# Is there a metric compactification that doesn't create new paths?

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.

• 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. Commented Jul 3 at 21:03
• 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/…. Commented Jul 3 at 21:05
• I would like to see a construction for the rationals. Commented Jul 3 at 21:18
• It is easy to construct for proper spaces (closed+bounded=compact). Commented Jul 4 at 2:37
• @HenrikRüping: Every countable dense subspace of the Cantor space is homeomorphic to $\mathbb Q$. So the Cantor space is a metrizable compactification of $\mathbb Q$. Commented Jul 4 at 11:08

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 $$X\setminus B$$.