Suppose I have a separable metric space $X$. I wanted to ask if there exists a separable metric compactification of this space $\overline{X}$, s.t. $X$ is considered open in $\overline{X}$.
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What about the Alexandroff extension? I'm not entirely sure that there always is a way to extend the metric to the compactification, though. – A.P. Dec 28 '15 at 14:36
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I forgot to mention I need the compactification to be metric – User666x Dec 28 '15 at 14:45
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This question might be relevant. – A.P. Dec 28 '15 at 14:45
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Yes, thank you. I noticed a few things and changed my question accordingly – User666x Dec 28 '15 at 15:04
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Do you require the metric on $\overline X$ to be an extension of the given metric on $X$? – Lee Mosher Dec 28 '15 at 16:10
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Yes it should be an extension – User666x Dec 28 '15 at 16:53
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This is possible iff $X$ is also locally compact.
Let $\beta X$ denote the Stone-Cech compactification of $X$. The following lemmas are standard.
Lemma. $X$ is open in $\beta X$ iff $X$ is locally compact.
Lemma. If $\gamma X$ is a compactification of $X$, then there is a continuous surjection $f:\beta X\to\gamma X$ such that $f[X]=X$ and $f[\beta X\setminus X]=\gamma X\setminus X$.
It easily follows that:
Theorem.
(1) If $X$ is open in some compactification, then $X$ is locally compact.
(2) If $X$ is locally compact, then $X$ is open in every compactification.
Back to your question...
By (1), locally compact is necessary. So $\mathbb Q$, for instance, would be a counterexample. (Actually, if $\gamma \mathbb Q$ is a compactification of $\mathbb Q$, then $\mathbb Q$ contains no open subset of $\gamma \mathbb Q$!)
Now let $X$ be separable metric and locally compact. As $X$ is separable metric, it embeds into the Hilbert cube $[0,1]^\omega$. The closure of $X$ in this cube is a separable metric compactification, and $X$ is open in it by (2).
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