The StoneČech compactification of Tychonoff spaces
1 Completely regular spaces and Tychonoff spaces
A topological space $X$ is said to be completely regular if whenever $F$ is a nonempty closed set and $x\in X\setminus F$, there is a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $f(F)=\{1\}$. A completely regular space need not be Hausdorff. For example, if $X$ is any set with more than one point, then the trivial topology, in which the only closed sets are $\mathrm{\varnothing}$ and $X$, is vacuously completely regular, but not Hausdorff. A topological space is said to be a Tychonoff space if it is completely regular and Hausdorff.
Lemma 1.
A topological space $X$ is completely regular if and only if for any nonempty closed set $F$, any $x\mathrm{\in}X\mathrm{\setminus}F$, and any distinct $a\mathrm{,}b\mathrm{\in}\mathrm{R}$, there is a continuous function $f\mathrm{:}X\mathrm{\to}\mathrm{R}$ such $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}a$ and $f\mathit{}\mathrm{(}F\mathrm{)}\mathrm{=}\mathrm{\{}b\mathrm{\}}$.
Theorem 2.
If $X$ is a Hausdorff space and $A\mathrm{\subset}X$, then $A$ with the subspace topology is a Hausdorff space. If $\mathrm{\{}{X}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ is a family of Hausdorff spaces, then ${\mathrm{\prod}}_{i\mathrm{\in}I}{X}_{i}$ is Hausdorff.
Proof.
Suppose that $a,b$ are distinct points in $A$. Because $X$ is Hausdorff, there are disjoint open sets $U,V$ in $X$ with $a\in U,b\in V$. Then $U\cap A,V\cap A$ are disjoint open sets in $A$ with the subspace topology and $a\in U\cap A,b\in V\cap A$, showing that $A$ is Hausdorff.
Suppose that $x,y$ are distinct elements of ${\prod}_{i\in I}{X}_{i}$. $x$ and $y$ being distinct means there is some $i\in I$ such that $x(i)\ne y(i)$. Then $x(i),y(i)$ are distinct points in ${X}_{i}$, which is Hausdorff, so there are disjoint open sets ${U}_{i},{V}_{i}$ in ${X}_{i}$ with $x(i)\in {U}_{i},y(i)\in {V}_{i}$. Let $U={\pi}_{i}^{1}({U}_{i}),V={\pi}_{i}^{1}({V}_{i})$, where ${\pi}_{i}$ is the projection map from the product to ${X}_{i}$. $U$ and $V$ are disjoint, and $x\in U,y\in V$, showing that ${\prod}_{i\in I}{X}_{i}$ is Hausdorff. ∎
We prove that subspaces and products of completely regular spaces are completely regular.^{1}^{1} 1 Stephen Willard, General Topology, p. 95, Theorem 14.10.
Theorem 3.
If $X$ is Hausdorff and $A\mathrm{\subset}X$, then $A$ with the subspace topology is completely regular. If $\mathrm{\{}{X}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ is a family of completely regular spaces, then ${\mathrm{\prod}}_{i\mathrm{\in}I}{X}_{i}$ is completely regular.
Proof.
Suppose that $F$ is closed in $A$ with the subspace topology and $x\in A\setminus F$. There is a closed set $G$ in $X$ with $F=G\cap A$. Then $x\notin G$, so there is a continuous function $f:X\to [0,1]$ satisfying $f(x)=0$ and $f(F)=\{1\}$. The restriction of $f$ to $A$ with the subspace topology is continuous, showing that $A$ is completely regular.
Suppose that $F$ is a closed subset of $X={\prod}_{i\in I}{X}_{i}$ and that $x\in X\setminus F$. A base for the product topology consists of intersections of finitely many sets of the form ${\pi}_{i}^{1}({U}_{i})$ where $i\in I$ and ${U}_{i}$ is an open subset of ${X}_{i}$, and because $X\setminus F$ is an open neighborhood of $x$, there is a finite subset $J$ of $I$ and open sets ${U}_{j}$ in ${X}_{j}$ for $j\in J$ such that
$$x\in \bigcap _{j\in J}{\pi}_{j}^{1}({U}_{j})\subset X\setminus F.$$ 
For each $j\in J$, ${X}_{j}\setminus {U}_{j}$ is closed in ${X}_{j}$ and $x(j)\in {U}_{j}$, and because ${X}_{j}$ is completely regular there is a continuous function ${f}_{j}:{X}_{j}\to [0,1]$ such that ${f}_{j}(x(j))=0$ and ${f}_{j}({X}_{j}\setminus {U}_{j})=\{1\}$. Define $g:X\to [0,1]$ by
$$g(y)=\underset{j\in J}{\mathrm{max}}({f}_{j}\circ {\pi}_{j})(y),y\in X.$$ 
In general, suppose that $Y$ is a topological space and denote by $C(Y)$ the set of continuous functions $Y\to \mathbb{R}$. It is a fact that $C(Y)$ is a lattice with the partial order $F\le G$ when $F(y)\le G(y)$ for all $y\in Y$. Hence, the maximum of finitely many continuous functions is also a continuous functions, hence $g:X\to [0,1]$ is continuous. Because $({f}_{j}\circ {\pi}_{j})(x)=0$ for each $j\in J$, $g(x)=0$. On the other hand, $F\subset X\setminus {\bigcap}_{j\in J}{\pi}_{j}^{1}({U}_{j})$, so if $y\in F$ then there is some $j\in J$ such that ${\pi}_{j}(y)\in {X}_{j}\setminus {U}_{j}$ and then $({f}_{j}\circ {\pi}_{j})(y)=1$. Hence, for any $y\in F$ we have $g(y)=1$. Thus we have proved that $g:X\to [0,1]$ is a continuous function such that $g(x)=0$ and $g(F)=\{1\}$, which shows that $X$ is completely regular. ∎
Therefore, subspaces and products of Tychonoff spaces are Tychonoff.
If $X$ is a normal topological space, it is immediate from Urysohn’s lemma that $X$ is completely regular. A metrizable space is normal and Hausdorff, so a metrizable space is thus a Tychonoff space. Let $X$ be a locally compact Hausdorff space. Either $X$ or the onepoint compactification of $X$ is a compact Hausdorff space $Y$ of which $X$ is a subspace. $Y$ being a compact Hausdorff space implies that it is normal and hence completely regular. But $X$ is a subspace of $Y$ and being completely regular is a hereditary property, so $X$ is completely regular, and therefore Tychonoff. Thus, we have proved that a locally compact Hausdorff space is Tychonoff.
2 Initial topologies
Suppose that $X$ is a set, ${X}_{i}$, $i\in I$, are topological spaces, and ${f}_{i}:X\to {X}_{i}$ are functions. The initial topology on $X$ induced by $\mathrm{\{}{f}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ is the coarsest topology on $X$ such that each ${f}_{i}$ is continuous. A subbase for the initial topology is the collection of those sets of the form ${f}_{i}^{1}({U}_{i})$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$.
If ${f}_{i}:X\to {X}_{i}$, $i\in I$, are functions, the evaluation map is the function $e:X\to {\prod}_{i\in I}{X}_{i}$ defined by
$$({\pi}_{i}\circ e)(x)={f}_{i}(x),i\in I.$$ 
We say that a collection $\{{f}_{i}:i\in I\}$ of functions on $X$ separates points if $x\ne y$ implies that there is some $i\in I$ such that ${f}_{i}(x)\ne {f}_{i}(y)$. We remind ourselves that if $X$ and $Y$ are topological spaces and $\varphi :X\to Y$ is a function, $\varphi $ is called an embedding when $\varphi :X\to \varphi (X)$ is a homeomorphism, where $\varphi (X)$ has the subspace topology inherited from $Y$. The following theorem gives conditions on when $X$ can be embedded into the product of the codomains of the ${f}_{i}$.^{2}^{2} 2 Stephen Willard, General Topology, p. 56, Theorem 8.12.
Theorem 4.
