Club, Stationary Sets, and Pressing Down

In the regular case, given $\beta <\kappa$, set $a_0=\beta +1$ and $a_{n+1}=\max (a_n,\sup f[a_n])+1$. Regularity bounds $f[a_n]$ below $\kappa$; recursion defines the increasing sequence, and $a={\sup }_n{a}_n<\kappa$ because $\omega <\mathrm{cf}(\kappa )$. For $x<a$ take $n$ with $x<a_n$; then $f(x)<a_{n+1}<a$. Thus $a\in C_f$ and $a>\beta$.

In the nondecreasing case use $a_{n+1}=\max (a_n,f(a_n))+1$ instead. Each term stays below the limit $\theta$, and the omega supremum stays below $\theta$. If $x<a_n$, monotonicity gives $f(x)\le f(a_n)<a_{n+1}<a$. Again this proves unboundedness.

In either case, if a nonzero limit $\delta$ is a limit point of $C_f$, then for every $x<\delta$ some $\gamma \in C_f\cap \delta$ exceeds $x$. Hence $f(x)<\gamma <\delta$. This proves closure. Zero itself belongs to $C_f$ vacuously, but is not a required closure limit.

Define $f(\beta )=\min \{x\in X:x>\beta \}$. This exists by unboundedness, is nondecreasing, and satisfies $f(\beta )>\beta$. Its closure points form a club. A nonzero closure point cannot be a successor $\gamma +1$, since $f(\gamma )\ge \gamma +1$; and for every $\beta <\delta$ a nonzero closure point has $\beta <f(\beta )<\delta$. Thus it is in $\mathrm{acc}(X)$. Conversely each nonzero limit point of $X$ is closed under $f$.

Removing zero from the closure-point club preserves unboundedness and closure at nonzero limits. Equivalently, closure of acc follows directly: below a limit of limit points, first choose a limit point above a given bound, then a point of $X$ above that bound. Hence acc is club with exactly the stipulated nonzero-limit convention.

For $\mu =0$, the intersection is $\theta$, which is closed and unbounded in itself. For $\mu >0$, let $f_i(\beta )=\min (C_i\setminus (\beta +1))$ and $g(\beta )={\sup }_{i<\mu }{f}_i(\beta )$. The supremum is below $\theta$, and $g$ is nondecreasing and strictly above its argument.

The nonzero closure points $\delta$ of $g$ are unbounded by the closure lemma. For every $i$ and $\beta <\delta$, $\beta <f_i(\beta )\le g(\beta )<\delta$. Thus $C_i\cap \delta$ is unbounded in $\delta$; $\delta$ is a nonzero limit and belongs to each $C_i$. This proves unboundedness of the intersection.

If $\delta$ is a nonzero limit point of the intersection, it is a limit point of each $C_i$, hence lies in each. This proves closure; the case $\mu =1$ is included. Specializing $\theta =\kappa$ gives the last assertion.

Every tail $[\beta ,\theta )$ is closed and unbounded: for any bound take a larger ordinal above $\beta$, and a limit of tail points is still at least $\beta$. A bounded set is disjoint from a suitable tail, so cannot be stationary. This also excludes the empty set and all singletons.

Two clubs intersect in a club and hence nontrivially, so each club is stationary. Supersets preserve intersections with every club. For stationary $S$ and clubs $C,D$, the club $C\cap D$ meets $S$, so $S\cap C$ meets every $D$ and is stationary.

For a small family of nonstationary sets, their union is in the dual ideal by its completeness. Explicitly choose an avoiding club for each member and intersect those clubs; the resulting club avoids the union. For the empty family the union is empty, avoided by $\theta$.

Define $g(\beta )={\sup }_{\xi \le \beta }\min (C_{\xi }\setminus (\beta +1))$. Regularity keeps this below $\kappa$. By the closure-point lemma, there are unboundedly many nonzero closure points $\alpha$ of $g$; these are limits since $g(\beta )>\beta$. Fix $\xi <\alpha$ and any $\eta <\alpha$, then take $\beta <\alpha$ with $\beta \ge \xi ,\eta$. The least $C_{\xi }$ point above $\beta$ is below $\alpha$. Thus $\alpha \in C_{\xi }$, proving $\alpha \in D$.

