Given a set $X \subset R^n$ with $m$ points. We can find it's Convex Hull and together with set of extreme points $E(X)$. And none of any points are linear multiplier of each other.
Under a linear projection of $f: R^n \to R^{n-1}$, we can find extreme points of $Y:=\{f(x)| x \in X\}$, denote $E(Y)$. which would again be extreme points of $X$, since linear projection preserves convexity.
$PI((E(Y))) \cap E(X) \neq \emptyset$, the pre-image of extreme points of $Y$ contains a subset of extreme points of $X$
There are ${m \choose n }$ such linear projection defined by the data points. Namely picking $n$ data points, looking at the affine subspace they span, and then projecting orthogonally onto that affine subspace
Would the union of ${m \choose n}$ linear projection and find extreme points in the lower dimension be sufficient to recover all the extreme points in the original set?
would this hold $E(X)\subseteq \cup_{i \in {m \choose n}}PI(E(Y_i))$ ?
For example, if in 2d space(n=2) and 10 points (m=10). Any two points can define a line, we would have ${10 \choose 2}$ lines defined by the data. If we project the 10 points to any line, the convex hull would form a line segment, we would detect 2 extreme points. If we project towards all the lines and collect all the extreme points there would be $2*{10 \choose 2}$ points, which contains duplicates of course. but if we deduplicate them, would all the extreme points of the 2d space being captured by these procedures?
I don't understand how: do you mean that you are picking some data points, looking at the affine subspace they span, and projecting orthogonally onto that affine subspace? If, so $n-1$ points will define an $(n-2)$-dimensional affine subspace, so you probably want to choose $n$ data points to project onto an $(n-1)$-dimensional space.
Apologies if this isn't what you meant — your question is pretty imprecise as stated so this was my best attempt at interpreting it. Also, what do you mean by an "extreme point of $X$", since $X$ is supposed to be finite (hence not convex unless $\lvert X \rvert \leq 1$)? Finally, how is an extreme point of $f(X)$ supposed to be the same as an extreme point of $X$, since these live in different spaces? I think this question is answerable but needs to be clarified first. I'll edit this answer once the question is clear and precise :)
Edit The question has been clarified, and the answer is no. Indeed, this fails immediately in $\mathbb{R}^2$ (and in all higher dimensions):
Let $X$ be the convex hull of $p_1 = (0,0)$, $p_2 = (0,1)$, $p_3 = (1,3)$, and $p_4 = (1,2)$ (these four points are the extreme points of $X$). Then $p_2$ is not sent to an extreme point of $f(X)$ for any orthogonal projection $f$ onto the affine span of any two of the $p_i$'s.