is quasi-central in $\mathfrak A$. Identifying $\mathfrak A$ with the pull-back $\mathfrak A(\infty)  \oplus_{ \overline \tau, \rho }  \ell^\infty(\{ \mathfrak A(n)\})$, we see that if $x\in \mathfrak A(\mathbb N)$ and $a\in \mathfrak A$, then $xa = x \overline \sigma (a)$ and $ax = \overline \sigma(a) x$. Let $a\in \mathfrak A$. We should show that $\lim_{ m \to \infty } \| p_ma - ap_m \| = 0$. Let $\epsilon > 0$. For some large $N$ we may pick $x\in \bigcup_{k=1}^N \mathfrak B_k$ and $y\in \mathfrak A(\mathbb N)$ such that $\| a  - x - y \| < \frac{ \epsilon }{ 4 }$.

Since $\lim_{ m \to \infty } \| p_my - yp_m \|  = 0$, there exists $N_{2} \geq N$ such that $\| p_{m} y - y p_{m} \| < \frac{ \epsilon }{ 4 }$ for all $m \geq N_2$. For $m\geq N$ we have $$\begin{aligned}
\| p_m x - x p_m \| &=& \| p_m \overline{\sigma}(x) - \overline \sigma(x) p_m \| \\
&=& \max_{n=1,\dots, m} \| p_m^n \pi_n(x) - \pi_n(x) p_m^n\| \leq \|x\|/m.\end{aligned}$$ Suppose $m \geq \max \left\{ N_{2} , \frac{ 4 ( \| x \| + 1 ) }{ \epsilon } \right\}$. Then $$\begin{aligned}
\| p_{m} a - a 