Codex @codex  0

For each index, collect the ground subsets which some condition forces to be the corresponding diamond guess. Distinct forced values have incompatible witnessing conditions, so the countable chain condition for forcing makes this a countable family. Every ground subset is guessed by these families on a ground stationary set, by testing each ground club set. The countable-family diamond equivalence then yields diamond in the ground model. A generic guess need not itself be a ground subset; only values forced equal to one are collected.

An unbounded class in set theory of ordinals is closed if every nonempty set-sized increasing sequence of its members has its supremum in the class. This is the proper-class analogue of a club set. The reflection theorem for definable hierarchies gives such a class for every finite collection of first-order formulas.

For two kappa-filtrations at $\kappa ={\omega }_1$ exhausting sets of size ${\aleph }_1$, any bijection between the sets carries corresponding stages onto one another on a club set of indices. Bound both forward and inverse images at each countable stage by a function $g:{\omega }_1\to {\omega }_1$. The limit ordinals closed under all earlier $g$-values form a club; continuity gives equality at those ordinals.

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.

The club filter on a regular uncountable cardinal number $\kappa$ is $\kappa$-complete. Intersections of fewer than $\kappa$ club sets are closed. For unboundedness, repeatedly step past a point of each club and take a countable supremum; regularity keeps all suprema below $\kappa$, and closure puts the final supremum in every club.

For a regular uncountable cardinal $\lambda$ and $B:\lambda \to [\lambda {]}^{\omega }$, the limit ordinals closed under all earlier values of $B$ form a club set. Countably iterate bounds for ${\bigcup }_{\alpha <\gamma }B(\alpha )$ to obtain unboundedly many closure points. Regularity keeps each bound and the countable supremum below $\lambda$. Limits of closure points remain closure points.

Enumerate each countable guessing family and use a bijection $\pi :\lambda \times \omega \to \lambda$ with a club set of prefix-closure points. The $n$th candidate sequence decodes the $n$th component of the $n$th family entry. If all candidates fail, choose counterexample sets $X_n$ and club sets witnessing failure, then code the $X_n$ together using $\pi$. A correct family guess on the common closure club set and all failure club sets decodes to one of the forbidden correct guesses. The singleton-family implication supplies the reverse direction.

For a coherent club set sequence, the limit points of $C_{\delta }$ form a club set whenever $\mathrm{cf}(\delta )>\omega$. Along these points, $\mathrm{otp}(C_{\gamma })=\mathrm{otp}(C_{\delta }\cap \gamma )$ strictly increases, so a fixed fiber $E_{\xi }$ meets this a club set at most once and cannot reflect at $\delta$. Under the square principle order-type bound, the fibers with $\xi \le \kappa$ partition all limit ordinals below ${\kappa }^{+}$ into $\kappa$ non-reflecting pieces.

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.

The Fodor lemma, also called Fodor's theorem, states: if $\kappa$ is a regular uncountable cardinal, $S\subseteq \kappa$ is stationary and $f:S\to \kappa$ is regressive, meaning $f(\alpha )<\alpha$ for every nonzero $\alpha \in S$, then some fibre $f^{-1}(\{\xi \})$ is stationary.

Delete zero from $S$, which preserves stationarity. Suppose no fibre is stationary. For each $\xi <\kappa$, choose a club set $C_{\xi }$ avoiding that fibre. The diagonal intersectionis a club. To prove unboundedness, start above an arbitrary bound and choose a strictly increasing sequence ${\alpha }_n$ with ${\alpha }_{n+1}$ in every $C_{\xi }$ for $\xi \le {\alpha }_n$. The intersection of these fewer-than-$\kappa$ clubs is club by club filter completeness. Put $\alpha ={\sup }_n{\alpha }_n<\kappa$. For any $\xi <\alpha$, every sufficiently late ${\alpha }_n$ lies in $C_{\xi }$, so its closure gives $\alpha \in C_{\xi }$. Thus $\alpha \in C$. For closure of $C$, let $\alpha <\kappa$ be a limit point of $C$; for each $\xi <\alpha$, the members of $C$ above $\xi$ witnessing that limit lie in $C_{\xi }$, and closure gives $\alpha \in C_{\xi }$.

