Now assume that $C_b(X)$ is separable. Thus $({\rm ball}\,C_b(X)^*,{\rm wk}^*)$ is metrizable (5.1). Since $X$ is homeomorphic to a subset of ${\rm ball}\,C_b(X)^*$ (6.1), $X$ is metrizable. It also follows that $\beta X$ is metrizable. It must be shown that $X=\beta X$.

Suppose there is a $\tau$ in $\beta X\setminus X$. Let $\{x_n\}$ be a sequence in $X$ such that $x_n\to\tau$. It can be assumed that $x_n\neq x_m$ for $n\neq m$. Let $A=\{x_n:n\,{\rm is\, even}\}$ and $B=\{x_n:n\,{\rm is\, odd}\}$. Then $A$ and $B$ are disjoint closed subsets of $X$ (not closed in $\beta X$, but in $X$) since $A$ and $B$ contain all of their limit points in $X$. Since $X$ is normal, there is a continuous function $f:X\to[0,1]$ such that $f=0$ on $A$ and $f=1$ on $B$. But then $f^\beta(\tau)=\lim f(x_{2n})=0$ and $f^\beta(\tau)=\lim f(x_{2n+1})=1$, a contradiction. Thus $\beta X\setminus X=\varnothing$. $\hspace{10cm}\blacksquare$

The quote above comes from the end of p. 140 on Conway's A course in functional analysis. The assumption is only that $X$ is completely regular, plus the one in the first sentence of course. My problem is: how does he deduce $\beta X$ is metrizable? He indicates no reference for this fact, and I can't seem to be able to do it myself. I thought of the completion theorem, so if I prove the completion of the metric space $X$ is compact then it is $\beta X$ by uniqueness and $\beta X$ is therefore metrizable. Indeed, if the completion $X'$ is compact, supposing $f:X\to\mathbb{R}$ is continuous, let $y\in X'\smallsetminus X$, then we have $x_n\to y$ a sequence in $X$, and by continuity I can prove $f(x_n)$ converges for any such sequence to a unique limit which I call $f(y)$, and this extends $f$ to $X'$, right? So $X'$ has the extension property for continuous functions which characterizes $\beta X$, hence $X'=\beta X$. But is it true that $X'$ is compact? And how can I prove it? And if otherwise, why is $\beta X$ metrizable?


OK the above argument works for continuous functions to complete metric spaces, but not any compact Hausdorff space, so I'd have to chuck it away since it doesn't get to $\beta X$. I was thinking maybe the weak-* closure of $X$ in $C_b(X)^\ast$ is a candidate? But again, how do I extend functions to prove that thing is $\beta X$?


The last question in the update seems a bit cryptic, so let me expand. $\beta X$ is characterized by being compact, having $X$ as a dense subspace, and being such that for any $f:X\to K$, with $K$ ha compact hausdorff space, there exists $\tilde f:\beta X\to K$ an extension of $f$ to a continuous map. So if we take the weak-* closure of $X$ in $C_b(X)^\ast$, it is compact because the ball by Banach-Alaoglu is and that closure is closed in the ball, it has $X$ as a dense subspace by definition, but what about that extension property?

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    $\begingroup$ One might think of $\beta X$ as the weak-* closure of $X$ in the closed unit ball of $C_b(X)^*$, since this is metrizable it follows easily that $\beta X$ is also metrizable. I don't follow what you mean in your update. The space $C_b(X)$ consists of continuous, bounded functions from $X$ into $\mathbb{C}$? $\endgroup$ – SamM May 7 '16 at 12:51
  • $\begingroup$ @SamM see added clarification. $C_b(X)$ I guess is indeed the space of continuous bounded functions $X\to\mathbb{C}$, with the uniform norm. Or maybe real-valued… let me look at what Conway says before this proof. $\endgroup$ – MickG May 7 '16 at 12:56
  • $\begingroup$ I've just had a look myself: it seems that information you require is contained in Theorem 6.2 on page 138. $\endgroup$ – SamM May 7 '16 at 13:01
  • $\begingroup$ @SamM Hm. Seems I have a different notion of Stone-Čech compactification than Conway. He only requires that maps in $C_b(X)$ extend to continuous maps $\beta X\to\mathbb{F}$ (what is $\mathbb{F}$?). I thought it was necessary to require all continuous maps from $X$ to any compact Hausdorff space $K$ extended to continuous maps from $\beta X$ to the same $K$. If $\beta X$ only requires extensions of maps in $C_b(X)$, then that theorem proves $\beta X$ is the weak-* closure of $X$. And it seems this closure is all I need for the last paragraph, which really answers my question. $\endgroup$ – MickG May 7 '16 at 13:16
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    $\begingroup$ To MickG. Re: a sentence in your 1st paragraph ",,,if the completion of the metric space $X$ is compact it is $\beta X$..." which is wrong, The completion of the real interval $(0,1)$ with the usual metric is $[0,1]$, which is compact. But $(0,1)$ is homeomorphic to $R$ and $\beta R$ is not metrizable. $\endgroup$ – DanielWainfleet May 8 '16 at 17:04

