Proposition 13.7.2.label Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_{b}(E; F)$.
Proof. Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_{F}(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_{E}(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_{E}(0, 1)) \subset Y + U$. In which case, $T(B_{E}(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_{E}(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$.$\square$
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