You are here
Homeway below
Primary tabs
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. 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 $D=\{x\mid p\leq x<b\}$ 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. 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. 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 twosided ideals of a ring, and the lattice of subspaces (projective geometry) of a vector space.
4. 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 $P=C_{1}\times C_{2}$. 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. 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).
Mathematics Subject Classification
06B35 no label found06A99 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections