Is the unit ball of $B(H)$ a Baire space (with the SOT)?

Question

Let $H$ be a Hilbert space, and let $B(H)$ be the set of bounded linear operators $t \colon H \to H$. Recall that we say $t_i \to t$ in the strong operator topology if $t_i \xi \to t \xi$ for every $\xi \in H$.

Question: Let $B := (B(H)_1, SOT)$ be the unit ball of $B(H)$ equipped with the strong operator topology. Is $B$ a Baire space, that is, is it the case that any intersection of countably many open dense sets is still dense?

The answer is positive if $H$ is separable. Indeed, see here for a proof of the fact that $B$ is then completely metrizable, and therefore a Baire space by the Baire Category Theorem. I'm hence mostly interested in the case $H$ is not separable, which, in turn, implies that $B(H)_1$ is much larger, and hence (roughly speaking) it is easier for a set to be nowhere dense.

Answer 1

I would say it is. Below is a sketch to more-or-less reduce to the separable case.

Given a finite set $V$ of unit vectors and some $\epsilon>0$, let $$ U_{V,\epsilon}=\{t\in B(H)_1\mid \|tv\|\leq\epsilon \text{ for every }v\in V\}. $$ The sets $\bar t + U_{V,\epsilon}$ form a basis neighborhood for the strong operator topology.

Let now $\mathcal U_n$ be a sequence of dense open sets in $(B(H)_1, SOT)$ and $U_0$ any open. We wish to show that $U_0\cap \bigcap_{n\in\mathbb N}\mathcal U_n$ is not empty. We start the argument as for the usual proof of the Baire theorem. Since $\mathcal U_1$ is dense, we may find $t\in B(H)_1$, $V_1$ and $\epsilon_1$ so that the open set $U_1= t_1+U_{V_1,\epsilon_1}$ has closure contained in $U_0\cap \mathcal U_1$. In turn, $U_1$ intersects $\mathcal U_2$ etc... This way we construct a sequence of nested open sets $U_n= t_n+ U_{V_n,\epsilon_n}$. Moreover, we may also assume that $V_n\subseteq V_{n+1}$, and that $\epsilon_{n+1} \ll \epsilon_n$, if necessary.

Since each $V_n$ is finite, their union $V=\bigcup_{n\in\mathbb N}V_n$ is countable. Let $$ H'=\overline{<t_n(V)\mid n\in\mathbb N>}^{\|\cdot\|} $$ be the norm closure of the span of $V$ and its image under any of the $t_n$. Observe that $H'$ now is separable and hence completely metrizeable. Let $p$ the projection onto $H'$ and observe that for every $n\in\mathbb N$ the operator $pt_np$ still belongs to $U_n$. It follows that the restrictions of $U_n$ to $B(H')_1$ (i.e. the sets $U_{n}'=\{t\in B(H')_1\mid \|(t-pt_np)(v)\|\leq\epsilon_n\ \forall v\in V_n \in U_{n}\}$) still are non-empty nested open sets, each containing the closure of all the following ones. Since they shrink very fast by construction, it should now be easy to deduce that they contain a Cauchy sequence in $B(H')_1$. Extending the resulting limit operator by $0$ on the orthogonal complement of $H'$ yields the desired element in $B(H)_1$.

(The sketch is inspired by the proof that arbitrary products of complete metric spaces are Baire. See Exercise 17 in Bourbaki's General Topology part 2, pp 254.)