Lemma 5.12.4. Let $X$ be a Hausdorff topological space.

1. If $E \subset X$ is quasi-compact, then it is closed.

2. 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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