nLab countable chain condition




The countable chain condition, appearing in various guises (for topological spaces, for posets, for Boolean algebras), frequently recurs in discussions on forcing in set theory, particularly in discussions about preservation of cardinals and cofinalities in forcing extensions.

It is something of a misnomer because it is really a condition about antichains. But in the topological context it is equivalent to a condition on chains, and the name has stuck.



Suppose PP is a poset with a bottom element 00. A (strong) antichain in PP is a subset APA \subseteq P such that for all a,bAa, b \in A, we have either a=ba = b or 00 is their meet (we need not assume general meets exist in PP). If PP is a poset without a bottom, we define APA \subseteq P to be an antichain if it becomes an antichain in P +P^+, the poset obtained by freely adjoining a bottom to PP; this is equivalent to saying a,ba, b have no lower bound in PP.


A poset PP satisfies the countable chain condition if every antichain in PP is a countable set (i.e., at most denumerable).

A topological space XX satisfies the countable chain condition if its frame of open sets Op(X)Op(X) does (as a poset).

For future reference, we record a result on the countable chain condition in Heyting algebras HH. Let Reg(H)Reg(H) denote the Boolean algebra of regular elements (xHx \in H such that x=¬¬xx = \neg \neg x).


A Heyting algebra HH satisfies the countable chain condition iff Reg(H)Reg(H) does.


The inclusion Reg(H)HReg(H) \hookrightarrow H preserves meets and 00, so if every antichain in HH is countable, then the same is true in Reg(H)Reg(H).

The quotient map ¬¬:HReg(H)\neg \neg: H \to Reg(H) also preserves meets and 00 (details are at Heyting algebra), and moreover the restriction of this map to an antichain AHA \subseteq H is an injection (if ¬¬a=¬¬b\neg \neg a = \neg \neg b for a,bAa, b \in A, then a=ba = b or ab=0a \wedge b = 0; in the latter case a¬¬a=¬¬a¬¬b=¬¬(ab)=0a \leq \neg \neg a = \neg \neg a \wedge \neg \neg b = \neg \neg (a \wedge b) = 0 and similarly b=0b = 0, so a=ba = b). So if every antichain in Reg(H)Reg(H) is countable, the same is true in HH.

Chains, not antichains


Let BB be a complete Boolean algebra. For cardinals κ\kappa, there is a bijective correspondence between

  • maps g:κB{0}g: \kappa \to B \setminus \{0\} such that {g(α):α<κ}\{g(\alpha): \alpha \lt \kappa\} is an antichain in BB;

  • injective colimit-preserving poset maps f:κBf: \kappa \to B,

here regarding κ\kappa as an ordinal.


Given an antichain g:κB{0}g: \kappa \to B \setminus \{0\}, define a colimit-preserving chain f:κBf: \kappa \to B by f(β)= α<βg(α)f(\beta) = \sum_{\alpha \lt \beta} g(\alpha) (so f(0)=0f(0) = 0, as mandated by colimit-preservation). Given an injective colimit-preserving chain f:κBf: \kappa \to B, define an antichain g:κB{0}g: \kappa \to B \setminus \{0\} by g(α)=f(α+1)f(α)f(α+1)¬f(α)g(\alpha) = f(\alpha + 1) \setminus f(\alpha) \coloneqq f(\alpha + 1) \wedge \neg f(\alpha). It is straightforward to verify that these assignments are inverse to one another.


BB satisfies the countable chain condition iff every chain b 0<b 1<b_0 \lt b_1 \lt \ldots in BB is countable.

This corollary is stated in terms of ascending chains, but by self-duality of Boolean algebras, it could equally well be stated in terms of descending chains.

For topological spaces XX, the frame H=Op(X)H = Op(X) satisfies the countable chain condition iff Reg(H)Reg(H), the complete Boolean algebra of regular open sets, satisfies the countable chain condition (Lemma ). Now a regular element is just an element of the form ¬U\neg U, which for an open set UU is the complement of its closure. Thus for the topological case we deduce the following formulation in terms of chains:


A space XX satisfies the countable chain condition if and only if every collection of open sets {U α} α<κ\{U_\alpha\}_{\alpha \lt \kappa} that satisfies the condition

  • U α¯U β¯\widebar{U_\alpha} \subset \widebar{U_\beta} (strict inclusion between closures) whenever α<β\alpha \lt \beta

is countable.

Products of separable spaces

We show in this section that arbitrary products of separable spaces satisfy the countable chain condition.


A separable space satisfies the countable chain condition.


Let DD be a countable dense set, and suppose given a collection of pairwise disjoint inhabited open sets {U α:αA}\{U_\alpha: \alpha \in A\}. Each U αU_\alpha contains at least one dDd \in D. Thus the map DAD \to A sending dDd \in D to the unique αA\alpha \in A such that dU αd \in U_\alpha is surjective, and so AA is at most countable.

Next we will need a result in combinatorial set theory. A Δ\Delta-system, also called a sunflower, is a collection of finite sets A iA_i such that any two intersect in the same set: there is some “common core” SS such that A iA j=SA_i \cap A_j = S when iji \neq j.


(Sunflower lemma) Let 𝒜\mathcal{A} be an uncountable collection of finite sets. Then there is some uncountable subset 𝒜\mathcal{B} \subseteq \mathcal{A} that is a sunflower.


