Proof. Let $I$ be a set and for $i \in I$ let $X_ i$ be a quasi-compact topological space. Set $X = \prod X_ i$. Let $\mathcal{B}$ be the set of subsets of $X$ of the form $U_ i \times \prod _{i' \in I, i' \not= i} X_{i'}$ where $U_ i \subset X_ i$ is open. By construction this family is a subbasis for the topology on $X$. By Lemma 5.12.15 it suffices to show that any covering $X = \bigcup _{j \in J} B_ j$ by elements $B_ j$ of $\mathcal{B}$ has a finite refinement. We can decompose $J = \coprod J_ i$ so that if $j \in J_ i$, then $B_ j = U_ j \times \prod _{i' \not= i} X_{i'}$ with $U_ j \subset X_ i$ open. If $X_ i = \bigcup _{j \in J_ i} U_ j$, then there is a finite refinement and we conclude that $X = \bigcup _{j \in J} B_ j$ has a finite refinement. If this is not the case, then for every $i$ we can choose an point $x_ i \in X_ i$ which is not in $\bigcup _{j \in J_ i} U_ j$. But then the point $x = (x_ i)_{i \in I}$ is an element of $X$ not contained in $\bigcup _{j \in J} B_ j$, a contradiction. $\square$

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