$A$ has a countable dense subset. How to describe a possible countable dense subset in $M_n(A)$? Let $A$ be a C$^*$-algebra and $M$ be a countable dense subset in $A$. Let $M_n(A)$ be the $C^*$-algebra of $n\times n$-matrices with entries in $A$, $n\in\mathbb{N}$. Then $M_n(A)$ should have a countable dense subset as well. Is $L=\{(a_{ij})_{i,j=1,\ldots,n}\in M_n(A):\;  a_{ij}\in M\; \text{for all}\; i,j \}$ such a subset? 
It is clear that $L$ is countable, but is $L$ dense in $M_n(A)$?
 A: Yes, because entrywise convergence in $M_n(A)$ implies convergence. Indeed, if $a^{(t)}=(a^{(t)}_{kj})$, $a=(a_{kj})$, and $a_{kj}^{(t)}\to a_{kj}$ for each pair $k,j$, then $a^{(t)}\to a$. Below is an argument. 
The norm in $M_n(A)$ is given by the operator norm of the action of $M_n(A)$ on $H^n$, when we represent $A\subset B(H)$. Then, for any $b\in M_n(A)$, we have 
\begin{align}
\|b\|&=\sup\left\{|\langle b\xi,\eta\rangle|:\ \xi,\eta\in H^n,\ \|\xi\|=\|\eta\|=1 \right\}\\ \ \\
&=\sup\left\{\left| \sum_{k,j}\langle b_{kj}\xi_j,\eta_k\rangle\right|:\ \xi,\eta\in H^n,\ \|\xi\|=\|\eta\|=1 \right\}\\ \ \\
&\leq\sup\left\{ \sum_{k,j}|\langle b_{kj}\xi_j,\eta_k\rangle|:\ \xi,\eta\in H^n,\ \|\xi\|=\|\eta\|=1 \right\}\\ \ \\
&\leq\sup\left\{ \sum_{k,j}\| b_{kj}\|\,\|\xi_j\|\,\|\eta_k\|:\ \xi,\eta\in H^n,\ \|\xi\|=\|\eta\|=1 \right\}\\ \ \\
&\leq \sum_{k,j}\| b_{kj}\| \\ \ \\
&\leq n^2\,\max\{\|b_{kj}\|:\ k,j=1,\ldots,n\}\\ \ \\
\end{align}
Then
$$
\|a^{(t)}-a\|\leq n^2\,\max\{\|a^{(t)}_{kj}-a_{kj}\|:\ k,j=1,\ldots,n\}\to0.
$$
