13
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 5
    $\begingroup$ @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

1 Answer 1

Reset to default
9
$\begingroup$

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 $X\setminus B$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.