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 operatorPlanetmathPlanetmath. This will be done by checking that the defining properties are satisfied. To begin, recall the definition of our operationMathworldPlanetmath:

Definition 1.

Let P be a poset and A a subset of P. The upper set of A is defined to be the set


Now, we verify each of the properties which is required of a closure operator.

Theorem 1.



Any statement of the form “(a)P(a)” is identically false no matter what the predicateMathworldPlanetmath P (i.e. it is an antitautology) and the set of objects satsfying an identically false condition is empty, so =. ∎

Theorem 2.



This follows from reflexivityMathworldPlanetmath — for every aA, one has aa, hence aA. ∎

Theorem 3.



By the previous result, AA. Hence, it only remains to show that AA. This follows from transitivity. In order for some a to be an element of A, there must exist b and c such that abC and CA. By transitivity, AC, so aA, hence AA as well. ∎

Theorem 4.

If A and B are subsets of a partially ordered setMathworldPlanetmath, then


On the one hand, if a(AB), then ab for some bAB. It then follows that either bA or bB. In the former case, aA, in the latter case, aB so, either way a(A)(B). Hence (AB)(A)(B).

On the other hand, if a(A)(B), then either a(A) or a(B). In the former case, there exists b such that ab and bA. Since AAB, we also have bAB, hence a(AB). Likewise, in the second case, we also conclude that a(AB). Therefore, we have (A)(B)(AB). ∎

Theorem 5.


Theorem 6.


Title upper set operation is a closure operator
Canonical name UpperSetOperationIsAClosureOperator
Date of creation 2013-03-22 16:41:43
Last modified on 2013-03-22 16:41:43
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 8
Author rspuzio (6075)
Entry type TheoremMathworldPlanetmath
Classification msc 06A06