# 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 $\leq$ on $X$ as follows: for any $x,y\in X$ we have $x\leq y$ if and only if $x\in\{y\}^{o}$. This relation will be called the induced partial order on $X$.

$(X,\leq)$ is a poset.

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

Assume now that $x\leq y$ and $y\leq x$ for some $x,y\in X$. Assume that $x\neq y$. Then, since $X$ is a $\mathrm{T}_{0}$ space, there is an open set $U$ such that $x\in U$ and $y\not\in U$ or there is an open set $V$ such that $y\in V$ and $x\not\in V$. 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 $\leq$ is antisymmetric.

Finally assume that $x\leq y$ and $y\leq z$ 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 $\leq$ is transitive, which completes the proof. $\square$

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 $x\leq y$. We wish to show that $f(x)\leq f(y)$, so assume this is not the case. Let $A=\{f(y)\}^{o}$. Then $f(x)\not\in A$. 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 $x\leq y$, 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 $y\leq x$ we have $f(y)\leq f(x)$ (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

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

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

Title induced partial order on an Alexandroff space InducedPartialOrderOnAnAlexandroffSpace 2013-03-22 18:45:55 2013-03-22 18:45:55 joking (16130) joking (16130) 4 joking (16130) Derivation msc 54A05