Let $A$ be a unital $C^*$-algebra. $RR(A) = 0$ iff for any $\epsilon > 0$ and for any pair of orthogonal positive elements $a, b \in A_+$ there is a projection $p \in A$ such that $||pa - p || < \epsilon$ and $(1 - p) b = b$.
Is there any idea how to prove both sides? I'm stucked.
$\textbf{(ii)}$The elements in $A_{sa}$ with finite spectra are dense in $A_{sa}$, then $\textbf{(iii)}$ every hereditary $C^*$-subalgebra of $A$ has an approximate unit consisting of projections.
$\textbf{proof}$:This is the implication proved in G. K, PEDERSEN, The linear span of projections in simple $C^*$-algebras.
By the hypothesis we can have $\textbf{(v)}$ there is a projection $p \in A$ such that $||(1-p)a||\leq \epsilon$ and $||pb|| \leq \epsilon$.
We have $\textbf{(v)} \Rightarrow \textbf{(i)} = RR(A) = 0$. Since, if $a \in A_{sa}$ and $\epsilon >0$, consider the orthogonal elements $a_+$ and $a_-$ in $A_+$. Choose a projection $p$ in $A$ such that $||(1-p)a_+|| \leq \epsilon$ and $||pa_- || \leq \epsilon$. We get $|| a - (pap + (1-p)a(1-p))|| 2\epsilon - \epsilon p \leq pap$. Thus the element $y = pap + 2\epsilon p + (1-p)a(1-p)-2\epsilon(1-p)$ is invertable in $A_{sa}$ and $||a - y|| \leq 4 \epsilon$.
Also, with a little effort we can show that $ \textbf{(v)} \Rightarrow \textbf{(ii)}$ and $ \textbf{(v)} \Rightarrow \textbf{(iii)}$
And of course it is easy to see that $ \textbf{(i)} \Rightarrow \textbf{(vi)}$ and $\textbf{(vi)}$: for each pair $a, b \in A_{+}$ and $\epsilon >0$ such that $||ab|| < \epsilon^{2}$, there is a projection $p$ in $A$, such that $||(1-p)a|| < \epsilon$ and $||pb||<\epsilon$