Timeline for When the boundary of any subset is compact?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 28, 2016 at 9:07 | comment | added | Mathieu Baillif | If there are only finitely many of them, take the members of the cover that contain the corresponding $U$s, this gives a finite cover of $X$. If there are infinitely many, either there is a infinite subset such that $\{x_U\}$ is open, so there is a clopen discrete subset, or infinitely many points are non-isolated, so they form an infinite discrete closed nowhere dense set, which is thus not compact. | |
| Sep 28, 2016 at 9:03 | comment | added | Mathieu Baillif | If $X$ is a $T_1$ space each of whose nowhere dense subsets is compact, then either $X$ is compact, or there is a clopen infinite discrete subset in $X$. Let me sketch the proof (following Blair). Take a cover $C$ of $X$. Take a maximal disjoint family $F$ of open sets such that each one is contained in a member of $C$. Take one point $x_U$ in each $U$ in $F$. Then $X-\cup F$ is closed and nowhere dense, cover it with finitely members of $C$. The $x_U$ not covered form a closed discrete set. (cont.) | |
| Sep 28, 2016 at 8:44 | vote | accept | Lisa_K | ||
| Sep 28, 2016 at 8:43 | comment | added | Lisa_K | +1 Thank you. Any informatin for general Tychonoff spaces? I mean not necessary spaces without isolated points. | |
| Sep 28, 2016 at 6:56 | comment | added | Mathieu Baillif | You can find all the references in Blair's paper, which can be found here, for instance: eudml.org/doc/17268?lang=cs&limit=10 | |
| Sep 28, 2016 at 6:42 | comment | added | Nate Eldredge | Can you give the full citations of the papers you mention here? | |
| Sep 28, 2016 at 6:40 | history | answered | Mathieu Baillif | CC BY-SA 3.0 |