# Bijection between ideals of a poset and order-preserving maps from the poset to $\{0,1\}$.

Show that there is a bijection between the ideals of a poset $$P$$ and the order-preserving functions from $$P$$ to $$\{0,1\}$$ with the relation $$0\leq 1$$.

A nonempty subset $$I\subset P$$ is called an ideal (just a "lower set" in some texts) if $$y\leq x$$ for any $$x\in I$$, then $$y\in I$$.

Let $$\mathcal{I}$$ be the set of ideals of $$P$$ and $$\text{Hom}_{\mathcal{P}os}(P,\{0,1\})$$ be the set of order preserving maps from $$P\to \{0,1\}$$. Define $$\Phi:\mathcal{I}\to\text{Hom}_{\mathcal{P}os}(P,\{0,1\})$$ by $$I\mapsto f(x)=\begin{cases} 1 & \text{ if } x \text{ is maximal in }I, \\ 0 & \text{ else} \end{cases}$$ Is this the correct map to use here? The map $$f$$ defined in this way is order-preserving and $$\Phi$$ should be one-to-one since if $$\Phi(I)=\Phi(J)$$ then $$I$$ and $$J$$ have all the same maximal elements and would thus be the same ideal(I think elements in $$I$$ and $$J$$ that are not comparable to anything else would be problematic for this potentially...). Now if $$f\in\text{Hom}_{\mathcal{P}os}(P,\{0,1\})$$, let $$I$$ be the ideal generated by all $$x\in P$$ such that $$f(x)=1$$. Then $$\Phi(I)=f$$. Thus $$\Phi$$ is a bijection. Is this correct? Thanks in advance.

• An ideal is not just a lower set. It is a upward directed lower set. It is the order dual of a filter - an downward directed upper set. Sep 22 '18 at 2:22
• Does "x∈I is maximal" mean x is a maximal element of I or does it mean x is in a maximal ideal I. Sep 22 '18 at 2:27
• Is set of lower sets of P given the subset order? Sep 22 '18 at 2:33
• @WilliamElliot I did say that in some texts this definition is just called a lower set, but I guess you missed that? $x\in I$ is maximal in $I$, will edit accordingly. Does the set of lower sets of $P$ need an order? Sep 22 '18 at 2:36

I think your bijection does not work because more than one ideal may not have a maximal element. For example, for the poset $$\mathbb{R}$$, the ideals $$(-\infty,0)$$ and $$(-\infty,1)$$ will cause a problem. I propose defining the map $$\Phi:\mathcal{I}\to\text{Hom}_\text{Pos}\big(P,\{0,1\}\big)$$ by setting $$\Phi(I):=f_I\,,\text{ where }f_I(x):=\left\{\begin{array}{ll}0\,,&\text{if }x\in I\,,\\ 1\,,&\text{if }x\in P\setminus I\,, \end{array}\right.$$ for all $$I\in\mathcal{I}$$. Then, the inverse map $$\Phi^{-1}:\text{Hom}_\text{Pos}\big(P,\{0,1\}\big)\to\mathcal{I}$$ is given by $$\Phi^{-1}(f):=f^{-1}\big(\{0\}\big)\text{ for all }f\in\text{Hom}_{\text{Pos}}\big(P,\{0,1\}\big)\,.$$
We can say more about $$\Phi$$. Equip $$\mathcal{I}$$ with an order induced by reverse inclusion: for all $$I,J\in\mathcal{I}$$, $$I\leq J$$ iff $$I\supseteq J$$. Order $$\text{Hom}_{\text{Pos}}\big(P,\{0,1\}\big)$$ by setting $$f\leq g$$ if $$f(x)\leq g(x)$$ for all $$x\in P$$, where $$f,g\in\text{Hom}_{\text{Pos}}\big(P,\{0,1\}\big)$$. Then, $$\Phi$$ is an isomorphism of partially ordered sets.
B = [0,oo)×{0} $$\cup$$ {(0,1)}.