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A remainder of a Tychonoff space is a $bX\setminus X$, where $bX$ is a compactification of $X$. Does every separable metrizable space has a separable metrizable remainder?

thanks,

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    Do you mean: Does every separable, metrizable space have at least one compactification with a separable, metrizable remainder? (Use `\setminus` to get the set difference symbol.)2012-10-22
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    Crossposted: http://mathoverflow.net/questions/110291/remainder-of-metrizable-seperable-space2012-10-22

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The answer is yes. Let $D=\{x_n:n\in\omega\}$ be a countable dense subset of $X$, and let $d$ be a compatible metrix bounded by $1$. Define

$$\varphi:X\to[0,1]^\omega:x\mapsto\left\langle d(x,x_n):n\in\omega\right\rangle\;;$$

it’s not hard to check that $\varphi$ is a homeomorphism of $X$ onto $\varphi[X]$. Now let $bX=\operatorname{cl}\varphi[X]$, where the closure is taken in $[0,1]^\omega$; clearly this is a compactification of $X$, and since $[0,1]^\omega$ is a separable metrizable space, so is $bX\setminus X$.

Added: The space $[0,1]^\omega$ is known as the Hilbert cube. It is universal for separable metrizable spaces or, equivalently, for second countable $T_3$-spaces.

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    May be clear, but how can we say since $bX$ is separable metrizable, so is $bX\X$. Is it just true for Hilbert Cube?2012-10-22
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    metrizability is hereditary but separability is not.2012-10-22
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    @ege: A metrizable space is separable iff it is second countable, and second countability *is* hereditary.2012-10-22
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    @Emil: Thanks for answering: I was a bit slow getting back to this.2012-10-22
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    Again sorry but, compatible metrix bounded by 1 is the same with dicsrete metric or not? I could not find the definition of compatible metric.2012-10-23
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    @ege: If $\langle X,\tau\rangle$ is a metrizable space, a metric $d$ on $X$ is *compatible* if it generates the topology $\tau$; *compatible* is short for *compatible with the given topology*. By definition there is at least one compatible metric for a metrizable space, and given any metric $d$ on a space, it’s a basic exercise to show that there is a bounded metric that generates the same topology. Thus, every metrizable space has a compatible bounded metric, and by scaling we can take the bound to be $1$.2012-10-23
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    I think, $d_{1}=minimum\{1,d\}$ is a bounded metric by 1 and generates the same topology. And also, these are equivalent metrics. Is it right?2012-10-23
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    @ege: Yes, that’s right. Another way to do it is to define $$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\;.$$2012-10-23
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    Again sorry, but how we can say $\phi$ is one to one and continuous. I think to show to one to one, we need some properties of the metric $d$. Take any $x and y$ which are different.Then there exists a $n$ such that $d(x,x_{n})$ not equal to $d(y,x_{n}) ( but how can I show)2012-10-29
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    @ege: Since $D$ is dense in $X$, there is a sequence $\langle x_{n_k}:k\in\Bbb N\rangle\to x$. If $d(x,x_m)=d(y,x_m)$ for all $m\in\Bbb N$, then $\langle x_{n_k}:k\in\Bbb N\rangle\to y$, and therefore $x=y$.2012-10-29
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    @Brian: Sorry I didn’t notice this earlier, but in your answer, “second-countable $T_4$” and “separable metrizable” are the same thing, one is not more general than the other.2012-10-30
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    @Emil: Of course. I must have been asleep.2012-10-30
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    thanks so much for your help.2012-11-01