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Jul 7 at 2:02 vote accept Jeremy Brazas
Jul 6 at 18:53 history edited LSpice CC BY-SA 4.0
`\backslash` -> `\setminus`
Jul 6 at 15:45 answer added KP Hart timeline score: 10
Jul 4 at 11:08 comment added Will Brian @HenrikRüping: Every countable dense subspace of the Cantor space is homeomorphic to $\mathbb Q$. So the Cantor space is a metrizable compactification of $\mathbb Q$.
Jul 4 at 2:37 comment added Anton Petrunin It is easy to construct for proper spaces (closed+bounded=compact).
Jul 3 at 21:18 comment added HenrikRüping I would like to see a construction for the rationals.
Jul 3 at 21:05 comment added Will Brian 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/….
Jul 3 at 21:03 comment added Will Brian 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.
Jul 3 at 20:18 history asked Jeremy Brazas CC BY-SA 4.0