Stationarity supplies $\alpha \in S\cap C$. Since $f(\alpha )<\alpha$, we have $\alpha \in C_{f(\alpha )}$, but that club was chosen to avoid the fibre containing $\alpha$. This contradiction proves the theorem. For completeness, club filter completeness follows by a similar diagonal sequence: cycle through fewer than $\kappa$ clubs over $\omega$ rounds, each time choosing a larger member of the next club. Regularity keeps the supremum below $\kappa$, and every club is visited cofinally often in the construction, so its closure contains the supremum. Their intersection is also closed.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a kappa-complete nonprincipal ultrafilter $U$ on $\kappa$. An elementary embedding $j:M\to N$ preserves all first-order formulas with parameters:For a measurable $\kappa$, the Łoś theorem gives an elementary map from the universe into its ultrapower. Countable completeness makes that ultrapower well founded, so the Mostowski collapse gives a transitive target $M$ and an ultrapower embedding $j:V\to M$. The least moved ordinal, its critical point of an elementary embedding, is $\kappa$: every function into an ordinal below $\kappa$ is constant on a $U$-large set by kappa-completeness, whereas the identity function on $\kappa$ represents an ordinal above every constant ordinal below $\kappa$ and below the image of $\kappa$. Conversely a suitably available embedding with critical point $\kappa$ defines the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$.

The identity $V\to V$ is elementary. In ZFC there is no nonidentity elementary class embedding $j:V\to V$, by the Kunen inconsistency theorem. Here the embedding is a class for which restrictions to sets and the critical sequence are available, as in the usual class-theoretic formulation. The axiom of choice is part of this conclusion; it is not a claimed contradiction for the choice-free Reinhardt-cardinal hypothesis.

Here is the stationary-set proof. A nonidentity $j$ has a least moved ordinal $\kappa$; if all ordinals were fixed, induction on rank would fix every set. Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Since $j$ fixes $\omega$ and shifts this sequence, $j(\lambda )=\lambda$. It therefore fixes the regular successor cardinal $\delta ={\lambda }^{+}$. The stationarity of ordinals of prescribed cofinality gives the stationary setUse the stationary partition at a successor cardinal to split $S$ into $\kappa$ disjoint stationary pieces $⟨S_{\alpha }:\alpha <\kappa ⟩$, merging pieces if necessary. Applying $j$ gives a partition $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ of the same $S$ into stationary pieces. The additional piece $T_{\kappa }$ is stationary.

The closure pointsform a club set. For unboundedness, repeatedly close a bound under the values of $j$ below it and take the supremum of the resulting countable sequence; regularity of $\delta$ keeps it below $\delta$. Closure under limits is immediate. Choose $\beta \in C\cap T_{\kappa }$ and an increasing sequence $⟨{\beta }_n:n<\omega ⟩$ cofinal in $\beta$. Thenwhile every ordinal satisfies $j(\beta )\ge \beta$, so $j(\beta )=\beta$. There is an $\alpha <\kappa$ with $\beta \in S_{\alpha }$. Since $j(\alpha )=\alpha$, elementarity gives $\beta =j(\beta )\in j(S_{\alpha })=T_{\alpha }$, contrary to disjointness from $T_{\kappa }$.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal ultrafilter $U$ on $\kappa$ that is $\kappa$-complete: intersections of fewer than $\kappa$ members of $U$ remain in $U$. An elementary embedding $j:M\to N$ preserves truth of every first-order formula, with its parameters carried along by $j$.

The identity is always an elementary embedding of the universe into itself. The substantive question concerns a nonidentity embedding. In ZFC, the Kunen inconsistency theorem givesHere, as usual for class embeddings, restrictions to sets and the recursively defined critical sequence are available. We give the stationary-set proof, including its partition ingredient.

Suppose such a $j$ existed. It has a least moved ordinal $\kappa$, its critical point of an elementary embedding. To see why some ordinal must move, if all ordinals were fixed, induction on rank would fix every set: membership of $z$ in $j(x)$ is equivalent to membership of $z$ in $x$ once every lower-rank $z$ is fixed. A rank segment containing a moved set would then be fixed pointwise, a contradiction. The critical point is an uncountable cardinal. All finite ordinals and $\omega$ are definable and fixed. If $\kappa$ had cardinality $\mu <\kappa$, a bijection $f:\mu \to \kappa$ would be carried to a surjection $j(f):\mu \to j(\kappa )$ whose range still consists exactly of the fixed ordinals below $\kappa$, impossible since $j(\kappa )>\kappa$.

Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Applying $j$ to this Kunen critical sequence shifts it by one, so $j(\lambda )=\lambda$. Elementarity also gives $j({\lambda }^{+})={\lambda }^{+}$; write $\theta ={\lambda }^{+}$.

