# Existence of the Cech-Stone compactification

I've been reading about compactifications, and I encountered a theorem about the Cech-Stone compactification. I've seen the proof of this theorem in other textbooks, but difference definitions and constructions were used. In this text, they give the definition of a compactification, and then the Cech-Stone compactification of a completely regular space $X$ is defined as the compactification which is maximal in the $\leq$ ordering. The $\leq$ ordering on the family of compactifications on $X$ is defined as: $cX \leq dX$ if and only if there is a continuous and onto $f:dX \rightarrow cX$ which fixes all points in $X$.

Theorem: The Cech-Stone compactification exists for every completely regular space $X$.

They provide a sketch of the proof, but I am having trouble filling in the gaps.

Sketch of Proof: Let $\mathcal{C}_{X}$ be the set of all compactifications on $X$, then $\prod \mathcal{C}_{X}$ is a compact space. Embed $X$ into $\prod \mathcal{C}_{X}$ by $\phi$: $x \mapsto \{ c(x) : cX \in \mathcal{C}_X \}$. Then, the closure of this image of $X$ is as required.

I don't understand why the mapping is an embedding. Also, I'm having trouble understanding why the closure of the image of $X$ ends up being the Cech-Stone compactification.

Any help would be greatly appreciated!

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There’s a bit of sloppiness right off the bat in the definition of $\mathscr{C}_X$ as the set of all compactifications of $X$: they really mean that $\mathscr{C}_X$ is a set containing one representative of each homeomorphism class of compactifications of $X$. That is, each member of $\mathscr{C}_X$ is a compactification of $X$, and every compactification of $X$ is homeomorphic to exactly one member of $\mathscr{C}_X$. It actually takes a bit of set-theoretic finesse to justify the existence of this set, but never mind; we’ll take all that as read and worry about the topology.

Let $Y=\prod\mathscr{C}_X$, and let $\varphi:X\to Y:x\mapsto\big\langle c(x):cX\in\mathscr{C}_X\big\rangle$. For each $cX\in\mathscr{C}_X$, $c:X\to cX$ is an embedding, so it’s clear that $\varphi$ is an injection. To show that $\varphi$ is continuous, it suffices to show that for each $cX\in\mathscr{C}_X$ and each open set $U$ in $cX$, $\varphi^{-1}\left[\{y\in Y:y_{cX}\in U\}\right]$ is open in $X$, since sets of the form $\{y\in Y:y_{cX}\in U\}$ are a subbase for the product topology on $Y$. But

\begin{align*} \varphi^{-1}\left[\{y\in Y:y_{cX}\in U\}\right]&=\{x\in X:c(x)\in U\}\\ &=c^{-1}[U]\\ &=c[X]\cap U\;, \end{align*}

which is open in $X$ simply because $cX$ is a compactification of $X$.

(Here $c[X]=\{c(x):x\in X\}\subseteq cX$.)

To show that $\varphi$ is also open, suppose that $U$ is an open set in $X$. then

\begin{align*} \varphi[U]&=\{\varphi(x):x\in U\}\\ &=\{\langle c(x):cX\in\mathscr{C}_X\rangle:x\in U\}\\ &=\{\langle c(x):cX\in\mathscr{C}_X\rangle:c(x)\in cU\}\;. \end{align*}

If $cX,c'X\in\mathscr{C}_X$ and $x\in X$, $c(x)\in cU$ iff $c'(x)\in c'U$, so $\varphi[U]=\varphi[X]\cap\{y\in Y:y_{cX}\in cU\}$, where $cX$ is a single, fixed compactification of $X$. And $\{y\in Y:y_{cX}\in cU\}$ is a basic open set in the product $Y$, so $\varphi[U]$ is open in $\varphi[X]$. Thus, $\varphi$ maps $x$ homeomorphically to $\varphi[X]$.

Since $Y$ is compact, $\operatorname{cl}_Y\varphi[X]$ is certainly a compactification of $X$; call it $K$. All that’s needed to finish the argument is to show that for each $cX\in\mathscr{C}_X$ there is a continuous surjection $f:K\to cX$ that fixes $X$ pointwise. (Actually, that last bit is verbal sloppiness: what’s really meant is that for each $x\in X$, $f(\varphi(x))=c(x)$.) We can simply take $f$ to be the projection map from $Y$ to the factor $cX$, restricted to the subspace $K$: $f=\pi_{cX}\upharpoonright K$. Projections are always continuous, and $f(\varphi(x))=c(x)$ by the definition of $\varphi$, so we’re done: $K$ is the Čech-Stone compactification of $X$.

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