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Let $X$ be a topological space and $\mathscr{U}$ a locally finite and disjoint collection of closed sets in $X$. We need to show $\mathscr{U}$ is a discrete collection.

So pick some $x \in X$. Then by local finiteness, there is a neighborhood $U$ that touches $U_1,..., U_n \in \mathscr{U}$. If $x$ is in none of the $U_i$, then for each $U_i$, we have a neighborhood $G_i$ of $x$ that is disjoint from $U_i$ by definition of closed set. Then consider the neighborhood $U \cap G_1 \cap ... \cap G_n$. This will touch no elements from $\mathscr{U}$. WLOG assume $x$ is in $U_1$. Then since $\mathscr{U}$ is disjoint, $x$ is not in $U_2,..., U_n$. Then we construct those same $G_i$ as above and consider the neighborhood $U \cap G_2 \cap ... \cap G_n$. This will be a neighborhood of $x$ that only touches $U_1 \in \mathscr{U}$. Thus, we conclude that $\mathscr{U}$ is discrete.

Any feedback would be greatly appreciated!

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Your argument is fine.

It is also possible to prove the result without any division into cases by showing that the union of a locally finite family of closed sets is closed. Once you’ve done that, you can argue as follows. Let $x\in X$ be arbitrary, and let $\mathscr{C}=\{U\in\mathscr{U}:x\notin U\}$; then $\mathscr{C}$ is locally finite, so $X\setminus\bigcup\mathscr{C}$ is an open nbhd of $x$ disjoint from every member of $\mathscr{U}$ that does not contain $x$, and the result follows immediately.

The argument that $\bigcup\mathscr{C}$ is closed is just your first case applied to $\mathscr{C}$ instead of $\mathscr{U}$.

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  • $\begingroup$ Oh great, thank you for the other insight! $\endgroup$ – Good Morning Captain Apr 3 at 19:22
  • $\begingroup$ @GoodMorningCaptain: You’re welcome! $\endgroup$ – Brian M. Scott Apr 3 at 19:23

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