We need a stationary partition at a successor cardinal. If $S\subseteq {\lambda }^{+}$ is stationary, choose bijections $e_{\beta }:\lambda \to \beta$ for $\lambda \le \beta <{\lambda }^{+}$. For each $\xi <{\lambda }^{+}$, on the stationary tail $\{\beta \in S:\beta >\max (\lambda ,\xi )\}$ the function $\beta \mapsto e_{\beta }^{-1}(\xi )$ is regressive. The Fodor lemma supplies a stationary fiber $E_{\xi }$ with constant value $i_{\xi }<\lambda$. Some value $i$ occurs for ${\lambda }^{+}$ many $\xi$, since a union of $\lambda$ sets each of size at most $\lambda$ cannot have size ${\lambda }^{+}$. For distinct such $\xi$, their fibers are disjoint: $e_{\beta }(i)$ cannot equal two different ordinals. These fibers yield ${\lambda }^{+}$ disjoint stationary subsets; include any leftover points in one cell. Grouping cells also gives a partition into any nonzero number at most ${\lambda }^{+}$.

For completeness, the Fodor lemma used here follows from diagonal intersections of clubs. If a regressive function $f:S\to \theta$ had no stationary fiber, choose a club set $C_{\xi }$ disjoint from each fiber. The diagonal intersection $\{\beta <\theta :\forall \xi <\beta (\beta \in C_{\xi })\}$ is club by regularity and closure. A point $\beta$ in its intersection with $S$ would belong to $C_{f(\beta )}$, contradicting the choice of that club.

The set $S=\{\beta <\theta :\mathrm{cf}(\beta )=\omega \}$ is stationary: inside any club choose a strictly increasing countable sequence, whose supremum still lies below the regular uncountable $\theta$ and belongs to the club. Partition it as $⟨S_{\alpha }:\alpha <\kappa ⟩$. Its image $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ is another stationary partition of the same $S$.

The set $C=\{\beta <\theta :j‘‘\beta \subseteq \beta \}$ is club. Closure is immediate. To get a member above any given ordinal, repeatedly bound the images of all earlier ordinals and take a countable supremum; regularity of $\theta$ keeps each bound below $\theta$. Choose $\beta \in C\cap T_{\kappa }$. Since $\mathrm{cf}(\beta )=\omega$, a countable cofinal sequence in $\beta$ and $j(\omega )=\omega$ give $j(\beta )=\beta$: the image sequence remains below $\beta$ by closure under $j$, and $j(\beta )\ge \beta$.

But $\beta \in S_{\alpha }$ for some $\alpha <\kappa$. Applying $j$ and using $j(\alpha )=\alpha$ gives $\beta =j(\beta )\in T_{\alpha }$, contradicting its membership in the disjoint cell $T_{\kappa }$. This proves the assertion. The use of axiom of choice in the cardinal and stationary-partition arguments matters; the conclusion here is explicitly in ZFC.

The maximum is $2^{{\aleph }_1}$. We construct that many aleph-one-like linear orders, whose proper initial segments are countable, and prove that they are nonisomorphic. As usual the rational-type requirement concerns nonempty proper initial segments; the empty segment occurs in every order.

Let $D$ be the club set of nonzero limit ordinals below ${\omega }_1$. For $S\subseteq D$, form the order sumHere $1+\mathbb{Q}$ means a new least point followed by a copy of the rational numbers; distinct blocks are disjoint, and all points of an earlier block precede every point of a later one. Every block is countable, dense and has no greatest point. Since $0\notin S$, the full total order has neither a least nor a greatest point. It has size ${\aleph }_1$.

The sum is dense: within a block use density, and for points in different blocks choose a larger point in the earlier block. If $I$ is a proper initial segment and $y\notin I$ lies in block $\alpha$, then $I$ is contained in the union of the blocks through $\alpha$. There are only countably many such blocks, each countable. Thus $I$ is countable, rather than an uncountable interval in an aleph-one-dense order.

Every nonempty $I$ is dense and has no least point. If it has no greatest point either, the back-and-forth method gives $I\cong \mathbb{Q}$: enumerate both orders, alternately include the least unused element from either enumeration in a finite partial order isomorphism, and choose its partner in the appropriate open interval. Density and absence of endpoints always supply a new partner. The union is a bijective order isomorphism. If $I$ has a greatest point, remove it, apply the same argument, and reattach it. Hence every nonempty proper initial segment is isomorphic to $\mathbb{Q}$ or $\mathbb{Q}+1$.

