Question:
Suppose $X$ is a closed convex nonempty set in $\mathbb{R}^n$. Define $W := \{x \in \mathbb{R}^n \mid x_i = x_i' \; \forall i \in I\}$ for some $x' \in \mathbb{R}^n$ and $I \subseteq \{1,\dots,n\}$.
If $X \cap W = \emptyset$, is there a hyperplane $H := \{x\in \mathbb{R}^n \mid v^\intercal x = \alpha \}$ such that $$ X \subseteq \{x\in \mathbb{R}^n \mid v^\intercal x \leq \alpha \} $$ and $$ W \subseteq \{x\in \mathbb{R}^n \mid v^\intercal x > \alpha \} $$ ?
Where this comes from:
I'm looking at an algorithm that may run into this type of situation in each iteration. If it does, one needs to find a polyhedron $P$ that such that $X \subseteq P$ and $P \cap W = \emptyset$. I'm trying to show that this is guaranteed.
If there is such a hyperplane $H$ as in the question, I'd just set $P := \{x \in \mathbb{R}^n \mid v^\intercal x \leq \alpha\}$.
Thoughts:
I'm not sure what to call this type of separation, but it's not covered by any of the typical separating hyperplane theorems that I've seen.
If $X$ were also bounded (and therefore compact), strict separation would hold. But I can't make that assumption about $X$.
I would have just asked this about generic closed convex nonempty sets $X$ and $W$, but then I realized that if $X = \{ x \in \mathbb{R}^2 \mid x_2 \geq \textrm{e}^{x_1} \}$ and $W = \{x \in \mathbb{R}^2 \mid x_2 \leq 0\}$ then there would be no such hyperplane. (Though there would be if the definitions of $X$ and $W$ were switched.)
Here is a counter-example to the statement as given. Define: $$ X = \{(x,y) \in \mathbb{R}^2 : y \geq e^x\}$$ $$ W = \{(x,0) \in \mathbb{R}^2 : x \in \mathbb{R}\}$$ Suppose there is a nonzero vector $(u,v)$ and real number $\alpha$ such that \begin{align} &xu + yv \leq \alpha \quad \forall (x,y) \in X \quad (Eq 1)\\ &xu + 0v > \alpha \quad \forall (x,0) \in W \quad (Eq 2) \end{align} Then (Eq 2) means $$ xu > \alpha \quad \forall x \in \mathbb{R} $$ Now if $u\neq 0$ we can make $x = M(-u)$ for large $M$ to make $xu$ as small as we like, violating the fact that $xu > \alpha$ for all $x \in \mathbb{R}$. So we know that $u=0$. This means that (Eq 2) reduces to $$ 0 > \alpha $$ On the other hand (Eq 1) and $u=0$ gives $$ yv \leq \alpha \mbox{ whenever $y \geq e^x$ for some $x \in \mathbb{R}$} $$ In particular: $$ e^x v \leq \alpha \quad \forall x \in \mathbb{R}$$ Taking a limit as $x\rightarrow -\infty$ gives $$ 0 \leq \alpha$$ a contradiction.