Collection $\mathcal B=${$\bar A:A\in \mathcal A$} is locally finite,given that $\mathcal A$ is locally finite.

Question

Let $\mathcal A$ be a locally finite collection of subsets of X.Then the collection $\mathcal B=${$\bar A:A\in \mathcal A$} of the closures of the element of $\mathcal A$ is locally finite.

Proof:

Let $x\in X$ and $\mathcal A$ is locally finite, so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $\mathcal A$.

More precisely,$A_i\cap U\neq \phi$,where $1\le i \le n $----------------$(1)$

Since,$A_i\subset \bar A_i$,so from $(1)$,we have $\bar A_i\cap U\neq \phi$,where $1\le i \le n $.

Hence,$U$ can intersect an atmost the same number of elements of $\mathcal B$ as that of $\mathcal A$.

Is the proof correct?

Answer 1

Essential is here that for $U$ you can always choose an open neighborhood.

In that situation we have the implication:$$U\cap\overline A\neq\varnothing\implies U\cap A\neq\varnothing\tag1$$

Also the converse if this is true, but that is not relevant here.

$(1)$ assures us that the cardinality of the set $\{\overline A\mid A \in\mathcal A\text{ and } U\cap \overline A\neq\varnothing\}$ will not exceed the set the cardinality of the set $\{A\in\mathcal A\mid U\cap A\neq\varnothing\}$.

So if $\{A\in\mathcal A\mid U\cap A\neq\varnothing\}$ is a finite set then so is $\{\overline A\mid A\in\mathcal A\text{ and } U\cap \overline A\neq\varnothing\}$ (QED).