# Alexandroff space is T1 if and only if it is discrete

Proposition^{}. Let $X$ be an Alexandroff space. Then $X$ is ${\mathrm{T}}_{1}$ if and only if $X$ is discrete.

Proof. ,,$\Leftarrow $” It is easy to see, that every discrete space is Alexandroff and ${\mathrm{T}}_{1}$.

,,$\Rightarrow $” Recall that topological space^{} is ${\mathrm{T}}_{1}$ if and only if every subset is equal to the intersection^{} of all its open neighbourhoods. So let $x\in X$. Then the intersection of all open neighbourhoods ${\{x\}}^{o}$ of $x$ is equal to $\{x\}$. But since $X$ is Alexandroff, then ${\{x\}}^{o}=\{x\}$ is open and thus points are open. Therefore $X$ is discrete. $\mathrm{\square}$

Title | Alexandroff space is T1 if and only if it is discrete |
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Canonical name | AlexandroffSpaceIsT1IfAndOnlyIfItIsDiscrete |

Date of creation | 2013-03-22 18:46:08 |

Last modified on | 2013-03-22 18:46:08 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Derivation |

Classification | msc 54A05 |