$a)$ Prove that a convex polytope has finitely many extreme points.
$b)$ Prove that the unit disc $S:=\{x\in\mathbb{R}^2:x_1^2+x_2^2\le1\}$ is not a convex polytope.
Hint : what are the extreme points of $S?$
My thoughts:
"A polytope is a bounded polyhedron", then formalize it we have:
$$\exists a_1^t\dots a_n^t\in\mathbb{R}^n,c_1\dots c_n\in\mathbb{R}, r\in\mathbb{R}\cap(0,\infty),s.t.$$
$$\{x\in\mathbb{R}^n:a_1^tx\le c_1\}\cap\dots\cap\{x\in\mathbb{R}^n:a_n^tx\le c_n\}\subseteq B(0,r)$$
$$\text{iff } \{x\in\mathbb{R}^n:a_1^tx\le c_1\}\cap\dots\cap\{x\in\mathbb{R}^n:a_n^tx\le c_n\}\text{ is a polytope.}$$
Using "Elementary Linear Programming With Application" by Bernard Kolman ⋅ Robert E. Beck
Intuition:
$a)$ Since those extreme points must located on intersections of finitely many half-spaces which implies extreme points are finite
$b)$ This is just a closed $\mathbb{R}^2$ circle, which has infinitely many extreme points, so it can not be formed by finitely many half-spaces, which can be bounded but can't be a polytope, so certainly not a convex polytope.
Intuitively, $a),b)$ are both almost trivial, but how do I write the proof formally$?$
Thanks for your help.
More definitions
$0.$Definition of polyhedron
A polyhedron is the intersection of finitely many halfspaces
$1.$Definition of halfspaces $P$ in $\mathbb{R}^n$ $$P=\{x∈\mathbb{R}^n,a^tx\le(\ge)b\}$$
$2.$A point $x\in\mathbb{R^n}$ is a $\underline{\text{convex combination}}$ of the points $x_1,x_2,\dots,x_r$ in $\mathbb{R}^n$ if for some real numbers $c_1,c_2,\dots,c_r$ which satisfy $$\sum_{i=1}^r c_i=1\text{ and }c_i\ge0,\space1\le i\le r,$$ we have $$x=\sum_{i=1}^rc_ix_i$$
$3.$ The $\underline{\text{convex hull}}$ of a finite point set $S$ is the set of all convex combinations of its points.
$4.$ a $\underline{\text{convex polytope}}$ is the convex hull of a finite set of points
$5.$ A point $u$ in a convex set $S$ is called an $\underline{\text{extreme point}}$ of $S$ if it is not an interior point of any line segment in $S$. That is, $u$ is an extreme point of $S$ if there are no distinct points $x_1$ and $x_2$ in $S$ such that $$u=\lambda x_1+(1-\lambda)x_2,\space0<\lambda<1.$$
Rough work
$a)$Proof.
Do this by contradiction
Assume $S$ is such convex polytope and has infinitely many extreme points, then
$$\forall n\in \mathbb{N},\exists^{\ge n} u\in S, s.t. \forall x_1\neq x_2\in S,\lambda\in(0,1)\cap\mathbb{R},u=\lambda x_1+(1-\lambda)x_2$$
Which contradict with $S$ is a convex hull of a finite set so it's not a convex polytope$ (\Rightarrow\Leftarrow)$ $\tag*{$\square$}$
$b)$ Proof.
Define $B(a;r):=\{x∈\mathbb{R}^n:|x−a|<r\}$ where $a\in\mathbb{R}^n$
Let $\overline{S}:=\{{x∈\mathbb{R}^n:∀ε>0,B(x;ε)∩S≠\varnothing}\}$. then we have:
$$\forall u\in\overline{S},x_1\neq x_2\in S,\lambda\in(0,1)\cap\mathbb{R},u\neq\lambda x_1+(1-\lambda)x_2$$
And also $\overline{S}\subseteq S$
But $\overline{S}$ is not a finite set, which means S has infinitley many extreme points, by contrapositive of $a)$
That implies $S$ is not a convex polytope. $\tag*{$\square$}$
Yet I'm trying to write a proof using def $2,3,4,5$,
Which I think is the correct definition that I suppose to use.
(keep updating$\dots$)
Any help or hint or suggestion would be appreciated.