# A strongly open set which is not measurable in the weak operator topology

Let $H$ be a non-separable Hilbert space and $\{e_i\}_{i\in I}$ be an orthonormal basis for $H$. Let $J$ be a uncountable proper subset in $I$.

Let us put $$E=\{x\in B(H): \lVert xe_j\rVert <1: \exists j\in J\}$$

One may check that $E$ is an open set in the strong operator topology but not in the weak operator topology.

Question1: I feel $E$ is not in the sigma algebra generated by the weak operator topology but have no evidence to prove it.

Question2: Let us assume that dimension of $H$ is of $c$. It seems that $E$ is WOT measurable if and only if every SOT open set is WOT measurable.

• Check your proof. The set $E$ you describe isn't SOT open. Apr 4, 2016 at 7:40
• Dear Nik, Thanks for your pay attention. I have corrected . Please look at again.
– ABB
Apr 4, 2016 at 11:26
• I guess you can just see this by counting. More precisely, fixing the cardinality of $J$ and $I-J$ (both uncountable) and vary $J$ you get too many sets $E_J$, more than you have in the $\sigma$-algebra, which are all distinct, but isomorphic by unitaries. Now, the $\sigma$-algebra is invariant under unitaries, so if of one is measurable then they all are. Apr 4, 2016 at 20:07
• @user89334, Thanks for your nice comment. It seems both SOT and WOT on $B(H)$ have the same cardinal $2^{B(H)}$, which implies that both corresponding sigma algebras have the same cardinal too. It means that by counting argument, no result will be obtained!
– ABB
Apr 5, 2016 at 5:06
• @user89334: Depending on the size of $I$, a counting argument won't work; for instance, if $I$ has cardinality $\aleph_1$, then you get $2^{\aleph_1}$-many sets $E_J$, but it's consistent with ZFC that $2^{\aleph_1} = 2^{\aleph_0}$. Apr 5, 2016 at 13:42