Intersection between orthogonal complement of a subspace and a set 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?

Based on the comments of my previous answer, I realise that one can simplify the argument considerably. First of all, if $M$ is a hyperplane, then the result is easily proved as follows. One can define $M$ by a linear equation $M=\{(x_1,\ldots,x_n)\in\mathbb R^n\mid x_1a_1+\cdots+a_nx_n=0 \}$ where one can take at least one of the coefficients $a_i$ to be positive. This being so, no other $a_j$ can be non-positive, since otherwise taking $x_i=-a_j$, $x_j=a_i$ and all other coordinates $x_k=0$ would give a non-zero vector in $M\cap P$. But if all $a_j$ are positive, one has $(a_1,\ldots,a_n)\in M^\perp\cap \mathrm{Int}(P)$.
Now what remains is to show that any subspace $M$ intersecting $P$ in $\vec0$ only can be extended to a hyperplane $H$ with the same property, because it then follows that $M^\perp\supseteq H^\perp$ which as we have seen meets $\mathrm{Int}(P)$. Projecting onto $M^\perp$, we need to find an appropriate image for $H$; this needs to be a hyperplane $H'$ of $M^\perp$ that meets the image $\cal C$ of $P$ (by orthogonal projection) only in $\vec0$. But since $P$ meets $M$ only in $\vec0$, its image $\cal C$ is a convex cone that is disjoint from $\mathrm{Int}(-\cal C)$. But the Hahn-Banach separation theorem (on OP's request) provides a linear form that separates $\cal C$ and $\mathrm{Int}(-\cal C)$, and its kernel provides the requested $H'$ (this could also be done using the first part of the lemma in my other answer). Finally take $H=H'\oplus M$.
