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Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$?

I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result.

2 Answers 2

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You can actually get by with a little less: If $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, the result holds.

As you have noticed, we have that $\partial ( A \cup B ) \subseteq \partial A \cup \partial B$. Suppose that $x \notin \partial ( A \cup B )$. There are two cases:

  1. $x \notin \overline{ A \cup B } = \overline{A} \cup \overline{B}$. In this case it can easily be shown that $x \notin \partial A \cup \partial B$.
  2. $x \notin \overline{ X \setminus ( A \cup B ) }$. Then $x \in X \setminus \overline{ X \setminus ( A \cup B ) } = \mathrm{Int} ( A \cup B )$. Without loss of generality assume that $x \in A$, and since $A\cap \overline B=\emptyset$, then $x\in X\setminus\overline B$, which implies that $x\notin \partial (B)$. Furthermore, it can be shown that $U = \mathrm{Int} ( A \cup B ) \setminus \overline{B}$ is a neighbourhood of $x$ that is contained in $A$, and so $x \notin \partial A$. Thus $x \notin \partial A \cup \partial B$.