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](https://math.stackexchange.com/questions/2515140/metrizability-of-the-strong-operator-topology-on-the-set-of-unitary-operators-of) 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.



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


