Lemma 5.13.6. Let $X$ be a Hausdorff and locally quasi-compact space. Let $Z \subset X$ be a quasi-compact (hence closed) subset. Suppose given an integer $p \geq 0$, a set $I$, for every $i \in I$ an open $U_ i \subset X$, and for every $(p + 1)$-tuple $i_0, \ldots , i_ p$ of $I$ an open $W_{i_0 \ldots i_ p} \subset U_{i_0} \cap \ldots \cap U_{i_ p}$ such that

$Z \subset \bigcup U_ i$, and

for every $i_0, \ldots , i_ p$ we have $W_{i_0 \ldots i_ p} \cap Z = U_{i_0} \cap \ldots \cap U_{i_ p} \cap Z$.

Then there exist opens $V_ i$ of $X$ such that we have $Z \subset \bigcup V_ i$, for all $i$ we have $\overline{V_ i} \subset U_ i$, and we have $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$.

**Proof.**
Since $Z$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). Because $X$ is locally quasi-compact and $Z$ is quasi-compact, we can find a neighbourhood $Z \subset E$ which is quasi-compact, i.e., $E$ is quasi-compact and contains an open neighbourhood of $Z$ in $X$. If we prove the result after replacing $X$ by $E$, then the result follows. Hence we may assume $X$ is quasi-compact.

We prove the result in case $I$ is finite and $X$ is quasi-compact by induction on $p$. The base case is $p = 0$. In this case we have $X = (X \setminus Z) \cup \bigcup W_ i$. By Lemma 5.13.4 we can find a covering $X = V \cup \bigcup V_ i$ by opens $V_ i \subset W_ i$ and $V \subset X \setminus Z$ with $\overline{V_ i} \subset W_ i$ for all $i$. Then we see that we obtain a solution of the problem posed by the lemma.

Induction step. Assume the lemma proven for $p - 1$. Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of all $W_{i_0 \ldots i_ p}$ with $\{ j_0, \ldots , j_{p - 1}\} = \{ i_0, \ldots , i_ p\} $ (equality of sets). By induction there exists a solution for these opens, say $V_ i \subset U_ i$. It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ for all $(p + 1)$-tuples $i_0, \ldots , i_ p$ where $i_ a = i_ b$ for some $0 \leq a < b \leq p$. Thus we only need to modify our choice of $V_ i$ if $V_{i_0} \cap \ldots \cap V_{i_ p} \not\subset W_{i_0 \ldots i_ p}$ for some $(p + 1)$-tuple $i_0, \ldots , i_ p$ with pairwise distinct elements. In this case we have

\[ T = \overline{V_{i_0} \cap \ldots \cap V_{i_ p} \setminus W_{i_0 \ldots i_ p}} \subset \overline{V_{i_0}} \cap \ldots \cap \overline{V_{i_ p}} \setminus W_{i_0 \ldots i_ p} \]

is a closed subset of $X$ contained in $U_{i_0} \cap \ldots \cap U_{i_ p}$ not meeting $Z$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$ to “fix” the problem. After repeating this finitely many times for each of the problem tuples, the lemma is proven.
$\square$

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