Closure of the intersection of an open halfspace with a convex set

Question

Let $H$ be an open halfspace and $C$ be a convex subset of $\mathbb{R}^n$, and suppose that $H\cap C\neq \varnothing$.

Is it true that the closure of the intersection of $H$ with $C$ is equal to the intersection of the closure of $H$ with the closure of $C$ ?

Answer 1

Since $C\cap H \subset \overline{C} \cap \overline{H}$ it follows that $\overline{C\cap H} \subset \overline{C} \cap \overline{H}$ for any sets $C,H$.

We are given that $C \cap H$ is non empty and since $H$ is open, it follows that there is a point $x \in H \cap \operatorname{ri} C $.

For any convex set $K$, if $x \in \operatorname{ri} K$ and $y \in \overline{K}$, then $t y + (1-t) x \in \operatorname{ri} K$ for all $t \in [0,1)$. (To some degree, this is the essence of the almost 'topological' nature of convexity.)

Suppose $y \in \overline{C} \cap \overline{H}$, then we have $t y + (1-t) x \in C$ and $t y + (1-t) x \in H$ for all $t \in [0,1)$. Letting $ t \to 1$ shows that $y \in \overline{C\cap H}$.