Lemma 5.13.7. Let $X$ be a topological space. Let $Z \subset X$ be a quasi-compact subset such that any two points of $Z$ have disjoint open neighbourhoods in $X$. 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

1. $Z \subset \bigcup U_ i$, and

2. 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

1. $Z \subset \bigcup V_ i$,

2. $V_ i \subset U_ i$ for all $i$,

3. $\overline{V_ i} \cap Z \subset U_ i$ for all $i$, and

4. $V_{i_0} \cap \ldots \cap V_{i_ p} \subset W_{i_0 \ldots i_ p}$ 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). We prove the result in case $I$ is finite by induction on $p$.

The base case is $p = 0$. For $z \in Z \cap U_ i$ and $z' \in Z \setminus U_ i$ there exist disjoint opens $z \in V_{z, z'}$ and $z' \in W_{z, z'}$ of $X$. Since $Z \setminus U_ i$ is quasi-compact (Lemma 5.12.3), we can choose a finite nunber $z'_1, \ldots , z'_ r$ such that $Z \setminus U_ i \subset W_{z, z'_1} \cup \ldots \cup W_{z, z'_ r}$. Then we see that $V_ z = V_{z, z'_1} \cap \ldots \cap V_{z, z'_ r} \cap U_ i$ is an open neighbourhood of $z$ contained in $U_ i$ with the property that $\overline{V_ z} \cap Z \subset U_ i$. Since $z$ and $i$ where arbitrary and since $Z$ is quasi-compact we can find a finite list $z_1, i_1, \ldots , z_ t, i_ t$ and opens $V_{z_ j} \subset U_{i_ j}$ with $\overline{V_{z_ j}} \cap Z \subset U_{i_ j}$ and $Z \subset \bigcup V_{z_ j}$. Then we can set $V_ i = W_ i \cap (\bigcup _{j : i = i_ j} V_{z_ j})$ to solve the problem for $p = 0$.

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 $X$ not meeting $Z$ by our property (3) of the opens $V_ i$. 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. $\square$

Comment #4234 by Kazuki Masugi on

"Then we see that $V_z=U_{z,z′1}\cap …\cap U_{z,z′r}\cap U_i$ " should be "Then we see that $V_z=V_{z,z′1}\cap …\cap V_{z,z′r}\cap U_i$".

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