# Transitivity of the Cuntz sub-equivalence

Let $A$ be a $C^*$-algebra and $a,b \in A$ positive elements. We define a relation (Cuntz sub-equivalence) by saying $$a\lesssim b: \Leftrightarrow \exists\, (r_n)_{n\in\mathbb{N}}\subset{A}\text{ such that } \|r_n^* b r_n-a\|\rightarrow 0.$$ I want to show, that this relation is transitive. The only proof I can come up with requires the sequence $(r_n)_n$ to be bounded.

Thank you

Note that $a\precsim b$ if and only if for every $\varepsilon>0$ there exists $r\in A$ such that $\|a-rbr^*\|\leq\varepsilon$.
Assume that $a\precsim b\precsim c$.
To show that $a\precsim c$, let $\varepsilon>0$. First, using that $a\precsim b$, choose $r\in A$ such that $\|a-rbr^*\|\leq\tfrac{\varepsilon}{2}$. We may assume that $r\neq 0$. Next, using that $b\precsim c$, choose $s\in A$ such that $\|b-scs^*\|\leq \tfrac{\varepsilon}{2}\|r\|^{-2}$. Then $$\|a-rsc(rs)^*\| \leq \|a-rbr^*\| + \|rbr^*-rsc(rs)^*\| \\ \leq \tfrac{\varepsilon}{2} + \|r(b-scs^*)r^*\| \\ \leq \tfrac{\varepsilon}{2} + \|r\| \|b-scs^*\| \|r^*\| \leq \varepsilon.$$