If $\delta$ is a nonzero limit point of $D$, then for each $\xi <\delta$ the points of $D\cap \delta$ above $\xi$ belong to $C_{\xi }$ and are unbounded in $\delta$. Its closure gives $\delta \in C_{\xi }$ for every $\xi <\delta$, hence $\delta \in D$. This proves closure. Zero is in $D$ by definition.

If every fibre were nonstationary, ambient AC would select a club $C_{\xi }$ avoiding that fibre for every $\xi <\kappa$. Let $D={<mi\; mathvariant="normal">△</mi>}_{\xi <\kappa }{C}_{\xi }$, a club.

Take $\alpha \in S\cap D$. Then $\alpha >0$ and $\xi =f(\alpha )<\alpha$, so diagonal membership gives $\alpha \in C_{\xi }$. But this club avoids the fibre containing $\alpha$, a contradiction. Therefore a stationary fibre exists.

If $F$ is normal and every fibre of a regressive $f:S\to \kappa$ is small, all their complements belong to $F$. Their diagonal belongs to $F$ and must meet $S$. At an intersection point $\alpha$, its value $f(\alpha )<\alpha$ forces it into the complement of its own fibre, a contradiction.

Conversely let $A_{\xi }\in F$ and suppose their diagonal $D$ is not in $F$. Then $S=\kappa \setminus D$ is positive and excludes zero. For $\alpha \in S$, take the least $\xi <\alpha$ with $\alpha \mathrm{\notin }{A}_{\xi }$. This defines a regressive map. A positive fibre would be disjoint from the corresponding filter member $A_{\xi }$, impossible. Hence $D\in F$.

For $(A_i{)}_{i<\mu }$ in $F$, $\mu <\kappa$, pad by $\kappa$ at all remaining indices. Its diagonal, intersected with $[\mu ,\kappa )$, is in $F$ and is contained in ${\bigcap }_{i<\mu }{A}_i$. Upward closure proves completeness. For $\mu =0$ the intersection is $\kappa$.

For a sequence of club-filter members choose a witnessing club inside each. Their diagonal is a club contained in the diagonal of the original members, so that diagonal belongs to the club filter. Tails are clubs, giving normality and tail containment.

Let $F$ be a proper normal tail-containing filter and $C$ a club. If $C\mathrm{\notin }F$, then $S=(\kappa \setminus C)\cap [1,\kappa )$ is positive: intersecting a positive set with a filter member preserves positivity, as every further filter intersection remains in $F$. On $S$ put $f(\alpha )=\sup (C\cap \alpha )$. At a successor this is below $\alpha$; at a nonzero limit equality would put $\alpha$ in the closed $C$. Hence $f$ is regressive.

For any $\xi$, take $c\in C$ above $\xi$. A point $\alpha >c$ has $f(\alpha )\ge c>\xi$, so the fibre of $\xi$ is bounded by $c+1$ and is disjoint from a tail in $F$. All fibres are small, contradicting positive pressing down. Thus $C\in F$, and upward closure includes the entire club filter in $F$.

The inherited ordinal order enumerates $C$ by an ordinal $\eta \le \kappa$, successively taking the least unused point. Unboundedness and regularity force $\eta \ge \kappa$, hence $\eta =\kappa$. The least-unused rule also proves uniqueness.

If $C$ is closed and $0<\lambda <\kappa$ is limit, $\delta ={\sup }_{\xi <\lambda }e(\xi )<\kappa$ is a limit point of $C$, hence lies in $C$. Strict increase and least-unused enumeration force $e(\lambda )=\delta$. Thus $e$ is normal.

