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

Let {An}\{A_n\} be a properly nested sequence of non empty open convex subsets of R2\mathbb R^2 such that ∩∞n=1An\cap_{n=1}^\infty A_n is non empty and bounded ; then is it possible that ∩∞n=1An\cap_{n=1}^\infty A_n is open ? I only know that ∩∞n=1An\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

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Possible duplicate of Intersection of a properly nested sequence of convex sets , if nonempty and bounded , can never be open?
– Henrik
yesterday

@Henrik : Okay , so you are the possibly third person to mark it as a duplicate without even realizing the fundamental difference between the questions
– Saun Dev
yesterday

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@SaunDev Since this issue has come up more than once, I suggest that you edit your question to make it clear what the difference between the two questions is.
– arjafi♦
yesterday

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Here’s an example.

Consider {(x,y)∈R2∣y>x2}\{(x,y) \in\mathbb{R}^2 \mid y>x^2\}. Now, consider a sequence of integers xn=nx_n = n and construct the sets
S1={(x,y)∈R2∣y>x2}∩{(x,y)∈R2∣y>12+[(1+1)2−12](x−1)}\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
Sn=Sn−1∩{(x,y)∈R2∣y>n2+[(n+1)2−n2](x−n)}.\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.

I meant projective transformation because lines are preserved under projective transformations.
– Jacky Chong
2 days ago

I could have constructed the example on the disk by pushing back portion of the boundary at a time, but it’s hard to write it out so I took the parabola on the upper plane instead and map it back to the disk.
– Jacky Chong
2 days ago

Let us continue this discussion in chat.
– Jacky Chong
2 days ago

What about the following example: For k≥2k\ge 2 denote
xk=(cos(π/k),sin(π/k)).
x_k =(\cos(\pi/k), \sin(\pi/k)).

Let AnA_n be the interior of the convex hull of (0,0)(0,0), x2…xnx_2\dots x_n, and the arc {(cos(t),sin(t)):t∈(0,π/n)}\{ (\cos(t),\sin(t)) : t\in (0,\pi/n)\}. So it is a quarter circle with some segments removed.

Let AA be the interior of the convex hull of (0,0)(0,0), x2,…x_2,\dots, and (1,0)(1,0), hence open. Clearly, A⊂AnA\subset A_n holds for all nn.

Take x∈∩Anx\in \cap A_n. Write x=r(cost,sint)x=r(\cos t,\sin t) with r∈[0,1]r\in[0,1]. If t∈[π/(k+1),π/k]t\in [\pi/(k+1),\pi/k], k≥2k\ge 2, then x∈Anx\in A_n for all n≥k+1n\ge k+1. If t=0t=0 then r∈(0,1)r\in(0,1) and x∈Anx\in A_n for all nn.

This shows A=∩n≥2AnA=\cap_{n\ge 2} A_n.