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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?

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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 $\mathrm{int}(Q)$.

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