Let $X$ be a topological space, let ${X}_{i}$, $i\mathrm{\in}I$, be topological spaces, and let ${f}_{i}\mathrm{:}X\mathrm{\to}{X}_{i}$ be functions. The evaluation map $e\mathrm{:}X\mathrm{\to}{\mathrm{\prod}}_{i\mathrm{\in}I}{X}_{i}$ is an embedding if and only if both (i) $X$ has the initial topology induced by the family $\mathrm{\{}{f}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ and (ii) the family $\mathrm{\{}{f}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ separates points in $X$.
Proof.
Write $P={\prod}_{i\in I}{X}_{i}$ and let ${p}_{i}:e(X)\to {X}_{i}$ be the restriction of ${\pi}_{i}:X\to {X}_{i}$ to $e(X)$. A subbase for $e(X)$ with the subspace topology inherited from $P$ consists of those sets of the form ${\pi}_{i}^{1}({U}_{i})\cap e(X)$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$. But ${\pi}_{i}^{1}({U}_{i})\cap e(X)={p}_{i}^{1}({U}_{i})$, and the collection of sets of this form is a subbase for $e(X)$ with the initial topology induced by the family $\{{p}_{i}:i\in I\}$, so these topologies are equal.
Assume that $e:X\to e(X)$ is a homeomorphism. Because $e$ is a homeomorphism and ${f}_{i}={\pi}_{i}\circ e={p}_{i}\circ e$, $e(X)$ having the initial topology induced by $\{{p}_{i}:i\in I\}$ implies that $X$ has the initial topology induced by $\{{f}_{i}:i\in I\}$. If $x,y$ are distinct elements of $X$ then there is some $i\in I$ such that ${p}_{i}(e(x))\ne {p}_{i}(e(y))$, i.e. ${f}_{i}(x)\ne {f}_{i}(y)$, showing that $\{{f}_{i}:i\in I\}$ separates points in $X$.
Assume that $X$ has the initial topology induced by $\{{f}_{i}:i\in I\}$ and that the family $\{{f}_{i}:i\in I\}$ separates points in $X$. We shall prove that $e:X\to e(X)$ is a homeomorphism, for which it suffices to prove that $e:X\to P$ is onetoone and continuous and that $e:X\to e(X)$ is open. If $x,y\in X$ are distinct then because the ${f}_{i}$ separate points, there is some $i\in I$ such that ${f}_{i}(x)\ne {f}_{i}(y)$, and so $e(x)\ne e(y)$, showing that $e$ is onetoone.
For each $i\in I$, ${f}_{i}$ is continuous and ${f}_{i}={\pi}_{i}\circ e$. The fact that this is true for all $i\in I$ implies that $e:X\to P$ is continuous. (Because the product topology is the initial topology induced by the family of projection maps, a map to a product is continuous if and only if its composition with each projection map is continuous.)
A subbase for the topology of $X$ consists of those sets of the form $V={f}_{i}^{1}({U}_{i})$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$. As ${f}_{i}={p}_{i}\circ e$ we can write this as
$$V={({p}_{i}\circ e)}^{1}({U}_{i})={e}^{1}({p}_{i}^{1}({U}_{i})),$$ 
which implies that $e(V)={p}_{i}^{1}({U}_{i})$, which is open in $e(X)$ and thus shows that $e:X\to e(X)$ is open. ∎
We say that a collection $\{{f}_{i}:i\in I\}$ of functions on a topological space $X$ separates points from closed sets if whenever $F$ is a closed subset of $X$ and $x\in X\setminus F$, there is some $i\in I$ such that ${f}_{i}(x)\notin \overline{{f}_{i}(F)}$, where $\overline{{f}_{i}(F)}$ is the closure of ${f}_{i}(F)$ in the codomain of $f$.
Theorem 5.
Assume that $X$ is a topological space and that ${f}_{i}\mathrm{:}X\mathrm{\to}{X}_{i}$, $i\mathrm{\in}I$, are continuous functions. This family separates points from closed sets if and only if the collection of sets of the form ${f}_{i}^{\mathrm{}\mathrm{1}}\mathit{}\mathrm{(}{U}_{i}\mathrm{)}$, $i\mathrm{\in}I$ and ${U}_{i}$ open in ${X}_{i}$, is a base for the topology of $X$.
Proof.
Assume that the family $\{{f}_{i}:i\in I\}$ separates points from closed sets in $X$. Say $x\in X$ and that $U$ is an open neighborhood of $x$. Then $F=X\setminus U$ is closed so there is some $i\in I$ such that ${f}_{i}(x)\notin \overline{{f}_{i}(F)}$. Thus ${U}_{i}={X}_{i}\setminus \overline{{f}_{i}(F)}$ is open in ${X}_{i}$, hence ${f}_{i}^{1}({U}_{i})$ is open in $X$. On the one hand, $f({x}_{i})\in {U}_{i}$ yields ${x}_{i}\in {f}_{i}^{1}({U}_{i})$. On the other hand, if $y\in {f}_{i}^{1}({U}_{i})$ then ${f}_{i}(y)\in {U}_{i}$, which tells us that $y\notin F$ and so $y\in U$, giving ${f}_{i}^{1}({U}_{i})\subset U$. This shows us that the collection of sets of the form ${f}_{i}^{1}({U}_{i})$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$, is a base for the topology of $X$.
Assume that the collection of sets of the form ${f}_{i}^{1}({U}_{i})$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$, is a base for the topology of $X$, and suppose that $F$ is a closed subset of $X$ and that $x\in X\setminus F$. Because $X\setminus F$ is an open neighborhood of $x$, there is some $i\in I$ and open ${U}_{i}$ in ${X}_{i}$ such that $x\in {f}_{i}^{1}({U}_{i})\subset X\setminus F$, so ${f}_{i}(x)\in {U}_{i}$. Suppose by contradiction that there is some $y\in F$ such that ${f}_{i}(y)\in {U}_{i}$. This gives $y\in {f}_{i}^{1}({U}_{i})\subset X\setminus F$, which contradicts $y\in F$. Therefore ${U}_{i}\cap {f}_{i}(F)=\mathrm{\varnothing}$, and hence ${X}_{i}\setminus {U}_{i}$ is a closed set that contains ${f}_{i}(F)$, which tells us that $\overline{{f}_{i}(F)}\subset {X}_{i}\setminus {U}_{i}$, i.e. $\overline{{f}_{i}(F)}\cap {U}_{i}=\mathrm{\varnothing}$. But ${f}_{i}(x)\in {U}_{i}$, so we have proved that $\{{f}_{i}:i\in I\}$ separates points from closed sets. ∎
A ${T}_{1}$ space is a topological space in which all singletons are closed.
Theorem 6.
If $X$ is a ${T}_{\mathrm{1}}$ space, ${X}_{i}$, $i\mathrm{\in}I$, are topological spaces, ${f}_{i}\mathrm{:}X\mathrm{\to}{X}_{i}$ are continuous functions, and $\mathrm{\{}{f}_{i}\mathrm{:}i\mathrm{\in}I\mathrm{\}}$ separates points from closed sets in $X$, then the evaluation map $e\mathrm{:}X\mathrm{\to}{\mathrm{\prod}}_{i\mathrm{\in}I}{X}_{i}$ is an embedding.
Proof.
