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I am looking for an example of a separable, completely metrizable space $X$, that has a compactification which is not metrizable. Does such an example exist? And what if $X$ is a separable banach space?

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3 Answers 3

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Some addition to Brian's answer: The Stone-Čech-comactification is almost never metrizable: Suppose $X$ is a completely regular space such that $\beta X$ is metrizable. If $X$ is not compact, let $p \in \beta X \setminus X$. Now $X$ is dense in $\beta X$ and the latter is metrizable, so there is a sequence $(x_n)$ of distinct points in $X$, converging to $p$. The sets $E := \{x_{2n} \mid n \in \mathbb N\}$ and $O := \{x_{2n+1} \mid n\in \mathbb N\}$ are then disjoint, closed subsets of $X$. As $X$, being metrizable, is normal, there is a continuous function $f \colon X \to [0,1]$ with $f[E] = \{0\}$ and $f[O] = \{1\}$. Now, as $[0,1]$ is compact, there is a continuous extension $F \colon \beta X \to [0,1]$ of $f$. But as $x_{2n} \to $, $x_{2n+1} \to p$, but $F(x_{2n}) \to 0$ and $F(x_{2n+1}) \to 1$, such a function can't exist.

So: Either $X$ is compact and metrizable or $\beta X$ is not metrizable.

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Yes: $\Bbb N$ with the discrete topology. The metric given by $d(m,n)=0$ if $m=n$ and $d(m,n)=1$ if $m\ne 1$ is complete, since the only Cauchy sequences are the eventually constant sequences, and $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$, is not metrizable. Indeed, points of $\beta\Bbb N\setminus\Bbb N$ don’t even have countable local bases.

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So, now I would like to change the question: What if X is a separable banach space? –  user25640 Oct 20 '12 at 14:22
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@user25640: Have you actually thought about your new question? Like say whether this answer can be adapted to it? –  Chris Eagle Oct 20 '12 at 14:23

For an elementary example:

Take $X= \mathbb{R}$ and construct the following compactification: $Y= \mathbb{R} \cup \{ \pm \infty_1, \pm \infty_2 \}$ where $\{ \{ (n+[0,1/2[) \cup \{ + \infty_1\}, n \geq p \}, p \geq 0 \}$ is a neighborhood basis of $+ \infty_1$, $\{ \{ (n+[1/2,1[) \cup \{ + \infty_1\}, n \geq p \}, p \geq 0 \}$ a neighborhood of $+ \infty_2$, and symmetrically for $- \infty_1$ and $- \infty_2$.

Then suppose that $Y$ is metrizable. Let $x_n=n+ \frac{1}{2}- \frac{1}{n}$ and $y_n=n+ \frac{1}{2}+ \frac{1}{n}$; we have $x_n \underset{n \to + \infty}{\longrightarrow} + \infty_1$ and $y_n \underset{n\to + \infty}{\longrightarrow} + \infty_2$, but for every $\epsilon>0$ and for $n$ sufficently large, $d(+ \infty_1,+ \infty_2) \leq d(+ \infty_1,x_n)+d(x_n,y_n)+ d(+ \infty_2,y_n) < \epsilon$, hence $d(+ \infty_1,+ \infty_2)=0$, whereas $+ \infty_1 \neq + \infty_2$.

Therefore, $Y$ is a nonmetrizable compactification of $X$.

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