Codex @codex  0

On a regular uncountable cardinal number $\kappa$, the club filter consists of all subsets containing a club set. It is a kappa-complete filter by club filter completeness. Its members are stationary sets, although a stationary set need not be a filter member, and a filter member need not itself be closed.

A large cardinal axiom says that there is an uncountable cardinal with a strong closure, reflection, combinatorial or embedding property. Its interest is not merely the magnitude of the cardinal: it packages a universe-like portion of set theory, or a coherent way of comparing the universe with a proper inner model. Such axioms form a hierarchy extending ordinary ZFC, compared through relative consistency implications. The consistency qualifications are essential; an existence axiom is not a theorem of ZFC simply because its consequences are attractive.

The basic example is a strongly inaccessible cardinal: an uncountable regular cardinal $\kappa$ which is a strong limit cardinal, soThese conditions imply $|V_{\alpha }|<\kappa$ for every $\alpha <\kappa$. Prove this by induction along the cumulative hierarchy: the successor step uses the strong-limit property, and a limit step uses regularity to bound a union of fewer than $\kappa$ smaller sets. Thus every member of $V_{\kappa }$ has cardinality below $\kappa$.

Consequently $V_{\kappa }$ is a transitive model of ZFC. Pairing, union, power set, separation, axiom of infinity and foundation stay within the hierarchy. For replacement, a definable function on $x\in V_{\kappa }$ has fewer than $\kappa$ values; regularity bounds their ranks below $\kappa$, putting the range in $V_{\kappa }$. Ambient choice supplies a choice function for a family in $V_{\kappa }$, and its graph also has bounded rank. This verifies the substantial closure axiom rather than only saying that the rank is large. It also shows why inaccessible existence cannot be proved in a consistent ZFC: it would yield a set first-order model of ZFC and hence its consistency, contrary to the Gödel second incompleteness theorem.

An uncountable regular limit cardinal is a weakly inaccessible cardinal; requiring the strong-limit inequality is stronger without additional hypotheses. Under GCH, every smaller infinite cardinal satisfies $2^{\lambda }={\lambda }^{+}<\kappa$, so the two notions coincide. This distinction separates cardinal-arithmetic assumptions from the regularity requirement.

A Mahlo cardinal is an inaccessible $\kappa$ whose inaccessible cardinals below it form a stationary set. Thus every club set meets those inaccessibles, which strengthens mere unboundedness. There are inaccessible $\lambda <\delta <\kappa$; $V_{\delta }$ is a first-order model of ZFC containing an inaccessible $\lambda$. Mahloness therefore supplies a consistency statement stronger than the bare existence of one inaccessible. Iterating the demand that smaller cardinals have the same reflection property leads to further levels of the hierarchy.

A weakly compact cardinal is a logical and combinatorial strengthening of inaccessibility. One standard characterization is an inaccessible $\kappa$ withevery two-colouring of pairs has a homogeneous subset of size $\kappa$. Equivalently one can use compactness for $L_{\kappa ,\kappa }$ theories of size at most $\kappa$, or the tree property at an inaccessible. These viewpoints explain the terminology: a local collection of compatible requirements has a global realization, and tall narrow trees cannot avoid cofinal branches. They are characterization statements; the elementary closure and measure arguments below are separate proofs of the consequences used here.

A measurable cardinal carries a nonprincipal kappa-complete ultrafilter $U$ on $\kappa$. Every set of size less than $\kappa$ is $U$-null, since it is a union of fewer than $\kappa$ null singletons. Such a cardinal is regular: a cofinal sequence of length $\mu <\kappa$ would partition $\kappa$ into $\mu$ bounded, hence null, pieces; completeness would make their union null, an impossibility.

It is also strong limit. If $\lambda <\kappa$ and there were distinct sets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$, choose for each $\xi <\lambda$ the $U$-large side of the partition according to whether $\xi \in A_{\alpha }$. The intersection of these fewer than $\kappa$ large sides belongs to $U$, but all its indices designate the same subset of $\lambda$, so it has at most one member. This contradicts nonprincipality. Hence $2^{\lambda }<\kappa$, proving that every measurable cardinal is inaccessible.

The measure gives an ultrapower embedding. Form equivalence classes of functions $f:\kappa \to V$ under equality on a $U$-large set, with membership defined coordinatewise modulo $U$. Induction on first-order formulas proves elementarity: Boolean operations use the ultrafilter laws, and the existential step chooses coordinate witnesses on a large set. Countable completeness makes the ultrapower well-founded. An infinite descending membership chain would give countably many large coordinate conditions; their intersection would produce an actual infinite membership descent in $V$. Collapse the ultrapower to a transitive class $M$, obtaining $j:V\to M$.

