# 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 |