Timeline for $X$ Polish geodesic implies $(P_2(X), W_2)$ geodesic
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9 events
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| Oct 21, 2013 at 10:10 | comment | added | Tapio Rajala | @User11111: Take the example from my first comment and denote the (open) lines of length 1 by $l_n$. For each positive integer $n$ take a geodesics of length $1-\frac1n$ inside $l_n$. The collection of these geodesics is closed but the multi-preimage is not (points $x$ and $y$ are missing). | |
| Oct 21, 2013 at 9:20 | comment | added | User11111 | @TapioRajala Right! Now it should work, however I didn't check the details. Let's call the multi-preimage of $U$ the set $\lbrace x \in X: S(x) \cap U \neq \emptyset \rbrace$. The multi-preimage of a closed set is closed (this requires some care). Since $G(X)$ is metric every open ball is countable union of closed sets. We get the thesis since the multi-preimage of a union is the union of the multi-preimages. | |
| Oct 21, 2013 at 6:26 | comment | added | Tapio Rajala | In the link you gave, this is not the definition of a multifunction to be Borel (see page 2). A multifunction $f \colon X \mapsto Y$ is Borel if for every open set $U \subset Y$ the set $\{x \in X : S(x) \cap U \ne \emptyset\}$ is Borel. | |
| Oct 20, 2013 at 20:19 | comment | added | User11111 | Ok, this should work. What I was trying to do was to use Kuratowski Ryll-Nardzewski theorem (for the statement see for example thm 3.3 in ls10-ftp.cs.uni-dortmund.de/pub/techreports/…), which requires a multifunction to be Borel. What I understand is a Borel multifunction is the following: for every open set $U$ in $G(X)$ the preimage of ${U}$ through $F$ (seen as a point in $2^{G(X)}$) is an open set in $X^2$. | |
| Oct 20, 2013 at 20:17 | history | edited | Otis Chodosh | CC BY-SA 3.0 |
deleted 17 characters in body
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| Oct 20, 2013 at 20:11 | history | edited | Otis Chodosh | CC BY-SA 3.0 |
corrected mistake
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| Oct 20, 2013 at 20:06 | comment | added | Otis Chodosh | Oh, oops! I'm used to thinking about the case when $X$ is compact. | |
| Oct 20, 2013 at 19:37 | comment | added | Tapio Rajala | The set of geodesics between $x$ and $y$ in a Polish space need not be compact. As an easy example consider the geodesic space consisting of two points ($x$ and $y$) and a countably infinite collection of paths of length 1 joining them. | |
| Oct 20, 2013 at 18:52 | history | answered | Otis Chodosh | CC BY-SA 3.0 |