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,
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,
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.