By Theorem 5, there is a base for the topology of $X$ consisting of sets of the form ${f}_{i}^{1}({U}_{i})$, $i\in I$ and ${U}_{i}$ open in ${X}_{i}$. Since this collection of sets is a base it is a fortiori a subbase, and the topology generated by this subbase is the initial topology for the family of functions $\{{f}_{i}:i\in I\}$. Because $X$ is ${T}_{1}$, singletons are closed and therefore the fact that $\{{f}_{i}:i\in I\}$ separates points and closed sets implies that it separates points in $X$. Therefore we can apply Theorem 4, which tells us that the evaluation map is an embedding. ∎
3 Bounded continuous functions
For any set $X$, we denote by ${\mathrm{\ell}}^{\mathrm{\infty}}(X)$ the set of all bounded functions $X\to \mathbb{R}$, and we take as known that ${\mathrm{\ell}}^{\mathrm{\infty}}(X)$ is a Banach space with the supremum norm
$${\parallel f\parallel}_{\mathrm{\infty}}=\underset{x\in X}{sup}f(x),f\in {\mathrm{\ell}}^{\mathrm{\infty}}(X).$$ 
If $X$ is a topological space, we denote by ${C}_{b}(X)$ the set of bounded continuous functions $X\to \mathbb{R}$. ${C}_{b}(X)\subset {\mathrm{\ell}}^{\mathrm{\infty}}(X)$, and it is apparent that ${C}_{b}(X)$ is a linear subspace of ${\mathrm{\ell}}^{\mathrm{\infty}}(X)$. One proves that ${C}_{b}(X)$ is closed in ${\mathrm{\ell}}^{\mathrm{\infty}}(X)$ (i.e., that if a sequence of bounded continuous functions converges to some bounded function, then this function is continuous), and hence with the supremum norm, ${C}_{b}(X)$ is a Banach space.
The following result shows that the Banach space ${C}_{b}(X)$ of bounded continuous functions $X\to \mathbb{R}$ is a useful collection of functions to talk about.^{3}^{3} 3 Stephen Willard, General Topology, p. 96, Theorem 14.12.
Theorem 7.
Let $X$ be a topological space. $X$ is completely regular if and only if $X$ has the initial topology induced by ${C}_{b}\mathit{}\mathrm{(}X\mathrm{)}$.
Proof.
Assume that $X$ is completely regular. If $F$ is a closed subset of $X$ and $x\in X\setminus F$, then there is a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $f(F)=\{1\}$. Then $f\in {C}_{b}(X)$, and $f(x)=0\notin \{1\}=\overline{f(F)}$. This shows that ${C}_{b}(X)$ separates points from closed sets in $X$. Applying Theorem 5, we get that $X$ has the initial topology induced by ${C}_{b}(X)$. (This would follow if the collection that Theorem 5 tells us is a base were merely a subbase.)
Assume that $X$ has the initial topology induced by ${C}_{b}(X)$. Suppose that $F$ is a closed subset of $X$ and that $x\in U=X\setminus F$. A subbase for the initial topology induced by ${C}_{b}(X)$ consists of those sets of the form ${f}^{1}(V)$ for $f\in {C}_{b}(X)$ and $V$ an open ray in $\mathbb{R}$ (because the open rays are a subbase for the topology of $\mathbb{R}$), so because $U$ is an open neighborhood of $x$, there is a finite subset $J$ of ${C}_{b}(X)$ and open rays ${V}_{f}$ in $\mathbb{R}$ for each $f\in J$ such that
$$x\in \bigcap _{f\in J}{f}^{1}({V}_{f})\subset U.$$ 
If some ${V}_{j}$ is of the form $(\mathrm{\infty},{a}_{f})$, then with $g=f$ we have ${f}^{1}(\mathrm{\infty},{a}_{f})={g}^{1}({a}_{f},\mathrm{\infty})$. We therefore suppose that in fact ${V}_{f}=({a}_{f},\mathrm{\infty})$ for each $f\in J$. For each $f\in J$, define ${g}_{f}:X\to \mathbb{R}$ by
$${g}_{f}(x)=sup\{f(x){a}_{f},0\},$$ 
which is continuous and $\ge 0$, and satisfies ${f}^{1}({a}_{f},\mathrm{\infty})={g}_{f}^{1}(0,\mathrm{\infty})$, so that
$$x\in \bigcap _{f\in J}{g}_{f}^{1}(0,\mathrm{\infty})\subset U.$$ 
Define $g={\prod}_{f\in J}{g}_{f}$, which is continuous because each factor is continuous. This function satisfies $g(x)={\prod}_{f\in J}{g}_{f}(x)>0$ because this is a product of finitely many factors each of which are $>0$. If $y\in {g}^{1}(0,\mathrm{\infty})$ then $y\in {\bigcap}_{f\in J}{g}_{f}^{1}(0,\mathrm{\infty})\subset U$, so ${g}^{1}(0,\mathrm{\infty})\subset U$. But $g$ is nonnegative, so this tells us that $g(X\setminus U)=\{0\}$, i.e. $g(F)=\{0\}$. By Lemma 1 this suffices to show that $X$ is completely regular. ∎
A cube is a topological space that is homeomorphic to a product of compact intervals in $\mathbb{R}$. Any product is homeomorphic to the same product without singleton factors, (e.g. $\mathbb{R}\times \mathbb{R}\times \{3\}$ is homeomorphic to $\mathbb{R}\times \mathbb{R}$) and a product of nonsingleton compact intervals with index set $I$ is homeomorphic to ${[0,1]}^{I}$. We remind ourselves that to say that a topological space is homeomorphic to a subspace of a cube is equivalent to saying that the space can be embedded into the cube.
Theorem 8.
A topological space $X$ is a Tychonoff space if and only if it is homeomorphic to a subspace of a cube.
Proof.
Suppose that $I$ is a set and that $X$ is homeomorphic to a subspace $Y$ of ${[0,1]}^{I}$. $[0,1]$ is Tychonoff so the product ${[0,1]}^{I}$ is Tychonoff, and hence the subspace $Y$ is Tychonoff, thus $X$ is Tychonoff.
Suppose that $X$ is Tychonoff. By Theorem 7, $X$ has the initial topology induced by ${C}_{b}(X)$. For each $f\in {C}_{b}(X)$, let ${I}_{f}=[{\parallel f\parallel}_{\mathrm{\infty}},{\parallel f\parallel}_{\mathrm{\infty}}]$, which is a compact interval in $\mathbb{R}$, and $f:X\to {I}_{f}$ is continuous. Because $X$ is Tychonoff, it is ${T}_{1}$ and the functions $f:X\to {I}_{f}$, $f\in {C}_{b}(X)$, separate points and closed sets, we can now apply Theorem 6, which tells us that the evaluation map $e:X\to {\prod}_{f\in {C}_{b}(X)}{I}_{f}$ is an embedding. ∎
4 Compactifications
In §1 we talked about the onepoint compactification of a locally compact Hausdorff space. A compactification of a topological space $X$ is a pair $(K,h)$ where (i) $K$ is a compact Hausdorff space, (ii) $h:X\to K$ is an embedding, and (iii) $h(X)$ is a dense subset of $K$. For example, if $X$ is a compact Hausdorff space then $(X,{\mathrm{id}}_{X})$ is a compactification of $X$, and if $X$ is a locally compact Hausdorff space, then the onepoint compactification ${X}^{*}=X\cup \{\mathrm{\infty}\}$, where $\mathrm{\infty}$ is some symbol that does not belong to $X$, together with the inclusion map $X\to {X}^{*}$ is a compactification.
Suppose that $X$ is a topological space and that $(K,h)$ is a compactification of $X$. Because $K$ is a compact Hausdorff space it is normal, and then Urysohn’s lemma tells us that $K$ is completely regular. But $K$ is Hausdorff, so in fact $K$ is Tychonoff. A subspace of a Tychonoff space is Tychonoff, so $h(X)$ with the subspace topology is Tychonoff. But $X$ and $h(X)$ are homeomorphic, so $X$ is Tychonoff. Thus, if a topological space has a compactification then it is Tychonoff.
