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Aug 22, 2018 at 10:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 23, 2018 at 9:12 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jun 23, 2018 at 8:48 answer added Ljubomir Cukic timeline score: 1
Apr 29, 2018 at 0:33 comment added Yemon Choi I don't understand the votes to close...
Apr 26, 2018 at 21:20 comment added Pietro Majer Sorry, I though I saw a closure- In fact, yes, in this example $X$ itself is second countable, so I don't quite understand the role of the $X_n$ in the question
Apr 26, 2018 at 21:12 comment added Tomasz Kania Can't you take $X_1=X=L_p$ for $0<p<1$?
Apr 26, 2018 at 21:10 comment added Tomasz Kania @PietroMajer your cover is not countable.
Apr 26, 2018 at 20:54 review Close votes
Apr 29, 2018 at 10:21
Apr 26, 2018 at 20:33 comment added Pietro Majer For instance $X:=L^p([0,1])$, for $0<p<1$, is a TVS whose only convex open set is $X$ itself; it is a separable complete metric space with the distance $d(f,g):=\int_0^1|f-g|^pdt$, and you can take the $X_n$ to be subspaces of dimension $n$
Apr 26, 2018 at 20:22 history edited ABB CC BY-SA 3.0
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Apr 26, 2018 at 20:17 history asked ABB CC BY-SA 3.0