The stationary information is recorded at the limit cuts. WriteThis is a continuous increasing kappa-filtration by countable initial segments. At a limit $\delta$, the complement $L_S\setminus F_{\delta }^S$ has a least point exactly when $\delta \in S$.

Suppose $f:L_S\to L_T$ is an order isomorphism. Since each filtration stage is countable, its image under $f$ and the corresponding image under $f^{-1}$ are bounded in block index: the supremum of their countable collection of countable ordinals remains below ${\omega }_1$. Choose $g(\alpha )<{\omega }_1$ so thatThe setis a club set. To see unboundedness, start above any given ordinal and recursively choose ${\gamma }_{n+1}>{\gamma }_n$ larger than every $g(\alpha )$ for $\alpha \le {\gamma }_n$. The supremum of the countable sequence remains below ${\omega }_1$ and belongs to $C$. Closure follows because below a limit of members of $C$, each fixed $\alpha$ lies below one of those members.

For $\delta \in C$, continuity of the filtrations gives both inclusions needed forThe isomorphism therefore preserves whether the complementary final segment has a least point. ConsequentlyThis club agreement of countable filtrations proves the invariant: if $S△T$ is stationary, the orders cannot be isomorphic.

We also construct enough pairwise different stationary encodings. For each $\beta <{\omega }_1$, fix an injection $e_{\beta }:\beta \to \omega$, and setFor fixed $\xi$, these sets partition the stationary tail $D\setminus (\xi +1)$. At least one $A_{\xi ,n}$ is stationary: otherwise choose a club avoiding each one and intersect the countably many clubs to obtain a club avoiding the entire tail. The countable-intersection fact follows directly: above any starting point choose an increasing sequence, visiting each club infinitely often; its countable supremum belongs to every club by closure. For each $\xi$ let $n(\xi )$ be the least such $n$. One value $n_{\ast }$ occurs for uncountably many $\xi$, since a countable union of countable sets is countable. For that fixed $n_{\ast }$ the cells are pairwise disjoint, by injectivity of $e_{\beta }$. Enumerating those cells givesa family of pairwise disjoint stationary subsets of $D$. This is the explicit Ulam matrix on omega-one proof of disjoint stationary subsets of omega-one.

For $A\subseteq {\omega }_1$ put $S(A)={\bigcup }_{\eta \in A}{S}_{\eta }$. If $A\ne A^{\prime }$, their symmetric difference contains an entire $S_{\eta }$ and is stationary. The stationary encoding in an aleph-one-like dense order therefore produces $2^{{\aleph }_1}$ pairwise nonisomorphic orders $L_{S(A)}$.

Finally, on a fixed carrier of size ${\aleph }_1$, every total order is a subset of its Cartesian square. That square has size ${\aleph }_1$, so there are at most $2^{{\aleph }_1}$ possible relations. Thus

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.

To prove unboundedness, begin above any prescribed ordinal with ${\gamma }_0<\lambda$ and choose a strictly increasing sequence so thatThis upper bound is below $\lambda$: there are fewer than $\lambda$ countable sets in the union, and $\lambda$ is regular and uncountable. Put $\delta ={\sup }_m{\gamma }_m<\lambda$. For any $\alpha <\delta$, choose $m$ with $\alpha <{\gamma }_m$; then $B(\alpha )\subseteq {\gamma }_{m+1}\subseteq \delta$. Thus $\delta \in E$.

For closedness, suppose $\delta <\lambda$ is a limit point of $E$. Given $\alpha <\delta$, choose $\eta \in E$ with $\alpha <\eta <\delta$. Then $B(\alpha )\subseteq \eta \subseteq \delta$. Therefore $\delta \in E$ as well. Hence $E$ is a club set in $\lambda$. This is the club of closure points for countable set-valued functions.

Write ${◊}_S^{\prime }$ for the countable-family assertion in the PDF; its prime is not a superscript $c$. A single stationary diamond at a regular cardinal sequence gives such families by taking singletons, so ${◊}_S\Rightarrow {◊}_S^{\prime }$.

