Separating hyperplane when $E_1 \cap \mbox{int}(E_2) = \emptyset$

Question

I need to prove that given $E_1, E_2 \subset \mathbb{R}^n, E_1 \ne \emptyset, E_2 \ne \emptyset$ convex set and $\mbox{int}(E_2) \ne \emptyset, E1 \cap \mbox{int}(E_2) = \emptyset$, there exists a hyperplane H that separates $E_1$ from $E_2$. I've seen how to prove it for $E_1 \cap E_2 = \emptyset$, but, what about this case?

Answer 1

To elaborate on my comment, since $\text{int}(E_2)$ is convex, we know there is a separating hyperplane $h = \{x \in \mathbb{R}^n : \langle x, 1 \rangle = b \}$ with half spaces $h^+ = \{x \in \mathbb{R}^n : \langle x, 1 \rangle \geqslant b \}$ and $h^- = \{x \in \mathbb{R}^n : \langle x, 1 \rangle \leqslant b \}$ so that $E_1 \subseteq h^+$ and $\text{int}(E_2) \subseteq h^-$.

Recall that closure is monotone in set inclusion. Then $\overline{\text{int}(E_2)} \subseteq \overline{h^-}$, i.e., $E_2 \subseteq h^-$. $\square$

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I am using the definition of separation that one set lies in $h^+$ and another in $h^-$. If you need that $E_1 \cap E_2 = \varnothing$, you don't sufficient condition to conclude that.

It's not hard to come up with an example of an open convex set $O$ and closed convex set $C$ so $O$ and $C$ can be separated by a hyperplane so that $h \cap O = \varnothing$, but $h \cap C \neq \varnothing$ (hint: can you construct something of this form from two half spaces?) In this case, $h$ separates $\overline{O}$ and $C$, even though $\overline{O} \cap C \neq \varnothing$.