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!
Best Answer
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.