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Let $H$ be a Hilbert space over $\mathbb C$ , I know that the space of all compact operators on $H$ ($\mathcal K(H)$ ) is closed in $\mathcal L(H)$ in norm topology . I want to ask , what is the closure of $\mathcal K(H)$ in $\mathcal L(H)$ in strong operator topology ? Is the space of compact operators still closed then ?

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Compact operators are dense in all bounded operators in the strong operator topology (that is, the closure is $\mathcal{L}(H)$). Indeed, let $T:H\to H$ be a bounded operator and let $U$ be a basic strong operator neighborhood of $T$. This means there are finitely many vectors $v_1,\dots,v_n\in H$ and $\epsilon>0$ such that $U$ is the set of all operators $S$ such that $\|Sv_i-Tv_i\|<\epsilon$ for all $i$. Now let $V$ be the span of the $v_i$, which is a finite-dimensional subspace of $H$. Let $P$ be the orthogonal projection onto $V$ and let $S=TP$. Then $S$ has finite rank and hence is compact, but $Sv_i=TPv_i=Tv_i$ for all $i$, so $S\in U$.

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