Q: We consider the Linear Program
$$ min\ \ \mathbf{c}^T\mathbf{x} $$
$$s.t. \ \mathbf{x}\in P$$
where $P$ is a polyhedron.
Let $Q$ be the set of optimal solutions. Show that any extreme point of $Q$ is an extreme point of $P$.
In my proof, I made use of the following theorems:
$\underline{\mathbf{Theorem \ 1}}: \\ $
Let $P$ be a nonempty polyhedron and $x\in P$. Then
$x$ is a vertex $\iff$ $x$ is an extreme point $\iff$ $x$ is a BFS
$\underline{\mathbf{Theorem \ 2}}: \\ $
Consider the Linear Program
$$min \ \mathbf{c}^T\mathbf{x}\\ s.t. \ \mathbf{x}\in P$$
where $P$ is a polyhedron.
Suppose $P$ has at least one BFS and that there is an optimal solution. Then there is an optimal solution which is a BFS.
Now, heres my proof for this question using the above theorems:
$Proof.$ Suppose that $P$ is a nonempty polyhedron and $\mathbf x \in P$, and that $P$ has at least one BFS.
Since $Q$ is the set containing the optimal solutions of $P$, by Theorem 2, there is an optimal solution which is a BFS.
Let $\mathbf x$ be this BFS. Then by Theorem 1, $\mathbf x$ is an extreme point.
Since $\mathbf x \in P$ and $\mathbf x \in Q$, it follows that any extreme point of $Q$ is an extreme point of $P$. $_{Q.E.D.}$
Is my proof correct or is it lacking something? I.e. wrong assumptions and/or applications of theorems. Some help and clarification will be appreciated. Thanks!