# completely normal

Let $X$ be a topological space^{}. $X$ is said to be if whenever $A,B\subseteq X$ with $A\cap \overline{B}=\overline{A}\cap B=\mathrm{\varnothing}$, then there are disjoint open sets $U$ and $V$ such that $A\subseteq U$ and $B\subseteq V$.

Equivalently, a topological space $X$ is if and only if every subspace^{} is normal.

Title | completely normal^{} |
---|---|

Canonical name | CompletelyNormal |

Date of creation | 2013-03-22 12:13:51 |

Last modified on | 2013-03-22 12:13:51 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 7 |

Author | Mathprof (13753) |

Entry type | Definition |

Classification | msc 54-00 |

Synonym | complete normality |

Related topic | NormalTopologicalSpace |