In Theorem 8 we proved that any Tychonoff space can be embedded into a cube. Here review our proof of this result. Let $X$ be a Tychonoff space, and for each $f\in {C}_{b}(X)$ let ${I}_{f}=[{\parallel f\parallel}_{\mathrm{\infty}},{\parallel f\parallel}_{\mathrm{\infty}}]$, so that $f:X\to {I}_{f}$ is continuous, and the family of these functions separates points in $X$. The evaluation map for this family is $e:X\to {\prod}_{f\in {C}_{b}(X)}{I}_{f}$ defined by $({\pi}_{f}\circ e)(x)=f(x)$ for $f\in {C}_{b}(X)$, and Theorem 6 tells us that $e:X\to {\prod}_{f\in {C}_{b}(X)}{I}_{f}$ is an embedding. Because each interval ${I}_{f}$ is a compact Hausdorff space (we remark that if $f=0$ then ${I}_{f}=\{0\}$, which is indeed compact), the product ${\prod}_{f\in {C}_{b}(X)}{I}_{f}$ is a compact Hausdorff space, and hence any closed subset of it is compact. We define $\beta X$ to be the closure of $e(X)$ in ${\prod}_{f\in {C}_{b}(X)}{I}_{f}$, and the StoneČech compactification of $X$ is the pair $(\beta X,e)$, and what we have said shows that indeed this is a compactification of $X$.
The StoneČech compactification of a Tychonoff space is useful beyond displaying that every Tychonoff space has a compactification. We prove in the following that any continuous function from a Tychonoff space to a compact Hausdorff space factors through its StoneČech compactification.^{4}^{4} 4 Stephen Willard, General Topology, p. 137, Theorem 19.5.
Theorem 9.
If $X$ is a Tychonoff space, $K$ is a compact Hausdorff space, and $\varphi \mathrm{:}X\mathrm{\to}K$ is continuous, then there is a unique continuous function $\mathrm{\Phi}\mathrm{:}\beta \mathit{}X\mathrm{\to}K$ such that $\varphi \mathrm{=}\mathrm{\Phi}\mathrm{\circ}e$.
Proof.
$K$ is Tychonoff because a compact Hausdorff space is Tychonoff, so the evaluation map ${e}_{K}:K\to {\prod}_{g\in {C}_{b}(K)}{I}_{g}$ is an embedding. Write $F={\prod}_{f\in {C}_{b}(X)}{I}_{f}$, $G={\prod}_{g\in {C}_{b}(K)}{I}_{g}$, and let ${p}_{f}:F\to {I}_{f}$, ${q}_{g}:G\to {I}_{g}$ be the projection maps.
We define $H:F\to G$ for $t\in F$ by $({q}_{g}\circ H)(t)=t(g\circ \varphi )={p}_{g\circ \varphi}(t)$. For each $g\in G$, the map ${q}_{g}\circ H:F\to {I}_{g\circ \varphi}$ is continuous, so $H$ is continuous.
For $x\in X$, we have
$({q}_{g}\circ H\circ e)(x)$  $=$  $({q}_{g}\circ H)(e(x))$  
$=$  ${p}_{g\circ \varphi}(e(x))$  
$=$  $({p}_{g\circ \varphi}\circ e)(x)$  
$=$  $(g\circ \varphi )(x)$  
$=$  $g(\varphi (x))$  
$=$  $({q}_{g}\circ {e}_{K})(\varphi (x))$  
$=$  $({q}_{g}\circ {e}_{K}\circ \varphi )(x),$ 
so
$$H\circ e={e}_{K}\circ \varphi .$$  (1) 
On the one hand, because $K$ is compact and ${e}_{K}$ is continuous, ${e}_{K}(K)$ is compact and hence is a closed subset of $G$ ($G$ is Hausdorff so a compact subset is closed). From (1) we know $H(e(X))\subset {e}_{K}(K)$, and thus
$$\overline{H(e(X))}\subset \overline{{e}_{K}(K)}={e}_{K}(K).$$ 
On the other hand, because $\beta X$ is compact and $H$ is continuous, $H(\beta X)$ is compact and hence is a closed subset of $G$. As $e(X)$ is dense in $\beta X$ and $H$ is continuous, $H(e(X))$ is dense in $H(\beta X)$, and thus
$$\overline{H(e(X))}=\overline{H(\beta X)}=H(\beta X).$$ 
Therefore we have
$$H(\beta X)\subset {e}_{K}(K).$$ 
Let $h$ be the restriction of $H$ to $\beta X$, and define $\mathrm{\Phi}:\beta X\to K$ by $\mathrm{\Phi}={e}_{K}^{1}\circ h$, which makes sense because ${e}_{K}:K\to {e}_{K}(K)$ is a homeomorphism and $h$ takes values in ${e}_{K}(K)$. $\mathrm{\Phi}$ is continuous, and for $x\in X$ we have, using (1),
$$(\mathrm{\Phi}\circ e)(x)=({e}_{K}^{1}\circ h\circ e)(x)=({e}_{K}^{1}\circ H\circ e)(x)=\varphi (x),$$ 
showing that $\mathrm{\Phi}\circ e=\varphi $.
If $\mathrm{\Psi}:\beta X\to K$ is a continuous function satisfying $f=\mathrm{\Psi}\circ e$, let $y\in e(X)$. There is some $x\in X$ such that $y=e(x)$, and $f(x)=(\mathrm{\Psi}\circ e)(x)=\mathrm{\Psi}(y)$, $f(x)=(\mathrm{\Phi}\circ e)(x)=\mathrm{\Phi}(y)$, showing that for all $y\in e(X)$, $\mathrm{\Psi}(y)=\mathrm{\Phi}(y)$. Since $\mathrm{\Psi}$ and $\mathrm{\Phi}$ are continuous and are equal on $e(X)$, which is a dense subset of $\beta X$, we get $\mathrm{\Psi}=\mathrm{\Phi}$, which completes the proof. ∎
If $X$ is a Tychonoff space with StoneČech compactification $(\beta X,e)$, then because $\beta X$ is a compact space, $C(\beta X)$ with the supremum norm is a Banach space. We show in the following that the extension in Theorem 9 produces an isometric isomorphism ${C}_{b}(X)\to C(\beta X)$.
Theorem 10.
If $X$ is a Tychonoff space with StoneČech compactification $\mathrm{(}\beta \mathit{}X\mathrm{,}e\mathrm{)}$, then there is an isomorphism of Banach spaces ${C}_{b}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\to}C\mathit{}\mathrm{(}\beta \mathit{}X\mathrm{)}$.
Proof.
Let $f,g\in {C}_{b}(X)$, let $\alpha $ be a scalar, and let $K=[\alpha \parallel f\parallel \parallel g\parallel ,\alpha \parallel f\parallel +\parallel g\parallel ]$, which is a compact set. Define $\varphi =\alpha f+g$, and then Theorem 9 tells us that there is a unique continuous function $F:\beta X\to K$ such that $f=F\circ e$, a unique continuous function $G:\beta X\to K$ such that $g=G\circ e$, and a unique continuous function $\mathrm{\Phi}:\beta X\to K$ such that $\varphi =\mathrm{\Phi}\circ e$. For $y\in e(X)$ and $x\in X$ such that $y=e(x)$,
$$\mathrm{\Phi}(y)=\varphi (x)=\alpha f(x)+g(x)=\alpha F(y)+G(y).$$ 
Since $\mathrm{\Phi}$ and $\alpha F+G$ are continuous functions $\beta X\to K$ that are equal on the dense set $e(X)$, we get $\mathrm{\Phi}=\alpha F+G$. Therefore, the map that sends $f\in {C}_{b}(X)$ to the unique $F\in C(\beta X)$ such that $f=F\circ e$ is linear.
Let $f\in {C}_{b}(X)$ and let $F$ be the unique element of $C(\beta X)$ such that $f=F\circ e$. For any $x\in X$, $f(x)=(F\circ e)(x)$, so
$${\parallel f\parallel}_{\mathrm{\infty}}=\underset{x\in X}{sup}f(x)=\underset{x\in X}{sup}(F\circ e)(x)=\underset{y\in e(X)}{sup}F(y).$$ 
Because $F$ is continuous and $e(X)$ is dense in $\beta X$,
$$\underset{y\in e(X)}{sup}F(y)=\underset{y\in \beta X}{sup}F(y)={\parallel F\parallel}_{\mathrm{\infty}},$$ 
so ${\parallel f\parallel}_{\mathrm{\infty}}={\parallel F\parallel}_{\mathrm{\infty}}$, showing that $f\mapsto F$ is an isometry.
