Contents

# Contents

## Idea

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.

## Definitions

###### Definition

Suppose $P$ is a poset with a bottom element $0$. A (strong) antichain in $P$ is a subset $A \subseteq P$ such that for all $a, b \in A$, we have either $a = b$ or $0$ is their meet (we need not assume general meets exist in $P$). If $P$ is a poset without a bottom, we define $A \subseteq P$ to be an antichain if it becomes an antichain in $P^+$, the poset obtained by freely adjoining a bottom to $P$; this is equivalent to saying $a, b$ have no lower bound in $P$.

###### Definition

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

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

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

###### Lemma

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

###### Proof

The inclusion $Reg(H) \hookrightarrow H$ preserves meets and $0$, so if every antichain in $H$ is countable, then the same is true in $Reg(H)$.

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

### Chains, not antichains

###### Proposition

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

• maps $g: \kappa \to B \setminus \{0\}$ such that $\{g(\alpha): \alpha \lt \kappa\}$ is an antichain in $B$;

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

here regarding $\kappa$ as an ordinal.

###### Proof

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

###### Corollary

$B$ satisfies the countable chain condition iff every chain $b_0 \lt b_1 \lt \ldots$ in $B$ 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 $X$, the frame $H = Op(X)$ satisfies the countable chain condition iff $Reg(H)$, 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 $\neg U$, which for an open set $U$ is the complement of its closure. Thus for the topological case we deduce the following formulation in terms of chains:

###### Proposition

A space $X$ satisfies the countable chain condition if and only if every collection of open sets $\{U_\alpha\}_{\alpha \lt \kappa}$ that satisfies the condition

• $\widebar{U_\alpha} \subset \widebar{U_\beta}$ (strict inclusion between closures) whenever $\alpha \lt \beta$

is countable.

## Products of separable spaces

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

###### Proposition

A separable space satisfies the countable chain condition.

###### Proof

Let $D$ be a countable dense set, and suppose given a collection of pairwise disjoint inhabited open sets $\{U_\alpha: \alpha \in A\}$. Each $U_\alpha$ contains at least one $d \in D$. Thus the map $D \to A$ sending $d \in D$ to the unique $\alpha \in A$ such that $d \in U_\alpha$ is surjective, and so $A$ is at most countable.

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

###### Lemma

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

###### Proof

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

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

###### Proposition

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

###### Proof

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

Suppose instead $u \in \pi_S(U_\alpha) \cap \pi_S(U_{\alpha'})$. Regard $u \in \prod_{i \in S} X_i$ as a section of the canonical projection $\sum_{i \in S} X_i \to S$. That it belongs to $\pi_S(U_\alpha)$ simply means it extends to some section $\sigma$ of the canonical map $\sum_{i \in A} X_i \to A$ such that the restriction of $\sigma$ to $F_\alpha$ factors through the inclusion $\sum_{i \in F_\alpha} U_{\alpha, i} \hookrightarrow \sum_{i \in A} X_i$, as on the left half of the following diagram:

$\array{ \sum_{i \in F_\alpha} U_{\alpha, i} & \hookrightarrow & \sum_{i \in A} X_i & \hookleftarrow & \sum_{i \in F_{\alpha'}} U_{\alpha', i} \\ \mathllap{\exists v} \uparrow & & \mathllap{\sigma} \uparrow \uparrow \mathrlap{\sigma'} & & \uparrow \mathrlap{\exists w} \\ F_\alpha & \hookrightarrow & A & \hookleftarrow & F_{\alpha'}\\ & \nwarrow & & \nearrow & \\ & & S & & }$

Similarly there is a section $\sigma': A \to \sum_{i \in A} X_i$ and a partial section $w$, as shown on the right half. With the partial sections $v$ and $w$ in hand, amalgamate them to a partial section over the union $F_\alpha \cup F_{\alpha'}$ (we can do this, as $v: F_\alpha \to \sum_{i \in A}$ and $w: F_{\alpha'} \to \sum_{i \in A} X_i$ both restrict to the same map $u$ on the intersection $S = F_\alpha \cap F_{\alpha'}$). Then extend the amalgamation $v \cup w$ however you please to a full section $\sigma'': A \to \sum_{i \in A} X_i$. This $\sigma''$ belongs to both $U_\alpha$ and $U_{\alpha'}$, contradicting their disjointness, and proving the claim.

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

###### Corollary

An arbitrary product of separable spaces $X_i$ satisfies the countable chain condition.

###### Proof

Any finite product of separable spaces $X_i$ 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).

### Implications

• a metric space has a $\sigma$-locally discrete base

• Nagata-Smirnov metrization theorem

• a second-countable space has a $\sigma$-locally finite base: take the the collection of singeltons of all elements of a countable cover of $X$.

• second-countable spaces are separable: use the axiom of countable choice to choose a point in each set of a countable cover.

• separable spaces satisfy the countable chain condition: given a dense set $D$ and a family $\{U_\alpha : \alpha \in A\}$, the map $D \cap \bigcup_{\alpha \in A} U_\alpha \to A$ assigning $d$ to the unique $\alpha \in A$ with $d \in U_\alpha$ is surjective.

• separable spaces are weakly Lindelöf: given a countable dense subset and an open cover choose for each point of the subset an open from the cover.

• Lindelöf spaces are trivially also weakly Lindelöf.

• a space with a $\sigma$-locally finite base is first countable: obviously, every point is contained in at most countably many sets of a $\sigma$-locally finite base.

• a first-countable space is obviously Fréchet-Urysohn.

• a Fréchet-Uryson space is obviously sequential.

• a sequential space is obviously countably tight.

Last revised on March 2, 2020 at 11:17:34. See the history of this page for a list of all contributions to it.