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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.