Question

To determine \(\left( {{Z^ - }, \ge } \right)\)where \({Z^ - }\)is the set of negative integers and poset is well-defined.

Short answer

This poset is well ordered

Step-by-step solution

Given data

A poset \((R, \le )\) is well founded if there is no infinite decreasing sequence of elements in the poset, that is elements \({x_1},{x_2},{x_3}, \ldots ..{x_{n,}}\) such that \( - < {x_{n, \ldots }}, \ldots {x_2} < {x_{1,*}}\). A poset \((R, \le )\) is dense if for all \(x \in S\) and \(y \in S\) with \(x < y\). there is an element \(z \in R\) such that \(x < z < y\).

Concept used rule of ordering

A linear or simple ordering is reflexive, antisymmetric, transitive, and connected, as less than or equal to on the integers. A partial ordering is reflexive, anti symmetric, and transitive, as set inclusion.

Prove for well ordered set

We have \(\left( {{Z^ - }, \ge } \right)\)where \({Z^ - }\)the set of negative integers. If we expand \({Z^ - }\)then it will be like this, \({Z^ - } = \{ - \infty , - 4, - 3, - 2, - 1\} \). Here, \({Z^ - }\)is again well ordered because the partial ordering is \( \ge \)instead of \( \le \), So that \( - 1\) is no less in poset terms than \( - 10\).

Hence, the poset is well ordered.