Combination of [I, p. 75, Lemme 1, Bourbaki] and [I, p. 76, Corrolaire 1, Bourbaki].

Lemma 5.17.3. A topological space $X$ is quasi-compact if and only if the projection map $Z \times X \to Z$ is closed for any topological space $Z$.

Proof. (See also remark below.) If $X$ is not quasi-compact, there exists an open covering $X = \bigcup _{i \in I} U_ i$ such that no finite number of $U_ i$ cover $X$. Let $Z$ be the subset of the power set $\mathcal{P}(I)$ of $I$ consisting of $I$ and all nonempty finite subsets of $I$. Define a topology on $Z$ with as a basis for the topology the following sets:

1. All subsets of $Z\setminus \{ I\}$.

2. For every finite subset $K$ of $I$ the set $U_ K := \{ J\subset I \mid J \in Z, \ K\subset J \} )$.

It is left to the reader to verify this is the basis for a topology. Consider the subset of $Z \times X$ defined by the formula

$M = \{ (J, x) \mid J \in Z, \ x \in \bigcap \nolimits _{i \in J} U_ i^ c)\}$

If $(J, x) \not\in M$, then $x \in U_ i$ for some $i \in J$. Hence $U_{\{ i\} } \times U_ i \subset Z \times X$ is an open subset containing $(J, x)$ and not intersecting $M$. Hence $M$ is closed. The projection of $M$ to $Z$ is $Z-\{ I\}$ which is not closed. Hence $Z \times X \to Z$ is not closed.

Assume $X$ is quasi-compact. Let $Z$ be a topological space. Let $M \subset Z \times X$ be closed. Let $z \in Z$ be a point which is not in $\text{pr}_1(M)$. By the Tube Lemma 5.17.1 there exists an open $U \subset Z$ such that $U \times X$ is contained in the complement of $M$. Hence $\text{pr}_1(M)$ is closed. $\square$

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