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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.

Mathematics
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3 Answers3

15

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 $x \notin \partial A \cup \partial B$.
Community
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user642796
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5

Here is a proof using only set operations, to complement the other answers.

Proposition. Suppose $\overline{A} \cap B = A \cap \overline{B} = \varnothing$. Then $\mathrm{Int}(A \cup B) = \mathrm{Int}(A) \cup \mathrm{Int}(B)$.

Proof. Since $\mathrm{Int}(A) \cup \mathrm{Int}(B)$ is an open set contained in $A \cup B$, we have $$ \mathrm{Int}(A) \cup \mathrm{Int}(B) \subseteq \mathrm{Int}(A \cup B). $$ For the converse, note that $A \subseteq \overline{B}^c = \mathrm{Int}(B^c)$ and $B \subseteq \overline{A}^c = \mathrm{Int}(A^c)$, hence $\mathrm{Int}(A \cup B) \subseteq A \cup B \subseteq \mathrm{Int}(B^c) \cup \mathrm{Int}(A^c)$. Therefore $$ \mathrm{Int}(A \cup B) = \big(\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c)\big) \cup \big(\mathrm{Int}(A \cup B) \cap \mathrm{Int}(A^c)\big). \tag*{$(1)$} $$ Note that $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c) \subseteq (A \cup B) \cap B^c = A \setminus B = A$. Since the LHS is an open set contained in $A$, it follows that $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c) \subseteq \mathrm{Int}(A)$. Analogously, $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(A^c) \subseteq \mathrm{Int}(B)$, so it follows from $(1)$ that $\mathrm{Int}(A \cup B) \subseteq \mathrm{Int}(A) \cup \mathrm{Int}(B)$.$\quad\Box$

To prove the same for the boundary, recall that $\partial S = \overline{S} \setminus \mathrm{Int}(S)$. We get the following:

Corollary. Suppose $\overline{A} \cap B = A \cap \overline{B} = \varnothing$. Then $\partial(A \cup B) = \partial A \cup \partial B$.

Proof. Recall that $\overline{A \cup B} = \overline{A} \cup \overline{B}$, even for arbitrary sets. It follows that \begin{align*} \partial(A \cup B) &= \overline{A \cup B} \setminus \mathrm{Int}(A \cup B) \\[1ex] &= (\overline{A} \cup \overline{B}) \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B)) \\[1ex] &= \big(\overline{A} \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B))\big) \cup \big(\overline{B} \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B))\big) \\[1ex] &= (\overline{A} \setminus \mathrm{Int}(A)) \cup (\overline{B} \setminus \mathrm{Int}(B)) \tag*{$(2)$}\\[1ex] &= \partial A \cup \partial B, \end{align*} where to deduce $(2)$ we use that $\overline{A} \cap \mathrm{Int}(B) \subseteq \overline{A} \cap B = \varnothing$ and $\overline{B} \cap \mathrm{Int}(A) \subseteq \overline{B} \cap A = \varnothing$.$\quad\Box$

Under OP's stronger assumption that $\overline{A} \cap \overline{B} = \varnothing$, we have the following stronger result.

Theorem. Suppose $\overline{A} \cap \overline{B} = \varnothing$. Then $\partial(A \cup B)$ is the topological disjoint union of $\partial A$ and $\partial B$.

Proof. Since $\partial A \cap \partial B \subseteq \overline{A} \cap \overline{B} = \varnothing$, it follows from the preceding Corollary that $\partial(A \cup B)$ is the disjoint set union of $\partial A$ and $\partial B$. To see that it is also a topological disjoint union, note that $\partial A$ and $\partial B$ are closed sets, so both are clopen in $\partial(A \cup B)$.$\quad\Box$

Josse van Dobben de Bruyn
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2

I'd like to slightly strengthen the accepted answer's result to that which I was looking for when I came across this question:

If $A \cap B = \emptyset$, then $(\partial A \cup \partial B) \backslash ((\partial A \cap B) \cup (A \cap \partial B)) \subseteq \partial (A \cup B) \subseteq \partial A \cup \partial B$.

The second inclusion is noted by the OP, so I'll prove the first. Write the first set in the above chain as $S_{AB}$. If $x \in S_{AB}$, then by symmetry between $A$ and $B$ we may assume WLOG that $x \in \partial A$ (implying $x \notin B$ by the definition of $S_{AB}$), so every neighborhood of $x$ intersects both $A$ and $A^c$ (the complement of $A$). We consider two cases:

  1. $x \notin A$. In this case, $x \in A^c \cap B^c = (A \cup B)^c$, so any neighborhood of $x$ intersects both $A \subset A \cup B$ and $(A \cup B)^c$, showing $x \in \partial (A \cup B)$.

  2. $x \in A$, which then implies $x \notin \partial B$. In this case, we may find a neighborhood $U_0$ of $x$ contained entirely in either $B$ or $B^c$, and since $x \in B^c \cap U_0$, we must have $U_0 \subset B^c$. Thus for any neighborhood $U$ of $x$, $U \cap U_0 \subset B^c$ intersects $A^c$, and hence $U$ intersects both $A \subset A \cup B$ and $(A \cup B)^c = A^c \cap B^c$, showing $x \in \partial (A \cup B)$.

In both cases, then, $x \in \partial (A \cup B)$, showing $S_{AB} \subseteq \partial (A \cup B)$.

jawheele
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