# a connected normal space with more than one point is uncountable

The proof of the following result is an application of the generalized intermediate value theorem (along with Urysohn’s lemma):

###### Proposition.

###### Proof.

Let $X$ be a http://planetmath.org/node/941connected http://planetmath.org/node/1532normal space^{} with at least two distinct points ${x}_{1}$ and ${x}_{2}$. As the sets $\{{x}_{1}\}$ and $\{{x}_{2}\}$ are http://planetmath.org/node/2739closed and disjoint, Urysohn’s lemma furnishes a continuous function^{} $f:X\to [0,1]$ such that $f({x}_{1})=0$ and $f({x}_{2})=1$. Because $X$ is connected, the generalized intermediate value theorem implies that $f$ is surjective^{}. Thus $f$ may be suitably to give a bijection between a subset of $X$ and the uncountable set $[0,1]$, from which it follows that $X$ is uncountable.
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Title | a connected normal space with more than one point is uncountable |
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Canonical name | AConnectedNormalSpaceWithMoreThanOnePointIsUncountable |

Date of creation | 2013-03-22 17:17:46 |

Last modified on | 2013-03-22 17:17:46 |

Owner | azdbacks4234 (14155) |

Last modified by | azdbacks4234 (14155) |

Numerical id | 10 |

Author | azdbacks4234 (14155) |

Entry type | Theorem |

Classification | msc 54D05 |

Related topic | UrysohnsLemma |

Related topic | NormalTopologicalSpace |

Related topic | Uncountable |

Related topic | Bijection |

Related topic | ConnectedSpace |