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.
Combination of [I, p. 75, Lemme 1, Bourbaki] and [I, p. 76, Corrolaire 1, Bourbaki].
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:
All subsets of Z\setminus \{ I\} .
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
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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