induced Alexandroff topology on a poset

Let $(X,\le )$ be a poset. For any $x\in X$ define following subset:

$$(-\mathrm{\infty },x]=\{y\in X|y\le x\}.$$

The induced Alexandroff topology $\tau$ on $X$ is defined as a topology generated by ${\{(-\mathrm{\infty },x]\}}_{x\in X}$.

Proposition 1. $(X,\tau )$ is a ${\mathrm{T}}_0$, Alexandroff space.

Proof. Let $x,y\in X$ be such that $x\ne y$. Note that this implies that $x\nleqq y$ or $y\nleqq x$ (because $\le$ is antisymmetric). Therefore $x\notin (-\mathrm{\infty },y]$ or $y\notin (-\mathrm{\infty },x]$. Thus $(X,\tau )$ is ${\mathrm{T}}_0$.

Now in order to show that $(X,\tau )$ is Alexandroff it is enough to show that an arbitrary intersection of base sets is open. So assume that ${\{x_i\}}_{i\in I}$ is a subset of $X$ such that

$$A=\bigcap _{i\in I}(-\mathrm{\infty },x_i]\ne \mathrm{\varnothing }$$

and let $y\in A$. Then (since $\le$ is transitive) it is clear that

$$(-\mathrm{\infty },y]\subseteq A$$

and thus

$$\bigcap _{i\in I}(-\mathrm{\infty },x_i]=A=\bigcup _{y\in A}(-\mathrm{\infty },y]$$

and therefore the intersection is open, which completes the proof. $\mathrm{\square }$

Proposition 2. Let $X,Y$ be posets and $f:X\to Y$ a function. Then $f$ preserves order if and only if $f$ is continuous in induced Alexandroff topologies.

Proof. ,,$\Rightarrow$” Assume that $f$ preserves order and let $A=(-\mathrm{\infty },y]\subseteq Y$ be an open base set. We wish to show that $f^{-1}(A)$ is open in $X$. So take any $x\in f^{-1}(A)$. Now if $y\le x$, then $f(y)\le f(x)$ (since $f$ preserves order) and thus $f(y)\in A$. Therefore $y\in f^{-1}(A)$. Since $y$ was arbitrary we obtain that for any $x\in f^{-1}(A)$ we have $(-\mathrm{\infty },x]\subseteq f^{-1}(A)$ and thus

$$f^{-1}(A)=\bigcup _{x\in f^{-1}(A)}(-\mathrm{\infty },x],$$

which implies that $f^{-1}(A)$ is open.

,,$\Leftarrow$” Assume that $f$ is continuous and let $y\le x$ for some $x,y\in X$. Assume that $f(y)\nleqq f(x)$. Let $A=(-\mathrm{\infty },f(x)]$. Therefore $f(y)\notin A$, but $A$ is open, so $f^{-1}(A)$ is open (because $f$ is continuous). Thus $(-\mathrm{\infty },x]\subseteq f^{-1}(A)$. But $y\le x$, so $y\in f^{-1}(A)$. But this implies that $f(y)\in A$. Contradiction. $\mathrm{\square }$

Title	induced Alexandroff topology on a poset
Canonical name	InducedAlexandroffTopologyOnAPoset
Date of creation	2013-03-22 18:46:01
Last modified on	2013-03-22 18:46:01
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	4
Author	joking (16130)
Entry type	Derivation
Classification	msc 54A05