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

Question

Let $\{A_n\}$ be a properly nested sequence of non empty open convex subsets of $\mathbb R^2$ such that $\cap_{n=1}^\infty A_n$ is non empty and bounded ; then is it possible that $\cap_{n=1}^\infty A_n$ is open ? I only know that $\cap_{n=1}^\infty A_n$ is convex . Also see Intersection of a properly nested sequence of convex sets , if nonempty and bounded , can never be open? . Please help .Thanks in advance

Answer 1

Here's an example.

Consider $\{(x,y) \in\mathbb{R}^2 \mid y>x^2\}$. Now, consider a sequence of integers $x_n = n$ and construct the sets \begin{align} S_1 =\{(x,y) \in \mathbb{R}^2 \mid y> x^2\} \cap\{(x, y) \in \mathbb{R}^2 \mid y> 1^2+[(1+1)^2-1^2](x-1)\} \end{align} and \begin{align} S_n = S_{n-1}\cap \{(x, y) \in \mathbb{R}^2 \mid y> n^2+[(n+1)^2-n^2](x-n)\}. \end{align}

It's easier to just draw the picture, which is just the area above the parabola being push back.

Edit: To get boundedness you can use a projective transformation to map the upper half plane to the disk which will preserve the convexity and openness.

Note: The key here is that you don't push any part of the boundary of the original set infinitely often.

Answer 2

What about the following example: For $k\ge 2$ denote $$ x_k =(\cos(\pi/k), \sin(\pi/k)). $$ Let $A_n$ be the interior of the convex hull of $(0,0)$, $x_2\dots x_n$, and the arc $\{ (\cos(t),\sin(t)) : t\in (0,\pi/n)\}$. So it is a quarter circle with some segments removed.

Let $A$ be the interior of the convex hull of $(0,0)$, $x_2,\dots$, and $(1,0)$, hence open. Clearly, $A\subset A_n$ holds for all $n$.

Take $x\in \cap A_n$. Write $x=r(\cos t,\sin t)$ with $r\in[0,1]$. If $t\in [\pi/(k+1),\pi/k]$, $k\ge 2$, then $x\in A_n$ for all $n\ge k+1$. If $t=0$ then $r\in(0,1)$ and $x\in A_n$ for all $n$.

This shows $A=\cap_{n\ge 2} A_n$.