Let $X$ be a topological space, and let $A$ and $B$ be any subsets of $X$. How to show that if $\overline{A} \cap \overline{B} = \emptyset$, then $b(A \cup B) = b(A) \cup b(B)$?

Here $b(A)$ denotes the boundary of $A$. Thus $p \in b(A)$ if and only if, for every open set $G$ containing $p$, we have $G \cap A \neq \emptyset$ and $G \cap (X \setminus A) \neq \emptyset$.

Moreover, we have $$ \operatorname{Int} (A) \cap b(A) \cap \operatorname{Ext} (A) = \emptyset $$ and $$ \operatorname{Int} (A) \cup b(A) \cup \operatorname{Ext} (A) = X. $$ And we also have $$ \overline{A} =\operatorname{Int} (A) \cup b(A). $$

What next?

  • 3
    $\begingroup$ One notation for boundary is $\partial A$. $\endgroup$
    – copper.hat
    May 14 at 4:13
  • $\begingroup$ Does one inclusion hold all the time? Can you start there? $\endgroup$ May 14 at 4:58

3 Answers 3


For any $x\in b(A)\cup b(B), x\in b(A)$ or $x\in b(B)$. WLOG, let $x\in b(A)$. It follows that $x\in \overline A.$ So $x\notin \overline B$ (hence $x\in B^c$). Let $G_x$ be any open set containing $x$. Then, $G_x\cap (A\cup B)\ne \emptyset$ since $G_x\cap A$ is non empty; and $G_x\cap (A\cup B)^c=G_x\cap A^c\cap B^c\ne \emptyset$ (if not, then $G_x\subset A\cup B$. So if $V_x\subset G_x$ is any open set containing $x$ then $V_x\cap B$ can't be non empty else $V_x\subset A$ implying that $x$ is an interior point of $A$. So $x$ must be a limit point of $B$,i.e. $x\in \overline B$, which is a contradiction. ) It follows that $x\in b(A\cup B)$. So $$b(A)\cup b(B)\subset b(A\cup B)\tag 1$$

For any $x\in b(A\cup B), \text{ x is neither an interior point nor an exterior point of } A\cup B$ and $x\in \overline {A\cup B}=\overline A\cup \overline B.$ So wlog, let $x\in \overline A$. It follows that $x\notin \overline B$. Let $G_x$ be any open set containing $x$. $G_x\cap A\ne \emptyset$ holds because $x\in \overline A$.

Since $x$ is a boundary point of $A\cup B$, the following holds: $G_x\cap (A\cup B)^c=G_x\cap A^c\cap B^c\ne \emptyset$, whence it follows that $G_x\cap A^c\ne \emptyset.$

It follows that $$x\in b(A)\subset b(A)\cup b(B)\implies b(A\cup B)\subset b(A)\cup b(B)\tag 2$$

By $(1)$ and $(2)$, one gets the desired equality.

  • $\begingroup$ @SaaqibMahmood: I have made the following modification (rather a correction) in the post: In the first para, the reason for $G_x\cap (A\cup B)^c\ne \emptyset$ was earlier given as "since $\color{red}{x\in G_x\cap A^c}$ and $x\in B^c$" wherein the red highlighted part doesn't always have to hold. Therefore, I have corrected the same and added the explanation for the same. $\endgroup$
    – Koro
    May 16 at 8:10

One inclusion holds without the condition $\overline{A}\cap\overline{B}=\emptyset$:

\begin{align} & \ \ \ \partial(A\cup B) \\ &= \left( \overline{A\cup B} \right) \cap \overline{(A\cup B)^\complement} \\ &= \left( \overline{A\cup B} \right) \cap \overline{A^\complement\cap B^\complement} \qquad [ \mbox{ by one of DeMorgan's laws } ] \\ &\subset \left( \overline{A}\cup\overline{B} \right) \cap \left( \overline{A^\complement}\cap \overline{B^\complement} \right) \\ &= \left(\overline{A}\cap\overline{A^\complement}\cap\overline{B^\complement}\right) \cup\left(\overline{A^\complement}\cap\overline{B}\cap\overline{B^\complement}\right) \\ &= \left(\partial A\cap\overline{B^\complement}\right) \cup\left(\overline{A^\complement}\cap\partial B\right) \\ &\subset \partial A\cup\partial B. \end{align}

The other inclusion also doesn't need that condition. In fact you only need the weaker condition $\overline{A}\cap B=A\cap\overline{B}=\emptyset$:

Let $x \in \partial A \cup \partial B$. Assume without loss of generality that $x\in\partial A$. Then $x\notin A^\circ$ and because of $x\in\overline{A}$, we have $x\notin B$ (due to the condition) and $x\in\overline{A}\cup\overline{B}=\overline{A\cup B}$. If $x\notin(A\cup B)^\circ$, we are finished. Assume $x\in(A\cup B)^\circ$ and let $U\subseteq A\cup B$ be an open neighboorhood of $x$, then $U\not\subseteq A$ (as otherwise $x\in A^\circ$) and therefore $U\cap B\neq\emptyset$, then because of $x\in\overline{B}$, we have $x\notin A$ (due to the condition). Because of $x\notin A\cup B$, we get a contradiction with the assumption. (Or it would also directly imply the desired result $x\in\partial(A\cup B)$.)

  • $\begingroup$ how do we know that $x \in \overline{B}$ in the last paragraph? After all, we only that for any open set $U$ such that $x \in U$ and $U \subset A\cup B$, we have $U \cap B \neq \emptyset$. You see, there could be open sets containing $x$ which are not contained in $A \cup B$. Can we also show that such open sets too intersect $B$? $\endgroup$ May 16 at 4:56
  • 1
    $\begingroup$ If there is such a subset, I can consider the intersection with an open subset of $A\cup B$ which is open again. $\endgroup$ May 16 at 7:06

Showing that $\partial (A\cup B) \subset \partial A \cup \partial B$ is straightforward and is always true.

Suppose $\overline{A} \cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$.

Choose $x \in \partial A \cup \partial B$. We would like to show that $x \in \partial (A\cup B)$.

We have $x \in \overline{A} \cup \overline{B} = \overline{A \cup B}$.

If $x \in \overline{(A \cup B)^c}$ then $x \in \partial (A\cup B)$ and we are finished.

If $x \notin \overline{(A \cup B)^c}$ then $x \in W = (A \cup B)^\circ$. Note that $A^c \cap W =B \cap W$ and $B^c \cap W = A \cap W$. Since $x \in \partial A \cup \partial B$, every open neighourhood $U$ of $x$ (which must intersect $W$) contains elements of $A,B$. Hence $x \in \overline{A} \cap \overline{B}$. However, since $x \in W$ we have $x \in A$ or $x \in B$ which gives a contradiction and hence no such $x$ exists.


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