It is easy to check that for a C*-algebra $A$ in $\mathfrak{B}(H)$ and a (norm-)closed, left ideal $L$ in $A$ there exists a projection $p$ such that the closure $L_s$ of $L$ in the strong topology on $\mathfrak{B}(H)$ can be written as $$L_s=A_s p$$ where $A_s$ denotes the strong closure of $A$. Also obvious is that $$A_sp\cap A$$ is a closed, left ideal in $A$ and that $$L\subset A_s p\cap A$$ holds. My Question is whether the converse inclusion is also true, i.e. whether $$L=A_s p \cap A.$$ Equivalently one could ask whether for a hereditary C*-subalgebra $B\subset A$ $$B=B_s\cap A=pA_sp \cap A$$ is true. I would be grateful for answers and if possible a proof. Thanks a lot.


Let $A=B(H)$ and $L$ be the compact operators $K(H)$.

The strong closure of $L$ is $B(H)$: Let $\mathcal F$ be the directed set of finite dimensional subspaces of $H$ and for $V\in\mathcal F$ let $p_V\in K(H)$ be the orthogonal projection onto $V$. You then have $\lim_{V\in\mathcal F} p_Vx=x$ for any $x\in H$, so $\lim_{V\in\mathcal F}p_V = \Bbb1$ in the strong topology.

So $p$ must be $\Bbb1$, it follows that

$$L=K(H)\subsetneq B(H) = A_sp\cap A$$

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