For $\mathrm{\Phi}\in C(\beta X)$, define $\varphi =\mathrm{\Phi}\circ e$. $\mathrm{\Phi}$ is bounded so $\varphi $ is also, and $\varphi $ is a composition of continuous functions, hence $\varphi \in {C}_{b}(X)$. Thus $\varphi \mapsto \mathrm{\Phi}$ is onto, completing the proof. ∎
5 Spaces of continuous functions
If $X$ is a topological space, we denote by $C(X)$ the set of continuous functions $X\to \mathbb{R}$. For $K$ a compact set in $X$ (in particular a singleton) and $f\in C(X)$, define ${p}_{K}(f)={sup}_{x\in K}f(x)$. The collection of ${p}_{K}$ for all compact subsets of $K$ of $X$ is a separating family of seminorms, because if $f$ is nonzero there is some $x\in X$ for which $f(x)\ne 0$ and then ${p}_{\{x\}}(f)>0$. Hence $C(X)$ with the topology induced by this family of seminorms is a locally convex space. (If $X$ is $\sigma $compact then the seminorm topology is induced by countably many of the seminorms, and then $C(X)$ is metrizable.) However, since we usually are not given that $X$ is compact (in which case $C(X)$ is normable with ${p}_{X}$) and since it is often more convenient to work with normed spaces than with locally convex spaces, we shall talk about subsets of $C(X)$.
For $X$ a topological space, we say that a function $f:X\to \mathbb{R}$ vanishes at infinity if for each $\u03f5>0$ there is a compact set $K$ such that $$ whenever $x\in X\setminus K$, and we denote by ${C}_{0}(X)$ the set of all continuous functions $X\to \mathbb{R}$ that vanish at infinity.
The following theorem shows first that ${C}_{0}(X)$ is contained in ${C}_{b}(X)$, second that ${C}_{0}(X)$ is a linear space, and third that it is a closed subset of ${C}_{b}(X)$. With the supremum norm ${C}_{b}(X)$ is a Banach space, so this shows that ${C}_{0}(X)$ is a Banach subspace. We work through the proof in detail because it is often proved with unnecessary assumptions on the topological space $X$.
Theorem 11.
Suppose that $X$ is a topological space. Then ${C}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}$ is a closed linear subspace of ${C}_{b}\mathit{}\mathrm{(}X\mathrm{)}$.
Proof.
If $f\in {C}_{0}(X)$, then there is a compact set $K$ such that $x\in X\setminus K$ implies that $$. On the other hand, because $f$ is continuous, $f(K)$ is a compact subset of the scalar field and hence is bounded, i.e., there is some $M\ge 0$ such that $x\in K$ implies that $f(x)\le M$. Therefore $f$ is bounded, showing that ${C}_{0}(X)\subset {C}_{b}(X)$.
Let $f,g\in {C}_{0}(X)$ and let $\u03f5>0$. There is a compact set ${K}_{1}$ such that $x\in X\setminus {K}_{1}$ implies that $$ and a compact set ${K}_{2}$ such that $x\in X\setminus {K}_{2}$ implies that $$. Let $K={K}_{1}\cup {K}_{2}$, which is a union of two compact sets hence is itself compact. If $x\in X\setminus K$, then $x\in X\setminus {K}_{1}$ implying $$ and $x\in X\setminus {K}_{2}$ implying $$, hence $$. This shows that $f+g\in {C}_{0}(X)$.
If $f\in {C}_{0}(X)$ and $\alpha $ is a nonzero scalar, let $\u03f5>0$. There is a compact set $K$ such that $x\in X\setminus K$ implies that $$, and hence $$, showing that $\alpha f\in {C}_{0}(X)$. Therefore ${C}_{0}(X)$ is a linear subspace of ${C}_{b}(X)$.
Suppose that ${f}_{n}$ is a sequence of elements of ${C}_{0}(X)$ that converges to some $f\in {C}_{b}(X)$. For $\u03f5>0$, there is some ${n}_{\u03f5}$ such that $n\ge {n}_{\u03f5}$ implies that $$, that is,
$$ 
For each $n$, let ${K}_{n}$ be a compact set in $X$ such that $x\in X\setminus {K}_{n}$ implies that $$; there are such ${K}_{n}$ because ${f}_{n}\in {C}_{0}(X)$. If $x\in X\setminus {K}_{{n}_{\u03f5}}$, then
$$ 
showing that $f\in {C}_{0}(X)$. ∎
If $X$ is a topological space and $f:X\to \mathbb{R}$ is a function, the support of $f$ is the set
$$\mathrm{supp}f=\overline{\{x\in X:f(x)\ne 0\}}.$$ 
If $\mathrm{supp}f$ is compact we say that $f$ has compact support, and we denote by ${C}_{c}(X)$ the set of all continuous functions $X\to \mathbb{R}$ with compact support.
Suppose that $X$ is a topological space and let $f\in {C}_{c}(X)$. For any $\u03f5>0$, if $x\in X\setminus \mathrm{supp}f$ then $$, showing that $f\in {C}_{0}(X)$. Therefore
$${C}_{c}(X)\subset {C}_{0}(X),$$ 
and this makes no assumptions about the topology of $X$.
We can prove that if $X$ is a locally compact Hausdorff space then ${C}_{c}(X)$ is dense in ${C}_{0}(X)$.^{5}^{5} 5 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, p. 132, Proposition 4.35.
Theorem 12.
If $X$ is a locally compact Hausdorff space, then ${C}_{c}\mathit{}\mathrm{(}X\mathrm{)}$ is a dense subset of ${C}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}$.
Proof.
Let $f\in {C}_{0}(X)$, and for each $n\in \mathbb{N}$ define
$${C}_{n}=\{x\in X:f(x)\ge \frac{1}{n}\}.$$ 
For $n\in \mathbb{N}$, because $f\in {C}_{0}(X)$ there is a compact set ${K}_{n}$ such that $x\in X\setminus {K}_{n}$ implies that $$, and hence ${C}_{n}\subset {K}_{n}$. Because $x\mapsto {f}_{n}(x)$ is continuous, ${C}_{n}$ is a closed set in $X$, and it follows that ${C}_{n}$, being contained in the compact set ${K}_{n}$, is compact. (This does not use that $X$ is Hausdorff.)
Let $n\in \mathbb{N}$. Because $X$ is a locally compact Hausdorff space and ${C}_{n}$ is compact, Urysohn’s lemma^{6}^{6} 6 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, p. 131, Lemma 4.32. tells us that there is a compact set ${D}_{n}$ containing ${C}_{n}$ and a continuous function ${g}_{n}:X\to [0,1]$ such that ${g}_{n}({C}_{n})=\{1\}$ and ${g}_{n}(X\setminus {D}_{n})\subset \{0\}$. That is, ${g}_{n}\in {C}_{c}(X)$, $0\le {g}_{n}\le 1$, and ${g}_{n}({C}_{n})=\{1\}$. Define ${f}_{n}={g}_{n}f\in {C}_{c}(X)$. (A product of continuous functions is continuous, and because $f$ is bounded and ${g}_{n}$ has compact support, ${g}_{n}f$ has compact support.) For $x\in {C}_{n}$, ${f}_{n}(x)f(x)=({g}_{n}(x)1)f(x)=0$, and for $x\in X\setminus {C}_{n}$, ${f}_{n}(x)f(x)={g}_{n}(x)1f(x)\le 1\cdot \frac{1}{n}$. Therefore
$${\parallel {f}_{n}f\parallel}_{\mathrm{\infty}}\le \frac{1}{n},$$ 
and hence ${f}_{n}$ is a sequence in ${C}_{c}(X)$ that converges to $f$, showing that ${C}_{c}(X)$ is dense in ${C}_{0}(X)$. ∎
If $X$ is a Hausdorff space, then we prove that ${C}_{c}(X)$ is a linear subspace of ${C}_{0}(X)$. When $X$ is a locally compact Hausdorff space then combined with the above this shows that ${C}_{c}(X)$ is a dense linear subspace of ${C}_{0}(X)$.
