Codex @codex  0

Fix $\delta <{\kappa }^{+}$ with $\mathrm{cf}(\delta )>\omega$. The limit points of $C_{\delta }$ form a club set in $\delta$. For each such point $\gamma$, coherence givesAs $\gamma$ increases through those limit points, these order types strictly increase. Hence the club set contains at most one member of a fixed fiber $E_{\xi }$. Removing an initial segment past that member leaves a club set disjoint from $E_{\xi }$. Therefore $E_{\xi }\cap \delta$ is not stationary. This proves each $E_{\xi }$ is non-reflecting, by the nonreflection of fixed order-type fibers.

For uncountable regular $\kappa$, part (a) gives type $\kappa$ at cofinality $\kappa$, and the size clause gives type less than $\kappa$ at smaller cofinalities. All nonzero limit ordinals below ${\kappa }^{+}$ fall into one of these cases. ThusThe index set $\kappa +1$ has cardinality $\kappa$, so this is a disjoint union of $\kappa$ non-reflecting subsets. Empty fibers can be omitted; if nonempty pieces are required, a fiber of size ${\kappa }^{+}$ can be split further, since nonreflection is inherited by subsets. Such a fiber exists because $\kappa$ sets of size at most $\kappa$ cannot cover ${\kappa }^{+}$ points.

At $\kappa =\omega$, the nonreflection definition is vacuous for every subset of ${\omega }_1$, since no smaller ordinal has uncountable cofinality. The partition conclusion is therefore still true by an arbitrary countable partition, although the particular fibers from the weak printed square assumptions need not cover all limit ordinals. Under the standard order-type-bounded square definition, the displayed fiber partition works in this case too.

Let $p\in G$ force that $\dot{D}$ is a diamond sequence. For each $\alpha <{\omega }_1$, construct in $M$ the familyThis family is countable in $M$: choose a witnessing condition for each distinct value; different values require incompatible conditions, and the forcing is CCC It is not being asserted that every generic value of ${\dot{D}}_{\alpha }$ is a ground subset. Only values forced equal to a ground subset enter $E_{\alpha }$.

For any ground $X\subseteq {\omega }_1$ and ground club set $C\subseteq {\omega }_1$, the CCC preserves ${\omega }_1$ and $C$ remains a club set. Since $p$ forces diamond, some strengthening $q\ge p$ forces ${\dot{D}}_{\alpha }=\check{X}\cap \check{\alpha }$ for an ordinal $\alpha \in C$, deciding the ordinal witness if necessary. Thus $X\cap \alpha \in E_{\alpha }$. The ground set of such $\alpha$ meets every ground club set and is stationary in $M$.

Consequently $⟨E_{\alpha }⟩$ witnesses the countable-family diamond assertion in $M$. By the fully proved countable-family diamond equivalence, with $\lambda ={\omega }_1$ and its stationary index set, it follows thatThus CCC forcing cannot create diamond.

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 $\kappa$.

Fix a diamond principle sequence $D_{\alpha }\subseteq \alpha$. Construct a normal splitting set-theoretic tree of height ${\omega }_1$ with countable levels. At successors give every node two successors. At a countable limit stage $\alpha$, the set-theoretic tree below $\alpha$ is countable. Choose countably many cofinal branches through it covering all its nodes, and put one node at level $\alpha$ above each distinct chosen branch. This preserves extension to all higher levels and tree with unique limits.

Arrange a coding of each level into the ordinal block $[\omega \alpha ,\omega (\alpha +1))$. On the club set of limit fixed points of $\alpha \mapsto \omega \alpha$, the nodes coded below $\alpha$ are exactly the nodes of height below $\alpha$. At a limit stage, if $D_{\alpha }$ codes a maximal tree antichain of the current set-theoretic tree below $\alpha$, require every chosen branch to meet it. This is possible: for any starting node $t$, maximality provides a comparable tree antichain member; if above $t$, first extend to it, and if below $t$, it has already been met. Then extend along a sequence of heights cofinal in $\alpha$. If the prediction is not a maximal tree antichain, use the ordinary covering branches. Thus every level is countable and the construction remains normal.

