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Jan 8, 2023 at 11:43 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question was bumped anyway)
Jul 14, 2014 at 19:21 answer added user12713 timeline score: 3
Feb 4, 2013 at 8:35 history edited Dirk CC BY-SA 3.0
added 250 characters in body
Feb 4, 2013 at 8:33 vote accept Dirk
Jan 30, 2013 at 22:42 answer added R W timeline score: 14
Jan 30, 2013 at 16:49 comment added Dirk Thanks a lot for the enlightening answers so far! Since I am off for a long weekend, I will have a closer look in a few days.
Jan 30, 2013 at 16:48 comment added Dirk Indeed, I wanted metric spaces - question edited. Thanks for pointing this out.
Jan 30, 2013 at 16:45 history edited Dirk CC BY-SA 3.0
Changed setup from Hausdorff to metric spaces.
Jan 30, 2013 at 16:26 answer added jbc timeline score: 1
Jan 30, 2013 at 14:13 answer added Dan timeline score: 9
Jan 30, 2013 at 11:40 history edited user9072
tag
Jan 30, 2013 at 11:32 answer added Jochen Wengenroth timeline score: 7
Jan 30, 2013 at 11:01 answer added Pietro Majer timeline score: 11
Jan 30, 2013 at 10:11 comment added Ian Morris Do you perhaps want $\Omega$ to be metric? It is not clear to me how the Prokhorov and Wasserstein metrics on $\mathcal{P}(\Omega)$ might be defined in the absence of a metric on $\Omega$, and I am suspicious of the suggestion that $\mathcal{P}(\Omega)$ is metrisable even when $\Omega$ is not (for example in the case where $\Omega$ is the Stone-Čech compactification of $\mathbb{N}$). In fact, can we not identify $\Omega$ with the subset of $\mathcal{P}(\Omega)$ which comprises the Dirac measures, and deduce that if $\mathcal{P}(\Omega)$ is metrisable then $\Omega$ must be metrisable also?
Jan 30, 2013 at 9:24 history asked Dirk CC BY-SA 3.0