Lemma 5.17.1 (Tube lemma). Let $X$ and $Y$ be topological spaces. Let $A \subset X$ and $B \subset Y$ be quasi-compact subsets. Let $A \times B \subset W \subset X \times Y$ with $W$ open in $X \times Y$. Then there exists opens $A \subset U \subset X$ and $B \subset V \subset Y$ such that $U \times V \subset W$.
Proof. For every $a \in A$ and $b \in B$ there exist opens $U_{(a, b)}$ of $X$ and $V_{(a, b)}$ of $Y$ such that $(a, b) \in U_{(a, b)} \times V_{(a, b)} \subset W$. Fix $b$ and we see there exist a finite number $a_1, \ldots , a_ n$ such that $A \subset U_{(a_1, b)} \cup \ldots \cup U_{(a_ n, b)}$. Hence
\[ A \times \{ b\} \subset (U_{(a_1, b)} \cup \ldots \cup U_{(a_ n, b)}) \times (V_{(a_1, b)} \cap \ldots \cap V_{(a_ n, b)}) \subset W. \]
Thus for every $b \in B$ there exists opens $U_ b \subset X$ and $V_ b \subset Y$ such that $A \times \{ b\} \subset U_ b \times V_ b \subset W$. As above there exist a finite number $b_1, \ldots , b_ m$ such that $B \subset V_{b_1} \cup \ldots \cup V_{b_ m}$. Then we win because $A \times B \subset (U_{b_1} \cap \ldots \cap U_{b_ m}) \times (V_{b_1} \cup \ldots \cup V_{b_ m})$. $\square$
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