The Stone-Cech Compactification

In general, a compactification of a topological space $X$ is a pair $(K,i)$ where $K$ is a compact topological space and $i:X \rightarrow K$ is a homeomorphism onto its range. The one point compactification is the smallest. The Stone-Cech compactification is, in a suitable sense, the largest. Its defining property is that for all $f \in C_b(X)$, then $f$-extends to $g \in C(K)$.

Example: Consider $X = \mathbb{R}$. The one point compactification is homeomorphic to $S^1$.
Consider $\arctan$ in $\mathbb{R}$, a bounded continuous function. It does not extend to a continuous function on the one-point compactification since $\lim_{x \rightarrow -\infty} \arctan(x) \neq \lim_{x \rightarrow +\infty} \arctan(x)$. To extend $\arctan$, we need at least a two point compactification, which is homeomorphic to $[0,1]$.
Consider $\sin$ on $\mathbb{R}$. Call $g$ the extension of $\sin$ to the Stone-Cech compactification. Since $\sin$ vanishes on $\{n \pi : n \in \mathbb{N}\}$, $g^{-1}(0)$ is a compact subset of the compactification and so $(n \pi)_{n \in \mathbb{N}}$ has a limit point, $x$.
Consider $((2n+\frac12)\pi)_{n \in \mathbb{N}} \subset \sin^{-1}(1)$. Similarly, this has a limit point $y$. Since

$$g(x) = \lim_{n \rightarrow \infty} g(n\pi)$$

$$g(y) = \lim_{n \rightarrow \infty} g((2n+\frac12)\pi) = 1$$

and $x \neq y$. In this manner, we can create countably many new limit points ``at infinity." So instead of just adding one new point, we are adding at least a copy of $[-1,1]$.
Consider $\sin(x^2), \sin(x^3), \ldots$.
So, the Stone-Cech compactification will be very large, even for ``natural" spaces like $\mathbb{R}$. However, there are a few examples where the one-point compactification and the Stone-Cech compactification are the same.

We construct the Stone-Cech compactification using a weak topology on part of the dual of $C_b(X)$.

Let $(X,\tau)$ be a topological space. For $x \in X$, define $\delta_x \in C(X)^*$ by $\delta_X(f) = f(x)$. Clearly, $\Vert\delta_x\Vert = 1$, $\delta_x$ is linear. Define $\Delta:X \rightarrow C_b(X)^*$ by $\Delta(x) = \delta_x$.

Claim: $\Delta$ is $(\tau,$   wk $^*)$-continuous. If $x_i \rightarrow x_0$ in $X$, then $f(x_i) \rightarrow f(x_0)$ for all $f \in C_b(X)$. That is, $\delta_{x_i}(f) \rightarrow \delta_{x_0}(f)$ for all $f$. That is, $\delta_{x_i} \rightarrow \delta_{x_0}$ in the wk $^*$ topology.

Proposition 3.2.1   $\Delta$ is a homeomorphism onto $\Delta(X)$ iff $X$ is completely regular.

Proof.

$\Leftarrow$

$\Delta$ is injective.
If $x_1 \neq x_2$, then $\exists f \in C_b(X)$ such that $f(x_1) \neq f(x_2)$. Hence, $\delta_{x_1}(f) \neq \delta_{x_2}(f)$, so $\delta_{x_1} \neq \delta_{x_2}$.
$\Delta$ is an open map
Let $U \subset X$ be open, $x_0 \in U$. It suffices to find $M \subset \Delta(U)$ with $\Delta(x_0) \in M$ and $M$ open. By complete regularity, $\exists f \in C_b(X)$ such that $f(x_0) = 1$, supp $f \subset \bar U$. Let $M_1 = \{ \mu \in C_b(X)^* : \mu(f) > 0 \}$, a wk $^*$-open set. To see this, observe

$$M_1 = \hat f ( (0,+\infty) )$$

in $C_b(X)^{**}$. Since $\hat f$ is wk $^*$-cts, $M_1$ is wk $^*$-open.
Let $M = M_1 \cap \Delta(X)$, a relatively wk $^*$-open set. This set shows the desired properties noted above, so $\Delta$ is an open map.

