# a space is T1 if and only if every singleton is closed

Say $X$ is a http://planetmath.org/node/1852${\mathrm{T}}_{\mathrm{1}}$ topological space^{}. Let’s show that $\{x\}$ is closed for every $x\in X$:

The ${T}_{1}$ axiom (http://planetmath.org/T1Space) gives us, for every $y$ distinct from $x$, an open ${U}_{y}$ that contains $y$ but not $x$. Since we’re in a topological space, we can take the union of all these open sets to get a new open set,

$$U=\bigcup _{y\ne x}{U}_{y}.$$ |

$\{x\}$ is the complement of $U$, closed because $U$ is open: None of the ${U}_{y}$ contain $x$, so $U$ doesn’t contain $x$. But any $y\ne x$ is in $U$, since $y\in {U}_{y}\subset U$. That takes care of that.

Now let’s say we have a topological space $X$ in which $\{x\}$ is closed for every $x\in X$. We’d like to show that ${T}_{1}$ holds:

Given $x\ne y$, we want to find an open set that contains $x$ but not $y$. $\{y\}$ is closed by hypothesis^{}, so its complement is open, and our search is over.

Title | a space is T1 if and only if every singleton is closed |
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Canonical name | ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed |

Date of creation | 2013-03-22 14:20:15 |

Last modified on | 2013-03-22 14:20:15 |

Owner | waj (4416) |

Last modified by | waj (4416) |

Numerical id | 7 |

Author | waj (4416) |

Entry type | Proof |

Classification | msc 54D10 |

Related topic | ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA |