# How to prove normal cones are closed?

Let $$C$$ be a convex subset of $$\mathbb{R}^{n}$$ and let $$\bar{x} \in C$$. Then the normal cone $$N_{C}(\bar{x})$$ is closed and convex. Here, we're defining the normal cone as follows:

$$N_{C}(\bar{x}) = \{v \in \mathbb{R}^{n} \vert \langle v, x - \bar{x} \rangle \le 0, \forall x \in C \}.$$

Proving convexity is straightforward, as is proving $$N_{C}(\bar{x})$$ is closed when $$C$$ is open ($$i.e.$$ every $$x \in C$$ is an interior point). However, I'm not sure how to prove that $$N_{C}(\bar{x})$$ is closed more generally?

Write $$N_C(\bar{x}) = \cap_{x \in C} \{v | \langle v, x-\bar{x} \rangle \le 0 \}$$.
Hence $$N_C(\bar{x})$$ is the intersection of closed hyperplanes which is closed (the function $$v \mapsto \langle v, x-\bar{x} \rangle$$ is continuous).
This approach also shows that $$N_C(\bar{x})$$ is convex.