and $\overline{\cup_{n=1}^{\infty} F_n} = A$. Applying Definition \[D\_9719\_AppRep\_Phi\] for each $F_n$, we get $z_{n}(g) \in {{\operatorname{U}}}(A)$ for $g \in G$ such that:

1.  \[D\_9719\_AppRep.1’\] $\| {\alpha}_g (a) - z_{n}(g) a z_{n}(g)^* \| < \frac{1}{n}$ for all $a \in F_n$ and all $g \in G$.

2.  \[D\_9719\_AppRep.2’\] $\| z_{n}(g) z_{n}(h) - z_{n}(gh)\| < \frac{1}{n}$ for all $g, h \in G$.

3.  \[D\_9719\_AppRep.3’\] $\| {\alpha}_g (z_{n}(h)) - z_{n}(g h g^{-1})\| < \frac{1}{n}$ for all $g, h \in G$.

Now, for $g \in G$, set $w^{(0)}_{ g}= (z_{n}(g))_{n\in {{\mathbb{Z}}_{> 0}}}$ as an element in $l^{\infty} ({{\mathbb{Z}}_{> 0}}, A)$ and $w(g)$ be its image in $A_{\infty}$. Then, for all $a \in \cup_{m=1}^{\infty} F_m$, we have $$\label{Eq1.20200510}
\lim_{n \to \infty}
 \left\| \alpha_{g} (a) - z_{n}(g) a z_{n}(g)^*\right\|=0.$$ Let ${\varepsilon}>0$ and let $x \in A$. Choose $m \in {{\mathbb{Z}}_{> 0}}$ and $a \in F_m$ such that $$\label{Eq2.20200510}
\|x - a \|<\tfrac{{\varepsilon}}{3}.$$ By (\[Eq1.20200510\]), we can choose $N \in {{\mathbb{Z}}_{> 0}}$ such that $N \geq m$ and for all $n \geq N$ $$\label{Eq3.20200510}
\left\| \alpha_{g} (a) - z_{n}(g) a z_{n}(g)^*\right\| < \frac{{\varepsilon}}{3}.$$ For all $n \geq N$, we use (\[Eq2.20200510\]) and (\[Eq3.20200510\]) to get $$\begin{aligned}
\left\| \alpha_{g} (x) - z_{n}(g) x z_{n}(g)^*\right\|
&<
\left\| \alpha_{g} (x) - 