Lemma 5.12.4. Let X be a Hausdorff topological space.
If E \subset X is quasi-compact, then it is closed.
If E_1, E_2 \subset X are disjoint quasi-compact subsets then there exists opens E_ i \subset U_ i with U_1 \cap U_2 = \emptyset .
Lemma 5.12.4. Let X be a Hausdorff topological space.
If E \subset X is quasi-compact, then it is closed.
If E_1, E_2 \subset X are disjoint quasi-compact subsets then there exists opens E_ i \subset U_ i with U_1 \cap U_2 = \emptyset .
Proof. Proof of (1). Let x \in X, x \not\in E. For every e \in E we can find disjoint opens V_ e and U_ e with e \in V_ e and x \in U_ e. Since E \subset \bigcup V_ e we can find finitely many e_1, \ldots , e_ n such that E \subset V_{e_1} \cup \ldots \cup V_{e_ n}. Then U = U_{e_1} \cap \ldots \cap U_{e_ n} is an open neighbourhood of x which avoids V_{e_1} \cup \ldots \cup V_{e_ n}. In particular it avoids E. Thus E is closed.
Proof of (2). In the proof of (1) we have seen that given x \in E_1 we can find an open neighbourhood x \in U_ x and an open E_2 \subset V_ x such that U_ x \cap V_ x = \emptyset . Because E_1 is quasi-compact we can find a finite number x_ i \in E_1 such that E_1 \subset U = U_{x_1} \cup \ldots \cup U_{x_ n}. We take V = V_{x_1} \cap \ldots \cap V_{x_ n} to finish the proof. \square
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