# Assume $A \subset \mathbb{R}^n$ is connected such that $A^c$ is separated by $B,C$, then $A \cup B$ is connected.

The following problem showed up on a previous qualifying exam:

Assume $$A \subset \mathbb{R}^n$$ is connected such that $$\begin{equation}A^c = B \cup C, \text{ such that } \overline{B} \cap C = \overline{C} \cap B = \emptyset \end{equation}$$ show that $$A \cup B$$ is connected. (We take $$\mathbb{R}^n$$ with the usual metric topology.)

This is what I have so far:

Assume for the sake of contradiction that $$A \cup B$$ is disconnected, then $$A \cup B = D \cup E$$ where $$D,E$$ are separated. Then as $$A$$ is connected, we must have $$A \subset D$$ or $$A \subset E$$; WLOG assume $$A \subset D$$.

I'm unsure how to proceed from here because I want to write $$A$$ as a union of two non-empty separated sets, but from $$A \cup B = D \cup E$$, I cannot see how to get a non-empty separated sets since intersecting them by $$A$$ or $$B^c$$ ends up with $$E \cap A$$ and $$E \cap B^c$$ being empty since $$E \subset B$$ as $$A \subset D$$.

Any help would be appreciated!

• Do you make an hypothesis regarding the topology used on $\mathbb R^n$? Is it the one induced by the standard distance? Dec 30 '18 at 10:37
• @mathcounterexamples.net yes it is Dec 30 '18 at 16:12

In fact the result is more general and true for any connected topological space $$X$$, which is the case for $$\mathbb R^n$$ endowed with the topology induced by the usual distance.

Let’s suppose that $$A \neq \emptyset$$. We’ll deal with the case $$A = \emptyset$$ at the end of the answer.

Suppose that $$D,E$$ are open sets of $$X$$ with $$A \cup B \subseteq D \cup E$$ and $$D \cap E = \emptyset$$. As $$A$$ is supposed to be connected, we can suppose without loss of generality that $$A \subseteq D$$. We’ll prove that $$U=E \cap (A \cup B)$$ is open and closed in $$X$$ hence empty as in a connected space the only subspaces that are both closed and open are the empty set and the space itself. This will provide the conclusion.

Let’s first notice that $$U \subseteq B$$ as $$D \cap E= \emptyset$$ and therefore $$U= E \cap B$$.

$$U$$ is open

Take $$b \in U$$. We have $$b \notin \overline{A}$$ as $$b \in \overline{A}$$ would imply the contradiction $$E \cap A \neq \emptyset$$ as $$E$$ is open. The hypothesis $$B \cap \overline{C}= \emptyset$$ allows to state $$b \notin \overline{C}$$. We also have $$b \notin \overline{A \cup C}$$ as $$\overline{A \cup C} \subseteq \overline{A} \cup \overline{C}$$. So $$b$$ belongs to the open subset $$V =X \setminus \overline{A \cup C}$$ that is included in $$B$$. And also to the open subset $$E \cap V$$ that is included in $$U$$. This proves that $$b$$ belongs to the interior of $$U$$. As this is true for all $$b \in U$$, $$U$$ is equal to its interior and is open.

$$U$$ is closed

Take $$x \in \overline{U} = \overline{E \cap B}$$. As $$\overline{E \cap B} \subseteq \overline{E} \cap \overline{B}$$, we have $$x \in \overline{B}$$ and $$x \notin C$$ according to the hypothesis $$\overline{B} \cap C =\emptyset$$. Hence $$x \in A \cup B$$. $$x$$ cannot belong to $$A$$: if that would be the case, $$x$$ would belong to the open set $$D$$ and $$D$$ would intersect $$U$$ and therefore $$E$$, a contradiction. We get $$x \in B$$ and as $$x \notin D$$, we have $$x \in E$$ and finally $$x \in U$$. So $$U = \overline{U}$$, proving that $$U$$ is closed.

The case $$A = \emptyset$$

In that case, the space $$X$$ is the union of $$B$$ and $$C$$. $$B$$ and $$C$$ are open and closed subsets of the connected space $$X$$ and are therefore connected. We’re done!

The result doesn’t hold if $$X$$ is not connected. Take for example $$X=\{1,2,3\}$$ endowed with the topology induced by the metric distance and $$A=\{1\}$$, $$B=\{2\}$$ and $$C=\{3\}$$.

Note: this answer is based on following one of the excellent French math site les-mathematiques.net.

If $$A$$ is connected then the complementary of $$A$$ is connected, because $$\mathbb{R}^n$$ is for $$n>1$$. Suppose $$A\cup B$$ disconnected. Then the not connected part of the union has to be $$B$$, leading to a contradiction: $$B$$ is a not connected subspace of a connected one. On the real line is true by taking $$A$$ the open connected unit sphere without boundary and $$B$$ one of the connected components of boundary.

• The complement of $A$ is not connected. We're assuming $A^c$ is disconnected since $A$ can be written as the union of two separated sets. For illustration in $n > 1$, consider the punctured unit ball with punctured point being $\textbf{x}=0$. Then $A^c$ is the complement of the unit ball and the one point in the origin, so $A^c$ is disconnected. Dec 30 '18 at 16:47