Let $\mathcal{H}$ be a separable Hilbert space. A unitary system $\mathcal{U}$ is a subset of the unitary operators acting on $\mathcal{H}$ which contains the identity operator $I$. A vector $\eta\in \mathcal{H}$ is called a Bessel vector for $\mathcal{U}$ if $$\mathcal{U}\eta=\{U\eta;\quad U\in\mathcal{U}\}$$ is Bessel sequence; that is, $$\sum_{U\in\mathcal{U}}|\langle f, U\eta\rangle|^2<B\|f\|^2$$ for every $f\in\mathcal{H}$. If $ \mathcal{U}\eta $ is a Bessel sequence for $\mathcal{H}$, then the vector $\eta$ is called a Bessel vector for $\mathcal{U}$. The set of all Bessel vectors for $\mathcal{U}$ is denoted by $\mathcal{B(U)}$.
My question: Is $\mathcal{B(U)}$ closed with Hilbert space norm? It seems that $\mathcal{B(U)}$ is not closed with Hilbert space norm. Please give me a counterexample of it.