The Stone-Čech 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.

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

$$g(y)=\underset{j\in J}{\max }(f_j\circ {\pi }_j)(y),y\in X.$$

In general, suppose that $Y$ is a topological space and denote by $C(Y)$ the set of continuous functions $Y\to ℝ$. 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 one-point 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 $f_i$.

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 one-to-one 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 one-to-one.

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 $X\to ℝ$, 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 $X\to ℝ$. $C_b(X)\subset {\mathrm{\ell }}^{\mathrm{\infty }}(X)$, 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 $X\to ℝ$ is a useful collection of functions to talk about.

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 $ℝ$ (because the open rays are a subbase for the topology of $ℝ$), 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 $ℝ$ 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 $g_f:X\to ℝ$ 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 $ℝ$. Any product is homeomorphic to the same product without singleton factors, (e.g. $ℝ\times ℝ\times \{3\}$ is homeomorphic to $ℝ\times ℝ$) 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 $ℝ$, 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 one-point 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 one-point 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 Stone-Čech compactification of $X$ is the pair $(\beta X,e)$, and what we have said shows that indeed this is a compactification of $X$.

The Stone-Čech 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 Stone-Čech compactification.

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 Stone-Čech 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 Stone-Čech 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 $X\to ℝ$. 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 $f:X\to ℝ$ vanishes at infinity if for each $ϵ>0$ 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 $X\to ℝ$ 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 $ϵ>0$. 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 $ϵ>0$. 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 $ϵ>0$, there is some $n_{ϵ}$ such that $n\ge n_{ϵ}$ 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 $x\in X\setminus K_{n_{ϵ}}$, then

showing that $f\in C_0(X)$. ∎

If $X$ is a topological space and $f:X\to ℝ$ 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 $X\to ℝ$ with compact support.

Suppose that $X$ is a topological space and let $f\in C_c(X)$. For any $ϵ>0$, 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 $C_0(X)$.

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 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 $x\in X\setminus C_n$, $|f_n(x)-f(x)|=|g_n(x)-1||f(x)|\le 1\cdot \frac{1}{n}$. 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 ${\delta }_x:C_b(X)\to ℝ$ 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.

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 one-to-one. 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 $C_b{(X)}^{*}\to ℝ$ 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 Banach-Alaoglu 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.

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 $\mathrm{\{}{U}_{n\mathrm{,}k}\mathrm{:}\mathrm{1}\mathrm{\le }k\mathrm{\le }{N}_n\mathrm{\}}$. 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 $ϵ>0$. 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,

$$\sum _{k=1}^{N_n}|f(x)-{\alpha }_k|f_{n,k}(x)\le \sum _{k=1}^{N_n}ϵf_{n,k}(x)=ϵ,$$

showing that $|f(x)-g(x)|\le ϵ$. 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 Stone-Čech 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. (This is proved using the Stone-Weierstrass 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. 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 $\lambda \in ℂ$,

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 $C_b(X,ℂ)$ rather than $C_b(X,ℝ)$, 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 $ℝ$ is true when the codomain is $ℂ$, 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 $\parallel f\parallel \le 1$. 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 $A\to ℂ$. 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. 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 Banach-Alaoglu 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 $\mathrm{\Gamma }(a)\in C_0(\sigma (A))$. The Gelfand-Naimark theorem 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.

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 $M(A)=A$. It can be proved that for any $C^{*}$-algebra $A$, the multiplier algebra $M(A)$ is unital. For a locally compact Hausdorff space $X$, it can be proved that $M(C_0(X))=C_b(X)$. 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 Stone-Čech compactification of discrete topological spaces. This proof is presented in Carothers’ book.