Let $H$ be an infinite dimensional Hilbert space and let $(x_i)_1^\infty$ be an orthonormal basis for $H$.
Consider $U(H)$ the unitary group of the continuous unitary operators on $H$. Equip $U(H)$ with the topology in which the basic neighborhood of the identity I has the form $$V=B(F,\epsilon)=\{S \in U(H): ||y-S(y)||<\epsilon\mbox{ for all $y$ in $F$}\}$$ where $F$ is a finite subset of the orthonormal basis.
Let $C$ be in $U(H)$. We have that $CV$ is an open neighborhood of $C$. Now, I want to show that: $$CV=\{S \in U(H): || C(y)-C(S(y))||<\epsilon\mbox{ for all $y$ in $F$}\}.$$ Is this true? How can I prove it? An inclusion I think is clear, I show that $CV$ is contained in this set. Let $CS$ in $CV$. We have $||C(S(f))-C(f)||=||C(S(f)-f)||<||C||\epsilon=\epsilon$. Is this ok? thank you
By definition, unitaries preserve the norm. So $$ \|Cy-CSy\|=\|C(y-Sy)\|=\|y-Sy\|. $$ Thus, $CV=V$.