$\{x \in H: \overline{U(H)x}~ \text{ is compact}\}$ is a closed subset of $H$, where $U(H)$ denotes the space of unitary operators on $B(H)$

Question

Let $H$ be a Hilbert space and $U(H)$ denotes the space of unitary operators on $B(H)$. For $x \in H$, we define $U(H)x :=\{Ux: U \in U(H)\}$. Now consider the subset $\Omega \subseteq H$ as $\Omega :=\{x \in H: \overline{U(H)x}~ \text{ is compact}\}$, where $\overline{U(H)x}$ is the closure of the set $U(H)x$. I want to show that $\Omega$ is closed.
To prove that, let us consider a net $\{x_l\}_l$ in $\Omega$ such that $x_l \to x$. I have to show that $x \in \Omega$. Now $x_l \in \Omega$ implies $\overline{U(H)x_l}$ is compact, for all $l$. Now from here how can I show that $\overline{U(H)x}$ is compact? Please help me to solve this. Thank you for you time and help.

Answer 1

Hints: Let $\epsilon >0$ and choose $i$ such that $\|x-x_i\|<\epsilon/2$. $\overline {U(H)x_i}$ is covered by a finite number of open balls $B(y_j,\epsilon/2), 1 \leq j \leq m$ of radius $\epsilon /2$.

Using triangle inequality and the fact that unitary operators have norm $1$ verify that the balls $B(y_j,\epsilon), 1 \leq j \leq m$ cover $\overline {U(H)x}$. To finish the proof just recall that complete and totally bounded sets are compact.