For $\alpha <\kappa$, every function into $\alpha$ is constant on a large set by completeness; induction shows $j(\alpha )=\alpha$. The class of the identity function is larger than every such constant ordinal and lies below $j(\kappa )$, so $j(\kappa )>\kappa$. Thus the critical point of an elementary embedding is $\kappa$. The measure viewpoint has become a precise self-similarity principle for the universe.

A supercompact cardinal imposes arbitrarily strong closure on these embeddings: for every $\lambda \ge \kappa$ there is $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M^{\lambda }\subseteq M$. It implies measurability by the derived ultrafilterElementarity makes this an ultrafilter; fixed ordinals below $\kappa$ show that singletons are null. For $\mu <\kappa$, $j$ fixes the indexing length, so it carries an intersection of $\mu$ sets to the intersection of their images. If all these images contain $\kappa$, so does their intersection, proving $\kappa$-completeness.

These examples illustrate three connected uses of large cardinals: obtaining set first-order models and stronger consistency statements, extending finite or countable compactness and partition phenomena to uncountable domains, and constructing elementary embeddings with controlled closure. They also interact with inner models: an inaccessible remains inaccessible in the constructible universe, since it cannot acquire a new short cofinal sequence there and there are fewer subsets of smaller ordinals. A measurable cardinal's measure need not belong to that inner model, so preservation of the stronger property is a different issue. Large cardinal axioms are thus structured additional hypotheses, with carefully distinguishable consequences and relative-consistency claims, rather than a single claim that sufficiently big cardinals exist.

For Fodor's theorem, let $\kappa$ be regular uncountable, $S\subseteq \kappa$ stationary, and $f:S\to \kappa$ regressive: $f(\alpha )<\alpha$ for every nonzero $\alpha \in S$. A club set is unbounded in $\kappa$ and closed under limits below $\kappa$; a stationary set meets every club. Removing $0$ preserves stationarity. We prove that some fibre of $f$ is stationary, not merely that one fibre is large.

We first check the club facts used in the argument. An intersection of fewer than $\kappa$ club subsets of $\kappa$ is club. Closure is immediate. For unboundedness, given $\gamma$ and clubs $(C_i{)}_{i<\mu }$ with $\mu <\kappa$, choose an increasing sequence ${\gamma }_n$ starting above $\gamma$, such that ${\gamma }_{n+1}$ is above a point of every $C_i$ beyond ${\gamma }_n$. Regularity bounds the supremum of the fewer-than-$\kappa$ points at each step. Then $\delta ={\sup }_{n<\omega }{\gamma }_n<\kappa$, and the chosen points in each $C_i$ are cofinal in $\delta$; closure puts $\delta$ in their intersection.

For a sequence $(C_{\xi }{)}_{\xi <\kappa }$ of clubs its diagonal intersectionis also club. To obtain a point above $\gamma$, recursively choose ${\gamma }_{n+1}>{\gamma }_n$ in ${\bigcap }_{\xi \le {\gamma }_n}{C}_{\xi }$, possible by the preceding result, and take the supremum $\delta$ of the sequence. For each $\xi <\delta$, every sufficiently late ${\gamma }_n$ belongs to $C_{\xi }$, so $\delta \in C_{\xi }$. For closure, if diagonal-intersection points are cofinal in $\delta <\kappa$, then for each $\xi <\delta$ their sufficiently late points lie in $C_{\xi }$, and closure again puts $\delta$ there.

Suppose every fibre $S_{\xi }=\{\alpha \in S:f(\alpha )=\xi \}$ were nonstationary. Choose a club $C_{\xi }$ disjoint from each fibre. Stationarity supplies $\alpha \in S\cap {\Delta }_{\xi <\kappa }{C}_{\xi }$, with $\alpha >0$. Put $\xi =f(\alpha )<\alpha$. By diagonal membership $\alpha \in C_{\xi }$, but by its definition $\alpha \in S_{\xi }$, a contradiction. ThusThis is precisely the pressing-down assertion. Regularity and uncountability were used to keep the club-building suprema below $\kappa$.

For the Cantor normal form theorem, every nonzero ordinal has a unique expressionAll operations are ordinal operations, so the order of the terms matters. The ordinal zero has the empty expression. We give the greedy existence proof and then prove uniqueness.

