Let $K,L$ be convex bodies in $\mathbb{R}^n$ s.t. $(K \cup L)$ is convex. If $x \in K, y \in L$, then $[x,y] \cap (K \cap L) \neq \emptyset$.

Question

Let $K,L$ be convex bodies in $\mathbb{R}^n$ s.t. $(K \cup L)$ is convex. If $x \in K, y \in L$, then $[x,y] \cap (K \cap L) \neq \emptyset$.

Here convex bodies are defined to be non-empty, compact and convex subsets of $\mathbb{R}^n$.

I’m looking for a simple argument to verify this statement. The interesting case is of course when $x,y \notin (K \cap L)$. If the statement does not hold then any point on $[x,y]$ would be in either $K$ or $L$, but not both. I suppose then I should get a sequence of points on the line segment $[x,y]$ that tend to either $x$ or $y$, which would lead to a contradiction. But the argument becomes a bit too convoluted. I’d appreciate some help.

Answer 1

We know that $[x, y] \subset K \cup L$ because the union is convex.

Now assume that $[x,y] \cap (K \cap L) = \emptyset $. Then $$ [x,y] = \bigl([x,y] \cap K \bigr) \cup \bigl([x,y] \cap L \bigr) $$ is a partition of the segment into nonempty, disjoint, and closed subsets, which is a contradiction to $[x, y]$ being connected.