Munkres, Chapter 2, question on locally finite family of sets

Question

I've been working through the Munkres Topology text on my own, and I am not sure if the following argument is correct. Fishing around the internet a bit for some alternative answers and it looks like the approaches others have taken to this problem are not the same way that I approached it, so I'm concerned that I'm making some conceptual errors. The problem:

Let {A$_{\alpha}$} be a collection of subsets of X. Let X=$\cup_{\alpha}$A$_{\alpha}$. Let f: X$\rightarrow$Y; suppose that f$\mid$A$_{\alpha}$ is continuous for each $\alpha$. An indexed family of sets {A$_{\alpha}$} is said to be locally finite if each point of X has a neighborhood that intersects A$_{\alpha}$ for only finitely many values of $\alpha$. Show that if the family {A$_{\alpha}$} is locally finite and each A$_{\alpha}$ is closed, then f is continuous.

Attempt at solution:

Let x $\in$ X. Since there exists a neighborhood U$_{x}$ of x that intersects A$_{\alpha}$ for only finitely many $\alpha$, U$_{x}$ can be contained in a finite collection of A$_{\alpha}$'s. So x $\in$ U$_{x}$ $\subset$ $\cup_{i=1}^{n}$A$_{i}$ $\subset$ X. Then, since each A$_{i}$ is closed and f$\mid$A$_{i}$ continuous for each i, by repeated application of the Pasting Lemma, f$\mid$$\cup_{i=1}^{n}$A$_{i}$ is continuous. Then, since U$_{x}$ $\subset$ $\cup_{i=1}^{n}$A$_{i}$, by restricting the domain (Theorem 18.2d), f$\mid$U$_{x}$ continuous. Since it is given that such a neighborhood U$_{x}$ exists for each x $\in$ X, then X can be written as the union of open sets U$_{x}$, where f$\mid$U$_{x}$ is continuous for each x, and so by Theorem 18.2f (local formulation of continuity), f: X$\rightarrow$Y is continuous.

Feel like I'm missing something, or making some incorrect assumptions. Any help would be appreciated.

Answer 1

Your argument is fine. An alternative that you may have encountered is first to prove the very useful fact that locally finite families are closure-preserving, meaning that if $\mathscr{C}$ is any locally finite family of sets, $\operatorname{cl}\bigcup\mathscr{C}=\bigcup_{C\in\mathscr{C}}\operatorname{cl}C$.

Then let $F$ be any closed subset of $Y$, and for each $\alpha\in A$ let $C_\alpha=(f\upharpoonright A_\alpha)^{-1}[F]\subseteq A_\alpha$. Then $C_\alpha$ is closed in $A_\alpha$ (since $f\upharpoonright A_\alpha$ is continuous) and hence in $X$. Moreover, $\mathscr{C}=\{C_\alpha:\alpha\in A\}$ is easily seen to be locally finite, so

$$f^{-1}[F]=\bigcup_{\alpha\in A}C_\alpha=\bigcup_{\alpha\in A}\operatorname{cl}C_\alpha=\operatorname{cl}\bigcup_{\alpha\in A}C_\alpha=\operatorname{cl}F\;,$$

and $F$ is closed.