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

by xman
Tags: boundary, closure, proof, subset
 P: 97 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.
 P: 97 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.
 P: 97 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.