Intersection of a properly nested sequence of convex sets , if nonempty and bounded , can never be open?

Question

Let $\{A_n\}$ be a sequence of convex sets in $\mathbb R^2$ such that $A_{n+1}\subsetneq A_n,\forall n\in \mathbb N$ and $\cap_{n=1}^{\infty} A_n$ is non-empty and bounded , then is it possible that $\cap_{n=1}^{\infty} A_n$ is an open set ?

I can only see that $\cap_{n=1}^{\infty} A_n$ is always convex .

Please help . Thanks in advance

Answer 1

yes, it is possible. Suppose the sequence $S_n=B(0,1)\cup \{(cos 2\pi t, sin 2\pi t);t\in(0,1/n)\}.$ Where $B(0,1)$ denotes the open unit ball.

They are convex and the intersection is the open set $B(0,1)$.

Answer 2

This is not true. For, instace we can consider the sequence of balls $B(0,1+1/n)$. The intersection is the closed ball $\bar{B}(0,1)$, which is not open.