# Closed balls in $B(H)$ are totally bounded?!

Let $H$ be a separable Hilbert space. It is well-known that the strong operator topology is metrizable on bounded parts of $B(H)$. The metric is given by: $$d(x,y)=\sum \frac{\|(x-y)e_i\|}{2^i}$$ where $\{e_i\}$ is an orthonormal basis in $B(H)$.

Trivially the closed unit ball $B_1$ of $B(H)$ is $d$-bounded. Is $B_1$ $d$-totally bounded too?

If the space $B(H)$ is infinite dimensional then the unit ball under the metric $d$ which is induced by the norm on $H$ is not compact.
But in a metric space a set is compact $\iff$ is complete and totally bounded with the subspace topology induced by the metric on that set.
So the unit ball $B_1$ is not totally bounded because it is not compact.
• Could you please say why $B_1$ is $d$-complete? – Ali Bagheri Sep 17 '17 at 15:33
• Of course $B_1$ is $d$-complete. The unit ball is sot-closed: a sot limit of contractions is a contraction. Even a wot limit of contractions is a contraction. – Martin Argerami Sep 17 '17 at 19:00