= Solution
Let $\mathcal F=C(X,[0,1])$. Form the evaluation map into a <product space>
$$
e:X\longrightarrow[0,1]^{\mathcal F},\qquad e(x)=(f(x))_{f\in\mathcal F},\qquad \beta X=\overline{e(X)}.
$$
The cube is <compact> by <Tychonoff's theorem> and <Hausdorff>, so its closed subspace $\beta X$ is a <compact Hausdorff space>. A <completely regular Hausdorff space> has enough <continuous functions> to separate points and closed sets. Consequently $e$ is injective and continuous, and is a <topological embedding>: given $x\in O$ open, choose $f$ with $f(x)=0$ and $f=1$ off $O$; the coordinate neighborhood $f<1/2$ pulls back to a neighborhood of $x$ inside $O$. Identify $X$ with this dense embedded subspace. This constructs the <Stone-Čech compactification>.
Each $f\in\mathcal F$ extends as its coordinate function. Rescaling therefore extends every bounded real <continuous function>; real and imaginary parts extend bounded complex <continuous functions>. The extension is unique because $X$ is dense and the target is <Hausdorff>. Its <supremum norm> is unchanged by density, so restriction is an <isometric isomorphism of normed spaces> of <Banach spaces>
$$
\boxed{C(\beta X)\cong C_b(X).}
$$
These are the <space of continuous functions on a compact space> and the space of <bounded continuous functions>, respectively.
More generally, embed a <compact Hausdorff space> $K$ in its own evaluation cube $[0,1]^{C(K,[0,1])}$. This is an embedding by complete regularity, and its image is closed by compactness. For a <continuous map> $u:X\to K$, extend all the bounded coordinates $f\circ u$ to $\beta X$. The resulting <continuous map> into the cube takes its values in the closed copy of $K$: it does so on the dense subset $X$, and the inverse image of that closed copy is closed. Thus
$$
\boxed{u:X\to K\text{ extends uniquely to }\widetilde u:\beta X\to K.}
$$
Uniqueness again follows from density. This is the universal property of the <Stone-Čech compactification>. It implies uniqueness of the compactification up to a <homeomorphism> fixing $X$, by extending the identity in both directions. Every Hausdorff compactification of $X$ is the image of $\beta X$ under a continuous surjection fixing $X$, since that image is compact and contains the dense copy of $X$. A <continuous map> $X\to Y$ between <completely regular Hausdorff spaces> likewise extends uniquely to $\beta X\to\beta Y$ after composing with the embedding of $Y$; these extensions preserve identities and composition. If $X$ is already compact, its dense copy is closed, so $\beta X=X$.
For discrete $X$, every indicator $\mathbf1_D$, $D\subseteq X$, is continuous. Its extension to $\beta X$ takes only values $0,1$, since this closed condition holds on the dense subset $X$. Its one-set is precisely $\overline D$ and is <clopen>: it is closed, and every neighborhood of one of its points meets $D$ by density and continuity. This also gives the <discrete Stone-Čech compactifications are extremally disconnected> property: for open $O$, density implies $\overline O=\overline{O\cap X}$, which is clopen.
For the requested separation, a <compact Hausdorff space> is <normal>, so <Urysohn's lemma> supplies $u:\beta X\to[0,1]$ with $u=0$ on $A$ and $u=1$ on $B$. Put $D=\{x\in X:u(x)<1/2\}$ and
$$
U=\overline D,\qquad V=\beta X\setminus U.
$$
The preceding indicator argument makes both sets clopen. At a point of $A$, the neighborhood $u<1/2$ has its dense $X$-part inside $D$, so the point lies in $U$. At a point of $B$, the neighborhood $u>1/2$ misses $D$, so the point lies outside $U$. We have proved the stronger <clopen separation in discrete Stone-Čech compactifications> conclusion
$$
\boxed{\beta X=U\mathbin{\dot\cup}V,\qquad A\subseteq U,\quad B\subseteq V,\quad U,V\text{ clopen}.}
$$
Finally take \b[$\boxed{K=\beta\mathbb N}$], the <Stone-Čech compactification of the natural numbers>. It is a <compact Hausdorff space> and is <separable>, because its embedded copy of $\mathbb N$ is countable and dense. For every $D\subseteq\mathbb N$, the indicator extends to an element $h_D\in C(K)$. For distinct $D,D'$, evaluating at an integer in their symmetric difference gives $\|h_D-h_{D'}\|_\infty=1$. There are uncountably many such indicators. A separable <metric space> cannot contain an uncountable family at pairwise distance one: disjoint balls of radius $1/3$ would require distinct members of a countable dense set. Hence \b[$C(K)$ is not separable], despite separability of $K$.
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