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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.

nanshan
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    SOT topology is normed on a finite dimensional space. Infact all Hausdorff topologies on finite dimensional vectorspaces are equivalent. – s.harp Jun 18 '18 at 21:37
  • @s.harp Thank you so much, but may I ask where I can find such conclusion, or can you explain more why all Hausdorff topologies on finite-dimensional vector spaces are equivalent? – nanshan Jun 18 '18 at 22:03
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    These lecture notes seem to be about the result: http://www.math.uni-konstanz.de/~infusino/Lect8.pdf Also this answer from Daniel Fischer on this website is good: https://math.stackexchange.com/questions/445547/how-to-endow-topology-on-a-finite-dimensional-topological-vector-space – s.harp Jun 18 '18 at 23:08

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