Lemma 5.13.4. Let $X$ be a Hausdorff and quasi-compact space. Let $X = \bigcup _{i \in I} U_ i$ be an open covering. Then there exists an open covering $X = \bigcup _{i \in I} V_ i$ such that $\overline{V_ i} \subset U_ i$ for all $i$.

**Proof.**
Let $x \in X$. Choose an $i(x) \in I$ such that $x \in U_{i(x)}$. Since $X \setminus U_{i(x)}$ and $\{ x\} $ are disjoint closed subsets of $X$, by Lemmas 5.12.3 and 5.12.4 there exists an open neighbourhood $U_ x$ of $x$ whose closure is disjoint from $X \setminus U_{i(x)}$. Thus $\overline{U_ x} \subset U_{i(x)}$. Since $X$ is quasi-compact, there is a finite list of points $x_1, \ldots , x_ m$ such that $X = U_{x_1} \cup \ldots \cup U_{x_ m}$. Setting $V_ i = \bigcup _{i = i(x_ j)} U_{x_ j}$ the proof is finished.
$\square$

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