TITLE: Intersection between orthogonal complement of a subspace and a set
QUESTION [15 upvotes]: Consider the normed vector space $E=\mathbb{R}^n$. Define
$ P=\{x \in \mathbb{R}^n: x_i \geq 0, \forall i \}$.
Let $M$ be a subspace such that $M \cap P = \{0\}$. I want to see that $M^\perp \cap {\rm Int}(P) \neq \emptyset $. This seems obvious geometrically, any idea how a short proof would look like?

REPLY [2 votes]: This is an elementary proof I discovered using a geometric version of Hahn-Banach. Hope you like it. Thanks to all of you who contributed to the discussion. :)
Suppose $M^\perp \cap \rm{Int}P=\emptyset$. Then since both sets are convex and $\rm{Int}P$ is open, it follows that there exists $0 \neq f \in E^*$ that separates them. That is, there exists $\alpha \in \mathbb{R}$ such that 
$f(x) \leq \alpha \leq f(y) \text{ for all } x\in M^\perp, y \in \rm{Int} P. $
By Reisz's Lemma $\exists x^* \in E$, $x^*\neq0$, such that $f(x)=(x,x^*)$. Now it is easy to see that since $M^\perp$ is a subspace, the first inequality implies that $(x,x^*)=0$ for all $x \in M^\perp$, ie $x^* \in (M^\perp)^\perp=M$ (since $E$ is finite dimensional). 
Therefore, from the second inequality we have $(y,x^*) \geq 0$ for all $y\in \rm{Int}P$. Finally, one only needs to note that $e_i \in \overline{\rm{Int}P}$ for all $i=1,\cdots,n$. The continuity of the inner product now implies that $(e_i,x^*) \geq 0$ for all $i$. I.e., $x^* \in P \cap M$ and $x^* \neq 0$, a contradiction. 
$\square$