upper set operation is a closure operator
In this entry, we shall prove the assertion made in the main entry (http://planetmath.org/UpperSet) that is a closure operator. This will be done by checking that the defining properties are satisfied. To begin, recall the definition of our operation:
Let be a poset and a subset of . The upper set of is defined to be the set
Now, we verify each of the properties which is required of a closure operator.
Any statement of the form “” is identically false no matter what the predicate (i.e. it is an antitautology) and the set of objects satsfying an identically false condition is empty, so . ∎
This follows from reflexivity — for every , one has , hence . ∎
By the previous result, . Hence, it only remains to show that . This follows from transitivity. In order for some to be an element of , there must exist and such that and . By transitivity, , so , hence as well. ∎
If and are subsets of a partially ordered set, then
On the one hand, if , then for some . It then follows that either or . In the former case, , in the latter case, so, either way . Hence .
On the other hand, if , then either or . In the former case, there exists such that and . Since , we also have , hence . Likewise, in the second case, we also conclude that . Therefore, we have . ∎
|Title||upper set operation is a closure operator|
|Date of creation||2013-03-22 16:41:43|
|Last modified on||2013-03-22 16:41:43|
|Last modified by||rspuzio (6075)|