Timeline for Are countable dense subspaces of $\mathbb{R}^n$ homeomorphic to ${\mathbb Q}^n$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2017 at 9:54 | comment | added | Dominic van der Zypen | Amazing, thanks @WillBrian! And I guess we can make this result to hold even for all $n < {\frak t}$! :) | |
| Sep 15, 2017 at 8:36 | comment | added | Will Brian | @GeraldEdgar: The Stone-Cech remainder of the naturals, $\mathbb N^*$, has the interesting property that every nonempty $G_\delta$ set has nonempty interior. What about intersections of larger cardinality? The cardinal number $\mathfrak p$ is, by definition, the smallest cardinality of a collection of open subsets of $\mathbb N^*$ that has a nonempty intersection with empty interior. That's a formal definition. An informal definition is that $\mathfrak p$ is the cardinal where certain diagonalization-type constructions stop working. To me, that's what makes it a really interesting number. | |
| Sep 14, 2017 at 20:34 | comment | added | Gerald Edgar | And of corse everyone but me knows what $\mathfrak p$ is. | |
| Sep 14, 2017 at 17:43 | comment | added | Will Brian | Related cool fact: If $n$ is allowed to be an infinite cardinal, then all countable dense subsets of $\mathbb R^n$ are homeomorphic if and only if $n < \mathfrak p$. (The question has already been answered, but I thought you might like to know.) | |
| Sep 14, 2017 at 17:32 | answer | added | coudy | timeline score: 9 | |
| Sep 14, 2017 at 9:44 | review | Close votes | |||
| Sep 14, 2017 at 13:04 | |||||
| Sep 14, 2017 at 6:43 | vote | accept | Dominic van der Zypen | ||
| Sep 14, 2017 at 6:11 | answer | added | Bjørn Kjos-Hanssen | timeline score: 20 | |
| Sep 14, 2017 at 5:44 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |