The collection of closures of each set in a locally finite collection of subsets of a topological space is locally finite.

general-topology

Let $\mathcal{X}$ be a locally finite collection of subsets of a topological space $M$. Show that the collection $\{\overline{X}:X\in \mathcal{X}\}$ is also locally finite.

By definition of local finiteness, for each $p\in M$, we can find a neighborhood $U$ of $p$ such that $U$ intersects only finitely many $X$ in $\mathcal{X}$. We shall denote the collection of subsets that $U$ intersects by $\{X_{1},...,X_n\}$. We shall denote the collection of subsets that $U$ does not intersect by $\mathcal{C}$, where $\mathcal{C}$ is defined to be $\mathcal{X}-\{X_1,...,X_n\}$.

Since $X_i \subseteq \overline{X_i}$ for $1 \leq i \leq n$, it follows that $U$ intersects each set in $\{\overline{X_1},...,\overline{X_n}\}$. By definition of local finiteness, it follows that $U$ is disjoint with each set in $\mathcal{C}$.

For each $C\in \mathcal{C}$, the closure $\overline{C}$ is defined to be the intersection of all closed subsets of $M$ containing $C$. My trouble is that each closed subset of $M$ containing $C$ may include a point that is also in $U$. If we can find such a point for each $C\in \mathcal{C}$, then the collection $\{\overline{X}:X\in \mathcal{X}\}$ is not locally finite. If we can prove such a point exists only for a finite number of $C\in \mathcal{C}$, then $\{\overline{X}:X\in \mathcal{X}\}$ would be locally finite.

The closure $\overline{C}$ could also be defined in terms of limit points. We know that $C\cap U=\emptyset$ for all $C\in \mathcal{C}$, so we are only concerned with whether a limit point of $C$ is a point of $U$.

We assume for the sake of contradiction that some point $x$ is a limit point of $C$ and a point of $U$. This implies that $U$ is a neighborhood of $x$. It makes sense that $U$ must also contain at least one point of $C$, which would yield a contradiction. However, I am unsure how to make this reasoning mathematically rigorous.

If the open $U$ does not intersect $A$, then $U$ also does not intersect $\overline{A}$ (we know that $A \subseteq X\setminus U$, which is closed, so $\overline{A} \subseteq X\setminus U$ too, by your definition of closure of $A$, and so $\overline{A} \cap U = \emptyset$ too). So $U$ can only intersect those same $\overline{X} \in \mathcal{X}$ that $U$ already did.