Lemma 3.12.2 (Locally Finite Compact).label Let $X$ be a topological space, $\cali{U}\suf 2^{X}$ locally finite and $K\suf X$ compact then $\curl{U\in\cali{U}:U\cap K\neq \emp}$ is finite.
Proof. For each $x\in K$, there exists $N_{x}\in \cali{N}(x)$ such that $\curl{U\in\cali{U}:U\cap N_x\neq\emp}$ is finite. By compactness of $K$, there exists $X_{K}\suf X$ finite such that $K\suf \cups{}{x\in X_K}N_{x}$. In which case
\begin{align*}{U\in\cali{U}:U\cap K\neq \emp}\suf \cups{}{x\in X_K}\curl{U\in\cali{U}:U\cap N_x\neq\emp}\end{align*}
$\square$
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