Conversely, if $e$ is normal and $\delta <\kappa$ is a nonzero limit point of $C$, the indices of the points in $C\cap \delta$ form an initial segment $\lambda <\kappa$ with no last element. (They cannot be all kappa since C is unbounded.) Continuity gives $e(\lambda )=\delta$, so $C$ is closed. Finally, a strictly increasing map on kappa has unbounded range: a bounded range cannot contain kappa distinct ordinals. It is the increasing enumeration of its range, giving the last equivalence.

Transfinite induction gives $f(\alpha )\ge \alpha$: zero is automatic, successors use strict increase, and limits use continuity. Given $\beta <\kappa$, iterate $a_0=\beta +1$, $a_{n+1}=f(a_n)$. If some adjacent terms agree, that term is a fixed point above $\beta$. Otherwise the sequence is strictly increasing and its supremum $a<\kappa$ is a nonzero limit.

In the latter case continuity and cofinality of the $a_n$ in $a$ give $f(a)={\sup }_{n}f(a_n)={\sup }_n{a}_{n+1}=a$. This proves unboundedness in both cases.

If fixed points are unbounded in a nonzero limit $\delta <\kappa$, monotonicity and continuity give $f(\delta )=\sup \{f(\gamma ):\gamma <\delta ,f(\gamma )=\gamma \}=\delta$. Thus the fixed-point set is closed. Zero may or may not be fixed; no claim depends on it.

Given a club $C\subseteq \theta$, define a strictly increasing sequence $(c_{\xi }{)}_{\xi <\lambda }$ in $C$: start at $\min C$, take the least greater C point at successors, and at nonzero limits take the supremum. Every such supremum is below theta since its index is below $\mathrm{cf}(\theta )$, and closure places it in C. Likewise $\delta ={\sup }_{\xi <\lambda }{c}_{\xi }<\theta$ belongs to C.

This sequence gives $\mathrm{cf}(\delta )\le \lambda$. If a cofinal subset $B\subseteq \delta$ had size $\mu <\lambda$, assign to each $b\in B$ the least $\xi <\lambda$ with $b<c_{\xi }$. These indices would be unbounded in lambda: a bound would bound B below delta. Regularity of lambda rules this out. Therefore $\mathrm{cf}(\delta )=\lambda$, so C meets the stratum. Since C was arbitrary the stratum is stationary.

Suppose a club $C$ avoids $S\setminus \mathrm{Tr}(S)$. Since $\mathrm{acc}(C)$ is club, let $\alpha$ be the least point of $S\cap \mathrm{acc}(C)$. Closure of C gives $\alpha \in C$, so avoidance forces $\alpha \in \mathrm{Tr}(S)$. In particular $\mathrm{cf}(\alpha )>\omega$.

Now $C\cap \alpha$ is unbounded in $\alpha$, so ${\mathrm{acc}}_{\alpha }(C\cap \alpha )={\mathrm{acc}}_{\kappa }(C)\cap \alpha$ is a club in $\alpha$. By minimality of $\alpha$ it is disjoint from $S\cap \alpha$. This contradicts trace membership.

For each $\beta <\kappa$, apply Fodor to the stated stationary tail domain. Its stationary constant subset lies in a fibre $g^{-1}(\{\gamma \})$ with $\gamma \ge \beta$. Thus $B=\{\gamma <\kappa :g^{-1}(\{\gamma \})\text{ stationary}\}$ is unbounded. Regularity gives $\mathrm{\mid }B\mathrm{\mid }=\kappa$, so its increasing enumeration has domain kappa.

The fibres at values in B are pairwise disjoint stationary sets. Keep them all, and adjoin every remaining point of S to the fibre at the least value in B. That enlarged fibre remains stationary and disjoint from all the others, and the union is now S.

Use AC to choose for each $\alpha \in S$ an increasing cofinal sequence $c^{\alpha }:\lambda \to \alpha$. Suppose every coordinate $\xi <\lambda$ has a threshold $b_{\xi }<\kappa$ for which $\{\alpha \in S:c^{\alpha }(\xi )\ge b_{\xi }\}$ is nonstationary; choose an avoiding club $C_{\xi }$.