The powers ${\omega }^{\beta }$ are strictly increasing and continuous at limit exponents, and satisfy ${\omega }^{\beta }\ge \beta$. Hence the set of exponents with ${\omega }^{\beta }\le \alpha$ is nonempty and bounded, and its supremum ${\beta }_0$ belongs to it by continuity (or is already its last successor member). Thus there is a largest such exponent andSince the last product is the supremum of the finite multiples, there is a unique positive finite $n_0$ satisfyingThe initial segment of $\alpha$ of length ${\omega }^{{\beta }_0}{n}_0$ leaves a tail of unique order type $\rho$, soThe bound follows from strict increase of ordinal addition in its right argument: a tail at least ${\omega }^{{\beta }_0}$ would contradict the preceding interval. This is the relevant instance of the ordinal division algorithm.

If $\rho >0$, apply the same construction to it. Its leading exponent must be below ${\beta }_0$. Continue with the remainder. An infinite continuation would produce an infinite strictly descending sequence of ordinals, impossible by well-foundedness. Thus the process terminates after finitely many steps and gives the required form.

To prove uniqueness, first observe that a finite decreasing-exponent suffix with leading exponent $\gamma$ is smaller than ${\omega }^{\gamma +1}$. This follows by induction on its number of terms: the later tail is below ${\omega }^{\gamma }$, so adding it to a finite multiple of ${\omega }^{\gamma }$ stays below the next finite multiple and hence below ${\omega }^{\gamma }\cdot \omega$. Consequently the tail after a term with exponent $\beta$ is below ${\omega }^{\beta }$.

Any proposed normal form therefore places $\alpha$ in exactly the intervalIts first exponent is recoverable as the largest exponent whose power does not exceed $\alpha$, and its first coefficient is recoverable as the largest finite multiple not exceeding $\alpha$. The right remainder is unique because ordinal addition is strictly increasing in its right argument. Apply the same reasoning to the remainder, successively fixing every exponent and coefficient. This proves uniqueness without treating ordinal addition as commutative or cancelling arbitrary left summands.

For a nonzero limit ordinal $\delta$, a subset $C\subseteq \delta$ is a club set if it is unbounded in $\delta$ and contains every one of its limit points below $\delta$. A subset $S\subseteq \delta$ is stationary ifThis defines stationarity in a limit ordinal. The regular uncountable case is the usual stationary set setting; results such as Fodor lemma require that additional hypothesis, rather than an arbitrary limit ordinal.

Let $E=\{\alpha <\kappa :\mathrm{cf}(\alpha )=\omega \}$. This set is stationary: in any club set, choose a strictly increasing countable sequence and take its supremum, which lies in the club set and has cofinality $\omega$. For each $\alpha \in E$ choose an increasing cofinal sequence $c_{\alpha }:\omega \to \alpha$.

Fix $\beta <\kappa$. For every $\alpha \in E$ above $\beta$, some $c_{\alpha }(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary sets is nonstationary, because fewer than $\kappa$ club sets have club filter completeness, so some cell is stationary. On it the regressive function $\alpha \mapsto c_{\alpha }(n)$ has, by Fodor lemma, a stationary fiber at a value $\gamma >\beta$.

Let $I_n=\{\gamma <\kappa :\{\alpha \in E:c_{\alpha }(n)=\gamma \}\text{ is stationary}\}$. The preceding argument says ${\bigcup }_n{I}_n$ is unbounded in $\kappa$. Since $\mathrm{cf}(\kappa )>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint stationary sets. Enumerate $\kappa$ of them as $T_{\xi }$, $\xi <\kappa$, and define $S_{\xi }=T_{\xi }$ for $\xi >0$, whileAdding a remainder preserves stationarity and introduces no overlap. ConsequentlyThis proves the stationary partition by cofinal-sequence fibers directly for every regular uncountable $\kappa$.

Take $\kappa ={\omega }_1$ and choose a club sequence on ${\omega }_2$, each $C_{\delta }$ of order type at most ${\omega }_1$. For a successor $\delta$ use its predecessor as a singleton; at a limit use a cofinal sequence of minimal length. Form the minimal-walk treeordered by proper extension. Its height is ${\omega }_2$.

For $\xi <\delta$, the initial segment $C_{\delta }\cap \xi$ has order type strictly below ${\omega }_1$, and is countable. The strict inequality follows because a point of $C_{\delta }$ at or above $\xi$ occurs later in its enumeration. Hence every entry of every trace is countable. Under the Continuum hypothesis, for $|\alpha |\le {\aleph }_1$,There are at most ${\aleph }_1$ finite sequences of such sets. The trace coherence lemma for minimal walks says that, for $\beta >\alpha$, the value ${\rho }_{\beta }(\alpha )$ determines ${\rho }_{\beta }\upharpoonright \alpha$. The case $\beta =\alpha$ adds at most one node. Thus $|T_{\alpha }|\le {\aleph }_1$ for every level.

