# ideal completion of a poset

Let $P$ be a poset. Consider the set $\operatorname{Id}(P)$ of all order ideals of $P$.

###### Theorem 1.

$\operatorname{Id}(P)$ is an algebraic dcpo, such that $P$ can be embedded in.

###### Proof.

We shall list, and when necessary, prove the following series of facts which ultimately prove the main assertion. For convenience, write $P^{\prime}=\operatorname{Id}(P)$.

1. 1.

$P^{\prime}$ is a poset with $\leq$ defined by set theoretic inclusion.

2. 2.

For any $x\in P$, $\downarrow\!\!x\in P^{\prime}$.

3. 3.

$P$ can be embedded in $P^{\prime}$. The function $f:P\to P^{\prime}$ defined by $f(x)=\downarrow\!\!x$ is order preserving and one-to-one. If $x\leq y$, and $a\leq x$, then $a\leq y$, hence $\downarrow\!\!x\subseteq\downarrow\!\!y$. If $\downarrow\!\!x=\downarrow\!\!y$, we have that $x\leq y$ and $y\leq x$, so $x=y$, since $\leq$ is antisymmetric.

4. 4.

$P^{\prime}$ is a dcpo. Suppose $D$ is a directed set  in $P^{\prime}$. Let $E=\bigcup D$. For any $x,y\in E$, $x\in I$ and $y\in J$ for some ideals $I,J\in D$. As $D$ is directed, there is $K\in D$ such that $I\subseteq K$ and $J\subseteq K$. So $x,y\in K$ and hence there is $z\in K\subseteq E$ such that $x\leq z$ and $y\leq z$. This shows that $E$ is directed. Next, suppose $x\in E$ and $y\leq x$. Then $x\in I$ for some $I\in D$, so $y\in I\subseteq E$ as well. This shows that $E$ is a down set. So $E$ is an ideal of $P$: $\bigvee D=E\in P^{\prime}$.

5. 5.

For every $x\in P$, $\downarrow\!\!x$ is a compact element of $P^{\prime}$. If $\downarrow\!\!x\leq\bigvee D$, where $D$ is directed in $P^{\prime}$, then $\downarrow\!\!x\subseteq\bigcup D$, or $x\in\bigcup D$, which implies $x\in I$ for some ideal $I\in D$. Therefore $\downarrow\!\!x\subseteq I$, and $\downarrow\!\!x$ is way below itself: $\downarrow\!\!x$ is compact.

6. 6.

$P^{\prime}$ is an algebraic dcpo. Let $I\in P^{\prime}$. Let $C=\{\downarrow\!\!x\mid x\in I\}$. For any $x,y\in I$, there is $z\in I$ such that $x\leq z$ and $y\leq z$. This shows that $\downarrow\!\!x\leq\downarrow\!\!z$ and $\downarrow\!\!y\leq\downarrow\!\!z$ in $C$, so that $C$ is directed. It is easy to see that $I=\bigvee C$. Since $I$ is a join of a directed set consisting of compact elements, $P^{\prime}$ is algebraic.

Definition. $\operatorname{Id}(P)$ is called the ideal completion of $P$.

Remarks.

• In general, the ideal completion of a poset is not a complete lattice  . It is complete in the sense of being directed complete. This is different from another type of completion, called the MacNeille completion of $P$, which is a complete lattice.

• If $P$ is an upper semilattice  , then so is $\operatorname{Id}(P)$. In fact, the join of any non-empty family of ideals exists. Furthermore, if $P$ has a bottom element $0$, then $\operatorname{Id}(P)$ is a complete lattice.

###### Proof.

Let $S$ be a non-empty family of ideals in $P$. Let $A$ be the set of $P$ consisting of all finite joins of elements of those ideals in $S$, and $B=\downarrow\!\!A$. Clearly, $B$ is a lower set. For every $a,b\in B$, we have $c,d\in A$ such that $a\leq c$ and $b\leq d$. Since $c$ and $d$ are both finite joins of elements of those ideals in $S$, so is $c\vee d$. Since $a\leq c\vee d$ and $b\leq c\vee d$, $B$ is directed. If $I$ is any ideal larger than any of the ideals in $S$, clearly $A\subseteq I$, since $I$ is directed. So $B=\downarrow\!\!A\subseteq\downarrow\!\!I=I$. Therefore, $B=\bigvee S$.

If $0\in P$, then $\langle 0\rangle$, the bottom of $\operatorname{Id}(P)$, is the join of the empty family of ideals in $P$. By this entry (http://planetmath.org/CriteriaForAPosetToBeACompleteLattice), $\operatorname{Id}(P)$ is a complete lattice. ∎

• ###### Proof.

Let $I,J$ be two ideals in $P$ and $K=I\cap J$. By definition, $I$ and $J$ are non-empty, so let $a\in I$ and $b\in J$. As $P$ is a lower semilattice, $c:=a\wedge b$ exists and $c\leq a$ and $c\leq b$. So $c\in I\cap J$, and that $K=I\cap J$ is non-empty. If $x\leq y\in K$, then $x\leq y\in I$ or $x\in I$. Similarly $x\in J$. Therefore $x\in I\cap J=K$ and $K$ is a lower set. If $r,s\in K$, then there is $u\in I$ and $v\in J$ such that $r,s\leq u,v$. So $r,s\leq u\wedge v$ and $K$ is directed. This means that $I\cap J\in\operatorname{Id}(P)$. ∎

## References

• 1 G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).
Title ideal completion of a poset IdealCompletionOfAPoset 2013-03-22 17:03:01 2013-03-22 17:03:01 CWoo (3771) CWoo (3771) 8 CWoo (3771) Definition msc 06A12 msc 06A06 LatticeOfIdeals ideal completion