## 5.25 Stone-Čech compactification

The Stone-Čech compactification of a topological space $X$ is a map $X \to \beta (X)$ from $X$ to a Hausdorff quasi-compact space $\beta (X)$ which is universal for such maps. We prove this exists by a standard argument using the following simple lemma.

Lemma 5.25.1. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that $f(X)$ is dense in $Y$ and that $Y$ is Hausdorff. Then the cardinality of $Y$ is at most the cardinality of $P(P(X))$ where $P$ is the power set operation.

Proof. Let $S = f(X) \subset Y$. Let $\mathcal{D}$ be the set of all closed domains of $Y$, i.e., subsets $D \subset Y$ which equal the closure of its interior. Note that the closure of an open subset of $Y$ is a closed domain. For $y \in Y$ consider the set

$I_ y = \{ T \subset S \mid \text{ there exists } D \in \mathcal{D}\text{ with }T = S \cap D\text{ and }y \in D\} .$

Since $S$ is dense in $Y$ for every closed domain $D$ we see that $S \cap D$ is dense in $D$. Hence, if $D \cap S = D' \cap S$ for $D, D' \in \mathcal{D}$, then $D = D'$. Thus $I_ y = I_{y'}$ implies that $y = y'$ because the Hausdorff condition assures us that we can find a closed domain containing $y$ but not $y'$. The result follows. $\square$

Let $X$ be a topological space. By Lemma 5.25.1, there is a set $I$ of isomorphism classes of continuous maps $f : X \to Y$ which have dense image and where $Y$ is Hausdorff and quasi-compact. For $i \in I$ choose a representative $f_ i : X \to Y_ i$. Consider the map

$\prod f_ i : X \longrightarrow \prod \nolimits _{i \in I} Y_ i$

and denote $\beta (X)$ the closure of the image. Since each $Y_ i$ is Hausdorff, so is $\beta (X)$. Since each $Y_ i$ is quasi-compact, so is $\beta (X)$ (use Theorem 5.14.4 and Lemma 5.12.3).

Let us show the canonical map $X \to \beta (X)$ satisfies the universal property with respect to maps to Hausdorff, quasi-compact spaces. Namely, let $f : X \to Y$ be such a morphism. Let $Z \subset Y$ be the closure of $f(X)$. Then $X \to Z$ is isomorphic to one of the maps $f_ i : X \to Y_ i$, say $f_{i_0} : X \to Y_{i_0}$. Thus $f$ factors as $X \to \beta (X) \to \prod Y_ i \to Y_{i_0} \cong Z \to Y$ as desired.

Lemma 5.25.2. Let $X$ be a Hausdorff, locally quasi-compact space. There exists a map $X \to X^*$ which identifies $X$ as an open subspace of a quasi-compact Hausdorff space $X^*$ such that $X^* \setminus X$ is a singleton (one point compactification). In particular, the map $X \to \beta (X)$ identifies $X$ with an open subspace of $\beta (X)$.

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 $X = \varphi ^{-1}(X)$. $\square$

## Comments (2)

Comment #6004 by Chen-Yu Chi on

In the 5th line of the proof of Lemma 0909, it is said that "...we see that $S\cap D$ is dense in $D$. Hence, ..." How do we conclude the part after "Hence" from the density of $S\cap D$ \underline{in $D$}? Should one replace "$S\cap D$ is dense in $D$" by "$\overline{S\cap D}=D$"?

Comment #6008 by Fan Zheng on

They are equivalent.

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