# If $\overline{A}\cap\overline{B}=\emptyset$, then $b(A)\cup b(B) = b(A\cup B)$

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?

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

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.

• @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.
– 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 $$x\in\partial(A\cup B)$$.)

• 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 $B$? May 16 at 4:56
• If there is such a subset, I can consider the intersection with an open subset of $A\cup B$ which is open again. 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.