Is the relative interior stable under intersection

Question

Assume we have sets $S_1 ...S_n$ for which $\cap_{i=1...m} S_i \ne \varnothing$ holds. Every set $S_i$ is closed and convex.

Does $ x \in \mathrm{relint}( \cap_{i=1...m} S_i)$ implie that $x \in \cap_{i=1...m} \mathrm{relint}(S_i)$ holds?

I didn't fiend an answer to this in textbooks. For simple scatches, it seems to be true.

Thanks in advance! :-)

Answer 1

Not always true; here is a counterexample:

Let $S_1$ and $S_2$ be two closed balls who intersect at exactly one point $x$ on their boundaries. It follows that $\textrm{relint}(S_1\cap S_2)=\textrm{relint}\{x\}=\{x\}$.

However, $\textrm{relint}S_1=\textrm{int}S_1$ is the interior of $S_1$, i.e. the open ball with the same center and radius (and the same holds for $S_2$). Since the closed balls only intersect at a boundary point, this means their relative interiors do not intersect, i.e. $\textrm{relint}S_1\cap\textrm{relint}S_2=\varnothing\not\ni x$.

Here is an illustration

(note I'm using the standard convex analysis definition of relative interior -- $\textrm{relint}C = \{x\in C\, |\, \textrm{cone}(C-x)=\textrm{span}(C-x)\}$).