Lemma 3.12.3 (Locally Finite Closure).label Let $X$ be a topological space, $\cali{U}\suf 2^{X}$ locally finite then $\curl{\cl{U}:U\in\cali{U}}$ is locally finite.
Proof. For each $x\in X$, there exists $N_{x}\in\cali{N}^{o}(x)$ such that $\curl{U\in\cali{U}:N_x\cap U\neq\emp}$ is finite. Since $N_{x}$ is open, for any $U\in\cali{U}$ we have
\begin{align*}N_{x}\cap U=\emp\implies \cl{U}\suf N_{x}^{c}\implies N_{x}\cap\cl{U}=\emp\end{align*}
Thus $\curl{U\in\cali{U}:N_x\cap U\neq\emp}=\curl{U\in\cali{U}:N_x\cap\cl{U}\neq\emp}$ is finite.$\square$
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