi },x_{\eta }\})=b(x_{\xi })$ for $\xi <\eta$. One of the fewer-than-$\kappa$ colors occurs on $\kappa$ selected points, by regularity. These points are homogeneous, proving $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$. ConsequentlyA strictly increasing countable iteration has a supremum of countable cofinality, so it cannot meet this regular set-theoretic tree-property condition; no continuity of $f$ has been assumed.

The final printed assertion needs its convention specified. The usual tree property alone does not imply strong inaccessibility; it can consistently hold at successor cardinals. In the argument just given, strong inaccessibility follows from function closure together with that set-theoretic tree property. There is also a stronger branching form of the tree property for which the assertion holds literally: require a cofinal branch in every separated rooted height-$\kappa$ set-theoretic tree whose nodes each have fewer than $\kappa$ immediate successors, without restricting level sizes. The singular-chain example proves regularity under this convention too. If $2^{\lambda }\ge \kappa$ for $\lambda <\kappa$, take the full binary set-theoretic tree through level $\lambda$, choose $\kappa$ distinct nodes on that level, and attach to its $\xi$th chosen node a chain of length $\xi$, for $\xi <\kappa$. It has height $\kappa$ and at most two successors per node, but every branch has length below $\kappa$. Thus this branching property implies the strong-limit condition and hence strong inaccessibility. At a strongly inaccessible cardinal it agrees, for separated set-theoretic trees, with the usual level-size convention: regularity bounds the union of fewer than $\kappa$ successor sets, and strong-limit arithmetic bounds the possible predecessor sequences at limit levels.

Work with well-ordered cardinals, as usual with the axiom of choice. We first prove the infinite cardinal arithmetic identity ${\kappa }^2=\kappa$. If a least infinite counterexample $\kappa$ existed, order pairs of ordinals below $\kappa$ first by their maximum coordinate, then lexicographically within each level. Every initial segment ending at $(\xi ,\eta )$ lies inside $(\max (\xi ,\eta )+1{)}^2$. Its cardinal is less than $\kappa$, by the smaller-cardinal identity or finite counting. This well-order cannot have an element with $\kappa$ predecessors, so its order type is at most the initial ordinal $\kappa$. Thus $|\kappa \times \kappa |\le \kappa$, contrary to its being a counterexample; the reverse injection is $\xi \mapsto (\xi ,0)$. HenceA sum of finitely many finite cardinals is finite, so ${\aleph }_0$ is regular. A sum of fewer than ${\aleph }_1$ smaller cardinals is a countable union of countable sets and has size at most ${\aleph }_0^2={\aleph }_0$, so ${\aleph }_1$ is regular.

Likewise fewer than ${\aleph }_2$ summands means at most ${\aleph }_1$ summands, each of size at most ${\aleph }_1$. Their sum is at most ${\aleph }_1^2={\aleph }_1<{\aleph }_2$. Thus ${\aleph }_2$ is regular. In contrast,is a countable sum of smaller cardinals, with ${\aleph }_0<{\aleph }_{\omega }$. Therefore ${\aleph }_{\omega }$ is singular, of cofinality $\omega$.