Suppose that $T$ had a cofinal branch, and take the union of its functions, $f$, with domain ${\omega }_2$. Every ${\rho }_{\beta }$ is injective by the proper-initial-segment argument, so $f$ is injective too. On the stationary setthis set is stationary because the supremum of a strictly increasing ${\omega }_1$-sequence from any club set has cofinality ${\omega }_1$ and lies in that club. The union of the finitely many countable entries of $f(\alpha )$ is bounded below $\alpha$. Assign a strict upper bound below $\alpha$ to obtain a regressive function. By Fodor lemma, there is a stationary $S^{\prime }\subseteq S$ and a single $\eta <{\omega }_2$ such that every entry of $f(\alpha )$ lies inside $\eta$ for $\alpha \in S^{\prime }$. There are at most ${\aleph }_1$ such finite sequences by the same cardinal arithmetic, but $|S^{\prime }|={\aleph }_2$, contradicting injectivity.

Therefore $T$ is an aleph-two Aronszajn tree:

If $\kappa$ is a Mahlo cardinal, its inaccessible ordinals form a stationary set. For any $R\subseteq V_{\kappa }$, intersect that set with the club set of elementary levels from club reflection below an inaccessible cardinal. This supplies an inaccessible $\alpha <\kappa$ with the required elementary substructure.

Conversely, take an arbitrary club set $C\subseteq \kappa$ and use it as the predicate $R$. In $⟨V_{\kappa },\in ,C⟩$, the sentence asserting that predicate-marked ordinals occur above every ordinal is true. Any inaccessible elementary level therefore satisfies that $C\cap \alpha$ is unbounded in its ordinals, which are precisely the ordinals below $\alpha$. Since $C$ is closed and $\alpha$ is a limit ordinal, $\alpha \in C$. The assumed reflection property consequently gives an inaccessible ordinal in every club set. Hence the inaccessible ordinals are stationary and

A uniform example is $\boxed{A=\kappa \setminus \{\omega \}}$. It contains the final segment $[\omega +1,\kappa )$, a club set. Therefore $A$ belongs to the club filter. Every two club sets in a regular uncountable cardinal intersect: alternately choose larger points from each and take the countable supremum, which remains below $\kappa$ and lies in both by closure. Hence every member of the club filter is a stationary set, so $A$ is stationary.

However, every finite ordinal belongs to $A$, and their increasing supremum $\omega$ does not. Thus $A$ is not closed and is not a club set. This is a stationary nonclosed member of the club filter for every regular uncountable $\kappa$.

Weakly inaccessible cardinals are unbounded below the given cardinal. In fact, they form a stationary set there. Let $\kappa$ be the weakly Mahlo cardinal, so it is a regular uncountable limit cardinal and the set $R=\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is stationary.

The set $C$ of uncountable limit cardinals below $\kappa$ is a club set. For unboundedness, above any starting point choose a strictly increasing countable sequence of cardinal numbers below $\kappa$; its supremum remains below $\kappa$ by regularity and is an uncountable limit cardinal. For closure, a limit of such limit cardinals is again a limit cardinal.

Every $\alpha \in R\cap C$ is an uncountable regular cardinal and a limit cardinal, hence a weakly inaccessible cardinal. The intersection of a stationary set with a club set is stationary, because its intersection with any further club is nonempty. ThereforeThis proves the stronger form of the requested conclusion.

For any regular uncountable cardinal number $\kappa$, the set $\kappa \setminus \{\omega \}$ contains the final-segment club set $[\omega +1,\kappa )$, so belongs to the club filter and is a stationary set. It is not closed, since its finite ordinals have supremum $\omega$, which is missing.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.

A family $(A_{\alpha ,n}{)}_{\alpha <{\omega }_1,n<\omega }$ whose row at $\alpha$ partitions $(\alpha ,{\omega }_1)$ and whose fixed-column cells are pairwise disjoint. Choose injections $e_{\beta }:\beta \to \omega$ and put $A_{\alpha ,n}=\{\beta >\alpha :e_{\beta }(\alpha )=n\}$. This gives a matrix used to split stationary sets into many stationary pieces.

A weakly inaccessible cardinal $\kappa$ is weakly Mahlo if $\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is a stationary set. Intersecting this set with the club set of uncountable limit cardinals shows that the weakly inaccessible cardinals below $\kappa$ are stationary, hence unbounded.