Determine the vertices of $P^{'}=\{ c^{T}x: x \in P\}$ where $P$ is a polytope Let $P$ be a polytope and $c \in \mathbb R^{n}$, then determine the vertices of  $P^{'}=\{ c^{T}x: x \in P\}$
My idea: 
Note that since $P$ is a polytope there exists $r > 0$ so that $P\subseteq B_{r}(0)$, then note that: $\vert c^{T}x\vert =\vert \sum_{i =1}^{n}c_{i}x_{i}\vert\leq \sum_{i =1}^{n}\vert c_{i}x_{i}\vert\leq (\sum_{i =1}^{n}\vert c_{i}\vert ^{2})^{\frac{1}{2}}(\sum_{i =1}^{n}\vert x_{i}\vert ^{2})^{\frac{1}{2}}\leq(\sum_{i =1}^{n}\vert c_{i}\vert ^{2})^{\frac{1}{2}}r < \infty$ by Cauchy-Schwarz and thus $P^{'}$ is also bounded, and it is a polyhedron by the projection lemma. 
I am not sure on finding ALL the vertices. I know that looking at the two optimization problems: 
$\max c^{T}x$ s.t. $x \in P$ as well as $\min c^{T}x$ s.t. $x \in P$ have optimal point $x^{+}$ and $x^{-}$, respectively, since $P$ is compact. $x^{+}$ and $x^{-}$ are then also vertices of $P$, and note: 
$c^{T}x^{-} \leq d \leq c^{T}x^{+}$ for any $d \in P'$. Note that $c^{T}x^{-}$ and $c^{T}x^{+}$ maximize/minimize the optimization problem: $1\times p^{'}$ s.t. $p^{'} \in P'$ and hence are vertices of $P^{'}$. 
I am not sure whether I went about this the right way. In my mind, I would still need to show that there are indeed no other vertices. How should I do this?
 A: Note that, for a fixed $x$, the quantity $c^T x$ is a scalar. Therefore, the set $P'=\{ c^{T}x: x \in P\}$ is a subset of the real line. I assume that, if you express $P'$ as a union of disjoint, closed intervals, then what you call "vertices of $P'$" are the edges of the intervals.
Clearly, $c^T x^+=\max_{x\in P} c^T x$ and $c^T x^-=\min_{x\in P} c^T x$ are vertices of $P'$. To prove that there are no other vertices, we prove that $P'$ is indeed one closed interval, i.e., that any $d$ with $c^T x^-\le d\le c^Tx^+$ is in $P'$.
The definition of polytope is not universal, but let us assume that your definition of polytope requires a polytope to be path-connected*. In this case, there exists a path in $P$ from $x^-$ to $x^+$. This is to say, there exists a continuous function $x(t):[0,1]\to\mathbb{R}^n$ with $x(0)=x^-$, $x(1)=x^+$, and $x(t)\in P$ for $t\in[0,1]$. The function
$$f(t)=c^T x(t)$$
satisfies $f(0)=c^Tx^-$, $f(1)=c^Tx^+$, and is continuous in $t\in[0,1]$, because it is a finite linear combination of continuous functions.  By the intermediate value theorem, $f(t)$ takes on all values in $[c^T x^-, c^Tx^+]$. Furthermore, $f(t)\in P'$ for $t\in[0,1]$. Therefore, all values in $[c^T x^-, c^Tx^+]$ belong to $P'$. No other $y\in \mathbb{R}$ belongs to $P'$, so we conclude
$$P'=[c^T x^-, c^Tx^+]$$
and therefore, $c^T x^-$ and $c^Tx^+$ are the only vertices of $P'$.
* If your definition of polytope allows non-connected polytopes (for example, a set of two polyhedra), then you can split your polytope into path-connected (sub-)polytopes, apply the reasoning in this answer to each sub-polytope, and combine the resulting intervals to find the vertices.
