6.1Local Finiteness

Updated 23 Aug 2026Chapter (PDF)

We only cover local finiteness and paracompactness to the extent necessary to generalize the mechanism of partitions of unity to spaces with a condition weaker than compactness. We also briefly state the major metrization theorems (Nagata-Smirnov and Smirnov) and their immediate consequences for manifolds, while blackboxing the highly technical proofs.

[6.1.1]Definition(Local Finiteness)#

Let $X$ be a topological space together with a collection of subsets $\mathcal{A}$ . We say $\mathcal{A}$ is locally finite if every point $x \in X$ has some open neighborhood $U$ of $x$ such that only finitely many members of $\mathcal{A}$ intersect $U$ .

[6.1.2]Example#

Note that $\mathcal{A} = \{(n, n + 1)\}_{n \in \bZ_+}$ is a locally finite collection with respect to $\bR$ . Indeed, for any $x \in X$ , the open neighborhood $(x - 0.5, x + 0.5)$ is intersected by, at most, three members of $\mathcal{A}$ . Also note that $\{(0, 1/n)\}_{n \in \bZ_+}$ is a locally finite collection on $(0, 1)$ , but not on $\bR$ (since $0$ , $1$ are limit points and thus admit countably many elements of this collection into any neighborhood of those points).

[6.1.3]Theorem(Subset and Closure Properties of Locally Finite Collections)#

Let $\mathcal{A}$ be a locally finite collection of subsets of $X$ .

Any subcollection of $\mathcal{A}$ is locally finite.
The collection $\mathcal{B} = \{\bar{A}\}_{A \in \mathcal{A}}$ is a locally finite collection.
$\bar{\bigcup_{A \in \mathcal{A}} A} = \bigcup_{A \in \mathcal{A}} \bar{A}$ .

Proof.

$(1)$ : Immediate.

$(2)$ : Let $x \in X$ be arbitrary. Choose an open neighborhood $U$ of $x$ such that the only members of $\mathcal{A}$ that intersect $U$ are $\{A_k\}_{k = 1}^n$ . Let $m \notin \{1, \dots, n\}$ such that $A_m \in \mathcal{A}$ , then note $U \cap \bar{A_m} = \emptyset$ since $U \cap A_m = \emptyset$ . Thus $U$ intersects finitely many elements of $\mathcal{B}$ .

$(3)$ : Let $Y = \bigcup_{A \in \mathcal{A}} A$ . It is a standard topological fact that $\bigcup_{A \in \mathcal{A}} \bar{A} \subseteq \bar{Y}$ . We prove the reverse inclusion. Suppose $x \in \bar{Y}$ . By local finiteness, there is some open neighborhood $U$ of $x$ that intersects only finitely many members of $\mathcal{A}$ , say $A_1, \dots, A_k$ . For the sake of contradiction, suppose $x \notin \bigcup_{A \in \mathcal{A}} \bar{A}$ . Then $x$ is not in any of $\bar{A_1}, \dots, \bar{A_k}$ . Thus $V = U \setminus (\bar{A_1} \cup \dots \cup \bar{A_k})$ is an open neighborhood of $x$ . But $V$ intersects no elements of $\mathcal{A}$ , meaning it does not intersect $Y$ , which contradicts $x \in \bar{Y}$ . Thus $x \in \bigcup_{A \in \mathcal{A}} \bar{A}$ , completing the proof.

[6.1.4]Definition(Refinement, Countably Locally Finite)#

Let $\mathcal{A}$ be a covering of $X$ . A collection $\mathcal{B}$ is a refinement of $\mathcal{A}$ if for each $B \in \mathcal{B}$ , there is some $A \in \mathcal{A}$ such that $B \subseteq A$ . If the elements of $\mathcal{B}$ are open, we call it an open refinement.

Furthermore, a collection $\mathcal{B}$ is countably locally finite (or $\sigma$ -locally finite) if it can be written as a countable union of locally finite collections.