If $\overline{A} \cap B \neq \emptyset, E = A \cup B$, prove that E is connected

Question

How can the following proposition be proved?

Proposition: If $A$ and $B$ are two connected subsets of $X$, and $\overline{A} \cap B \neq \emptyset$ (where $\overline{A}$ denotes the closure of $A$), then $A\cup B$ is connected.

The definitions of connected and separated sets are relevant:

Definition: A set $X$ is separated (or not connected) if there exist nonempty sets $A$ and $B$ such that $X = A \cup B$ and $$\overline{A} \cap B = A \cap \overline{B} = \emptyset.$$ A set is connected if it is not separated.

Intuitively, I feel that this statement makes sense because the only way $A$ and $B$ are connected but $A \cup B$ isn't is if there is some possible separate component of $A$ and $B$. It must be that some part outside $A$ (in this case, $\overline{A}$) allows a separation between $A$ and $B$.

Answer 1

We prove this by contradiction. Assume $A\cup B$ is disconnected, then there are two open sets $U$ and $V$ such that

• $A\cup B\subseteq U\cup V$ and $U\cap(A\cup B)\neq\emptyset, V\cap (A\cup B)\neq\emptyset$.
• $U'\cap V'=\emptyset$ where $U'=U\cap(A\cup B), V'=V\cap (A\cup B)$. Note that $A\cup B=U'\cup V'$. Moreover $U'$ and $V'$ are open in $A\cup B$ under the subspace topology.

Since $A$ is connected, it follows that either $A\subseteq U$ or $A\subseteq V$, let's say $A\subseteq U$. Now, as $B$ is also connected, it follows that $B\subseteq U$ or $B\subseteq V$. Then we have $B\subseteq V$ because both $U$ and $V$ have non-empty intersections with $A\cup B$.

Now take $b\in Cl(A)\cap B$. Then $V$ is an open neighbourhood of $b$ thus $V\cap A\neq\emptyset$. However as $A\subseteq U$, it follows that $V\cap A\subseteq U$, thus $V'\cap U'\neq\emptyset$, which is a contradiction.