Let $H$ be a complex Hilbert space with orthonormal basis $\{e_i:i\in I\}$ . Consider the $C^\ast$-algebra of the compact operators on $H$, $K(H)$. For a finite subset $F\subseteq I$, let $P_F$ be the orthogonal projection onto $span\{e_i:i\in F\}$. In an exercise one has to show, that $(P_F)$ is an approximate unit of $K(H)$. This is the background of the follwing questions:
$(P_F)$ has to be monotonically increasing, but what is a partial ordering on $K(H)$ ? If $F, S$ are finite subsets of $I$, such that $F\subseteq S$, it has to be $P_F\le P_S$. I know that $span\{e_i:i\in F\}\subset span\{e_i:i\in S\}$ but I dont know, what $P_F\le P_S$ means.
If you want to prove, that $\|TP_F-T\|\to 0$ and $\|P_FT-T\|\to 0$ for all $T\in K(H)$, it suffices to show, that $\|AP_F-A\|\to 0$ and $\|P_FA-A\|\to 0$ for all $A\in F(H)$, where $F(H)$ is the set of all finite rank operators on $H$. My question: I have a solution of this claim which says that is it possible to express $A$ as follows: $A=\sum_{k=1}^n\theta_{x_k,y_k}$ with $\theta_{x_k,y_k}(z)=x_k\langle y_k,z\rangle$ for all $x_k,y_k,z\in H$, see for example https://en.wikipedia.org/wiki/Finite-rank_operator . Then if you know that $\|P_FA-A\|\to 0$, how to prove $\|AP_F-A\|\to 0$? Is the following correct: $\|AP_F-A\|=\|(AP_F-A)^*\|=\|P_FA^*-A^*\|\to 0$? I think here we need, that $F(H)$ is closed under taking adjoints, is it true? Regards.