Without loss of generality, we may suppose all A𝒜A \in \mathcal{A} have the same cardinality nn. We argue by induction on nn. The case n=0n = 0 is trivial. Assume the result holds for nn, and suppose |A|=n+1{|A|} = n + 1 for all A𝒜A \in \mathcal{A}. If some aX=𝒜a \in X = \bigcup \mathcal{A} belongs to uncountably many A𝒜A \in \mathcal{A}, then these AA form a collection 𝒜\mathcal{A}', and by inductive hypothesis the collection {A{a}:A𝒜}\{A \setminus \{a\}: A \in \mathcal{A}'\} has a sunflower \mathcal{B}'. Then induction goes through by forming the sunflower {B+{a}:B}\{B + \{a\}: B \in \mathcal{B}'\}.

Otherwise, each aXa \in X belongs to at most countably many A𝒜A \in \mathcal{A}. In this case we may form an uncountable pairwise disjoint sequence {A α𝒜:αω 1}\{A_\alpha \in \mathcal{A}: \alpha \in \omega_1\} by transfinite induction: if A αA_\alpha have been chosen for all α:α<β<ω 1\alpha: \alpha \lt \beta \lt \omega_1, then only countably many AA have an inhabited intersection with α<βA α\bigcup_{\alpha \lt \beta} A_\alpha, and we simply choose an A βA_\beta belonging to the uncountable collection that remains after removing such AA. (So here the common core SS of the sunflower is the empty set.)


If {X i} iI\{X_i\}_{i \in I} is an arbitrary family of spaces such that any finite product of the X iX_i satisfies the countable chain condition, then the full product X= iIX iX = \prod_{i \in I} X_i also satisfies the countable chain condition.


Suppose otherwise, that {U α} αA\{U_\alpha\}_{\alpha \in A} is an uncountable collection of pairwise disjoint inhabited open sets of XX. Shrinking them if necessary, we may suppose each U αU_\alpha is a basic open of the form iF αU α,i× iF αX i\prod_{i \in F_\alpha} U_{\alpha, i} \times \prod_{i \notin F_\alpha} X_i, where F αIF_\alpha \subseteq I is a finite set. The collection {F α}\{F_\alpha\} has a sunflower by the Sunflower Lemma, say with common core SS, and without loss of generality we may suppose {F α}\{F_\alpha\} is that sunflower (i.e., that the α\alpha's used to index the sunflower are just the αA\alpha \in A: throw away any other αA\alpha \in A). Let π S:X iSX i\pi_S: X \to \prod_{i \in S} X_i be the obvious projection map. We first claim that the sets π S(U α)\pi_S(U_\alpha) are pairwise disjoint, a purely set-theoretic matter.

Suppose instead uπ S(U α)π S(U α)u \in \pi_S(U_\alpha) \cap \pi_S(U_{\alpha'}). Regard u iSX iu \in \prod_{i \in S} X_i as a section of the canonical projection iSX iS\sum_{i \in S} X_i \to S. That it belongs to π S(U α)\pi_S(U_\alpha) simply means it extends to some section σ\sigma of the canonical map iAX iA\sum_{i \in A} X_i \to A such that the restriction of σ\sigma to F αF_\alpha factors through the inclusion iF αU α,i iAX i\sum_{i \in F_\alpha} U_{\alpha, i} \hookrightarrow \sum_{i \in A} X_i, as on the left half of the following diagram:

iF αU α,i iAX i iF αU α,i v σσ w F α A F α S \array{ \sum_{i \in F_\alpha} U_{\alpha, i} & \hookrightarrow & \sum_{i \in A} X_i & \hookleftarrow & \sum_{i \in F_{\alpha'}} U_{\alpha', i} \\ \mathllap{\exists v} \uparrow & & \mathllap{\sigma} \uparrow \uparrow \mathrlap{\sigma'} & & \uparrow \mathrlap{\exists w} \\ F_\alpha & \hookrightarrow & A & \hookleftarrow & F_{\alpha'}\\ & \nwarrow & & \nearrow & \\ & & S & & }

Similarly there is a section σ:A iAX i\sigma': A \to \sum_{i \in A} X_i and a partial section ww, as shown on the right half. With the partial sections vv and ww in hand, amalgamate them to a partial section over the union F αF αF_\alpha \cup F_{\alpha'} (we can do this, as v:F α iAv: F_\alpha \to \sum_{i \in A} and w:F α iAX iw: F_{\alpha'} \to \sum_{i \in A} X_i both restrict to the same map uu on the intersection S=F αF αS = F_\alpha \cap F_{\alpha'}). Then extend the amalgamation vwv \cup w however you please to a full section σ:A iAX i\sigma'': A \to \sum_{i \in A} X_i. This σ\sigma'' belongs to both U αU_\alpha and U αU_{\alpha'}, contradicting their disjointness, and proving the claim.

Thus the π S(U α)\pi_S(U_\alpha) form an uncountable collection of open, inhabited, pairwise disjoint sets, contradicting the countable chain condition hypothesis on the finite product iSX i\prod_{i \in S} X_i. With this the proof is complete.


An arbitrary product of separable spaces X iX_i satisfies the countable chain condition.


Any finite product of separable spaces X iX_i is separable and thus satisfies the countable chain condition by Proposition , so the full product does as well by Proposition .

Axioms: axiom of choice (AC), countable choice (CC).



Last revised on March 2, 2020 at 16:17:34. See the history of this page for a list of all contributions to it.