Timeline for Is there a metric compactification that doesn't create new paths?
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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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`
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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 |