Theorem 41.7 in Munkres Topology

Question

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The only part I am having difficulty justifying is why there exists a $W_x \in \{W_\alpha\}$ that intersects only finitely many sets in $\{\mbox{Supp } \psi_\alpha \}$

Answer 1

I think this is a subtle point. Let $W_x$ be one of the sets in the collection $\{W_\alpha\}$ such that $x \in W_x$. Let $W_\alpha$ be another set in the same collection. For simplicity of notation, let $S_\alpha$ be the support of $\psi_\alpha$. The subtlety is that $S_\alpha$ is a closed set: it is the smallest closed set containing the open set $O_\alpha = \psi_\alpha^{-1}(\mathbb{R}\setminus\{0\})$.

We have the following relationships:

$$W_\alpha \subset \overline{W_\alpha} \subset O_\alpha \subset V_\alpha.$$

Suppose that $y \in W_x \cap S_\alpha$. If $y \in O_\alpha$, then $y \in V_x \cap V_\alpha$. If $y$ belongs to the frontier (sometimes called the boundary) of $O_\alpha$, then since $W_x$ is an open neighborhood of $y$ it must intersect both $O_\alpha$ (and its complement) and so $W_x \cap O_\alpha \neq \emptyset$; therefore $V_x \cap V_\alpha \neq \emptyset$ in this second case as well.

Since the collection $\{V_\alpha\}$ is locally finite, there can only be finitely many such pairs of indices $x$ and $\alpha$ such that $W_x \cap S_\alpha \neq \emptyset$.