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

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. $\square$

Title Alexandroff space is T1 if and only if it is discrete AlexandroffSpaceIsT1IfAndOnlyIfItIsDiscrete 2013-03-22 18:46:08 2013-03-22 18:46:08 joking (16130) joking (16130) 4 joking (16130) Derivation msc 54A05