# Covering Relation

• Jul 10th 2007, 08:30 AM
Discrete
Covering Relation
Let (S, $\preceq$) be a poset. We say that an element y $\in$ S covers an element x $\in$ S if x $\prec$ y and there is no element z $\in$ S such that $x \prec z \prec y$. The set of pairs (x,y) such that y covers x is called the covering relation of (S, $\preceq$)

What is the covering relation of the partial ordering {(a,b) $\vert$ a divides b} on {1,2,3,4,6,12}?

My answer so far is {(1,2), (1,3), (2,4), (2,6), (3,6), (4,12), (6,12)}

is this right?
• Jul 10th 2007, 11:21 AM
Plato
Quote:

Originally Posted by Discrete
Let (S, $\preceq$) be a poset. We say that an element y $\in$ S covers an element x $\in$ if x $\prec$ y and there is no element z $\in$ S such that $x \prec y \prec z$. The set of pairs (x,y) such that y covers x is called the covering relation of (S, $\preceq$) What is the covering relation of the partial ordering {(a,b) $\vert$ a divides b} on {1,2,3,4,6,12}?
My answer so far is {(1,2), (1,3), (2,4), (2,6), (3,6), (4,12), (6,12)}is this right?

NO indeed!
Look very carefully at the definition of covering elements.
Is this true, $2 \prec 4 \prec 12$?
If it is true, then can 4 cover 2?
Try again!
• Jul 10th 2007, 02:15 PM
Discrete
Sorry I revised the question! I misstyped it. Is it correct now?
• Jul 10th 2007, 02:41 PM
Plato
Yes now it is correct using the corrected definition.
The other definition give maximal elements in chains.