# proof that components of open sets in a locally connected space are open

###### Theorem.

A topological space^{} $X$ is locally connected if and only if each component of an open set
is open.

###### Proof.

First, suppose that $X$ is locally connected and that $U$ is an open set of $X$. Let $p\in C$, where $C$ is a component of $U$. Since $X$ is locally connected there is an open connected set, say $V$ with $p\in V\subset U$. Since $C$ is a component of $U$ it must be that $V\subset C$. Hence, $C$ is open. For the converse, suppose that each component of each open set is open. Let $p\in X$. Let $U$ be an open set containing $p$. Let $C$ be the component of $U$ which contains $p$. Then $C$ is open and connected, so $X$ is locally connected.

∎

As a corollary, we have that the components of a locally connected space are both open and closed.

Title | proof that components of open sets in a locally connected space are open |
---|---|

Canonical name | ProofThatComponentsOfOpenSetsInALocallyConnectedSpaceAreOpen |

Date of creation | 2013-03-22 17:06:07 |

Last modified on | 2013-03-22 17:06:07 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 6 |

Author | Mathprof (13753) |

Entry type | Theorem |

Classification | msc 54A99 |