Let $A$ be an abelian C-algebra, B is another abelian C-algebra. Let $D_{A}$ and $D_{B}$ be the dense subsets of $A$ and $B$. Consider $A\otimes B$, is $D_{A}\otimes D_{B}$ still dense in $A\otimes B$?
I suspect that the answer is no, but I cannot give a counter examples.
If $D_A\subset A$ and $D_B\subset B$ are dense subspaces, then the algebraic tensor product $D_A\odot D_B$ is dense in $A\otimes B$. First, $A\odot B$ is dense in $A\otimes B$ by definition. So it suffices to prove that every finite sum of the form $\sum_{k=1}^N a_k\otimes b_k$ can approximated by elements from $D_A\odot D_B$. But that's not hard: Simply approximate every $a_k$ by a sequence $(a_k^{(n)})$ in $D_A$ and every $b_k$ by a sequence $(b^{(n)}_k)$ in $D_B$. Since the norm on $A\otimes B$ is a cross norm, we have $a_k^{(n)}\otimes b_k^{(n)}\to a_k\otimes b_k$. Then we can linearly extend this to get $$ \sum_{k=1}^N a_k^{(n)}\otimes b_k^{(n)}\to\sum_{k=1}^N a_k\otimes b_k. $$ Note that I didn't use that $A$ and $B$ are abelian.