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

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.

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.

• This is good. I just worked your proof out as well. Thank you. Commented Jul 3, 2015 at 14:23
• @mrmingus: You’re welcome. Commented Jul 3, 2015 at 19:39
• @BrianM.Scott I hope you don't mind, but I have a question about the OP that I would like to direct towards you, since I am not sure mrmingus still visits us. It isn't clear to me why $U_x \subseteq \bigcup_{i=1}^n A_i$. What guarantees this? How does one go from having $U_x$ intersect finitely many $A_i$'s to being entirely covered by finitely many of them? Commented Dec 30, 2016 at 13:04
• @user193319: It follows from the fact that $\bigcup_\alpha A_\alpha=X$. Thus, $U_x\subseteq\bigcup_\alpha A_\alpha$. Sets $A_\alpha$ that don’t intersect $U_x$ clearly don’t help to cover $U_x$, so $U_x\subseteq\bigcup\{A_\alpha:U_x\cap A_\alpha\ne\varnothing\}$. Commented Dec 31, 2016 at 20:55
• @BrianM.Scott I am still having trouble seeing why $\{A_\alpha ~|~ U_x \cap A_\alpha \neq \emptyset \}$ is finite. There is no disputing that $U_x$ is contained in $\bigcup \{A_\alpha ~|~ U_x \cap A_\alpha \neq \emptyset \}$, but I don't see what guarantees that this is a finite union, which is a crucial step in mrmingus' proof. I like mrmingus' proof. It is very nearly what I came up with, so it is very intuitive to me, but I am just having trouble with this one step. Commented Jan 2, 2017 at 13:40