# a space is T1 if and only if every subset A is the intersection of all open sets containing A

Say we have $X$, a ${T}^{1}$-space, and $A$, a subset of $X$. We aim to show that the intersection^{} of all open sets containing $A$ equals $A$. By de Morgan’s laws, that would be true if the complement of $A$, ${A}^{c}$, equalled the union of all closed sets^{} in ${A}^{c}$. Let’s call this union of closed sets $C$.

Each set that makes up $C$ is contained by ${A}^{c}$, so $C\subset {A}^{c}$. If we could show ${A}^{c}\subset C$, we’d be done.

Since $X$ is ${T}^{1}$, each singleton in ${A}^{c}$ is closed (http://planetmath.org/ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed). Their union, a subset of $C$, contains ${A}^{c}$, so we’re through.

Now suppose we know that in some topological space^{} $X$, any subset $A$ of $X$ is the intersection of all open sets containing $A$. Given $x\ne y$, we’re looking for an open set containing $x$ but not $y$, to show that $X$ is ${T}^{1}$.

$$\{x\}=\bigcap _{\begin{array}{c}U\text{open}\\ U\ni x\end{array}}U$$ |

by hypothesis^{}. If all open sets containing $x$ contained $y$, $y$ would be in the intersection; since $y$ isn’t in the intersection, $X$ must be ${T}^{1}$.

Title | a space is T1 if and only if every subset A is the intersection of all open sets containing A |
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Canonical name | ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA |

Date of creation | 2013-03-22 14:20:18 |

Last modified on | 2013-03-22 14:20:18 |

Owner | waj (4416) |

Last modified by | waj (4416) |

Numerical id | 4 |

Author | waj (4416) |

Entry type | Proof |

Classification | msc 54D10 |

Related topic | T1Space |

Related topic | ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed |