# About closed sets in the plane

1. Jul 19, 2013

### 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?

2. Jul 19, 2013

### lavinia

a closed disk is not a boundary

3. Jul 19, 2013

### micromass

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

4. Jul 19, 2013

### Jorriss

The result in the OP is infact true.

5. Jul 19, 2013

### lavinia

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.

6. Jul 19, 2013

### micromass

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

7. Jul 19, 2013

### Jorriss

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.

8. Jul 19, 2013

### lavinia

right.

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

Last edited: Jul 19, 2013
9. Jul 20, 2013

### jbunniii

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. Jul 20, 2013

### R136a1

Awesome! Thanks a lot!!