6

If sets $A, B$ in Euclidean space are closed sets, they have the same boundary and their interiors intersection is non-empty, can we say $A=B$? Any suggestions and comments are welcome!

user10354138
  • 33,239
longjian li
  • 143

2 Answers2

3

No. For example, $A=\overline{\bigcup_{n\in\mathbb{Z}} [\arctan(2n-1),\arctan(2n)]}\cup [10,11]$, $B=\overline{([-\pi/2,\pi/2]-\bigcup [\arctan(2n-1),\arctan(2n)]}\cup[10,11]$.

To come up with such an example, note we can WLOG forget the interior intersection is nonempty by adding a requirement $A,B$ bounded (since we can take union with a closed ball far far away) and $\mathbb{R}$ is homeomorphic to a bounded open interval. So we want to come up with two different nonempty closed sets in $\mathbb{R}$ with the same boundary, which is easy enough: $\bigcup[2n-1,2n]$ and $\bigcup [2n,2n+1]$ both have boundary $\mathbb{Z}$.

user10354138
  • 33,239
  • 1
    In your last example the two sets exhaust the real line and there is no 'ball far away' left to make the interiors intersect. – Kavi Rama Murthy Jul 15 '20 at 10:15
  • @KaviRamaMurthy Which is why there is the homeomorphism of $\mathbb{R}$ with a bounded open interval to make sure we can add the ball far away (we also need to add the endpoints of the intervals but that is OK) – user10354138 Jul 15 '20 at 10:18
  • Thank you. I got it now. +1 for the answer. – Kavi Rama Murthy Jul 15 '20 at 10:25
  • 1
    if these two sets have stronger condition: if set A,B are closed, connected, convex sets in Euclidean space and they have the same boundry and their interior's intersection is non-empty, can we say A=B? – longjian li Jul 15 '20 at 10:51
3

Counterexample in $\mathbb R^2$:

                                               Counterexample

\begin{align*} A&=\text{red disk}\bigcup\text{black vertical line}\\ B&=\text{red disk}\bigcup\text{black vertical line}\bigcup\text{blue area} \end{align*}

The red disk includes the circle on its boundary, and the blue area extends to infinity towards north, south, and east.

\begin{align*} \partial A&=\partial B=\text{red circle}\bigcup\text{black vertical line}\\ A^{\circ}&=\text{interior of red disk}\\ B^{\circ}&=\text{interior of red disk}\bigcup\text{blue area}\\ A^{\circ}\cap B^{\circ}&=\text{interior of red disk}\neq\varnothing\\ A&\neq B \end{align*}

triple_sec
  • 23,377