Why $B_\epsilon(d)\not\subset G'\not\rightarrow B_\epsilon(d)\subset G'^c$?

Question

Let $A$ be a $m\times n$ matrix and consider the cones $G_0=\{d\in\mathbb R^n:Ad<0\}$ and $G'=\{d\in\mathbb R^n:Ad\le0\}$

Prove that $G_0=int G'$

Here $Ad<0$ means that every component of the vector $Ad$ is negative.

Proof.

Let's see that $G_0\subset int G'$.

Let $d\in G_0$, thus $Ad<0.$

Suppose that $d\not\in intG'.$ Thus $\forall \epsilon>0,B_\epsilon(d)\not\subset G'$. Hence we must have $B_\epsilon(d)\subset G'^c$.

But this means $Ad>0\ $ !

---

My instructor told me the proof is wrong because $B_\epsilon(d)\not\subset G'\not\rightarrow B_\epsilon(d)\subset G'^c$ . 'The ball could be in the boundary'.

Answer 1

There is a possibility that $d$ in the boundary of $G'$.

The point $d$ in boundary of $G'$ satisfies every neighborhood of $d$ contains at least one point of $G'$ and at least one point not of $G'$, so $B_\epsilon (d)\not\subset G'$.