Some basic properties of spaces with countable chain condition

A space X is said to have countable chain condition (ccc) if every disjoint collection of open subsets of X is countable. When a space has ccc, for convenience we also say that it is a ccc space. We present some basic results about ccc space as well as an equivalent condition for ccc in terms of relatively locally finite open collection. A corollary of this equivalent condition is that in the class of ccc spaces, paracompactness coincides with the Lindelof property.

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Basic Properties

Let X be a space. We have the following simple results about ccc spaces.

  1. If X is separable, then X has ccc.
  2. If X is hereditarily Lindelof, then X has ccc.
  3. The property ccc is hereditary with respect to open subspaces.
  4. Let Y be a dense subspace of X. Then Y has ccc if and only if X has ccc.

We only prove result 4.

Proof of 4
Let Y be a dense subspace of the space X. Suppose that X does not have ccc. Then there is an uncountable disjoint collection \mathcal{U} of open subsets of X. Assuming that Y is an uncountable set, the collection of all U \cap Y (where U \in \mathcal{U}) is an uncountable disjoint collection of open subsets of Y. It follows that Y does not have ccc. Thus Y has ccc implies X has ccc.

Suppose that Y does not have ccc. Then there is an uncountable disjoint collection \mathcal{V} of open subsets of Y. For each V \in \mathcal{V}, there is some V^*, open subset of X, such that V=V^* \cap Y. Let \mathcal{V}^* be the collection of all V^*. Note that \mathcal{V}^* is uncountable and is a disjoint collection of open subsets of X. It follows that X does not have ccc. Thus X has ccc implies Y has ccc. \blacksquare

Theorem 1 and Corollary 2 (stated below) are discussed in a previous post (Product of Spaces with Countable Chain Condition). Theorem 1 implies that if the ccc property is preserved by taking product of any finite number of factors, then the ccc property is preserved by taking product of any number of factors. As a corollary, it follows that the product of any number of separable spaces has ccc.

Theorem 1
Suppose that \left\{X_\alpha: \alpha \in T \right\} is a family of spaces such that \prod \limits_{\alpha \in F} X_\alpha has countable chain condition for every finite F \subset T. Then \prod \limits_{\alpha \in T} X_\alpha has countable chain condition.

Corollary 2
Suppose that \left\{X_\alpha: \alpha \in T \right\} is a family of separable spaces. Then \prod \limits_{\alpha \in T} X_\alpha has countable chain condition.

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An Equivalent Condition for Countable Chain Condition

The next theorem is an alternative way of looking at countable chain condition. Corollary 3 is an application of this equivalent condition.

Theorem 3
Let X be a space. Then the following conditions are equivalent.

  1. If \mathcal{U} is a collection of open subsets of X such that \mathcal{U} is locally countable in the open subspace \bigcup \mathcal{U}, then \mathcal{U} is countable.
  2. If \mathcal{U} is a collection of open subsets of X such that \mathcal{U} is locally finite in the open subspace \bigcup \mathcal{U}, then \mathcal{U} is countable.
  3. X has ccc.

Note that when an open collection \mathcal{U} is locally finite (locally countable) in the open subspace \bigcup \mathcal{U}, \mathcal{U} is said to be a relatively locally finite (locally countable) open collection.

Proof
The directions 1 \Rightarrow 2 and 2 \Rightarrow 3 are clear.

3 \Rightarrow 1
Suppose that X has ccc. Let \mathcal{U} be a collection of open subsets of X such that \mathcal{U} is locally countable in the open subspace \bigcup \mathcal{U}. For each U \in \mathcal{U}, choose a non-empty open set f(U) \subset U such that f(U) \cap O \ne \varnothing for at most countably many O \in \mathcal{U}. Let \mathcal{U}_f be the collection of all f(U) (over all U \in \mathcal{U}). Note that f is a countable-to-one mapping from \mathcal{U} into \mathcal{U}_f.

For H,K \in \mathcal{U}_f, a chain from H to K is any finite collection

    \left\{A_1,A_2,\cdots,A_n \right\} \subset \mathcal{U}_f

such that H=A_1, K=A_n, and for 1 \le j < n, A_j \cap A_{j+1} \ne \varnothing. For each V \in \mathcal{U}_f, let \mathcal{C}(V) be the following:

    \mathcal{C}(V)=\left\{W \in \mathcal{U}_f: \text{ there exists a chain from } V \text{ to } W \right\}

Note that every V \in \mathcal{U}_f meets only countably many sets in \mathcal{U}. Thus every V \in \mathcal{U}_f meets only countably many sets in \mathcal{U}_f. As a result, each \mathcal{C}(V) is countable. For each V \in \mathcal{U}_f, let \mathcal{E}(V)=\bigcup \mathcal{C}(V). For V_1,V_2 \in \mathcal{U}_f, if \mathcal{E}(V_1) \cap \mathcal{E}(V_2) \ne \varnothing, \mathcal{C}(V_1)=\mathcal{C}(V_2), and as a result \mathcal{E}(V_1)=\mathcal{E}(V_2). Thus the collection of all distinct \mathcal{E}(V) is a collection of disjoint open sets in X. Since X has ccc, there can be only countably many \mathcal{E}(V).

Each \mathcal{E}(V) is the union of countably many open sets, namely the open sets in \mathcal{C}(V), which is countable. Each set f(U) \in \mathcal{C}(V) is traced back to at most countably many sets in open sets U in the original collection \mathcal{U}. Since the mapping f is a countable-to-one map from \mathcal{U} to \mathcal{U}_f, \mathcal{U} is countable. \blacksquare

Corollary 4
Let X be a space with countable chain condition. Then X is a paracompact space if and only if X is a Lindelof space.

Proof
The direction \Leftarrow is the theorem that every regular Lindelof space is paracompact. See Theorem 3.8.11 and Theorem 5.1.2 in [1]. Also see Corollary 20.8 in [2].

\Rightarrow
This direction is a corollary of Theorem 3. Since X is paracompact, every open cover of the space X has a locally finite open refinement. By Theorem 3, the locally finite open refinement must be countable. \blacksquare

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

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