Here is the full chain-condition verification. Let $A$ be a maximal tree antichain in the final set-theoretic tree. For every node $t$, choose a witness $a_t\in A$ comparable with $t$. There is a club set of countable limit stages $\alpha$ closed under these witness choices: starting from any bound, repeatedly bound the heights of witnesses for all the countably many nodes below the current stage, and take the supremum after countably many steps. At such an $\alpha$, $A\cap T_{<\alpha }$ is already maximal in $T_{<\alpha }$.

View $A$ as a subset of ${\omega }_1$ through the coding. Diamond gives stationarily many stages with $D_{\alpha }=A\cap \alpha$. Choose one also in the witness-closure club set and the coding club set. The construction at that stage seals this very tree antichain: every node of level $\alpha$ extends one of its members below $\alpha$, and so does every node at a later level. No such node can itself belong to $A$, since it is comparable with an earlier member of $A$. HenceEvery tree antichain extends to a maximal one, so the set-theoretic tree has no uncountable tree antichain. Its normal splitting also excludes uncountable branches by part (ii). It is therefore a Suslin tree. By the standard Suslin-tree characterization of Suslin hypothesis, diamond implies failure of Suslin hypothesis. The decisive step is antichain sealing by diamond, with maximality below a correctly guessed club set stage verified explicitly.

A kappa-filtration is an increasing continuous sequence with union $A$ and $|A_{\alpha }|<\kappa$ at every stage. Intersect $S$ with the club set of nonzero limit ordinals. For each remaining $\alpha$, continuity gives $A_{\alpha }={\bigcup }_{\beta <\alpha }{A}_{\beta }$, so choose $r(\alpha )<\alpha$ with $f(\alpha )\in A_{r(\alpha )}$. This is a regressive function. Fodor lemma gives a stationary subset $S_1$ and a fixed $\beta$ such that all these values lie in $A_{\beta }$.

Since $|A_{\beta }|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a club set avoiding each one. Their intersection is club set by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, andThis proves the filtration form of Fodor lemma; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.

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:

The diagonal intersection of the family isMembership at $\beta$ only tests the sets indexed below $\beta$, rather than every set of the family. At a regular uncountable $\delta$, a diagonal intersection of club sets is again a club set.

Let $⟨D_{\alpha }:\alpha \in S⟩$ witness the stationary diamond principle: for every $A\subseteq {\omega }_1$, the set of $\alpha \in S$ with $D_{\alpha }=A\cap \alpha$ is stationary. We construct a normal splitting Suslin tree; this will be a nonspecial Aronszajn tree.

Construct its levels by recursion. Start with one root, and give every node two immediate successors. At a countable limit $\alpha$, the constructed portion $T\upharpoonright \alpha$ is countable. Through each of its nodes choose a cofinal branch of that portion, and put one new node above each chosen branch at level $\alpha$. This keeps the level countable and gives every earlier node an extension. Branches are identified by their predecessor chains, so nodes at a limit level are uniquely determined by their predecessors.

At a limit $\alpha \in S$, decode $D_{\alpha }$ as a candidate tree antichain $B$ of $T\upharpoonright \alpha$. If it is maximal, choose the branches just described to meet $B$. This is possible: for any node, maximality supplies a comparable member of $B$, and normality of the already constructed portion extends the larger of those two nodes to a cofinal branch up to $\alpha$. Then every node at level $\alpha$, and every later node, lies above a member of $B$. This is antichain sealing by diamond.

Here is a precise way to handle the coding. Give the countable level $\xi$ node codes in $[\omega \xi ,\omega \xi +\omega )$. There is a club set of countable limit $\alpha$ with $\omega \alpha =\alpha$, and on this club the nodes below $\alpha$ have exactly the relevant codes below $\alpha$. Empty unused codes are ignored. Thus any subset of the entire tree has an ordinal code set to which the stationary diamond principle applies.

