Is Every Closed Subset of ##\mathbb{R}^2## the Boundary of Some Set?

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Discussion Overview

The discussion revolves around the question of whether every closed subset of ##\mathbb{R}^2## can be considered the boundary of some set within ##\mathbb{R}^2##. Participants explore various examples and counterexamples, focusing on closed disks and the Cantor set, among others.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the truth of the statement, seeking a counterexample.
  • Another participant asserts that a closed disk is not a boundary.
  • Contrarily, a participant claims that a closed disk can be the boundary of the intersection of the closed disk with the rational points in the plane.
  • Further discussion emphasizes that the closed disk is not a boundary of a subset of the plane, despite the previous claim.
  • Another participant suggests that a closed subset of ##\mathbb{R}^2## can be the boundary of a countable dense subset with an empty interior.
  • The Cantor set is proposed as an example of a closed set that is its own boundary, with reasoning provided about its properties.

Areas of Agreement / Disagreement

Participants express disagreement regarding whether closed disks can be boundaries of subsets of the plane. Some argue in favor of the original statement while others provide counterexamples, indicating that the discussion remains unresolved.

Contextual Notes

There are unresolved details regarding the conditions under which certain sets can be considered boundaries, particularly concerning the definitions and properties of the sets discussed.

R136a1
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I'm wondering if the following is true: Every closed subset of ##\mathbb{R}^2## is the boundary of some set of ##\mathbb{R}^2##.

It seems false to me, does anybody know a good counterexample?
 
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a closed disk is not a boundary
 
lavinia said:
a closed disk is not a boundary

Let ##A## be the closed disk, then ##A## is the boundary of ##A\cap(\mathbb{Q}\times \mathbb{Q})##.
 
lavinia said:
a closed disk is not a boundary
The result in the OP is infact true.
 
micromass said:
Let ##A## be the closed disk, then ##A## is the boundary of ##A\cap(\mathbb{Q}\times \mathbb{Q})##.

No. The question was a boundary of a subset of the plane. The closed disk is not a boundary of a subset of the plane.
 
lavinia said:
No. The question was a boundary of a subset of the plane. The closed disk is not a boundary of a subset of the plane.

:confused: ##A\cap (\mathbb{Q}\times \mathbb{Q})## is a subset of the plane. ##A## is its boundary.
 
lavinia said:
No. The question was a boundary of a subset of the plane. The closed disk is not a boundary os a subset of the plane.
Consider a closed subset of R2, A. Let B be a countable dense subset of A. B has an empty interior so A is the boundary of B. There are some details to fill in, but that sketches the idea.
 
micromass said:
:confused: ##A\cap (\mathbb{Q}\times \mathbb{Q})## is a subset of the plane. ##A## is its boundary.

right.

So I guess the Cantor set is the boundary of itself.
 
Last edited:
lavinia said:
So I guess the Cantor set is the boundary of itself.
This is correct. The Cantor set ##C## is closed, so it contains its boundary. On the other hand, ##C## contains no intervals, so if ##x \in C##, then any neighborhood of ##x## contains points not in ##C##. Therefore ##x## is a boundary point of ##C##.
 
  • #10
Awesome! Thanks a lot!
 

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