Pedersen ideal is the minimal hereditary dense ideal. An element $a$ is full if the ideal generated by $a$ is dense.
Assume $a\in \text{Ped}(A)$ is a positive full element. Let $A_1=\overline{aAa}$. Is $\text{Ped}(A_1)=A_1$?
Brown's theorem: Assume $A$ is $\sigma$-unital and $a$ is a positive full element, let $A_1=\overline{aAa}$, then $A\otimes \mathcal K\simeq A_1\otimes \mathcal K$.
Let $I'$ be a dense hereditary ideal of $A_1$, the hereditary ideal generated by $I'$ in $\mathcal K$ is exactly $I'\otimes \mathcal K_0$, where $\mathcal K_0$ are the finite rank operators. Since $I'\otimes \mathcal K_0\cap A$ is a dense hereditary ideal of $A$, it contains $\text{Ped}(A)$ and thus contains $A_1$, which shows $I'$ contains $A_1$. (?)