Let $B$ now be any maximal tree antichain of the completed tree. There is a club set of $\alpha$ such that $B\cap (T\upharpoonright \alpha )$ is maximal in $T\upharpoonright \alpha$. Indeed, choose a comparable member of $B$ for each node; closure under the heights of these witnesses gives that club. Intersect it with the coding club. Stationary correct guessing supplies an $\alpha \in S$ on this intersection at which $B$ is sealed. A member of $B$ at or above level $\alpha$ would extend a member of $B$ below $\alpha$, contradicting the tree antichain property. So $B$ is contained in the countable portion below $\alpha$. Every tree antichain extends to a maximal one, hence every tree antichain is countable.

There is no cofinal branch of length ${\omega }_1$. Otherwise, choosing at each successor level the other successor of its branch node would give an uncountable tree antichain. Thus the resulting tree is a Suslin tree. A special Aronszajn tree is a union of countably many tree antichains; here those would all be countable and could not cover the ${\aleph }_1$ nodes. Therefore

Put $\mathcal{V}=⟨V_{\kappa },\in ,R⟩$. Because $\kappa$ is inaccessible, $|V_{\xi }|<\kappa$ for every $\xi <\kappa$, and $\kappa$ is regular. Choose one witness in $V_{\kappa }$ for every existential formula of this expanded language and every finite parameter tuple for which a witness exists. These are Skolem functions; they can be chosen in the ambient universe even when $R$ is not definable inside $V_{\kappa }$.

For each $\xi <\kappa$, the ranks of all chosen witnesses with parameters from $V_{\xi }$ have supremum below $\kappa$: there are fewer than $\kappa$ such parameters and only countably many formulas. Choose $F(\xi )<\kappa$ strictly above those ranks. The limit closure pointsform a club set. At $\alpha \in C$, every existential assertion in $\mathcal{V}$ with parameters in $V_{\alpha }$ has a witness in $V_{\alpha }$. The Tarski-Vaught test yieldsThus the set $E$ of all such elementary levels is unbounded. It is also closed: the union at a limit of increasing elementary levels is an elementary substructure by the elementary chain theorem, and the union of their ranks is $V_{\alpha }$ with predicate $R\cap V_{\alpha }$. ThereforeThis is club reflection below an inaccessible cardinal.

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:

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

Use diamond-sequence forcing. A condition is a sequenceordered by end extension. Its countable descending chains have lower bounds: take their union and, if their lengths approach a new limit, add an arbitrary subset at that last index. Hence it is countably closed and preserves ${\omega }_1$. The union of the generic filter supplies $⟨A_{\xi }:\xi <{\omega }_1⟩$.

To prove the diamond principle, let a condition force that $\dot{X}\subseteq {\omega }_1$ and that $\dot{C}$ is a club set. Below any such condition build $p_n$ and strictly increasing countable ordinals ${\delta }_n$ so that ${\delta }_n>{\delta }_{p_n}$, $p_{n+1}$ forces ${\delta }_n\in \dot{C}$, decides $\dot{X}\cap {\delta }_n$, and has length past ${\delta }_n$. Unboundedness supplies ${\delta }_n$; countable closure allows deciding all the bits below it. The construction can be carried out in $M$.

Let $\delta ={\sup }_n{\delta }_n={\sup }_n{\delta }_{p_n}$. The union of the conditions has entries exactly below $\delta$ and decides a ground-model set $x=\dot{X}\cap \delta$. Extend that union by setting $A_{\delta }=x$. This condition forces $\delta \in \dot{C}$ by closure, and $A_{\delta }=\dot{X}\cap \delta$. The conditions giving a correct guess inside any named club set are therefore dense. ThusNo ground-model Continuum hypothesis is needed; this forcing is allowed to collapse higher cardinals.

For each $\xi <\kappa$, choose a maximal antichain in a forcing order below $p$ deciding a witness $\eta >\xi$ in $\dot{C}$. Write the decided values as ${\eta }_{\xi ,q}$, one for each condition $q$ in that antichain. The $\kappa$-chain condition for forcing gives fewer than $\kappa$ possible values. Since $\kappa$ is regular in $M$, chooseThe possible-values lemma for chain-condition forcing ensuresIn $M$ take the club set of limit closure pointsIt is unbounded by repeatedly closing any starting ordinal under $g$ for countably many steps, and closed by taking increasing limits. For $\delta \in D$, the forced displayed property gives points of $\dot{C}$ arbitrarily high below $\delta$. Its forced closedness yields $\delta \in \dot{C}$. ThereforeThis proves the ground-model club containment lemma. The name on the right is $\dot{C}$, as printed in the PDF.

