# Problem with a proof on Conway's book

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?

Update

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 $$\beta X$$?

Clarification

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?

• 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 $\mathbb{C}$? – SamM May 7 '16 at 12:51
• @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. – MickG May 7 '16 at 12:56
• I've just had a look myself: it seems that information you require is contained in Theorem 6.2 on page 138. – SamM May 7 '16 at 13:01
• @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. – MickG May 7 '16 at 13:16
• 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. – 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.