Finite Rank Operator is Compact

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Let $H, K$ be Hilbert spaces.

Let $T \in B_{00} \left({H, K}\right)$ be a bounded finite rank operator.

Then $T \in B_0 \left({H, K}\right)$, i.e., $T$ is compact.


Since $T$ is bounded, it maps bounded sets to bounded sets. Because its range is finite dimensional, the closure of an image of a bounded set is compact. Therefore, $T$ is a compact linear operator.