Question

To determine a compatible total order for the poset with the Hasse diagram.

Short answer

One such total order compatible with the given partial order is given by

\(a{ < _t}b{ < _t}d{ < _t}e{ < _t}c{ < _t}f{ < _t}g{ < _t}h{ < _t}i{ < _t}j{ < _t}k{ < _t}m{ < _t}l.\)

Step-by-step solution

Given data

R is the partial order on the set V= {a, b, c} corresponding to the Hasse diagram shown below:

Concept used of maximal element rule

A maximal element of a subset S of some preordered set is an element of S that is not smaller than any other element in S.

Prove for maximal element

An element of a poset \((P, \le )\) is said to be a maximal element if there is no \(b \in P\) such that \(a < b\). Now we suppose \((P, \le )\) is a non empty finite poset. Let \({x_0} \in P\). where \({\rm{P}}\) is non empty. If \({x_0}\) is a maximal element, we are through, otherwise there exists some \({x_1} \in P\) such that \({x_0} < {x_1}\)If \({x_1}\) is a maximal element, we are through, otherwise there exists some \({x_2} \in P\). such that \({x_0} < {x_1} < {x_2}\).. Since \(P\) is finite, we cannot continue this process infinitely, it will be terminated after a finite stage, say

\({a_0} < {a_1} < {a_2} \ldots \ldots \ldots .{a_{n - 1}} < {a_n}\)

Then \({a_n}\) is a maximal element. Hence, a finite non empty poset has a maximal element.

Hence proved.