Timeline for Anti-symmetric mappability relation
Current License: CC BY-SA 3.0
15 events
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| Feb 2, 2017 at 23:10 | comment | added | Włodzimierz Holsztyński | I feel that the question was somewhat incomplete. I'd add a requirement that there exist $\ x\ y\in X\ $ such that $\ x\ $ is mappable onto y. One may even consider a stronger requirement (a narrower class): every $\ x\in X\ $ is either mappable onto another point or a point different from $\ x\ $ is mappable onto $\ x$. | |
| Feb 2, 2017 at 16:25 | vote | accept | Dominic van der Zypen | ||
| Feb 2, 2017 at 15:44 | answer | added | Will Brian | timeline score: 2 | |
| Feb 2, 2017 at 15:26 | comment | added | Joel David Hamkins | Great! Why don't you summarize it all in an answer? This space has the desired property vacuously. | |
| Feb 2, 2017 at 15:21 | comment | added | Will Brian | @JoelDavidHamkins: I just looked at the paper. I think Theorem 11 is the one that clearly answers the question (it says: the identity is the only mapping of M_2 onto a non-degenerate subcontinuum of M_2). Here "mapping" means continuous function and "non-degenerate subcontinuum" means a closed, connected subspace with more than one point. Since continuous functions on M_2 always have closed, connected images, we can interpret this theorem to say that any continuous function from M_2 to itself, other than the identity, is constant. | |
| Feb 2, 2017 at 15:01 | comment | added | Joel David Hamkins | Cook claims to answer my question in his preamble, but I am confused about whether his theorem statement (theorem 9) actually does answer it. | |
| Feb 2, 2017 at 14:56 | comment | added | Joel David Hamkins | I thought that there might be such a space amongst the continua. And I had seen Cook's theorem, but somehow thought it was only about homeomorphisms. Does "map" for Cook mean continuous function only? | |
| Feb 2, 2017 at 14:46 | comment | added | Will Brian | @JoelDavidHamkins: Good point. And yes, there are such spaces, like the one Ramiro de la Vega points to in this related question: mathoverflow.net/questions/188729/… | |
| Feb 2, 2017 at 14:09 | comment | added | Joel David Hamkins | Is there a Hausdorff space (with more than one point) having no non-constant non-identity continuous self-maps? So every continuous self-map is either constant or the identity. Such a space would be an example. | |
| Feb 2, 2017 at 10:01 | history | edited | Dominic van der Zypen | CC BY-SA 3.0 |
added 11 characters in body
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| Feb 2, 2017 at 8:22 | comment | added | Włodzimierz Holsztyński | The question seems about quasi-ordering not being a partial ordering(?). | |
| Feb 2, 2017 at 8:17 | comment | added | Włodzimierz Holsztyński | @Anton, in topology it's common to assume that maps or mappings stand for continuous functions. | |
| Feb 2, 2017 at 8:14 | comment | added | Włodzimierz Holsztyński | mappable?--how about "$x$ pays attention to $y$" or "$x$ appreciates $y$" or .... | |
| Feb 2, 2017 at 8:09 | comment | added | user1688 | Do you mean a non-constant continuous map? | |
| Feb 2, 2017 at 7:53 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |