# union of non-disjoint connected sets is connected

###### Theorem 1.

Suppose $A,B$ are connected sets in a topological space $X$. If $A,B$ are not disjoint, then $A\cup B$ is connected.

###### Proof.

By assumption, we have two implications. First, if $U,V$ are open in $A$ and $U\cup V=A$, then $U\cap V\neq\emptyset$. Second, if $U,V$ are open in $B$ and $U\cup V=B$, then $U\cap V\neq\emptyset$. To prove that $A\cup B$ is connected, suppose $U,V$ are open in $A\cup B$ and $U\cup V=A\cup B$. Then

 $\displaystyle U\cup V$ $\displaystyle=$ $\displaystyle((U\cup V)\cap A)\,\cup\,((U\cup V)\cap B)$ $\displaystyle=$ $\displaystyle(U\cap A)\cup(V\cap A)\cup(U\cap B)\cup(V\cap B)$

Let us show that $U\cap A$ and $V\cap A$ are open in $A$. To do this, we use this result (http://planetmath.org/SubspaceOfASubspace) and notation from that entry too. For example, as $U\in\tau_{A\cup B,X}$, $U\cap A\in\tau_{A,A\cup B,X}=\tau_{A,X}$, and so $U\cap A$, $V\cap A$ are open in $A$. Since $(U\cap A)\cup(V\cap A)=A$, it follows that

 $\emptyset\neq(U\cap A)\cap(V\cap A)=(U\cap V)\cap A.$

If $U\cap V=\emptyset$, then this is a contradition, so $A\cup B$ must be connected. ∎

Title union of non-disjoint connected sets is connected UnionOfNondisjointConnectedSetsIsConnected 2013-03-22 15:17:50 2013-03-22 15:17:50 matte (1858) matte (1858) 8 matte (1858) Theorem msc 54D05