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

[R136a1]

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?

 

[lavinia]

a closed disk is not a boundary

 

[micromass]

a closed disk is not a boundary

Let $A$ be the closed disk, then $A$ is the boundary of $A\cap(\mathbb{Q}\times \mathbb{Q})$.

 

[Jorriss]

a closed disk is not a boundary

The result in the OP is infact true.

 

[lavinia]

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.

 

[micromass]

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.

$A\cap (\mathbb{Q}\times \mathbb{Q})$ is a subset of the plane. $A$ is its boundary.

 

[Jorriss]

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.

 

[lavinia]

$A\cap (\mathbb{Q}\times \mathbb{Q})$ is a subset of the plane. $A$ is its boundary.

right.

So I guess the Cantor set is the boundary of itself.

 

[jbunniii]

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 $C$.

 

[R136a1]

Awesome! Thanks a lot!