# A locally finite, disjoint collection of closed subsets is discrete proof verification

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!

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 $$\mathscr{U}$$.