The strong topology on $U(\mathcal H)$ is metrisable The strong operator topology on a Banach space $X$ is usually defined via semi-norms:
For any $x \in X$, $|\cdot|_x: B(X) \to \mathbb R, A \mapsto \|A(x)\|$ is a semi-norm, the strong topology is the weakest/coarsest topology which makes these maps continuous.
Alternatively it is generated by the sub-base $\left\{B_\epsilon(A;x)=\{B\in X \mid |B-A|_x<\epsilon\}\phantom{\sum}\right\}$.
If we define this topology on a separable Hilbert space $\mathcal H$ and restrict it to the subset of unitary operators, a statement in a book I am reading is that this is then a metrisable topology.
My question is then, how can I prove this (especially, why do we need separability)? What does the metric look like (is it constructable)?
When can this result be extended to other bounded subsets of $B(\mathcal H)$?
 A: I figured I might write up an answer. Let us consider a bounded subset $U\subset B(\mathcal H)$ where $\mathcal H$ is a seperable Hilbert space. The comment by @AliBagheri gives metric
$$d(A,B):=\sum_{n\in \mathbb N}2^{-n}\,\|(A-B) e_n\|$$
Where $\{e_n\}_{n \in \mathbb N}$ is a Hilbert basis of $\mathcal H$. The existence of a countable basis follows from separability, but boundedness is not necessary for this to be a metric. The content that follows is a proof that this metric generates the same topology as the strong operator topology on $U$. Let $C$ be a bound of $U$.
To see that the strong operator topology is finer than the metric topology, consider a metric ball $B_\epsilon(A)$. Eventually for some $N$, the series $\sum_{n=N}^\infty 2^{-n+1}\, C$ will be smaller than $\frac12 \epsilon$. If you consider an element of the finite intersection
$$B \in V:= \bigcap_{n=1}^N B_{\epsilon/2}(A;e_n)$$
Then from $\|(A-B)e_n\|≤2C$:
$$d(A,B)=\sum_{n=1}^\infty 2^{-n}\| (A-B) e_n\| ≤\sum_{n=1}^N 2^{-n} \frac\epsilon2+\sum_{n=N+1}^\infty 2^{-n}\, 2C <\epsilon$$
You get $V \subset B_\epsilon(A)$, $V$ is open in the strong operator topology and $A \in V$. So any neighbourhood of $A$ in the metric topology contains a neighbourhood of $A$ in the strong operator topology, and the strong topology is finer than the metric topology.
To get the other direction we need first that for a dense subset $\{x_n\}_n$ of $\mathcal H$, the sets $B_{\epsilon}(A;x_n)$ also generate the strong operator topology.
Obviously the strong operator topology is finer than this topology, on the other hand for $x \in \mathcal H$, there exists a $x_n$ in the dense subset so that $\|x_n-x\|<\frac{\epsilon}{4\,C}$. Then
$$|A-B|_{x} =\|(A-B)(x-x_n+x_n)\|≤\|A-B\|\,\|x-x_n\|+|A-B|_{x_n}<\frac{\epsilon}2+|A-B|_{x_n}$$
So $B_{\epsilon/2}(A;x_n)\subset B_\epsilon(A;x)$. This implies that the topology generated by $B_{\epsilon}(A;x_n)$ is finer than the strong operator topology, so they are equal.
The space of finite linear combinations of $e_n$ is a dense subset of $\mathcal H$. If $x=\sum_n^N a_n e_n$ then:
$$\left\|(A-B)\sum_{n=1}^N a_n e_n \right\|≤\sum_{n=1}^N |a_n|\, \|(A-B)e_n\|≤2^N\,\max_n\{|a_n|\}\sum_{n=1}^\infty 2^{-n}\, \|(A-B)e_n\|$$
(Finite linear combinations are needed, otherwise the sum in the middle does not have to converge). So $$d(A,B)<\frac\epsilon{2^N\,\max_n\{|x_n|\}}=:\delta$$ implies $B \in B_\epsilon(A;x)$. Since these $B_\epsilon(A;x)$ are a neighbourhood subbase of the strong operator topology, this implies that the metric topology is finer than the strong operator topology.
