I want to show that the unitary group $U(\mathcal H)$ of a Hilbertspace $\mathcal H$ is a topological group wrt the strong operator topology. For the standard proof it is most convenient to use that the strong operator topology is generated by the subbase:
$$\big\{\ U_x(A;\epsilon):=\{B\in X \mid |B-A|_x<\epsilon\}\mid A \in U(\mathcal H), x \in \mathcal H, \epsilon \in \mathbb R_{>0}\big\}$$
My problem is with the proof that multiplication is jointly continuous, the proof shows that for every open set of the form $U=U_x(A_1A_2;\epsilon)$ there exist open sets $V_1, V_2$ so that:
$$M(V_1\times V_2) \subset U$$
Where $M$ is the multiplication. In the construction we have $A_1 \in V_1$ and $A_2 \in V_2$.
Apparently this statement (which is not hard to show) is enough to show that multiplication on $U(\mathcal H)$ jointly continuous, but I can't reach the result from here.
What we need of course is that $M^{-1}(U)$ is open, but all we get from this is that it contains an open subset for any open $U$.
We want to show $M^{-1}(U)$ is open. Let $(A_1, A_2) \in M^{-1}(U)$, and let's find a neighborhood of $(A_1, A_2)$ in $M^{-1}(U)$.
Now $A_1A_2$ is in $U$, which is open. So there's a neighborhood of $A_1A_2$ in $U$ of the form $\cap_1^n U_{x_i}(A_1A_2;\epsilon_i)$, by your subbase observation.
Let $V_1^i\times V_2^i$ be the sets in your construction, for $1\le i \le n$.
Then $(A_1, A_2) \in \cap_1^n V_1^i\times V_2^i \subset M^{-1}(\cap_1^n U_{x_i}(A_1A_2;\epsilon_i)) \subset M^{-1}(U)$. But $\cap_1^n V_1^i\times V_2^i$ is open. So we took an arbitrary point $(A_1, A_2)$ from $M^{-1}(U)$ and produced a neighborhood of it in $M^{-1}(U)$. Therefore $M^{-1}(U)$ is open.