Lemma 13.
Suppose that $X$ is a Hausdorff space. Then ${C}_{c}\mathit{}\mathrm{(}X\mathrm{)}$ is a linear subspace of ${C}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}$.
Proof.
If $f,g\in {C}_{c}(X)$ and $\alpha $ is a scalar, let $K=\mathrm{supp}f\cup \mathrm{supp}g$, which is a union of two compact sets hence compact. If $x\in X\setminus K$, then $f(x)=0$ because $x\notin \mathrm{supp}f$ and $g(x)=0$ because $x\notin \mathrm{supp}g$, so $(\alpha f+g)(x)=0$. Therefore $\{x\in X:(\alpha f+g)(x)\ne 0\}\subset K$ and hence $\mathrm{supp}(\alpha f+g)\subset \overline{K}$. But as $X$ is Hausdorff, $K$ being compact implies that $K$ is closed in $X$, so we get $\mathrm{supp}(\alpha f+g)\subset K$. Because $\mathrm{supp}(\alpha f+g)$ is closed and is contained in the compact set $K$, it is itself compact, so $\alpha f+g\in {C}_{c}(X)$. ∎
Let $X$ be a topological space, and for $x\in X$ define ${\delta}_{x}:{C}_{b}(X)\to \mathbb{R}$ by ${\delta}_{x}(f)=f(x)$. For each $x\in X$, ${\delta}_{x}$ is linear and ${\delta}_{x}(f)=f(x)\le {\parallel f\parallel}_{\mathrm{\infty}}$, so ${\delta}_{x}$ is continuous and hence belongs to the dual space ${C}_{b}{(X)}^{*}$. Moreover, the constant function $f(x)=1$ shows that $\parallel {\delta}_{x}\parallel =1$. We define $\mathrm{\Delta}:X\to {C}_{b}{(X)}^{*}$ by $\mathrm{\Delta}(x)={\delta}_{x}$. Suppose that ${x}_{i}$ is a net in $X$ that converges to some $x\in X$. Then for every $f\in {C}_{b}(X)$ we have $f({x}_{i})\to f(x)$, and this means that ${\delta}_{{x}_{i}}$ weak* converges to ${\delta}_{x}$ in ${C}_{b}{(X)}^{*}$. This shows that with ${C}_{b}{(X)}^{*}$ assigned the weak* topology, $\mathrm{\Delta}:X\to {C}_{b}{(X)}^{*}$ is continuous. We now characterize when $\mathrm{\Delta}$ is an embedding.^{7}^{7} 7 John B. Conway, A Course in Functional Analysis, second ed., p. 137, Proposition 6.1.
Theorem 14.
Suppose that $X$ is a topological space and assign ${C}_{b}\mathit{}{\mathrm{(}X\mathrm{)}}^{\mathrm{*}}$ the weak* topology. Then the map $\mathrm{\Delta}\mathrm{:}X\mathrm{\to}\mathrm{\Delta}\mathit{}\mathrm{(}X\mathrm{)}$ is a homeomorphism if and only if $X$ is Tychonoff, where $\mathrm{\Delta}\mathit{}\mathrm{(}X\mathrm{)}$ has the subspace topology inherited from ${C}_{b}\mathit{}{\mathrm{(}X\mathrm{)}}^{\mathrm{*}}$.
Proof.
Suppose that $X$ is Tychonoff. If $x,y\in X$ are distinct, then there is some $f\in {C}_{b}(X)$ such that $f(x)=0$ and $f(y)=1$, and then ${\delta}_{x}(f)=0\ne 1={\delta}_{y}(f)$, so $\mathrm{\Delta}(x)\ne \mathrm{\Delta}(y)$, showing that $\mathrm{\Delta}$ is onetoone. To show that $\mathrm{\Delta}:X\to \mathrm{\Delta}(X)$ is a homeomorphism, it suffices to prove that $\mathrm{\Delta}$ is an open map, so let $U$ be an open subset of $X$. For ${x}_{0}\in U$, because $X\setminus U$ is closed there is some $f\in {C}_{b}(X)$ such that $f({x}_{0})=0$ and $f(X\setminus U)=\{1\}$. Let
$$ 
This is an open subset of ${C}_{b}{(X)}^{*}$ as it is the inverse image of $(\mathrm{\infty},1)$ under the map $\mu \mapsto \mu (f)$, which is continuous ${C}_{b}{(X)}^{*}\to \mathbb{R}$ by definition of the weak* topology. Then
$$ 
is an open subset of the subspace $\mathrm{\Delta}(X)$, and we have both ${\delta}_{{x}_{0}}\in V$ and $V\subset \mathrm{\Delta}(U)$. This shows that for any element ${\delta}_{{x}_{0}}$ of $\mathrm{\Delta}(U)$, there is some open set $V$ in the subspace $\mathrm{\Delta}(X)$ such that ${\delta}_{{x}_{0}}\in V\subset \mathrm{\Delta}(U)$, which tells us that $\mathrm{\Delta}(U)$ is an open set in the subspace $\mathrm{\Delta}(U)$, showing that $\mathrm{\Delta}$ is an open map and therefore a homeomorphism.
Suppose that $\mathrm{\Delta}:X\to \mathrm{\Delta}(X)$ is a homeomorphism. By the BanachAlaoglu theorem we know that the closed unit ball ${B}_{1}$ in ${C}_{b}{(X)}^{*}$ is compact. (We remind ourselves that we have assigned ${C}_{b}{(X)}^{*}$ the weak* topology.) That is, with the subspace topology inherited from ${C}_{b}{(X)}^{*}$, ${B}_{1}$ is a compact space. It is Hausdorff because ${C}_{b}{(X)}^{*}$ is Hausdorff, and a compact Hausdorff space is Tychonoff. But $\mathrm{\Delta}(X)$ is contained in the surface of ${B}_{1}$, in particular $\mathrm{\Delta}(X)$ is contained in ${B}_{1}$ and hence is itself Tychonoff with the subspace topology inherited from ${B}_{1}$, which is equal to the subspace topology inherited from ${C}_{b}{(X)}^{*}$. Since $\mathrm{\Delta}:X\to \mathrm{\Delta}(X)$ is a homeomorphism, we get that $X$ is a Tychonoff space, completing the proof. ∎
The following result shows when the Banach space ${C}_{b}(X)$ is separable.^{8}^{8} 8 John B. Conway, A Course in Functional Analysis, second ed., p. 140, Theorem 6.6.
Theorem 15.
Suppose that $X$ is a Tychonoff space. Then the Banach space ${C}_{b}\mathit{}\mathrm{(}X\mathrm{)}$ is separable if and only if $X$ is compact and metrizable.
Proof.
Assume that $X$ is compact and metrizable, with a compatible metric $d$. For each $n\in \mathbb{N}$ there are open balls ${U}_{n,1},\mathrm{\dots},{U}_{n,{N}_{n}}$ of radius $\frac{1}{n}$ that cover $X$. As $X$ is metrizable it is normal, so there is a partition of unity subordinate to the cover $\mathrm{\{}{U}_{n\mathrm{,}k}\mathrm{:}\mathrm{1}\mathrm{\le}k\mathrm{\le}{N}_{n}\mathrm{\}}$.^{9}^{9} 9 John B. Conway, A Course in Functional Analysis, second ed., p. 139, Theorem 6.5. That is, there are continuous functions ${f}_{n,1},\mathrm{\dots},{f}_{n,{N}_{n}}:X\to [0,1]$ such that ${\sum}_{k=1}^{{N}_{n}}{f}_{n,k}=1$ and such that $x\in X\setminus {U}_{n,k}$ implies that ${f}_{n,k}(x)=0$. Then $\{{f}_{n,k}:n\in \mathbb{N},1\le k\le {N}_{n}\}$ is countable, so its span $D$ over $\mathbb{Q}$ is also countable. We shall prove that $D$ is dense in $C(X)={C}_{b}(X)$, which will show that ${C}_{b}(X)$ is separable.
