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Every separable metric space $A$ has a metrizable compactification, i.e. a compact metrizable space $X$ for which $A$ embeds topologically as a dense subspace of $X$. There are many approaches to constructing metrizable compactifications and probably several I am unaware of.

Question: Given an arbitrary separable metrizable space $A$, does there exist a metrizable compactification $X$ of $A$ such that there does not exist any path $\alpha:[0,1]\to X$ with $\alpha(0)\in A$ and $\alpha(1)\in X\setminus A$?

In other words, I'd like a metrizable compactification of $A$ so that every path component of $A$ is also a path component of the compactification. I can answer this question affirmatively when $A$ is locally compact (using a topologist's-sine modification of the one-point metric compactification) but this doesn't generalize and I only have ad-hoc approaches for non-locally compact cases.

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  • $\begingroup$ Nice question! I don't have time to check the details at the moment, but here is an idea that I think should work. First, prove the Stone-Cech compactification $\beta X$ has all the properties you want except for metrizability. Next, you can find a "metrizable reflection" of $\beta X$; it has many of the same properties as $\beta X$, and in particular this property of not giving new paths should reflect down. A metrizable reflection of $\beta X$ is obtained by taking the (countable!) lattice of open sets from $\beta X$ that are in a countable elementary submodel, then spacifying that lattice. $\endgroup$
    – Will Brian
    Commented Jul 3 at 21:03
  • $\begingroup$ The general idea is that if you see a property of $\beta X$ that can be expressed in a sufficiently "first-order" way, then you can reflect that property down to a metrizable compactification of $X$. You can see more about this kind of thing if you look at/around Lemma 3.2 here: wrbrian.wordpress.com/wp-content/uploads/2012/01/…. $\endgroup$
    – Will Brian
    Commented Jul 3 at 21:05
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    $\begingroup$ I would like to see a construction for the rationals. $\endgroup$
    – HenrikRüping
    Commented Jul 3 at 21:18
  • $\begingroup$ It is easy to construct for proper spaces (closed+bounded=compact). $\endgroup$
    – Anton Petrunin
    Commented Jul 4 at 2:37
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    $\begingroup$ @HenrikRüping: Every countable dense subspace of the Cantor space is homeomorphic to $\mathbb Q$. So the Cantor space is a metrizable compactification of $\mathbb Q$. $\endgroup$
    – Will Brian
    Commented Jul 4 at 11:08

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Here's a counterexample. Let $B$ be a Bernstein set in the plane, so $B$ and its complement intersect every uncountable closed subset of $\mathbb{R}^2$.

Let $X$ be a metric compactification of $B$, with embedding $f:B\to X$. Apply Lavrentieff's theorem to find $G_\delta$-subsets $G\subseteq\mathbb{R}^2$ and $H\subseteq X$ together with a homeomorphism $g:G\to H$ that extends $f$.

The complement of $G$ in $\mathbb{R}^2$ is countable and hence, by Cantor's theorem $G$ is pathwise connected. Since $B$ is a proper subset of $G$ there is a path in $G$ that connects a point from $B$ to a point in $G\setminus B$. The image of that path under $g$ connects a point of $B$ to a point in $X\setminus B$.

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