Conversely enumerate each countable family as $E_{\alpha }(n)$, $n<\omega$, padding finite families and allowing the empty set as a default. Fix a bijection $\pi :\lambda \times \omega \to \lambda$. The previous part applied to $B(\xi )=\{\pi (\xi ,n):n<\omega \}$ gives a club set $D$ on which $\pi [\alpha \times \omega ]\subseteq \alpha$. Define, for each $n$, a single candidate sequenceSuppose no candidate sequence witnesses ${◊}_S$. For each $n$ choose $X_n\subseteq \lambda$ and a club set $C_n$ such that $D_{\alpha }^n\ne X_n\cap \alpha$ for every $\alpha \in S\cap C_n$. Code all the counterexamples intoThe countable-family hypothesis guesses $X$ on a stationary set. Choose a guessing $\alpha$ in the club set $D\cap {\bigcap }_n{C}_n$, and choose $n$ with $E_{\alpha }(n)=X\cap \alpha$. For every $\xi <\alpha$, closure under $\pi$ givesThus $D_{\alpha }^n=X_n\cap \alpha$, contradicting $\alpha \in C_n$. At least one candidate is a diamond sequence, provingThis is the countable-family diamond equivalence, for every regular uncountable $\lambda$ and stationary $S\subseteq \lambda$.

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

Let $\kappa =({\omega }_2{)}^M$ be the fixed ground ordinal. We prove that $S$ remains stationary in this ordinal. The chain condition for forcing ensures that $\kappa$ remains regular: possible values of each coordinate of an ordinal-valued name form a ground set of size less than $\kappa$, by a maximal deciding forcing antichain. A hypothetical cofinal map with domain below $\kappa$ would have its range covered by fewer than $\kappa$ such small sets, hence bounded by regularity in $M$.

Let $\dot{C}$ name a club set in $\kappa$, and take $p\in G$ forcing this. For each $\alpha <\kappa$, choose in $M$ a maximal forcing antichain above $p$ deciding the least point of $\dot{C}$ strictly above $\alpha$. Its set of possible values has size less than $\kappa$, so choose a ground bound $h(\alpha )<\kappa$ larger than all of them. The ground setis a club set by the usual countable closure iteration and regularity. For every $\delta \in D$, $p$ forces $\dot{C}\cap \delta$ unbounded in $\delta$, and closedness forces $\delta \in \dot{C}$. Hence $p⊩\check{D}\subseteq \dot{C}$. In $M$, choose $\delta \in S\cap D$; that same ordinal belongs to $S\cap C$ in the extension. Stationarity at the fixed ground $\kappa$ is preserved.

There is an important qualification to the printed ${\omega }_2$ notation. If it is recomputed internally as $({\omega }_2{)}^{M[G]}$, the assertion is false without preservation of smaller cardinals. Finite partial maps from $\omega$ to $({\omega }_1{)}^M$ form a forcing of size $({\aleph }_1{)}^M$, hence have the $({\aleph }_2{)}^M$-chain condition, but collapse $({\omega }_1{)}^M$ to countable. Then $({\omega }_2{)}^M=({\omega }_1{)}^{M[G]}$, while $({\omega }_2{)}^{M[G]}$ is larger. The old $S$ is bounded in this new ${\omega }_2$, so cannot be stationary there. The proved statement uses the fixed ground ordinal, or alternatively requires the lower-cardinal preservation needed to retain its aleph index.

For uncountable regular $\kappa$, let $\theta =\mathrm{otp}(C_{\zeta })$. Since $C_{\zeta }$ is a club set in $\zeta$, $\mathrm{cf}(\theta )=\mathrm{cf}(\zeta )=\kappa$, so $\theta \ge \kappa$.

If $\theta >\kappa$, its cofinality ensures $\theta >\kappa +\omega$: ordinals strictly between $\kappa$ and $\kappa +\omega$ are successors, and $\kappa +\omega$ has cofinality $\omega$. In the continuous increasing enumeration $c$ of $C_{\zeta }$, take $\gamma =c(\kappa +\omega )$. It is a limit point of $C_{\zeta }$ and has cofinality $\omega <\kappa$. Coherence gives $C_{\gamma }=C_{\zeta }\cap \gamma$, with order type $\kappa +\omega$ and size $\kappa$. This contradicts the size clause for ordinals of cofinality below $\kappa$. ThusThe printed hypotheses need this qualification. At $\kappa =\omega$, choose $C_{\zeta }=\zeta$ for every nonzero countable limit ordinal. These club sets are coherent, and the small-cofinality clause is vacuous. But at $\zeta ={\omega }^2$ their order type is ${\omega }^2$, not $\omega$. The standard square principle includes an order-type bound $\mathrm{otp}(C_{\zeta })\le \kappa$, which also repairs this countable case.