The Stacks project

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

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. $\square$