induced partial order on an Alexandroff space

Let $X$ be a ${\mathrm{T}}_0$, Alexandroff space. For $A\subseteq X$ denote by $A^o$ the intersection of all open neighbourhoods of $A$. Define a relation $\le$ on $X$ as follows: for any $x,y\in X$ we have $x\le y$ if and only if $x\in {\{y\}}^o$. This relation will be called the induced partial order on $X$.

Proposition 1. $(X,\le )$ is a poset.

Proof. Of course $x\in {\{x\}}^o$ for any $x\in X$. Thus $\le$ is reflexive.

Assume now that $x\le y$ and $y\le x$ for some $x,y\in X$. Assume that $x\ne y$. Then, since $X$ is a ${\mathrm{T}}_0$ space, there is an open set $U$ such that $x\in U$ and $y\notin U$ or there is an open set $V$ such that $y\in V$ and $x\notin V$. Both cases lead to contradiction, because we assumed that $x\in {\{y\}}^o$ and $y\in {\{x\}}^o$. Thus every open neighbourhood of one element must also contain the other. Thus $\le$ is antisymmetric.

Finally assume that $x\le y$ and $y\le z$ for some $x,y,z\in X$. Since $y\in {\{z\}}^o$, then ${\{z\}}^o$ is an open neighbourhood of $y$ and thus ${\{y\}}^o\subseteq {\{z\}}^o$. Therefore $x\in {\{z\}}^o$, so $\le$ is transitive, which completes the proof. $\mathrm{\square }$

Proposition 2. Let $X,Y$ be two, ${\mathrm{T}}_0$, Alexandroff spaces and $f:X\to Y$ be a function. Then $f$ is continuous if and only if $f$ preserves the induced partial order.

Proof. ,,$\Rightarrow$” Assume that $f$ is continuous and suppose that $x,y\in X$ are such that $x\le y$. We wish to show that $f(x)\le f(y)$, so assume this is not the case. Let $A={\{f(y)\}}^o$. Then $f(x)\notin A$. But $A$ is open, so $f^{-1}(A)$ is also open (because we assumed that $f$ is continuous). Furthermore $y\in f^{-1}(A)$ and because $x\le y$, then $x\in f^{-1}(A)$, but this implies that $f(x)\in A$. Contradiction.

,,$\Leftarrow$” Assume that $f$ preserves the induced partial order and let $U\subseteq Y$ be an open subset. Let $x\in U$. Then for any $y\le x$ we have $f(y)\le f(x)$ (because $f$ preserves the induced partial order) and since ${\{f(x)\}}^o\subseteq U$ (because $U$ is open and ${\{f(x)\}}^o$ is the smallest open neighbourhood of $f(x)$) we have that $f(y)\in U$. Thus

$${\{x\}}^o=\{y\in X|y\le x\}\subseteq f^{-1}(U)$$

which implies that $f^{-1}(U)$ is open because $f^{-1}(A)$ contains a small neighbourhood of each point. This completes the proof. $\mathrm{\square }$