Commuting nets for commuting projections 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?
 A: 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. 
