Given a Polytope $X \subset R^n$ and assume we have nonempty extreme point set $P(X)$.
We can do have $n$ different $n-1$ coordinate projection (namely hiding one coordinate). We use $h(X, i)$ indicates $i$th coordinate get hidden.
Would it be possible For $n \geq 3$, $\exists x \in P(X), h(x, i) \notin P(h(X, i)), \forall i \in {1\cdots n} $ ?
For a $n =2$ case is kind of trivial to prove, take $X$ as a pentagon, and there are 5 extreme points. and project it to both X and Y axis would result in two line segments which in total there can be maximum 4 distinct extreme points. So that there must be 1 extreme point missing.
I'm having a trouble prove it for $n =3$ case.
Take the convex hull of $n+1$ points consisting of the standard basis vectors $e_1, \ldots, e_n$ together with $v_t = (t, t, \ldots, t)$ for some $t \in (0, 1/n)$. Then the projection of $v_t$ is in the interior of any coordinate projection.