In a linear programming proof we have the:
$\sup\{c^Tx: Ax \le b\}$
This supremem can be $\infty$, or defined as $-\infty$, if there are no vectors x such that $Ax \le b$. But it is stated that should the supremum be finite, then there is one vector that attains $x_0$ such that $c^Tx_0=\sup\{c^Tx: Ax \le b\}$. Why is this?