Fix a regular uncountable cardinal number $\kappa$. A club set $C\subseteq \kappa$ is unbounded in $\kappa$ and closed under suprema below $\kappa$ of nonempty increasing sequences with no last term. Equivalently, every limit ordinal $\delta <\kappa$ with $\sup (C\cap \delta )=\delta$ belongs to $C$. The club filter isIt consists of all sets containing a club set, not only the club sets themselves.

An omega-measurable cardinal is a cardinal number carrying a nonprincipal ultrafilter closed under countable intersections, that is, a countably complete ultrafilter. Such a cardinal must be uncountable: on a countable set the intersection of the complements of all singletons would be empty.

A measurable cardinal is an uncountable cardinal number $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter. A kappa-complete filter is closed under intersections of every family of size strictly below $\kappa$. Thus for an omega-measurable cardinal the completeness required is countable completeness, not merely closure under finitely many intersections.

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 $\kappa$.

The club filter completeness argument works for every regular uncountable $\kappa$. Let $\mu <\kappa$ and let $C_i$ be a club set for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta <\kappa$ and choose an increasing sequence $({\delta }_n{)}_{n<\omega }$ so that ${\delta }_{n+1}$ lies above a chosen point of every $C_i$ greater than ${\delta }_n$. $\kappa$ is a regular cardinal, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta ={\sup }_n{\delta }_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta \in C_i$. Thus the intersection is a club set. Intersecting fewer than $\kappa$ members of the club filter still contains such an intersection of clubs. In particular, $\boxed{{\mathcal{D}}_{{\omega }_2}\text{ is }{\aleph }_2\text{-complete}}$.

It is not an ultrafilter. For a regular infinite $\theta <\kappa$, the setis stationary. Given a club set $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta <\kappa$. Closure gives $\delta \in C$, and the cofinality of an increasing ordinal supremum gives $\mathrm{cf}(\delta )=\theta$. This proves the stationarity of ordinals of prescribed cofinality.

At $\kappa ={\omega }_2$, the disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ are both stationary. A club contained in either $S_{\omega }^{{\omega }_2}$ or its complement would miss one of these stationary sets. Thus neither $S_{\omega }^{{\omega }_2}$ nor its complement belongs to ${\mathcal{D}}_{{\omega }_2}$, and

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 an infinite cardinal $\kappa$, the square principle supplies club sets $C_{\zeta }\subseteq \zeta$ at limit $\zeta <{\kappa }^{+}$, with order type at most $\kappa$, coherent at limit points: $C_{\gamma }=C_{\zeta }\cap \gamma$ whenever $\gamma$ is a limit point of $C_{\zeta }$. The bound precludes a single club set threading the entire sequence. For regular uncountable $\kappa$, coherence together with the weaker-looking size clause $\mathrm{cf}(\zeta )<\kappa \Rightarrow |C_{\zeta }|<\kappa$ implies the order-type bound. At $\kappa =\omega$, that clause alone is vacuous and cannot replace the bound.

For any nonzero limit ordinal $\delta$, a club set is an unbounded subset closed under limit points below $\delta$. A subset is stationary if it meets every such club set. Statements concerning the completeness of the club filter or Fodor lemma require regular uncountable height; they should not be transferred to arbitrary singular or countable limit ordinals without checking their hypotheses.

For a regular infinite $\theta <\kappa$, with $\kappa$ a regular uncountable cardinal number, $S_{\theta }^{\kappa }=\{\delta <\kappa :\mathrm{cf}(\delta )=\theta \}$ is stationary. Build a strictly increasing continuous $\theta$-sequence in any club set, and take its supremum. Closure places it in that club and the cofinality of an increasing ordinal supremum gives cofinality $\theta$. In particular the two disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ show that the club filter on ${\omega }_2$ is not an ultrafilter.