Let $\Omega \subseteq \mathbb{R}^n$ be open. Suppose that $\chi \in C_0^\infty(\Omega \times \Omega \setminus \text{diag} (\Omega))$, i.e., $\chi$ is supported away from the diagonal in $\Omega \times \Omega$.
I would like to show that there exist functions $\varphi_n, \psi_n \in C_0^\infty(\Omega)$ such that $\text{supp}(\varphi_n) \cap \text{supp}(\psi_n) = \emptyset$ for each $n$ and the tensor products $ \varphi_n \otimes \psi_n$ converge to $\chi$ in the topology of $C_0^\infty(\Omega \times \Omega)$.
Here is my attempt so far. I already know that tensor products of the form $\tilde{\varphi} \otimes \tilde{\psi}$, $\tilde{\varphi}, \tilde{\psi} \in C_0^\infty(\Omega)$ are dense in the topology of $C_0^\infty(\Omega \times \Omega)$. So, viewing $\chi$ as a member of $C_0^\infty(\Omega \times \Omega)$, we can produce $\tilde{\varphi}_n \otimes \tilde{\psi}_n$ converging to $\chi$ in $C_0^\infty(\Omega \times \Omega)$.
If we can then produce $\chi_1, \chi_2 \in C_0^\infty(\Omega)$ such that $\text{supp}(\chi_1) \cap \text{supp}(\chi_2) = \emptyset$ and $(\chi_1 \otimes \chi_2 ) \chi = \chi$, we are done (can then take $\varphi_n = \chi_1 \tilde{\varphi}_n$, $\psi_n = \chi_2\tilde{\psi}_n$. However, I have yet to show that we can construct $\chi_1$, $\chi_2$ as desired.
Hints or solutions are greatly appreciated!