# Characterization of interior point of convex set using normal cones

There is a theorem saying that for any convex set $$Q$$, $$x\in \text{int } Q \Leftrightarrow N_Q(x)=\{0\}$$. I'm trying to prove the backward direction, and my argument is as follows: If $$N_Q(x)=\{0\}$$, then equivalently any nonzero vector cannot be in the normal cone at $$x$$, which means that any nonzero vector must make some acute angle with $$y-x$$ for some $$y\in Q$$. I'm trying to argue that this implies $$\exists \epsilon>0$$ such that $$B_\epsilon (x)\subset Q$$, but there seems to be a gap in the argument. How can I fill in this gap? Or are there other ways to prove this?

Assume that $$N_{Q}(x) = \{0\}$$. For the sake of contradiction, take $$x \in \mathbf{bd}(Q)$$, the set's boundary. Then, the supporting hyperplane theorem implies that $$\exists v \in \mathbb{R}^d, v \neq 0$$ such that
$$\langle v, x \rangle \geq \langle v, y \rangle, \; \forall y \in Q \implies \langle v, y - x \rangle \leq 0, \; \forall y \in Q \Leftrightarrow v \in N_Q(x).$$
This contradicts your assumption on $$N_Q$$, since $$v$$ is guaranteed to be nonzero. Hence $$x$$ must indeed be in $$\mathrm{int}(Q)$$.