$\Rightarrow$

As a compact Hausdorff space, $($ball $C_b(X)^*,$   wk $^*)$ is normal and normal implies completely regular.
Since supsets of completely regular topological spaces (with the relative topology) are completely regular, $\Delta(X)$ is completely regular. Since $\Delta$ is a homeomorphism, $X$ is completely regular.

$\qedsymbol$

Theorem 3.2.2   Stone-Cech Compactification
If $X$ is completely regular, then there is a unique (up to homeomorphism) compactification of $X$, $\beta X$ such that every $f \in C_b(X)$ extends to an element of $C(\beta X)$.

Proof. Existence
Let $\beta X =$   wk $^*-$   cl $\Delta(X)$. By Alaoglu, $\beta X$ is wk $^*$-compact. By the proposition, $\Delta :X \rightarrow \Delta(X)$ is a homeomorphism. Since $\beta X$ is the closure of $\Delta(X)$, the image of $X$ is dense in $\beta X$.
Let $f \in C_b(X)$. As $\beta X \subset C_b(X)^*$, define $g \in C(\beta X)$ by $g(\phi) = \phi(f)$. If $\phi_i \rightarrow \phi$ wk $^*$, $\phi_i(f) \rightarrow \phi(f)$, so $g(\phi_i) \rightarrow g(\phi)$, so $g$ is continuous. If $\phi = \delta_x$, then $g(\delta_x) = \delta_x(f) = f(x)$, so $g$ extends $f$.
Uniqueness
Let $\Omega$ be a compactification of $X$, i.e., $\Omega$ is compact, there is $\Pi:X \rightarrow \Omega$ a homeomorphism onto $\Pi(X)$, and $\Pi(X)$ is dense in $\Omega$ such that $\forall f \in C_b(X)$, there exists $\tilde f \in C(\Omega)$ such that $\tilde f \circ \Pi = f$.
Define $g: \Delta(X) \rightarrow \Pi(X)$ by $g = \Pi \circ \Delta^{-1}$, a homeomorphism.
Claim:
$g$ extends to a homeomorphism $G: \beta X \rightarrow \Omega$. The text proves this directly. We'll prove it using the Banach-Stone theorem, which says, in part, if $X, Y$ are compact Hausdorff spaces and there is an isometric isomorphism $\Phi$ from $C(X)$ to $C(Y)$, then $X$ is homeomorphic to $Y$ and there is a homeomorphism $\phi : X \rightarrow Y$ such that

$$\Phi(f) = \alpha f \circ \phi^{-1}, \forall f \in C(X)$$

where $\alpha : Y \rightarrow \mathbb{T}$ is continuous ( $\mathbb{T}:= \{ z \in \mathbb{F}: \vert z\vert = 1 \}$).
$g$ gives an isometric isomorphism $h:C(\Delta(X)) \rightarrow C(\Pi(X))$ by $h(f) = f \circ g^{-1}$ for all $f \in C(\Delta(X))$. By density, for $f \in C(\beta X)$, $f\vert _{\Delta(X)}$ determines $f$. I.e., given $f_1, f_2 \in C(\beta X)$ such that $f_1\vert _{\Delta(X)} = f_2\vert _{\Delta(X)}$ then $f_1 = f_2$ Similarly for $f_1, f_2 \in C(\Omega)$ and $f_1\vert _{\Pi(X)} = f_2\vert _{\Pi(X)}$.
Thus we have isometric isomorphisms:

$$\Phi : C(\beta X) \rightarrow C_b( \Delta(X) )$$

$$\Psi : C(\Omega) \rightarrow C_b( \Pi(X) )$$

To use Banach-Stone, we need an isomorphism between $C(\beta X)$ and $C(\Omega)$. (Commutative diagram). Thus, $\Lambda = \Psi^{-1} \circ h \circ \Phi$ is the isometric isomorphism we desire. By Banach-Stone, $\beta X$ and $\Omega$ are homeomorphic.
If you use the specific form of the homeomorphism from the Banach-Stone theorem, it is possible to show that the homeomorphism extends $g$.
$\qedsymbol$

Aside: For a completely regular topological space $X$, $C_b(X)$ is separable iff $X$ is a compact metric space.