# induced Alexandroff topology on a poset

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

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

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

$(X,\tau)$ is a $\mathrm{T}_{0}$, Alexandroff space.

Proof. Let $x,y\in X$ be such that $x\neq y$. Note that this implies that $x\not\leq y$ or $y\not\leq x$ (because $\leq$ is antisymmetric). Therefore $x\not\in(-\infty,y]$ or $y\not\in(-\infty,x]$. 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

 $A=\bigcap_{i\in I}\,(-\infty,x_{i}]\neq\emptyset$

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

 $(-\infty,y]\subseteq A$

and thus

 $\bigcap_{i\in I}\,(-\infty,x_{i}]=A=\bigcup_{y\in A}\,(-\infty,y]$

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

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 $A=(-\infty,y]\subseteq Y$ 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 $y\leq x$, then $f(y)\leq f(x)$ (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 $(-\infty,x]\subseteq f^{-1}(A)$ and thus

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

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

,,$\Leftarrow$” Assume that $f$ is continuous and let $y\leq x$ for some $x,y\in X$. Assume that $f(y)\not\leq f(x)$. Let $A=(-\infty,f(x)]$. Therefore $f(y)\not\in A$, but $A$ is open, so $f^{-1}(A)$ is open (because $f$ is continuous). Thus $(-\infty,x]\subseteq f^{-1}(A)$. But $y\leq x$, so $y\in f^{-1}(A)$. But this implies that $f(y)\in A$. Contradiction. $\square$

Title induced Alexandroff topology on a poset InducedAlexandroffTopologyOnAPoset 2013-03-22 18:46:01 2013-03-22 18:46:01 joking (16130) joking (16130) 4 joking (16130) Derivation msc 54A05