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. $\square$