Let $f\in C(X)$ and let $\u03f5>0$. Because $(X,d)$ is a compact metric space, $f$ is uniformly continuous, so there is some $\delta >0$ such that $$ implies that $$. Let $n\in \mathbb{N}$ be $>\frac{2}{\delta}$, and for each $1\le k\le {N}_{n}$ let ${x}_{k}\in {U}_{n,k}$. For each $k$ there is some ${\alpha}_{k}\in \mathbb{Q}$ such that $$, and we define
$$g=\sum _{k=1}^{{N}_{n}}{\alpha}_{k}{f}_{n,k}\in D.$$ 
Because ${\sum}_{k=1}^{{N}_{n}}{f}_{n,k}=1$ we have $f={\sum}_{k=1}^{{N}_{n}}f{f}_{n,k}$. Let $x\in X$, and then
$$f(x)g(x)=\left\sum _{k=1}^{{N}_{n}}(f(x){\alpha}_{k}){f}_{n,k}(x)\right\le \sum _{k=1}^{{N}_{n}}f(x){\alpha}_{k}{f}_{n,k}(x).$$ 
For each $1\le k\le {N}_{n}$, either $x\in {U}_{n,k}$ or $x\notin {U}_{n,k}$. In the first case, since $x$ and ${x}_{k}$ are then in the same open ball of radius $\frac{1}{n}$, $$, so
$$ 
In the second case, ${f}_{n,k}(x)=0$. Therefore,
$$\sum _{k=1}^{{N}_{n}}f(x){\alpha}_{k}{f}_{n,k}(x)\le \sum _{k=1}^{{N}_{n}}\u03f5{f}_{n,k}(x)=\u03f5,$$ 
showing that $f(x)g(x)\le \u03f5$. This shows that $D$ is dense in $C(X)$, and therefore that ${C}_{b}(X)=C(X)$ is separable.
Suppose that ${C}_{b}(X)$ is separable. Because $X$ is Tychonoff, by Theorem 10 there is an isometric isomorphism between the Banach spaces ${C}_{b}(X)$ and $C(\beta X)$, where $(\beta X,e)$ is the StoneČech compactification of $X$. Hence $C(\beta X)$ is separable. But it is a fact that a compact Hausdorff space $Y$ is metrizable if and only if the Banach space $C(Y)$ is separable.^{10}^{10} 10 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 353, Theorem 9.14. (This is proved using the StoneWeierstrass theorem.) As $\beta X$ is a compact Hausdorff space and $C(\beta X)$ is separable, we thus get that $\beta X$ is metrizable.
It is a fact that if $Y$ is a Banach space and ${B}_{1}$ is the closed unit ball in the dual space ${Y}^{*}$, then ${B}_{1}$ with the subspace topology inherited from ${Y}^{*}$ with the weak* topology is metrizable if and only if $Y$ is separable.^{11}^{11} 11 John B. Conway, A Course in Functional Analysis, second ed., p. 134, Theorem 5.1. Thus, the closed unit ball ${B}_{1}$ in ${C}_{b}{(X)}^{*}$ is metrizable. Theorem 14 tells us there is an embedding $\mathrm{\Delta}:X\to {B}_{1}$, and ${B}_{1}$ being metrizable implies that $\mathrm{\Delta}(X)$ is metrizable. As $\mathrm{\Delta}:X\to \mathrm{\Delta}(X)$ is a homeomorphism, we get that $X$ is metrizable.
Because $\beta X$ is compact and metrizable, to prove that $X$ is compact and metrizable it suffices to prove that $\beta X\setminus e(X)=\mathrm{\varnothing}$, so we suppose by contradiction that there is some $\tau \in \beta X\setminus e(X)$. $e(X)$ is dense in $\beta X$, so there is a sequence ${x}_{n}\in X$, for which we take ${x}_{n}\ne {x}_{m}$ when $n\ne m$, such that $e({x}_{n})\to \tau $. If ${x}_{n}$ had a subsequence ${x}_{a(n)}$ that converged to some $y\in X$, then $e({x}_{a(n)})\to e(y)$ and hence $e(y)=\tau $, a contradiction. Therefore the sequence ${x}_{n}$ has no limit points, so the sets $A=\{{x}_{n}:n\text{odd}\}$ and $B=\{{x}_{n}:n\text{even}\}$ are closed and disjoint. Because $X$ is metrizable it is normal, hence by Urysohn’s lemma there is a continuous function $\varphi :X\to [0,1]$ such that $\varphi (a)=0$ for all $a\in A$ and $\varphi (b)=1$ for all $b\in B$. Then, by Theorem 9 there is a unique continuous $\mathrm{\Phi}:X\to [0,1]$ such that $\varphi =\mathrm{\Phi}\circ e$. Then we have, because a subsequence of a convergent sequence has the same limit,
$\mathrm{\Phi}(\tau )$  $=$  $\mathrm{\Phi}\left(\underset{n\to \mathrm{\infty}}{lim}e({x}_{n})\right)$  
$=$  $\mathrm{\Phi}\left(\underset{n\to \mathrm{\infty}}{lim}e({x}_{2n+1})\right)$  
$=$  $\underset{n\to \mathrm{\infty}}{lim}(\mathrm{\Phi}\circ e)({x}_{2n+1})$  
$=$  $\underset{n\to \mathrm{\infty}}{lim}\varphi ({x}_{2n+1})$  
$=$  $0,$ 
and likewise
$$\mathrm{\Phi}(\tau )=\underset{n\to \mathrm{\infty}}{lim}\varphi ({x}_{2n})=1,$$ 
a contradiction. This shows that $\beta X\setminus e(X)=\mathrm{\varnothing}$, which completes the proof. ∎
6 𝐶^{*}algebras and the Gelfand transform
A ${C}^{*}$algebra is a complex Banach algebra $A$ with a map ${}^{*}:A\to A$ such that

1.
${a}^{**}=a$ for all $a\in A$ (namely, ${}^{*}$ is an involution),

2.
${(a+b)}^{*}={a}^{*}+{b}^{*}$ and ${(ab)}^{*}={b}^{*}{a}^{*}$ for all $a\in A$,

3.
${(\lambda a)}^{*}=\overline{\lambda}{a}^{*}$ for all $a\in A$ and $\lambda \in \u2102$,

4.
$\parallel {a}^{*}a\parallel ={\parallel a\parallel}^{2}$ for all $a\in A$.
We do not require that a ${C}^{*}$algebra be unital. If $a=0$ then $\parallel {a}^{*}\parallel =\parallel 0\parallel =\parallel a\parallel $. Otherwise,
$${\parallel a\parallel}^{2}=\parallel {a}^{*}a\parallel \le \parallel {a}^{*}\parallel \parallel a\parallel $$ 
gives $\parallel a\parallel \le \parallel {a}^{*}\parallel $ and
$${\parallel {a}^{*}\parallel}^{2}=\parallel {a}^{**}{a}^{*}\parallel =\parallel a{a}^{*}\parallel \le \parallel a\parallel \parallel {a}^{*}\parallel $$ 
gives $\parallel {a}^{*}\parallel \le \parallel a\parallel $, showing that ${}^{*}$ is an isometry.
We now take ${C}_{b}(X)$ to denote ${C}_{b}(X,\u2102)$ rather than ${C}_{b}(X,\mathbb{R})$, and likewise for $C(X),{C}_{0}(X)$, and ${C}_{c}(X)$. It is routine to verify that everything we have asserted about these spaces when the codomain is $\mathbb{R}$ is true when the codomain is $\u2102$, but this is not obvious. In particular, ${C}_{b}(X)$ is a Banach space with the supremum norm and ${C}_{0}(X)$ is a closed linear subspace, whatever the topological space $X$. It is then straightforward to check that with the involution ${f}^{*}=\overline{f}$ they are commutative ${C}^{*}$algebras.
A homomorphism of ${C}^{\mathrm{*}}$algebras is an algebra homomorphism $f:A\to B$, where $A$ and $B$ are ${C}^{*}$algebras, such that $f({a}^{*})=f{(a)}^{*}$ for all $a\in A$. It can be proved that $\parallel f\parallel \le 1$.^{12}^{12} 12 José M. GraciaBondía, Joseph C. Várilly and Héctor Figueroa, Elements of Noncommutative Geometry, p. 29, Lemma 1.16. We define an isomorphism of ${C}^{\mathrm{*}}$algebras to be an algebra isomorphism $f:A\to B$ such that $f({a}^{*})=f{(a)}^{*}$ for all $a\in A$. It follows that $\parallel f\parallel \le 1$ and because $f$ is bijective, the inverse ${f}^{1}$ is a ${C}^{*}$algebra homomorphism, giving $\parallel {f}^{1}\parallel \le 1$ and therefore $\parallel f\parallel =1$. Thus, an isomorphism of ${C}^{*}$algebras is an isometric isomorphism.
Suppose that $A$ is a commutative ${C}^{*}$algebra, which we do not assume to be unital. A character of $A$ is a nonzero algebra homomorphism $A\to \u2102$. We denote the set of characters of $A$ by $\sigma (A)$, which we call the Gelfand spectrum of $A$. We make some assertions in the following text that are proved in Folland.^{13}^{13} 13 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 12, §1.3. It is a fact that for every $h\in \sigma (A)$, $\parallel h\parallel \le 1$, so $\sigma (A)$ is contained in the closed unit ball of ${A}^{*}$, where ${A}^{*}$ denotes the dual of the Banach space $A$. Furthermore, one can prove that $\sigma (A)\cup \{0\}$ is a weak* closed set in ${A}^{*}$, and hence is weak* compact because it is contained in the closed unit ball which we know to be weak* compact by the BanachAlaoglu theorem. We assign $\sigma (A)$ the subspace topology inherited from ${A}^{*}$ with the weak* topology. Depending on whether $0$ is or is not an isolated point in $\sigma (A)\cup \{0\}$, $\sigma (A)$ is a compact or a locally compact Hausdorff space; in any case $\sigma (A)$ is a locally compact Hausdorff space.
The Gelfand transform is the map $\mathrm{\Gamma}:A\to {C}_{0}(\sigma (A))$ defined by $\mathrm{\Gamma}(a)(h)=h(a)$; that $\mathrm{\Gamma}(a)$ is continuous follows from $\sigma (A)$ having the weak* topology, and one proves that in fact $\mathrm{\Gamma}(a)\in {C}_{0}(\sigma (A))$.^{14}^{14} 14 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 15. The GelfandNaimark theorem^{15}^{15} 15 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 16, Theorem 1.31. states that $\mathrm{\Gamma}:A\to {C}_{0}(\sigma (A))$ is an isomorphism of ${C}^{*}$algebras.
It can be proved that two commutative ${C}^{*}$algebras are isomorphic as ${C}^{*}$algebras if and only if their Gelfand spectra are homeomorphic.^{16}^{16} 16 José M. GraciaBondía, Joseph C. Várilly and Héctor Figueroa, Elements of Noncommutative Geometry, p. 11, Proposition 1.5.
7 Multiplier algebras
An ideal of a ${C}^{\mathrm{*}}$algebra $A$ is a closed linear subspace $I$ of $A$ such that $IA\subset I$ and $AI\subset I$. An ideal $I$ is said to be essential if $I\cap J\ne \{0\}$ for every nonzero ideal $J$ of $A$. In particular, $A$ is itself an essential ideal.
Suppose that $A$ is a ${C}^{*}$algebra. The multiplier algebra of $A$, denoted $M(A)$, is a ${C}^{*}$algebra containing $A$ as an essential ideal such that if $B$ is a ${C}^{*}$algebra containing $A$ as an essential ideal then there is a unique homomorphism of ${C}^{*}$algebras $\pi :B\to M(A)$ whose restriction to $A$ is the identity. We have not shown that there is a multiplier algebra of $A$, but we shall now prove that this definition is a universal property: that any ${C}^{*}$algebra satisfying the definition is isomorphic as a ${C}^{*}$algebra to $M(A)$, which allows us to talk about “the” multiplier algebra rather than “a” multiplier algebra.
Suppose that $C$ is a ${C}^{*}$algebra containing $A$ as an essential ideal such that if $B$ is a ${C}^{*}$algebra containing $A$ as an essential ideal then there is a unique ${C}^{*}$algebra homomorphism $\pi :B\to C$ whose restriction to $A$ is the identity. Hence there is a unique homomorphism of ${C}^{*}$algebras ${\pi}_{1}:C\to M(A)$ whose restriction to $A$ is the identity, and there is a unique homomorphism of ${C}^{*}$algebras ${\pi}_{2}:M(A)\to C$ whose restriction to $A$ is the identity. Then ${\pi}_{2}\circ {\pi}_{1}:C\to C$ and ${\pi}_{1}\circ {\pi}_{2}:M(A)\to M(A)$ are homomorphisms of ${C}^{*}$algebras whose restrictions to $A$ are the identity. But the identity maps ${\mathrm{id}}_{C}:C\to C$ and ${\mathrm{id}}_{M(A)}:M(A)\to M(A)$ are also homomorphisms of ${C}^{*}$algebras whose restrictions to $A$ are the identity. Therefore, by uniqueness we get that ${\pi}_{2}\circ {\pi}_{1}={\mathrm{id}}_{C}$ and ${\pi}_{1}\circ {\pi}_{2}={\mathrm{id}}_{M(A)}$. Therefore ${\pi}_{1}:C\to M(A)$ is an isomorphism of ${C}^{*}$algebras.
One can prove that if $A$ is unital then $M(A)=A$.^{17}^{17} 17 Paul Skoufranis, An Introduction to Multiplier Algebras, http://www.math.ucla.edu/~pskoufra/OANotesMultiplierAlgebras.pdf, p. 4, Lemma 1.9. It can be proved that for any ${C}^{*}$algebra $A$, the multiplier algebra $M(A)$ is unital.^{18}^{18} 18 Paul Skoufranis, An Introduction to Multiplier Algebras, http://www.math.ucla.edu/~pskoufra/OANotesMultiplierAlgebras.pdf, p. 9, Corollary 2.8. For a locally compact Hausdorff space $X$, it can be proved that $M({C}_{0}(X))={C}_{b}(X)$.^{19}^{19} 19 Eberhard Kaniuth, A Course in Commutative Banach Algebras, p. 29, Example 1.4.13; José M. GraciaBondía, Joseph C. Várilly and Héctor Figueroa, Elements of Noncommutative Geometry, p. 14, Proposition 1.10. This last assertion is the reason for my interest in multiplier algebras. We have seen that if $X$ is a locally compact Hausdorff space then ${C}_{c}(X)$ is a dense linear subspace of ${C}_{0}(X)$, and for any topological space ${C}_{0}(X)$ is a closed linear subspace of ${C}_{b}(X)$, but before talking about multiplier algebras we did not have a tight fit between the ${C}^{*}$algebras ${C}_{0}(X)$ and ${C}_{b}(X)$.
8 Riesz representation theorem for compact Hausdorff spaces
There is a proof due to D. J. H. Garling of the Riesz representation theorem for compact Hausdorff spaces that uses the StoneČech compactification of discrete topological spaces. This proof is presented in Carothers’ book.^{20}^{20} 20 N. L. Carothers, A Short Course on Banach Space Theory, Chapter 16, pp. 156–165.