Let $A$ a $m\times n$ real matrix , $b\in \mathbb{R}^m$ and $c\in \mathbb{R}^n$. I want to prove that if the linear program $P=\min \{c^\top x: Ax=b, x\ge 0\}$ is feasible and bounded, then it has an optimal solution.
First of all, the LP is in the standard form, but we know that the problem can be reformulated in terms of canonical form. So, if we call $P=\{x\in \mathbb{R}^{n}: Ax=b, x\ge 0\}$, then $P$ is a polyhedron and by the Minkowski-Weyl theorem we know that $P$ can be expressed in the form $$ P=\mbox{conv}(V)+\mbox{cone}(Y)$$ where $V$ and $Y$ are set of vectors in $\mathbb{R}^n$, $\mbox{conv}(V)$ is the convex hull of vectors in $V$ and $\mbox{cone}(Y)$ is the conic hull or the recession cone of vectors in $Y$. So, I assume that the LP is bounded, which means that exists a constant $\alpha\in \mathbb{R}$ such that $c^\top x\ge \alpha$ for all $x\in P$. Intuitively I get that the recession cone must be or empty or formed only by the null vector; other way the polyhedron $P$ must contain a ray wich contains vector that are not in $P$, and that is absurd since we have that $P=\mbox{conv}(V)+\mbox{cone}(Y)$. However, I was wondering how can I proof this fact using the definition of bounded LP cited before (exists a constant $\alpha\in \mathbb{R}$ such that $c^\top x\ge \alpha$ for all $x\in P$). If I take an $u\in P$, then $$u=v+y$$ for $v \in \mbox{conv}(V)$ and $y\in \mbox{cone}(Y)$. By the definition of $\mbox{cone}(Y)$, then the vector $v+ty\in P$ for all $t\ge 0$. How can these facts conclude that $y$ must be $0$? Any help will be appreciated. I am using the definition $$\mbox{cone}(Y)=\left\{\sum_{i=1}^{k}t_iy_i: y_i\in Y, t_i\ge 0\right\}$$