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Timeline for Meagre sets of bounded operators

Current License: CC BY-SA 4.0

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Jun 8, 2021 at 7:31 history edited Matthias Ludewig CC BY-SA 4.0
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May 27, 2021 at 0:25 comment added Narutaka OZAWA It is well exemplified by the following. Let $P_t$ denote the orthogonal projection from $L^2[0,1]$ onto $L^2[0,t]$. Then, $[0,1]\ni t\mapsto P_t\in B(L^2[0,1])$ is a homeomorphism w.r.t. SOT, but $\{P_t\}$ is discrete in the norm topology and hence any subset is relatively open.
May 26, 2021 at 14:56 comment added Matthias Ludewig Oh right, because $\mathbb{B}(H)$ is not separable in the norm topology...
May 26, 2021 at 13:25 comment added Narutaka OZAWA Since a norm open subset of $B(H)$ is not necessarily a countable union of open balls, norm open subsets need not be Borel in SOT.
May 26, 2021 at 10:14 history asked Matthias Ludewig CC BY-SA 4.0