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I've been reviewing my old exams, and I came across a question that I was unable to answer. Can anyone help me out on this one?

Let $X = \mathbb{R}$, and let $cX$ be the compactification corresponding to the algebra $A \subset C_{b}(X)$ of all uniformly continuous bounded functions. Give an example of two disjoing closed subsets of $X$ which have non-disjoint closures in $cX$. Is the compactification $cX$ metrizable?

I have an idea what to do, but I'm having trouble coming up with an example to the first part of the problem. As for the second part, I'm stuck.

Thanks in advance for any help!

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How do you define compactification corresponding to $A \subseteq C_b(X)$? –  Tunococ Aug 31 '12 at 3:28
For the second part, you may be able to exhibit a sequence in $cX$ with no convergent subsequence, contradicting the Bolzano-Weierstrass theorem. –  Nate Eldredge Aug 31 '12 at 3:38
A hint for the first part: you want to take advantage of the non-compactness of $\Bbb R$. One way to do this is to find disjoint closed sets $H$ and $K$ such that $\inf\{|x-y|:x\in H\text{ and }y\in K\}=0$. –  Brian M. Scott Aug 31 '12 at 7:36
@Tunococ: If a closed unital subalgebra $A$ of $C_b(X)$ separates the points of $X$ (for $x \neq y$ there is $a \in A$ such that $a(x) \neq a(y)$) you can embed $X$ homeomorphically into the spectrum of $A$ by sending $x$ to the character $\hat{x} \colon A \to \mathbb{C}$ given by $\hat{x}(a) = a(x)$. Since the spectrum $\sigma(A)$ of $A$ is compact, the embedding $x \mapsto \hat{x}$ from $X$ into $\sigma(A)$ yields a compactification of $X$. See also Gelfand representation. –  t.b. Aug 31 '12 at 9:21

2 Answers 2

up vote 2 down vote accepted

The space in question is called the Samuel compactification. I suggest you use the closed subsets of the real line consisting of the integers and the set of elements of the form $n+\frac 1n$ ($n$ an integer) for the first part. A good way to get some results is to note that for the space of integers the Stone-Čech compactification and the Samuel compactification coincide. This, together with the embedding of the integers into the real line, allows the use of some well-known facts about this space (e.g. that it is not metrisable) to get corresponding negative results about the Samuel compactification of the line. (This is implicit in t.b.'s remark)

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Thanks for pointing out that name. Good to know! –  t.b. Aug 31 '12 at 8:37

Since $cX$ is compact Hausdorff, it is normal, so two closed sets $A,B \subset cX$ are disjoint iff there is a continuous $f : cX \to [0,1]$ which separates them (i.e. $f=0$ on $A$ and $f=1$ on $B$). Such $f$ is the extension of a bounded uniformly continuous $g : X \to [0,1]$.

Can you find two disjoint closed subsets of $X$ which cannot be separated by a bounded uniformly continuous function?

Their closures in $cX$ will be as desired.

For the second part, following my comment, you could find a sequence in $cX$ that has no convergent subsequence, showing that $cX$ is not sequentially compact. $cX$ is compact by construction, and by the Bolzano-Weierstrass theorem, every compact metrizable space is sequentially compact, so this would rule out metrizability.

Maybe we can even choose this sequence to lie in $X$.

Suppose we have a sequence $\{y_n\} \subset X$ that converges in $cX$. In particular, for any continuous $f : cX \to [0,1]$, $\lim f(y_n)$ exists. As before, any bounded uniformly continuous $g : X \to [0,1]$ extends to a continuous function on $cX$, so $\lim g(y_n)$ must exist as well.

So I propose the following:

Find a sequence $\{x_n\} \subset X$ such that, for every subsequence $\{x_{n_k}\}$, there is a bounded uniformly continuous $g: X \to [0,1]$ such that $\lim g(x_{n_k})$ does not exist.

This sequence will have no convergent subsequence in $cX$.

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For the second part it might be easier to note that $A$ isn't separable since $\ell^{\infty}$ embeds isometrically into it: for each $n \in \mathbb{N}$ choose a triangle function $f_n$ of width $1/2$ and height $1$ around $n$ and send $(a_n) \in \ell^{\infty}$ to $f = \sum a_n f_n$. –  t.b. Aug 31 '12 at 7:20
@t.b.: That’s practically the same as the way that I’d pick Nate’s sequence and functions $g$. –  Brian M. Scott Aug 31 '12 at 7:34
@Brian: I suppose so... Once you start playing around with $\mathbb{N}$ in $\mathbb{R}$ you're probably bound to end up with something like this. Does this compactification have a standard name among topologists? I've seen "the greatest ambit" (Brook) in some papers concerned with topological transformation groups, but I wondered how standard that was. –  t.b. Aug 31 '12 at 7:48
@t.b.: Not that I know of, but that doesn’t mean much: I don’t think that I’ve actually run into it before. –  Brian M. Scott Aug 31 '12 at 7:56

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