union of non-disjoint connected sets is connected

Theorem 1.

Suppose $A\mathrm{,}B$ are connected sets in a topologicalspace $X$. If $A\mathrm{,}B$ are not disjoint, then $A\mathrm{\cup }B$ is connected.

Proof.

By assumption, we have two implications.First, if $U,V$ are open in $A$ and $U\cup V=A$, then $U\cap V\ne \mathrm{\varnothing }$.Second, if $U,V$ are open in $B$ and $U\cup V=B$, then $U\cap V\ne \mathrm{\varnothing }$.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

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

$=$$(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

$$\mathrm{\varnothing }\ne (U\cap A)\cap (V\cap A)=(U\cap V)\cap A.$$

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