way below
Let be a poset and . is said to be way below , written , if for any directed set such that exists and that , then there is a such that .
First note that if , then since we can set , and if is finite, we have the converse (since ). So, given any element , what exactly are the elements that are way below ? Below are some examples that will throw some light:
Examples
- 1.
Let be the poset given by the Hasse diagram below:
where the dotted lines denote infinite chains between the end points. First, note that every element in is below () . However, only is way below . , for example, is not way below , because is a directed set such that and non of the elements in are above . This illustrates the fact that if has a bottom, it is way below everything else.
- 2.
Suppose is a lattice
. Then iff for any set such that exists and , there is a finite subset such that .
Proof.
. Suppose . Let be the set in the assumption
. Let be the set of all finite joins of elements of . Then . Also, every element of is bounded above by . If is an upper bound of elements of , then it is certainly an upper bound of elements of , and hence . So is the least upper bound of elements of , or . Furthermore, is directed. So there is an element such that . But for some finite subset of , and this completes
one side of the proof.
. Let be a directed set such that exists and . There is a finite subset of such that . Since is directed, there is an element such that is the upper bound of elements of . So , completing the other side of the proof.∎
- 3.
With the above assertion, we see that, for example, in the lattice of subgroups of a group , iff 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.
In particular, if is a chain, then implies that . If is a set such that exists and , then there is a such that (otherwise is an upper bound of elements of and ), so .
- 5.
Here’s an example where in but is not the bottom of . Take two complete infinite chains and with bottom and , and let be their product
(http://planetmath.org/ProductOfPosets) . What elements are way below ? First, take . Since is complete, , but every element of is stricly less than , so is not way below itself. What about elements of the form , ? If we take , then once again. But no elements of are above . So can not be way below . Similarly, neither can be way below . Finally, what about for and ? If is a set with , then and , where and . Since and are chains, implies that there is an such that . Similarly, there is a such that . Together, . So .
- 6.
Let be a topological space
and be the lattice of open sets in . Suppose and . If there is a compact subset such that , then .
Remarks.
- •
In a lattice , iff 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 be directed in the definition above, then is said to be way way below .
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).