Let $\mathcal{H}$ be a Hilbert space and $U(\mathcal{H})$ be the set of unitary operators on it. We define: \begin{equation} V_f(U_0,r):=\{U \in U(\mathcal{H})| \; \; ||U_0(f)-U(f)|| < r \}, \; \; f \in \mathcal{H}, \; r>0. \end{equation} We now claim a subset $W$ to be open in $\mathcal{O}_S$, the strong operator topology, iff \begin{equation} W \in \mathcal{O}_{S} \iff W=\bigcup_{U_0 \in W} \bigcap_{j=1}^{k} V_{f_j}(U_0,r_j). \end{equation} Why does the collection of such open sets define a topology?
My trial so far is this: $\varnothing \in \mathcal{O}_S$, as we would take empty union. And empty union is the empty set. Moreover, it is also true that: \begin{equation} U_0 \in V_f(U_0,r), \; \; \text{since} \; \;||U_0(f) - U_0(f)||=0 <r. \end{equation} To show that the set $U(\mathcal{H})$ also belongs to the topology, we have to show: \begin{equation} U(\mathcal{H})=\bigcup_{U_0 \in \mathcal{H}} \bigcap_{j=1}^{k} V_{f_j}(U_0,r_j), \end{equation} We do this by double inclusion. Let $A \in U(\mathcal{H})$. Then, we have by the above observation that \begin{equation} A \in V_{f_j}(A,r_j) \; \; \text{for all} \; \; f_j \in \mathcal{H}, \; \; r_j>0. \end{equation} This lets us conclude that: \begin{equation} A \in \bigcap_{j=1}^{k} V_{f_j}(A,r_j). \end{equation} Taking union over all $A\in U(\mathcal{H})$ shows us that: \begin{equation} A \in \bigcup_{A \in U(\mathcal{H})} \bigcap_{j=1}^{k} V_{f_j}(A,r_j), \; \; \text{i. e.} \; \; U(\mathcal{H}) \subseteq \bigcup_{A \in U(\mathcal{H})} \bigcap_{j=1}^{k} V_{f_j}(A,r_j) \end{equation} Now suppose $A \in \bigcup_{U_0 \in \mathcal{H}}\limits \bigcap_{j=1}^{k} \limits V_{f_j}(U_0,r_j).$ By our above observation, we know: \begin{equation} U_0 \in \bigcap_{j=1}^{k} \limits V_{f_j}(U_0,r_j). \end{equation} Taking union over all $U_0 \in U(\mathcal{H})$ lets us conclude $A \in U(\mathcal{H}).$ Thus we have successfully shown by double inclusion that \begin{equation} U(\mathcal{H})=\bigcup_{U_0 \in W} \bigcap_{j=1}^{k} V_{f_j}(U_0,r_j). \end{equation} How to check the other two axioms? There i'm a bit lost.
By definition the set $W$ is open iff for any $U_0\in W$ there is a finite collection of $f_j\in \mathcal{H}$ and a collection of positive numbers $r_j,$ $1\le j\le n,$ such that $$\bigcap_{j=1}^nV_{f_j}(U_0,r_j)\subset W$$ Therefore $$W\supset\bigcup_{U_0\in W}\bigcap_{j=1}^nV_{f_j}(U_0,r_j)$$ The reverse containment is obvious. All the elements $n,$ $f_j$ and $r_j$ depend on $W_0,$ so they should be denoted as $n_{W_0},\, f_{j,W_0},\, r_{j,W_0}.$
The set $U(\mathcal{H})$ is open by definition, as it is an entire collection of unitary operators. If we consider $U(\mathcal{H})$ as a subset of $B(\mathcal{H})$ then $U(\mathcal{H})$ is not open in $B(\mathcal{H}) $