union of non-disjoint connected sets is connected

Theorem 1.

Suppose $A\mathrm{,}B$ are connected sets in a topological space $X$. 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. ∎

Title	union of non-disjoint connected sets is connected
Canonical name	UnionOfNondisjointConnectedSetsIsConnected
Date of creation	2013-03-22 15:17:50
Last modified on	2013-03-22 15:17:50
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	8
Author	matte (1858)
Entry type	Theorem
Classification	msc 54D05