Let $H$ be a finite-dimensional Hilbert space, and $U$ be its unitary group. Let $B(H)$ be the space of all bounded linear operators on $H$.
Why $U$ is compact in the SOT(strong operator topology) of $B(H)$?
In Heine-Borel theorem, for a subset $S$ of $R^n$, $S$ is compact (in norm topology) if and only if it is closed and bounded. However, now $U$ is in the SOT (not norm topology) of $B(H)$. Can I still use Heine-Borel thm to prove it?
Thanks a lot.