Writing this answer to sum up the comment thread I had with @SamM, of which I post 1 and 2 screenshots ("it seems" link here), just in case someone decides to prune it as I have seen done on other posts. I credit him/her for the hints of those comments, and here is what I gathered out of them.

Firstly, it seems my doubt was from Conway having a different (but equivalent) definition of Stone-Čech compactification. My definition of $\beta X$ is by the universal property that if $K$ is a compact Hausdorff space (any one, not just real intervals) and $f:X\to K$ is continuous, then it extends to $\tilde f:\beta X\to K$, a continuous extension. Conway's definition, which I'll call $X'$, is that for any $f\in C_b(X)$ (which appears to be the space of continuous bounded real-valued functions on $X$) there exists an extension $\tilde f:X'\to\mathbb{F}$, where $\mathbb{F}$ is just a fancy notation to mean $\mathbb{R}$. Well, technically it is the field of scalars for a vector space, but here we are talking of $C_b$, which is a real Banach space.

It is theorem (6.1) on pp. 137 of Conway (screenshot) that $X$ is homeomorphic to a subset of the unit ball of $C_b(X)^\ast$, and it is Theorem (6.2) on pp. 137-138 of Conway (1 and 2) that the weak-* closure of that set is Conway's S-Č compactification $X'$. Long story short, once you have established that $\Delta:x\mapsto\delta_x$ is a homeomorphism onto its image (first theorem), the weak-* closure of $\Delta(X)$ is closed in the unit ball, which is weak-* compact by Banach-Alaoglu, hence that closure is compact, and if $f\in C_b$, then it defines a linear continuous functional on $C_b^\ast$, and in particular a continuous function on $X'$, and of course $\langle\delta_x,f\rangle=f(x)$ so this can be viewed as the required extension.

Exercise 4 on p. 141 is precisely showing that the more general property of $\beta X$ holds for $X'$. Of course, the property of $X'$ holds for $\beta X$, since $f\in C_b(X)$ is a continuous map with real values that is bounded, and thus can be seen as taking values in a compact interval of the real line, and the extension then follows from the property of $\beta X$.

Before pointing me to exercise 4, he suggested I take the inclusion o f$X$ into $X'$ and extend it to $\beta X$ by the universal property, since $X'$ is compact Hausdorff -- compact since it's closed in the compact ball, Hausdorff because the weak-* topology is always Hausdorff --, and show that the extension is a homeomorphism. I managed to prove it is continuous, surjective, open and closed, but I was stuck on injectivity, as is seen in the comment:

Let $i$ be the identity and $i'$ the inclusion of $X$ into $X'$ the weak-* closure of $X$. $f:=i'\circ i$ extends to $\tilde f:\beta X\to X'$. $\tilde f$ is continuous, hence has closed image. $X'$ is metrizable, hence Hausdorff, hence $\tilde f(\beta X)$ is closed. But $i'(X)=\tilde f(i_\beta(X))\subseteq\tilde f(\beta X)$, where $i_\beta$ is the inclusion of $X$ into $\beta X$. Then again, $i'(X)$ is dense in $X'$, so $\tilde f(\beta X)$, being closed and containing a dense subspace, must be $X'$. So $\tilde f$ is a continuous map from a compact space to a Hausdorff space, hence closed. …

To do exercise 4, I first of all need to know that for any $x\neq y$ in $X$ there is $f\in C_b(X)$ such that $f(x)\neq f(y)$. Compact Hausdorff implies Urysohn's lemma holds, so I can find $f\in C_b:f(x)=0,f(y)=1$.

