Showing that the dual of convex cone is closure of its open space

Question

I came across a quite obvious statement but I am stuck in proving that.

Let $\Lambda$ be a subset of $R^n$, the dual cone $\Lambda^*=\{w:\forall \lambda\in\Lambda, w\cdot\lambda\geq0\}$, and its open set $\Lambda^>=\{w:\forall \lambda\in\Lambda, w\cdot\lambda>0\}$. How can I show that $\Lambda^*$ is topologically closure of $\Lambda^>$? Assume that $\Lambda^>$ is non-empty.

Answer 1

The inclusion $\operatorname{cl}\Lambda^> \subset \Lambda^*$ is easy to show.

Now, let $w \in \Lambda^>$ be given. For an arbitrary $\hat w \in \Lambda^*$, you have $\hat w + \frac1n \, w \in \Lambda^>$ for all $n \in \mathbb N$. Then, it is easy to check that $\Lambda^* \subset \operatorname{cl}\Lambda^>$.