The intersection $C={\bigcap }_{\xi <\lambda }{C}_{\xi }$ is club, and $b={\sup }_{\xi <\lambda }{b}_{\xi }<\kappa$ by regularity. Take $\alpha \in S\cap C$ above b (the intersection is unbounded, by testing it against additional tails). Then $c^{\alpha }(\xi )<b_{\xi }\le b$ for every $\xi$, contradicting cofinality in $\alpha >b$.

Therefore some fixed coordinate $\xi$ has stationary tail domains at every threshold. The map $g(\alpha )=c^{\alpha }(\xi )$ is regressive on all S. The fibre-partition lemma applies.

Let $T=S\setminus \mathrm{Tr}(S)$, stationary. For each $\alpha \in T$ use AC to choose a club $D_{\alpha }\subseteq \alpha$ disjoint from $S\cap \alpha$, and let $c^{\alpha }:\alpha \to D_{\alpha }$ be its normal enumeration. Such clubs exist since alpha is regular uncountable and is not in the trace.

For each coordinate $\xi <\kappa$ consider the domain $T_{\xi }=\{\alpha \in T:\xi <\alpha \}$. Suppose no coordinate has stationary sets $\{\alpha \in T_{\xi }:c^{\alpha }(\xi )\ge b\}$ for every $b<\kappa$. Choose a failing threshold $b_{\xi }$ and avoiding club $C_{\xi }$, so $c^{\alpha }(\xi )<b_{\xi }$ whenever $\alpha \in T_{\xi }\cap C_{\xi }$.

Let $D={<mi\; mathvariant="normal">△</mi>}_{\xi <\kappa }{C}_{\xi }$ and let $E$ be the club of closure points of $\xi \mapsto b_{\xi }$. Choose $\alpha \in T\cap E$ and then $\gamma \in T\cap D$ above alpha. For every $\xi <\alpha$, diagonal membership of gamma and closure at alpha give $c^{\gamma }(\xi )<b_{\xi }<\alpha$. Since alpha is a nonzero limit, continuity gives $c^{\gamma }(\alpha )\le \alpha$. Strict increase implies $c^{\gamma }(\xi )\ge \xi$ by ordinal induction, so in fact $c^{\gamma }(\alpha )=\alpha$. This contradicts $D_{\gamma }\cap S=\mathrm{\varnothing }$, since $\alpha \in T\subseteq S$.

Consequently some $\xi$ has all stationary tail domains. Its coordinate map on $T_{\xi }$ is regressive and the fibre lemma partitions $T_{\xi }$ into kappa stationary sets. Add $S\setminus T_{\xi }$ to one piece; this preserves stationarity and disjointness and gives the desired partition of S.

The nonzero limit ordinals below kappa form a club: above any bound iterate successors omega times to find a larger limit below kappa, and a nonzero limit of such ordinals is a limit. Intersect S with this club and the tail above omega. Partition the resulting stationary T into $T_0=\{\alpha \in T:\mathrm{cf}(\alpha )<\alpha \}$ and $T_1=\{\alpha \in T:\mathrm{cf}(\alpha )=\alpha \}$. At least one is stationary.

If $T_0$ is stationary, its cofinality map is regressive. Fodor supplies a stationary subset of one cofinality $\lambda <\kappa$, which is infinite regular because its arguments are limits. The fixed-cofinality splitting lemma partitions this subset.

If $T_1$ is stationary, its members are regular uncountable cardinals: cofinalities of limits are regular cardinals, and these members equal their cofinalities and exceed omega. Apply the regular-cardinal splitting lemma. In either case adjoin every discarded point of S to one of the kappa pieces. Supersets preserve stationarity, and the pieces remain disjoint and exhaust S.

The whole cardinal is stationary because every club is nonempty. Split it into $(S_{\xi }{)}_{\xi <\kappa }$ by Solovay. Then $S_0$ is stationary and its complement contains the stationary $S_1$, so the complement is stationary as well.

Neither $S_0$ nor its complement contains a club, since such a club would be disjoint from the stationary set on the other side. Thus the club filter contains neither side of this partition and does not decide every subset.

