The Stacks project

Proof. Set $X^* = X \amalg \{ \infty \} $. We declare a subset $V$ of $X^*$ to be open if either $V \subset X$ is open in $X$, or $\infty \in V$ and $U = V \cap X$ is an open of $X$ such that $X \setminus U$ is quasi-compact. We omit the verification that this defines a topology. It is clear that $X \to X^*$ identifies $X$ with an open subspace of $X$.

Since $X$ is locally quasi-compact, every point $x \in X$ has a quasi-compact neighbourhood $x \in E \subset X$. Then $E$ is closed (Lemma 5.12.4 part (1)) and $V = (X \setminus E) \amalg \{ \infty \} $ is an open neighbourhood of $\infty $ disjoint from the interior of $E$. Thus $X^*$ is Hausdorff.

Let $X^* = \bigcup V_ i$ be an open covering. Then for some $i$, say $i_0$, we have $\infty \in V_{i_0}$. By construction $Z = X^* \setminus V_{i_0}$ is quasi-compact. Hence the covering $Z \subset \bigcup _{i \not= i_0} Z \cap V_ i$ has a finite refinement which implies that the given covering of $X^*$ has a finite refinement. Thus $X^*$ is quasi-compact.

The map $X \to X^*$ factors as $X \to \beta (X) \to X^*$ by the universal property of the Stone-Čech compactification. Let $\varphi : \beta (X) \to X^*$ be this factorization. Then $X \to \varphi ^{-1}(X)$ is a section to $\varphi ^{-1}(X) \to X$ hence has closed image (Lemma 5.3.3). Since the image of $X \to \beta (X)$ is dense we conclude that $X = \varphi ^{-1}(X)$. $\square$