Let $A$ be separable $C^*$-algebra and $B$ a arbitrary $C^*$-algebra and $\operatorname{CPC(A,B)}:=\{f:A\to B | f\;\text{ is completely positive contractive}\}$. Let $\{a_1,a_2,,\}$ a countable dense subset in $A^1=\{a\in A:\|a\|\le 1\}$. For $f,g\in \operatorname{CPC(A,B)}$ define a metric $$d_B(f,g):=\sum\limits_{k=1}^\infty 2^{-k}\|f(a_k)-g(a_k)\|.$$
Clearly $d_B(f,g)\le 2\sum\limits_{k=1}^\infty 2^{-k}<\infty$ and all the metric properties are easy to prove.
My first question: For the well-definedness of $d_B$, do I have to check that the definition of $d_B$ doesn't depend on the choice of the dense subset $\{a_1,a_2,,\}$ ? Furthermore I want to prove: for nets $(f_{\lambda})_\lambda\subseteq \operatorname{CPC(A,B)}$, it is
$$d_B(f_{\lambda},f)\to 0 \iff \|f_\lambda (a_k)-f(a_k)\|\to 0\ \text{ for all k}\in \mathbb{N}.$$
Since this implies that for all choices of dense subsets $\{a_1,a_2,,\}$, $d_B$ induces the same topology on $\operatorname{CPC(A,B)}$. However, here I'm stuck, my second question is: Why is $$\lim_\lambda \sum\limits_{k=1}^\infty 2^{-k}\|f_\lambda (a_k)-f(a_k)\|=\sum\limits_{k=1}^\infty \lim_\lambda 2^{-k} \|f_\lambda (a_k)-f(a_k)\|?$$
I appreciate your help
The definition does depend on the sequence $\{a_k\}$; even by reordering you would most likely change the metric.
Suppose that $d_B(f_\lambda,f)\to0$. Then, for any fixed $k$, $$ \|f_\lambda(a_k)-f(a_k)\|\leq 2^k d_B(f_\lambda,f)\to0. $$
Conversely, suppose that $\|f_\lambda(a_k)-f(a_k)\|\to0$ for all $k$. Fix $\varepsilon>0$. Choose $m$ such that $2^{-m+1}<\varepsilon/2$. Choose $\lambda_0$ such that $$\|f_\lambda(a_k)-f(a_k)\|<\varepsilon/2,\ \ \ k=1,\ldots,m$$ for all $\lambda\geq\lambda_0$ (this can be done, since we are using finitely many $k$). Now \begin{align} d_B(f_\lambda,f)&=\sum_{k=1}^m2^{-k}\|f_\lambda(a_k)-f(a_k)\|+\sum_{k=m+1}^\infty2^{-k}\|f_\lambda(a_k)-f(a_k)\|\\ \ \\ &\leq\frac\varepsilon2\,\sum_{k=1}^m2^{-k}+\sum_{k=m+1}^\infty2^{-k+1}\\ \ \\ &\leq\frac\varepsilon2+2^{-m+1}\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon. \end{align} In other words, if $\lambda\geq\lambda_0$, then $d_B(f_\lambda,f)<\varepsilon$.