4.18 Paracompact Spaces
Definition 4.18.1 (Locally Finite). Let $X$ be a topological space and $\mathcal{U}\subset 2^{X}$, then $\mathcal{U}$ is locally finite if for every $x \in X$, there exists $V \in \cn(x)$ such that $\bracs{U \in \mathcal{U}| V \cap U \ne \emptyset}$ is finite.
Lemma 4.18.2. Let $X$ be a topological space, $\mathcal{U}\subset 2^{X}$ be locally finite, and $K \subset X$ compact, then $\bracs{U \in \mathcal{U}|U \cap K \ne \emptyset}$ is finite.
Proof. For each $x \in K$, there exists $N_{x} \in \cn(x)$ such that $\bracs{U \in \mathcal{U}|U \cap N_x \ne \emptyset}$ is finite. By compactness of $K$, there exists $X_{K} \subset X$ finite such that $K \subset \bigcup_{x \in X_K}N_{x}$. In which case,
Lemma 4.18.3. Let $X$ be a topological space, $\mathcal{U}\subset 2^{X}$ be locally finite, then $\bracsn{\ol{U}|U \in \mathcal{U}}$ is also locally finite.
Proof. For each $x \in X$, there exists $N_{x} \in \cn^{o}(x)$ such that $\bracs{U \in \mathcal{U}|N_x \cap U \ne \emptyset}$ is finite. Since $N_{x}$ is open, for any $U \in \mathcal{U}$, $N_{x} \cap U = \emptyset$ implies that $N_{x}^{c} \supset \ol{U}$. Thus $\bracsn{U \in \mathcal{U}|N_x \cap \ol U \ne \emptyset}$ is finite as well.$\square$
Definition 4.18.4 (Refinement). Let $X$ be a topological space and $\mathcal{U}, \mathcal{V}\subset 2^{X}$ be open covers, then $\mathcal{V}$ is a refinement of $\mathcal{U}$ if for every $V \in \mathcal{V}$, there exists $U \in \mathcal{U}$ such that $V \subset U$.
Definition 4.18.5 (Paracompact). Let $X$ be a topological space, then $X$ is paracompact if every open cover of $X$ admits a locally finite refinement.