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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?


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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.) – Brian M. Scott Oct 22 '12 at 2:22
Crossposted:… – Qiaochu Yuan Oct 22 '12 at 2:35
up vote 3 down vote accepted

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? – ege Oct 22 '12 at 5:45
metrizability is hereditary but separability is not. – ege Oct 22 '12 at 6:10
@ege: A metrizable space is separable iff it is second countable, and second countability is hereditary. – Emil Jeřábek Oct 22 '12 at 15:57
@Emil: Thanks for answering: I was a bit slow getting back to this. – Brian M. Scott Oct 22 '12 at 18:07
thanks so much. – ege Oct 23 '12 at 1:01

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