Timeline for Existence of non-locally constant functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2014 at 18:51 | comment | added | Eric Wofsey | A slight variant of this argument does work: take $(x_n)$ to be any countable discrete subset, and instead of a single point $x$ take the set of all accumulation points. | |
| Feb 3, 2014 at 18:06 | comment | added | abx | Oops, sorry, you are right. | |
| Feb 3, 2014 at 17:59 | comment | added | Joseph Van Name | There are compact spaces that do not contain any non-trivial convergent sequences. For instance, $\beta\mathbb{N}$ has no convergent sequence. See the answers to this question mathoverflow.net/questions/138161/… for more details regarding $\beta\mathbb{N}$ and other spaces with no convergent sequences. | |
| Feb 3, 2014 at 17:58 | comment | added | Johannes Hahn | @abx: No, it doesn't. Any net has a convergent subnet, but that subnet need not necessarily be a sequence. | |
| Feb 3, 2014 at 17:48 | comment | added | abx | What I meant is that you don't need compactness to separate points. But you need it to find a convergent sequence -- in a compact (Hausdorff) space any infinite set contains a convergent subsequence. | |
| Feb 3, 2014 at 17:36 | comment | added | Joseph Van Name | Compactness is a necessary condition. | |
| Feb 3, 2014 at 17:29 | comment | added | Marten Wortel | How can you find a converging sequence? Don't you need some countability axiom for that? | |
| Feb 3, 2014 at 17:15 | history | answered | abx | CC BY-SA 3.0 |