# CCC + Paracompact => Lindelof

It is a well known fact in general topology that in the class of spaces with the countable chain condition (ccc), paracompactness equals the Lindelof property. I would like to write down a proof for this fact, proving that paracompact space with the ccc is Lindelof. A space has the countable chain condition (ccc) if every pairwise disjoint family of open sets in the space is countable. I would like to make an observation about two classes of ccc spaces. One is the class of separable spaces. The other is the product of separable spaces. In particular, the product of real lines has the ccc. So $\mathbb{R}^\mathcal{K}$ has the ccc for any cardinal $\mathcal{K}$. The countable chain condition is hereditary with respect to dense subspaces. The function space $C_p(X)$, the real-valued continuous function space with the pointwise convergence topology, is always a dense subspace of $\mathbb{R}^X$. Thus $C_p(X)$ always has the ccc. For this class of function spaces, there is no distinction between paracompactness and the Lindelof property (if it is paracompact, it is Lindelof).

Theorem. If a paracompact space $X$ has the ccc, then $X$ is paracompact.

This theorem is established after the following lemma is proved.

Lemma. In a space with the ccc, any locally finite open cover is countable.

Proof of Lemma. Let $X$ be a space with the countable chain condition (ccc). Let $\mathcal{V}$ be a locally finite open cover of $X$. For each $V \in \mathcal{V}$, let $O(V) \subset V$ be open such that $O(V)$ meets at most finitely many open sets in $\mathcal{V}$. Let $\mathcal{O}=\lbrace{O(V):V \in \mathcal{V}}\rbrace$.

For $E, F \in \mathcal{O}$, a finite collection $\lbrace{G_1,G_2,...,G_n}\rbrace \subset \mathcal{O}$ is said to be a chain from $E$ to $F$ if $E=G_1$, $F=G_n$, and $G_i \cap G_{i+1} \neq \phi$ for $1 \leq i < n$. For each $E \in \mathcal{O}$, let $T(E)=\lbrace{F \in \mathcal{O}:}$ there is a chain from $E$ to $F\rbrace$. Clearly each $T(E)$ is a countable collection. For each $E \in \mathcal{O}$, let $S(E)=\bigcup T(E)$. For $E_1,E_2 \in \mathcal{O}$, if $S(E_1) \cap S(E_2) \neq \phi$, then there is a chain from $E_1$ to $E_2$ and there is a chain from $E_2$ to $E_1$. This means $T(E_1)=T(E_2)$. Consider the distinct elements $S(E)$ where $E \in \mathcal{O}$. These open sets form a pairwise disjoint collection of open sets in $X$. By the ccc, there are only countably many distinct $S(E)$.

Let $\mathcal{W}$ be the countable collection of all distinct open sets $S(E)$. Each $S(E) \in \mathcal{W}$ is associated with countably many $E^* \in \mathcal{O}$. In turn, each $E^*=O(V)$ for some $V \in \mathcal{V}$. There is a one-to-countable mapping from $\mathcal{W}$ onto $\mathcal{V}$. Since $\mathcal{W}$ is countable, $\mathcal{V}$ is countable.