Am reading Wiki's proof of the Hyperplane Separation Theorem and am having trouble with the last part of the proof. Let me give you the structure of the argument and explain precisely where I have problem.
Theorem. Let $A$ and $B$ be two disjoint nonempty convex subsets of $\mathbb{R}^n$. Then there exist a nonzero vector $v$ and a real number $c$ such that
$$ \langle x,v\rangle \geq c,{\text{ and }}\langle y,v\rangle \leq c $$ for all $x \in A$ and $y \in B$.
The proof begins with a Lemma, the proof of which I understand:
Lemma. Let $K$ be a nonempty closed convex subset of $\mathbb{R}^n$. Then there exists a unique vector in $K$ of minimum norm (length).
Thus, given disjoint nonempty convex sets $A$, $B$ we can let
$$K=A-B$$
which is nonempty and convex. Hence so is its closure $\bar{K}$. We can therefore apply the Lemma to obtain a vector $v \in \bar{K}$ of minimum norm. One can then show that (see Wiki's article)
$$ \langle z,v\rangle \geq |v|^{2}$$
for all $z \in K$ or equivalently $\langle x-y,v\rangle \geq |v|^{2}$ for all $x \in A$ and $y \in B$. If $v$ is nonzero we are done because
$$\inf _{x\in A}\langle x,v\rangle \geq |v|^{2}+\sup _{y\in B}\langle y,v\rangle $$ so taking $c=|v|^{2}+\sup _{y\in B}\langle y,v\rangle$ does what is wanted.
Now, the last paragraph of the proof deals with the general case (where $v=0$ is possible) by dividing into two cases.
Case 1. We assume that the interior of $K$ is nonempty. The article then reads “ The interior can be exhausted by a nested sequence of nonempty compact convex subsets $K_{1}\subset K_{2}\subset K_{3}\subset \cdots$ ”. I suppose it means we can find such a sequence with $K^\circ=\cup_{n \in \mathbb{N}} K_n$.
My first question is: How can I construct such a sequence explicitly? Since $K^\circ$ is nonempty my idea is to let $K_1=\bar{B}(z_0)$ be a closed ball centered at some $z_0\in K^\circ$ with $\bar{B}(z_0) \subset K$. But I don't know how to define $K_2$ from there...
Case 2. The interior of $K$ is empty. The article then reads “The affine set that $K$ spans has dimension less than that of the whole space. Consequently $K$ is contained in some hyperplane $\langle \cdot ,v\rangle =c$ ”.
My second question is : How can I show that $\text{span}(K)$ has dimension less than $n$?
I can show that any subspace of $\mathbb{R}^n$ of dimension less than $n$ is included in a hyperplane of the form $\{ x \in \mathbb{R}^n : \langle x ,v\rangle =0 \}$ for some nonzero vector $v$. Hence in this second this case I believe we have $c=0$. Am I correct?
Thanks a lot for your help.
That's two questions.
The second is the easier one. We know $K$ is convex. If $K$ has $n+1$ affinely independent points, then $K$ contains the $n$-simplex $S$ they span. But $S$ has non-empty interior in $\Bbb R^n$, and therefore so has $K$. So every $n+1$ points in $K$ are affinely dependent, and so the affine hull of $K$ is a proper flat (translate of a linear subspace) in $\Bbb R^n$.
For the first, you need to prove that $K^\circ$ is convex. This requires a little work. Now let $A_n$ be the points of $K_n$ at distance $\ge 1/n$ from the complement of $K^\circ$ (I suppose let $A_n=\Bbb R^n$ if $K=\Bbb R^n$). Then each $A_n$ is closed and convex. Again this needs some work.
But you wanted an exhaustion by compact sets, not closed sets! Just take a sequence of compact convex sets $B_1\subset B_2\subset \cdots$ exhausting $\Bbb R^n$ and let $K_n=A_n\cap B_n$.