Proof that the boundary and the closure of a subset are closed.

1. Sep 11, 2006

xman

Hello,
I am currently working on proving the following theorem

The boundary $$\partial A$$ and the closure $$\overline{A}$$ of a subset A of $$\mathbb C$$ are closed sets.

Proof: Let $$A \subset \mathbb C.$$ We want to show the set $$\partial A \cap \overline{A}$$ is closed. To show that $$\partial A \cap \overline{A}$$ is closed, we will show that the complement of $$\partial A \cap \overline{A}$$ is open. So the complement of $$\partial A \cap \overline{A}$$ is the set of all points not in $$\partial A \cap \overline{A},$$ i.e. $$\mathbb C \sim (\partial A \cap \overline{A} ).$$ Since $$\overline{A} = A\cup \partial S$$ and the set $$\partial A \cap \overline{A}$$ contains all of its boundary points by definition of the intersection. So the complement of $$\partial A \cap \overline{A}$$ cannot contain any of its boundary points by the definition of the complement. Since the complement does not contain any points on its boundary, then the complement is open. Therefore, since the complement of $$\partial A \cap \overline{A}$$ is open then the set $$\partial A \cap \overline{A}$$ is closed.

I am really poor at doing proofs, any help, insight, or questions would be greatly appreciated. Please let me know. Thanks.

2. Sep 11, 2006

StatusX

The boundary and the closure are both closed. Their intersection is also closed, but their intersection is just the boundary. So what are you trying to prove?

3. Sep 11, 2006

xman

StatusX, thanks for the reply. I am trying to show the theorem:

Thm. The boundary $$\partial A$$ and the closure $$\overline{A}$$ of a subset $$A$$ of $$\mathbb C$$ are closed.

4. Sep 11, 2006

StatusX

Then why are you looking at their intersection? Also, what are you using as the definition of these terms?

5. Sep 11, 2006

xman

I misunderstood the statement, I was thinking and = intersection. I realized that the minute I read your first post. I'm using the following definitions.

"z is an interior point of A if there exists an r>0 such that the open disk is contained in A."

The open set consists of the set of all points of a set that are interior to to that set.

The boundary point is so called if for every r>0 the open disk has non-empty intersection with both A and its complement (C-A).

The closure of a set A is the union of A and its boundary.

So I need to show that both the boundary and the closure are closed sets. So, to show the sets are closed, do I consider the complement of the boundary for both cases? I can see how the closure of A is closed, since its complement is what I was essentially arguing in my first post. I am a little unsure about just the boundary part.

Am I at least on the right track now?

6. Sep 12, 2006

StatusX

Yes, the complement of the closure is the interior of the complement, so is open (prove this). Remember the intersection of closed sets is closed, so can you find a couple closed sets whose intersection is the boundary?

Last edited: Sep 12, 2006
7. Sep 12, 2006

xman

Thanks, StatusX. This is what I have. I know my proof isn't complete, I think I am having a problem with the boundary argument.

Let $$A \subset \mathbb C.$$ We want to show that $$\partial A$$ is closed and $$\overline{A}$$ is closed. To show that both the boundary and the closure are closed, we need to show that the complement of each is open. So first we will show that the boundary is closed and then we will show the closure is closed.

A point $$z\in \mathbb C$$ which for every r>0 there is an open disk $$\Delta(z,r)$$ which has a non-empty intersection with A and its complement $$\mathbb C\sim A$$ is said to be boundary point. Thus, the collection of all such points is the boundary. The complement of the boundary $$\mathbb C \sim \partial A$$ then cannot contain any points for which there exists an open disk who has a non-empty intersection with both A and $$\mathbb C\sim A.$$ Since the complement must have empty intersections with $$\mathbb C\sim A$$ and A this implies there exists an r>0 such that every open disk lies entirely inside the complement. Since for every point in the complement we can find an r>0 such that the open disk about that point lies entirely in the complement, then every point in the complement is an interior point to the complement. Since all points interior to the complement are interior points, then the complement is open. Since the complement of the boundary is open then the boundary is closed.

To show the closure of a subset of the complex plane is closed, we need to show that its complement is open. Since $$\overline{A}$$ contains all points in the union of A and its boundary $$\partial A$$ then the complement of the closure is all points outside of this union. Since the complement cannot contain any points in A then we need only worry about the boundary of A. By the previous paragraph we showed the boundary is closed. Since the complement cannot contain any points of A and the boundary is closed, then the complement of $$\overline{A}$$ is open. Since the complement of $$\overline{A}$$ is open then $$\overline{A}$$ is closed.

Let me know what you think.

Last edited: Sep 12, 2006