A subset $A$ of a geodesic space $X$ is called quasi convex if there exists a constant $k > 0$ so that if $x,y \in A$, the geodesic joining $x$ to $y$ is in the $k$ neighborhood of $A$. Is it true that if we intersect two quasi convex sets, we get another quasi convex set? The proof in the actual convex case is one line but this one doesn't seem so easy.
If $A_1,A_2$ are quasi convex with a common $k$, and $x,y \in A_1 \cap A_2$, we could write $$B(A_1 \cap A_2,k) \subset B(A_1,k)\cap B(A_2,k)$$ but it seems like we actually need the other inclusion. Is there a counter example?
The answer is no. Take $X = \mathbb{R}^2$ and $A_1,A_2$ to be combs with teeth exponentially far apart that intersect only on the teeth. Explicitly, let $$B = \{0,1,2,4,8,16, 2^5 \ldots\} \times [0,1]$$ $$A_1 = B \, \cup [0,\infty) \times \{0\}$$ $$A_2 = B \, \cup [0,\infty) \times \{1\}$$ so $A_1$ is a comb with teeth pointing up, $A_2$ is a comb with teeth pointing down, and $B$ is the teeth. Then $A_1 \cap A_2 = B$ and clearly this is not quasi convex since the line segments between teeth get longer and longer with no bound.