Timeline for Does every set $X$ have a topology for which the only continuous self-surjection is the identity map?
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 18, 2015 at 8:07 | answer | added | Dominic van der Zypen | timeline score: 3 | |
| Aug 14, 2015 at 7:15 | comment | added | Dominic van der Zypen | Very nice question! I think spaces $(X,\tau)$ with the property that the identity is the only continuous self-surjection must have some low separation property: if $x\neq y\in X$ have the property that ${\cal N}_x = {\cal N}_y$ then a continuous self-bijection that swaps $x, y$ can be constructed (if I'm not mistaken). | |
| Aug 13, 2015 at 5:49 | answer | added | Adam Przeździecki | timeline score: 12 | |
| Aug 13, 2015 at 5:05 | answer | added | Joseph Van Name | timeline score: 7 | |
| Aug 13, 2015 at 4:50 | comment | added | Eric Wofsey | This answer also gives an example whenever $|X|^{\aleph_0}=|X|$. Still, it seems like this sort of thing is massive overkill and there ought to be some fairly simple construction that always works. | |
| Aug 13, 2015 at 4:22 | comment | added | Eric Wofsey | This question also seems related. In particular, the answer there gives an example with $|X|=\mathfrak{c}$. | |
| Aug 13, 2015 at 4:04 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Aug 13, 2015 at 4:01 | comment | added | Joel David Hamkins | I have a feeling that the rigidity results mentioned in mathoverflow.net/a/6300/1946 will be relevant. | |
| Aug 13, 2015 at 3:47 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |