Fundamental group of the common boundary of two disjoint convex sets in $\mathbb{R}^n$

Question

Let $A,B\subseteq \mathbb{R}^n$ be two disjoint convex sets. In general the fundamental group of $\partial A$ ($\partial B$)(with the subspace topology) is not trivial (for example let $n=2$ and $A$ be the unit disc in $\mathbb{R}^2$). Now consider $\partial A\cap \partial B$ (also with the subspace topology). Being the common boundary of two convex sets that both have non-empty interior, it seems that it cannot encircle any two-dimension hole, so the fundamental group of it should be trivial. But I cannot prove it or give a counter-example, can someone help me?

Answer 1

Disjoint convex sets can be separated by a hyperplane $H$. Thus $\partial A \cap \partial B \subseteq H$. Also note that $a \in \overline A \cap H$ implies $a \in \partial A$ (if it were an interior point, you would find a ball, which extends to the other side of $H$, a contradiction). Since $\overline{A}$ and $\overline{B}$ are convex, the intersection $\overline{A} \cap \overline{B}\cap H = \partial A \cap \partial B \cap H = \partial A \cap \partial B$ is convex, in particular contractible.

Answer 2

The closure of a convex set is convex. And the intersection of two convex sets is convex. So the following set is convex: $$\overline A \cap \overline B = (A \cup \partial A) \cap (B \cup \partial B) = (A \cap \partial B) \cup (\partial A \cap B) \cup (\partial A \cap \partial B) $$ But in addition we have $A \cap \partial B \subset \partial A \cap \partial B$ because if $x \in A \cap \partial B$ the $x$ is a limit point of $B \subset \mathbb R^n - A$ and hence $x \in \overline{\mathbb R^n - A} \cap \overline A = \partial A$.

Similarly $\partial A \cap B \subset \partial A \cap \partial B$.

Putting this altogether, it follows that $\overline A \cap \overline B = \partial A \cap \partial B$, and since that set is convex, it is also simply connected, and so its fundamental group is trivial.