Ytukyg

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!

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 $Acup B$ 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.