Include in $\mathcal{H}$ every language function and constant, and the constant zero. For each formula $\mathrm{\exists }x\phi (x,\vec{y})$ with a specified finite list containing its other free variables, include $h_{\phi }(\vec{a})$ equal to the least ordinal witness in $M$ if one exists, and zero otherwise. The local structural recursion and satisfaction theorem make each displayed witness selector a set function.

Put $\mu =\max (\omega ,\mathrm{\mid }L\mathrm{\mid })<\kappa$. Finite strings over the alphabet of symbols, countably many variables and punctuation number at most $\mu$: induction from ${\mu }^2=\mu$ bounds each finite length, and recursion collects all finite lengths; the countable disjoint union has size at most $\omega \times \mu \le {\mu }^2=\mu$. Thus the formula/list pairs and the functions just included form a family of size at most $\mu$. Ambient AC suffices for these cardinal identifications.

Let nonzero alpha be closed under this family. Closure under language functions and constants makes it a substructure. Terms evaluated on parameters below alpha agree in both structures, by induction on terms. Equality and relation atoms therefore agree; induction on formulas preserves agreement under negation and conjunction. If M satisfies $\mathrm{\exists }x\phi (x,\vec{a})$, its selected witness is below alpha, and the induction hypothesis for $\phi$ proves truth in the restriction. Conversely a witness below alpha transfers to M by the same induction hypothesis. Thus every formula agrees.

The same induction proves the general witness criterion used below: any nonempty substructure in which every existential formula true in M with parameters in the substructure has some witness there is elementary. Conversely an elementary substructure has that witness property by the semantics of existential quantification.

Use the witness family $\mathcal{H}$. For $\beta <\kappa$, let $g(\beta )=\sup (\{h(\vec{a})+1:h\in \mathcal{H},\vec{a}\in (\beta +1{)}^{\mathrm{arity}(h)}\}\cup \{\beta +1\})$. There are fewer than kappa values: the finite-string cardinal count in the witness lemma bounds all tuples, and multiplying by $\mathrm{\mid }\mathcal{H}\mathrm{\mid }<\kappa$ still gives fewer than kappa. Regularity therefore gives $g(\beta )<\kappa$.

The nonzero closure points alpha of g form an unbounded set. Since $g(\beta )>\beta$, such alpha are limits. Every finite tuple below alpha is contained in some $\beta +1<\alpha$; hence every h value on it is below alpha. This includes the empty tuple for constants. The witness lemma gives $\alpha \in E_M$. Thus $E_M$ is unbounded.

If delta is a nonzero limit point of $E_M$, every finite tuple below delta lies below some $\alpha \in E_M\cap \delta$. Language-function closure at alpha makes the restriction to delta a substructure. Any existential formula true in M with such a tuple has a witness below alpha by elementarity there, hence below delta. The witness criterion proves elementarity at delta. Thus $E_M$ itself is closed, completing the club claim.

If S is stationary, intersect it with the club of elementary initial segments of any specified small-language structure. This proves (i) implies (ii), which immediately implies (iii), since countable languages have size below uncountable kappa.

Assume (iii) and let C be any club. Form the finite-language structure on kappa with ordinal order, constant zero, successor function $s(\beta )=\beta +1$, and next-club-point function $n_C(\beta )=\min (C\setminus (\beta +1))$. A nonzero elementary restriction at $\alpha \in S$ is in particular a substructure. Successor closure makes alpha a limit, and next-point closure gives a point of $C\cap \alpha$ strictly above every beta below alpha. Closedness of C forces $\alpha \in C$, so S meets C. This proves (iii) implies (i).

The related filter-base conclusion follows by combining fewer than kappa languages, renaming their nonlogical symbols to avoid collisions, and combining the corresponding structures on kappa. Regularity bounds the union language below kappa. Its elementary club is contained in the intersection of the original elementary clubs, because each original structure is a reduct and its formulas retain their interpretations. The empty collection has the whole cardinal as an upper containing set.