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 $b\le \bigvee D$, then there is a $d\in D$ such that $a\le d$.
First note that if $a\ll b$, then $a\le b$ 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:
$$\text{xymatrix}\mathrm{\&}\mathrm{\&}b\text{ar}\mathrm{@}.[d]\text{ar}\mathrm{@}.[dll]\text{ar}\mathrm{@}.[dr]\mathrm{\&}p\text{ar}\mathrm{@}[d]\text{ar}\mathrm{@}[dr]\mathrm{\&}\mathrm{\&}q\text{ar}\mathrm{@}[dl]\text{ar}\mathrm{@}[d]\mathrm{\&}r\text{ar}\mathrm{@}[dl]s\mathrm{\&}t\text{ar}\mathrm{@}[d]\mathrm{\&}u\text{ar}\mathrm{@}[dl]\mathrm{\&}v\mathrm{\&}\mathrm{\&}$$ where the dotted lines denote infinite chains between the end points. First, note that every element in $P$ is below ($\le $) $b$. However, only $v$ is way below $b$. $u$, for example, is not way below $b$, because $$ 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 $b\le \bigvee D$, there is a finite subset $F\subseteq D$ such that $a\le \bigvee F$.
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 $\bigvee D\le t$. 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 $a\le e$. 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 $b\le \bigvee D$. There is a finite subset $F$ of $D$ such that $a\le \bigvee F$. Since $D$ is directed, there is an element $d\in D$ such that $d$ is the upper bound of elements of $F$. So $a\le d$, 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.
In particular, if $P$ is a chain, then $a\le b$ implies that $a\ll b$. If $D$ is a set such that $\bigvee D$ exists and $b\le \bigvee D$, then there is a $d\in D$ such that $b\le d$ (otherwise $b$ is an upper bound of elements of $D$ and $\bigvee D\le b$), so $a\le d$.

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 $$. Since $P$ is complete, $\bigvee D=(1,1)$, but every element of $D$ is stricly less than $(1,1)$, so $(1,1)$ is not way below itself. What about elements of the form $(a,1)$, $a\ne 1$? If we take $$, 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 $$ and $$? 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, $a\le 1$ implies that there is an $s\in {D}_{1}$ such that $a\le s$. Similarly, there is a $t\in {D}_{2}$ such that $b\le t$. Together, $(a,b)\le (s,t)\in D$. 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 $U\le V$. 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  20130322 16:38:28 
Last modified on  20130322 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  waybelow 
Synonym  waywaybelow 
Defines  way below relation^{} 
Defines  way way below relation 