# Boundary of $A$ is closed iff $A$ is union of closed and open set?

This is part (x) of Exercise 8 in Section 2.2 of Topology and Groupoids, by Brown.

Exercise:

Prove that the boundary of $$A$$ is closed if and only if $$A$$ is the union of a closed and an open set.

Definitions:

$$\text{Bd } A = A \setminus \text{Int } A$$.

My attempt:

Assume the boundary of $$A$$ is closed. That means $$A \setminus \text{Int } A$$ is closed. Since $$A = (A \setminus \text{Int } A) \cup \text{Int } A$$, and $$\text{Int } A$$ is open, we see that $$A$$ is the union of a closed and an open set.

Conversely, assume $$A = C \cup O$$, where $$C$$ is closed and $$O$$ is open. I need to show that $$\text{Bd } A = (C \cup O) \setminus \text{Int } (C \cup O)$$ is closed.

We have

\begin{align*} (C \cup O) \setminus \text{Int } (C \cup O) &= (C \cup O) \cap \text{Int }(C \cup O)^c \text{ (complement relative to } C \cup O)\\ &= C \cap \text{Int }(C \cup O)^c \bigcup O \cap \text{Int }(C \cup O)^c\\ &= C \cap \text{Int }(C \cup O)^c \text{ (right side is empty)}. \end{align*}

According to the book, the intersection of any family of closed sets is closed. I know $$C$$ is closed. The problem I have is that I don't know if the right side is closed. It would be closed if the complement were relative to the space $$X$$, but the complement is relative to $$C \cup O$$.

Any help is appreciated.

Edit:

Try a proof by contradiction. Assume $$A$$ is the union of a closed and an open set, and $$A \setminus \text{Int }A$$ is open. That means

\begin{align*} X \setminus (A \setminus \text{Int }A) = (X \setminus A) \cup \text{Int }A \end{align*}

is closed. Under what circumstances could this be true? We know $$\text{Int }A$$ is open, and $$\text{Int }A \subseteq A$$, so $$\text{Int }A \nsubseteq X \setminus A$$. So we have the disjoint union of an open set and the complement of the union of a closed and open set.

• @Azif00 The OP has a different definition of boundary of $A$: specifically $A\setminus\operatorname{Int}A$ instead of $\overline A\setminus\operatorname{Int}A$.
– user239203
Aug 5 '20 at 20:38
• @Azif00 I think that what you are calling the boundary, Brown refers to as the frontier. It was shown earlier that the frontier of any set is closed, that $\text{Fr } A = \overline A \cap \overline{X \setminus A}$, and $\text{Bd } A = A \cap \overline{X \setminus A}$. Aug 5 '20 at 20:39
• @Gae.S. It is true. I had not noticed. Aug 5 '20 at 20:39
• Note that for all $S, T \subset X$ we have $S \cap (S\setminus T) = S \cap (X \setminus T)$, thus the first attempt is easily finished. Regarding the edit, note that "not closed" is not the same as "open". Aug 7 '20 at 17:51
• @DanielFischer If I understand you correctly, you are saying that 1) in my first attempt, it is equivalent to consider the complement relative to the space $X$ (which essentially completes the proof), and 2) in my second attempt I made the mistake of assuming that the boundary has to be open in order for there to be a contradiction when it just has to not be closed. Thanks for your help. If you submit an answer I will be happy to upvote and accept it. Aug 9 '20 at 19:28

Your first attempt works, in that situation it is immaterial whether you consider the complement with respect to $$C \cup O$$ or with respect to the ambient space $$X$$.
Note that for any sets $$S, T \subset X$$ we have $$S \cap (X \setminus T) = \{ x \in X : x \in S \land x \notin T\} = S \setminus T = S \cap (S \setminus T)\,.$$
Your second attempt does not work (at least not without major changes) since in general subsets of $$X$$ that aren't closed need not be open. In particular a set $$B = A \setminus \operatorname{Int} A$$ can only be open if it is empty (and then it is also closed). For $$B$$ is a subset of $$A$$, and if $$B$$ is open it is a subset of $$\operatorname{Int} A$$.