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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.

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    $\begingroup$ @Azif00 The OP has a different definition of boundary of $A$: specifically $A\setminus\operatorname{Int}A$ instead of $\overline A\setminus\operatorname{Int}A$. $\endgroup$
    – user239203
    Aug 5 '20 at 20:38
  • $\begingroup$ @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 $\text{Bd } A = A \cap \overline{X \setminus A}$. $\endgroup$
    – Novice
    Aug 5 '20 at 20:39
  • $\begingroup$ @Gae.S. It is true. I had not noticed. $\endgroup$
    – azif00
    Aug 5 '20 at 20:39
  • $\begingroup$ 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". $\endgroup$ Aug 7 '20 at 17:51
  • $\begingroup$ @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. $\endgroup$
    – Novice
    Aug 9 '20 at 19:28
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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 $\operatorname{Int} A$.

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