I'm trying to understand a part of this proof. In fact, I would like to know why $P \in f^{-1}(O Q)$. I understand the rest of the proof, but I have no idea where does this assertion comes from. Any help will be appreciated.
Theorem:
Let $X$ be a locally finite Alexandroff space which is not $T_{1}$. Then there exist two points $P \neq Q$ in $X$ such that for every two neighborhoods $U(P)$ and $U(Q)$ of these points any injection $f: U(P) \longrightarrow U(Q)$ mapping $P$ in $Q$ is not continuous.
Where $OQ$ is the maximal neighbourhood of $Q$ and $OP$ the maximal neighbourhood of $P$
Proof:
$X$ is not $T_{1}$, hence there exist points $P$ and $Q$ such that $Q \in O P$ and $P \notin O Q$. Thus $O Q \subset O P$ (strictly), and therefore $|O Q|<|O P|$. $O P$ is clearly an open subset of $U(P)$ and $O Q$ is an open subset of $U(Q)$. Let $f: U(P) \longrightarrow U(Q)$ be any continuous injection. Since $P \in f^{-1}(O Q)$ we obtain that $O P \subseteq f^{-1}(O Q) .$ Thus $|O P| \leq\left|f^{-1}(O Q)\right|=$ $O Q \mid$, and this contradicts the fact that $|O Q|<|O P|$.
Thanks for your time. :)
We also assume that $f$ maps $P$ to $Q$ (see the theorem statement!). So $f(P)=Q \in OQ$ so $P \in f^{-1}(OQ)$ by definition and as that set is open, minimality of $OP$ then gives $OP \subseteq f^{-1}(OQ)$ and we get the contradiction from $|f^{-1}(OQ)|=|OQ|$ from injectivity.