# way below

Let $P$ be a poset and $a,b\in P$. $a$ is said to be way below $b$, written $a\ll b$, if for any directed set $D\subseteq P$ such that $\bigvee D$ exists and that $b\leq\bigvee D$, then there is a $d\in D$ such that $a\leq d$.

First note that if $a\ll b$, then $a\leq b$ since we can set $D=\{b\}$, and if $P$ is finite, we have the converse (since $\bigvee D\in D$). So, given any element $b\in P$, what exactly are the elements that are way below $b$? Below are some examples that will throw some light:

Examples

1. 1.

Let $P$ be the poset given by the Hasse diagram below:

 $\xymatrix{&&b\ar@{.}[d]\ar@{.}[dll]\ar@{.}[dr]&\\ p\ar@{-}[d]\ar@{-}[dr]&&q\ar@{-}[dl]\ar@{-}[d]&r\ar@{-}[dl]\\ s&t\ar@{-}[d]&u\ar@{-}[dl]\\ &v&&}$

where the dotted lines denote infinite chains between the end points. First, note that every element in $P$ is below ($\leq$) $b$. However, only $v$ is way below $b$. $u$, for example, is not way below $b$, because $D=\{x\mid p\leq x is a directed set such that $\bigvee D=b$ and non of the elements in $D$ are above $u$. This illustrates the fact that if $P$ has a bottom, it is way below everything else.

2. 2.

Suppose $P$ is a lattice. Then $a\ll b$ iff for any set $D$ such that $\bigvee D$ exists and $b\leq\bigvee D$, there is a finite subset $F\subseteq D$ such that $a\leq\bigvee F$.

###### Proof.

$(\Rightarrow)$. Suppose $a\ll b$. Let $D$ be the set in the assumption. Let $E$ be the set of all finite joins of elements of $D$. Then $D\subseteq E$. Also, every element of $E$ is bounded above by $\bigvee D$. If $t$ is an upper bound of elements of $E$, then it is certainly an upper bound of elements of $D$, and hence $\bigvee D\leq t$. So $\bigvee D$ is the least upper bound of elements of $E$, or $\bigvee E=\bigvee D$. Furthermore, $E$ is directed. So there is an element $e\in E$ such that $a\leq e$. But $e=\bigvee F$ for some finite subset of $D$, and this completes one side of the proof.

$(\Leftarrow)$. Let $D$ be a directed set such that $\bigvee D$ exists and $b\leq\bigvee D$. There is a finite subset $F$ of $D$ such that $a\leq\bigvee F$. Since $D$ is directed, there is an element $d\in D$ such that $d$ is the upper bound of elements of $F$. So $a\leq d$, completing the other side of the proof. ∎

3. 3.

With the above assertion, we see that, for example, in the lattice of subgroups $L(G)$ of a group $G$, $H\ll K$ iff $H$ is finitely generated. Other similar examples can be found in the lattice of two-sided ideals of a ring, and the lattice of subspaces (projective geometry) of a vector space.

4. 4.

In particular, if $P$ is a chain, then $a\leq b$ implies that $a\ll b$. If $D$ is a set such that $\bigvee D$ exists and $b\leq\bigvee D$, then there is a $d\in D$ such that $b\leq d$ (otherwise $b$ is an upper bound of elements of $D$ and $\bigvee D\leq b$), so $a\leq d$.

5. 5.

Here’s an example where $a\ll b$ in $P$ but $a$ is not the bottom of $P$. Take two complete infinite chains $C_{1}$ and $C_{2}$ with bottom $0$ and $1$, and let $P$ be their product (http://planetmath.org/ProductOfPosets) $P=C_{1}\times C_{2}$. What elements are way below $(1,1)$? First, take $D=\{(a,1)\mid 0\leq a<1\}$. Since $P$ is complete, $\bigvee D=(1,1)$, but every element of $D$ is stricly less than $(1,1)$, so $(1,1)$ is not way below itself. What about elements of the form $(a,1)$, $a\neq 1$? If we take $D=\{(1,b)\mid 0\leq b<1\}$, then $\bigvee D=(1,1)$ once again. But no elements of $D$ are above $(a,1)$. So $(a,1)$ can not be way below $(1,1)$. Similarly, neither can $(1,b)$ be way below $(1,1)$. Finally, what about $(a,b)$ for $a<1$ and $b<1$? If $D$ is a set with $\bigvee D=(1,1)$, then $\bigvee D_{1}=1$ and $\bigvee D_{2}=1$, where $D_{1}=\{x\mid(x,1)\in D\}$ and $D_{2}=\{y\mid(1,y)\in D\}$. Since $C_{1}$ and $C_{2}$ are chains, $a\leq 1$ implies that there is an $s\in D_{1}$ such that $a\leq s$. Similarly, there is a $t\in D_{2}$ such that $b\leq t$. Together, $(a,b)\leq(s,t)\in D$. So $(a,b)\ll(1,1)$.

6. 6.

Let $X$ be a topological space and $L(X)$ be the lattice of open sets in $X$. Suppose $U,V\in L(X)$ and $U\leq V$. If there is a compact subset $C$ such that $U\subseteq C\subseteq V$, then $U\ll V$.

Remarks.

• In a lattice $L$, $a\ll a$ iff $a$ is a compact element. This follows directly from the assertion above. In fact, a compact element can be defined in a general poset as an element that is way below itself.

• If we remove the condition that $D$ be directed in the definition above, then $a$ is said to be way way below $b$.

## 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 way below Canonical name WayBelow Date of creation 2013-03-22 16:38:28 Last modified on 2013-03-22 16:38:28 Owner CWoo (3771) Last modified by CWoo (3771) Numerical id 10 Author CWoo (3771) Entry type Definition Classification msc 06B35 Classification msc 06A99 Synonym way way below Synonym way-below Synonym way-way-below Defines way below relation Defines way way below relation