Prooving that a point is not extreme

Question

I have a conceptual question (that looks simple) but I can't prove it mathematically :

Let $X:=\{x:Ax \leq b\}$ and let $x_0$ be such that $Ax_0 < b$. Prove that $x_0$ cannot be an extreme point of $X$.

My answer: -logically speaking- since $x_0$ doesn't satisfy $Ax=b$, then it doesn't fall on a hyperplane so it's not an extreme point ; but how can I prove it in a more formal way ?

Answer 1

Let $e$ be the all one vector.

Since $Ax_o < b$, $\exists \epsilon>0, A(x_0+\epsilon e) \leq b$ and $A(x_0-\epsilon e) \leq b$.

Then we have $x_0+\epsilon e \neq x_0$, $x_0-\epsilon e \neq x_0$ and$$\frac{(x_0+\epsilon e)+(x_0-\epsilon e)}{2}=x_0.$$

Hence $x_0$ is not an extreme point.

Answer 2

Such an $x_0$ is even an interior point. It is not hard to show (continuity!) that there is $\varepsilon > 0$, such that $$B_\varepsilon(x_0) := \{x : \|x - x_0\| \le \varepsilon\} \subset X.$$