Relative interior $\operatorname{relint}(C)$ of convex $C \subset \mathbb{R}^n$ set is again convex

Question

I'm trying to show that the relative interior $\operatorname{relint}(C)$ of a convex set $C \subset \mathbb{R}^n $ is again convex.

Using the following definition of the rel. interior

$\operatorname{relint}(X) := \{x \in X\mid \exists \epsilon > 0: \mathbb{B}(\epsilon, x) \cap \operatorname{aff}(X) \subset X\}$

I've got the steps:

let $x,y \in \operatorname{relint}(C)$, each with $\epsilon_x, \epsilon_y$. Then there must be $\lambda x + (1-\lambda)y \in \operatorname{relint}(C)$, i.e. their convex comb. must be in the interior as well, to show the above.

then the convex combination of $x,y$ is clearly in $\operatorname{aff}(C)$ but I don't know how to proceed with the ball.

I would have to show that there's a ball $\mathbb{B}(\epsilon', \lambda x + (1-\lambda)y)$ and that its intersection with the affine hull is in C again. Am I right?

Thanks!

Answer 1

Let $\varepsilon =\min(\varepsilon_1, \varepsilon_2)$, and let $\|r\|<\varepsilon$ with $x+r\in {\rm Aff}(C)$, then $y+r=(x+r)+y-x\in {\rm Aff}(C)$ as well, and $$\lambda x+(1-\lambda)y+r\ =\ \lambda(x+r)+(1-\lambda)(y+r)\ \in C$$ because both $x+r$ and $y+r$ are in $C$ by hypothesis, and $C$ is convex.