Timeline for Open set of geodesics implies the set of starting points is open
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2014 at 23:33 | vote | accept | User11111 | ||
| Oct 21, 2013 at 9:23 | history | edited | User11111 | CC BY-SA 3.0 |
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| Oct 21, 2013 at 7:24 | comment | added | Tapio Rajala | @Misha: It is an intereting question indeed. And the existence of a Borel selection is exactly where the current question was heading. If the multivalued mapping is Borel in the sense described above, then a Borel selection of geodesics immediately exists, see mathoverflow.net/questions/145351/… | |
| Oct 21, 2013 at 6:40 | comment | added | Misha | @TapioRajala: Dear Tapia, this makes more sense. Another interesting question is if for every $X$ there exists a Borel map $S: X^2\to G(X)$. For instance, this is the case for Riemannian manifolds. | |
| Oct 21, 2013 at 5:11 | comment | added | Tapio Rajala | @Misha. For a multivalued mappping $S$ in this context $S^{-1}(A)$ is defined as $\{x \in X : S(x) \cap A \ne \emptyset\}$. The unit circle is still a counterexample to preimages being open. Asking for Borel might make more sense. | |
| Oct 21, 2013 at 4:25 | comment | added | TaQ | @ User11111. If you have e.g. $X=\mathbb S^{\kern.5mm 1}$, for every open $A\subseteq G(X)$ the set of $(x,y)\in X^{\kern.5mm 2}$ with $S((x,y))\in\{A\}$, i.e $S((x,y))=A$ is the empty set, and hence is trivially open. | |
| Oct 21, 2013 at 3:58 | comment | added | TaQ | @ User11111. If you have $S:X^2\rightarrow 2^{G(X)}$, then what do you mean by "preimage through $S$ of an open subset of $G(X)$" ? | |
| Oct 21, 2013 at 3:44 | comment | added | Misha | Your map is $S: X^2\to 2^{G(X)}$, as you assign to $x,y$ the set of all geodesics connecting $x$ and $y$, just read what is written in your post. How are you planning to define continuity if you have no topology on the target space? | |
| Oct 20, 2013 at 19:56 | comment | added | User11111 | I'm not interested in a topology on $2^{G(X)}$, I'm taking $S$ open in $G(X)$. | |
| Oct 20, 2013 at 18:34 | comment | added | Misha | The question now makes less sense and should be edited: You have to explain which topology you put on the set of all subsets of $G(X)$ since your new map takes values not in $G(X)$ but in $2^{G(X)}$. If your topology is at all reasonable, the circle will provide a counter-example to the continuity question. | |
| Oct 20, 2013 at 18:18 | comment | added | Misha | Even if $X$ is the unit circle with path-metric then there is no continuous map $S$, therefore, preimages of open sets will not be in general open. | |
| Oct 20, 2013 at 18:10 | history | edited | User11111 | CC BY-SA 3.0 |
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| Oct 20, 2013 at 15:38 | vote | accept | User11111 | ||
| Oct 20, 2013 at 18:09 | |||||
| Oct 20, 2013 at 11:23 | answer | added | Tapio Rajala | timeline score: 2 | |
| Oct 20, 2013 at 10:57 | review | First posts | |||
| Oct 20, 2013 at 11:04 | |||||
| Oct 20, 2013 at 10:40 | history | asked | User11111 | CC BY-SA 3.0 |