Is it true that $F\cap \overset{\circ}{E}=\emptyset\implies F\cap \overset{\_}{E}=\emptyset$? Let $E,F\subset \mathbb{R}^n$ where $F$ is open and $E$ is arbitrary. Is it true that:
$$
F\cap \overset{\circ}{E}=\emptyset\implies F\cap \overset{\_}{E}=\emptyset
$$
Intuitively I think this is true, but my intuition on open and closed sets has proven to be not that good. How to prove it or how to construct a counter example?
Edit: How about 
$$
F\cap E=\emptyset\implies F\cap \overset{\_}{E}=\emptyset
$$

Simplest counterexample I can think of:
$E = \{(0, 0, \ldots, 0)\} \subseteq \mathbb{R}^n$, $F = (-1, 1)^n \subseteq \mathbb{R}^n$. Then $\mathring{E} = \emptyset \implies \mathring{E}\cap F = \emptyset$, but $\bar{E}\cap F = E\cap F = \{(0, 0, \ldots, 0)\}$.