Now of course if we want $f:X\to\Omega$ with $\Omega$ compact Hausdorff to extend to $\tilde f:X'\to\Omega$ continuous map, we will need to use nets. In particular, $X$ is dense in $X'$, by definition, so -- well first of all $\tilde f(x)=f(x)$ for $x\in X\subseteq X'$, or it isn't an extension of $f$ -- for $x\in X'$ we take a net $x_\alpha$ converging to $x$ from inside $X$. We need to prove $f(x_\alpha)$ converges in $\Omega$. For any $g\in C_b$ we know $g(f(x_\alpha))$ converges because $g\circ f$ extends by theorem (6.2) of Conway. If $f(x_\alpha)$ did not converge, it would need to have two subnets $f(x_{\alpha_\beta}),f(x_{\alpha_\gamma})$ converging to distinct points $y_1,y_2$. Take $g\in C_b:g(y_1)=0,g(y_2)=1$. Then $g(f(x_{\alpha_\beta}))\to g(y_1)=0$ and $g(f(x_{\alpha_\gamma}))\to g(y_2)=1$, because continuous maps preserve net convergence. But that is absurd since $g(f(x_\alpha))$ converges and so all its subnets must converge to the same number. Compactness is equivalent to every net having a convergent subnet, so I have at least one converging subnet for $f(x_\alpha)$, hence $f(x_\alpha)$ converges. Let $\tilde f(x_\alpha)=\lim_\alpha f(x_\alpha)$.

Next, we need to prove that any two nets $x_\alpha,x_\beta\to x$ give the same limit. Suppose $f(x_\alpha)\to y_1,f(x_\beta)\to y_2$. Then for any $g\in C_b$ we have $g(f(x_\alpha))\to g(y_1),g(f(x_\beta))\to g(y_2)$. Take a $g:g(y_1)=0,g(y_2)=1$. Then $g(f(x_\alpha))\to0,g(f(a_\beta))\to1$. But $g\circ f$ extends to $X'$ continuously, and $x_\alpha,x_\beta\to x$, so $g\circ f(x_\alpha),g\circ f(x_\beta)\to\widetilde{g\circ f}(x)$, impossible. Hence, the above extension is well-defined.

Now let us prove that if $g\in\mathcal{C}(\Omega)$ then $g\circ\tilde f=\widetilde{g\circ f}$. Suppose it isn't true. Then we have $x\in X'$ such that $g\circ\tilde f(x)\neq\widetilde{g\circ f}(x)$. If $x\in X$ this is impossible, since $\widetilde{g\circ f}(x)=g(f(x))=g(\tilde f(x))$. But then we have a net $x_\alpha\to x$ from inside $X$. $\tilde f$ is defined so that $\tilde f(x_\alpha)\to\tilde f(x)$, and $g$ is continuous, so $g\circ\tilde f(x_\alpha)\to g\circ\tilde f(x)$. $\widetilde{g\circ f}$ is defined as a continuous extension, hence $\widetilde{g\circ f}(x_\alpha)\to\widetilde{g\circ f}(x)$. But then:

$$\widetilde{g\circ f}(x)\leftarrow\widetilde{g\circ f}(x_\alpha)=g\circ\tilde f(x_\alpha)\to g\circ \tilde f(x),$$

which is again a contradiction. So $g\circ\tilde f=\widetilde{g\circ f}$ on all of $X'$.

Now let $x_\alpha$ be a net in $X'$ that converges to $x\in X'$. No idea where the $x_\alpha$s lie, they might bounce in and out of $X$ and converge to any point in $X$ or $X'\smallsetminus X$. We want to show $\tilde f(x_\alpha)\to\tilde f(x)$. We certainly know that:

$$g\circ\tilde f(x_\alpha)=\widetilde{g\circ f}(x_\alpha)\to\widetilde{g\circ f}(x)=g\circ\tilde f(x),$$

for any $g\in C_b$. Suppose we have $f(x_\alpha)\not\to f(x)$. Since it has a convergent subnet by compactness of $\Omega$, and if any subnet converges to $f(x)$ we know $f(x_\alpha)$ will have to converge to $f(x)$, we conclude there exists $f(x_{\alpha_\beta})$ which converges to a point $y\neq\tilde f(x)$. $g\circ\tilde f(x_\alpha)\to g\circ\tilde f(x)$ for any $g\in C_b$. But for $g\in C_b$ we have that $g\circ f(x_{\alpha)\beta})\to g(y)$. So we pick $g$ that separates $y$ and $f(x)$, as usual, and reach the same old contradiction, proving $\tilde f$ preserves net convergence, and this implies its continuity, ending the exercise.

Btw hidden in the above is that the topology on $\Omega$ is $\sigma(\Omega,C_b)$, or that $x_\alpha\to x$ if and only if $f(x_\alpha)\to f(x)$ for all $f\in C_b$.

As another extra, the theorem says $C_b$ is separable iff $X$ is a compact metric space, so $C_0=C_b$ will be nonseparable if $X$ is compact and nonmetrizable, as for example is $\{0,1\}^{\mathbb{R}}$, an example suggested by this post, a compact non-first-countable (hence non-metrizable) space.

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