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Let $\mathbb{H}$ be a complex infinite dimensional Hilbert space and $B \left( \mathbb{H} \right)$ bes the algebra of bounded linear operators on $\mathbb{H}$.

I would like to find a compact subset of $B \left( \mathbb{H} \right)$.

One of the candidate is the set of all compact operators since it is closed. Is the set of all compact operators on $B \left( \mathbb{H} \right)$ ?

Otherwise, could you please let me know another possibility?

Thank you in advance.

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No nonzero vector space can be compact because it is not bounded. So compact operators do not form a compact set. Any finite set is compact in any topological space. In particular the zero operator gives you a compact subset.

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