I think this should not be too difficult, but I am not an expert. I did not get an answer on stackexchange. 


Let $A$ be a $C$\*-algebra and let $p,q\in A^{\*\*}$ be two commuting projections. Then there exist self-adjoint nets $(x\_i)\_i$ and $(y\_j)\_j$ in $A$ such that $x\_i\to p$ and $y\_j\to q$ in the weak$^\*$-toplogy. Can these nets be chosen to be commutative, that is, $x\_iy\_j=y\_jx\_i$ for all $i,j$?


Extra question: Is this true if $A$ is a $JB$-algebra?



Commuting nets for commuting projections
I think the following provides a counterexample, though the bidual of a $C^\*$-algebra always makes me nervous.


Let $A=M\_2\otimes C[0,1]$. Any bounded, Borel measurable, $M\_2$ valued function on $[0,1]$ will give an element of $A^{\*\*}$; for the projection $p\in A^{\*\*}$ we take the function
\begin{equation}
p(t)=\begin{cases} \begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix} & 0\leq t\leq \frac12, \\ \begin{pmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{pmatrix} & \frac12 <t \leq 1\end{cases}
\end{equation}
and for $q$ we take the complementary projection $1-p$. Suppose by way of contradiction there are nets of selfadjoints $(x\_i), (y\_i)$ in $A$ with $x\_i\to p$ and $y\_i\to 1-p$ weak-$\*$, and $x\_iy\_j=y\_jx\_i$ for all $i,j$. Since the $y\_j$ converge to $1-p$ and commute with all the $x\_i$, it follows that all the $x\_i$ commute with $1-p$, and hence also with $p$. This now means that there are continuous $a\_i, b\_i, c\_i, d\_i$ so that $x\_i$ has the form
\begin{equation}
x\_i(t) = \begin{cases} \begin{pmatrix} a\_i(t) & 0\\ 0 & b\_i(t)\end{pmatrix} & 0\leq t\leq \frac12, \\ \begin{pmatrix} c\_i(t) & d\_i(t)\\ d\_i(t) & c\_i(t) \end{pmatrix} & \frac12 <t \leq 1\end{cases}.
\end{equation}
Continuity at $t=\frac12$ forces $x\_i(\frac12 )$ to be a scalar multiple of the identity for all $i$, but then $x\_i(\frac12 )$ cannot converge to $p(\frac12)$, which is a contradiction since weak-$\*$ convergence controls pointwise congvergence. 


