Question

To determine the poset \((Z, \prec )\) is well defined but is not a totally ordered set.

Short answer

Hence, the poset \((Z, < )\) is well defined but is not a totally ordered set.

Step-by-step solution

Given data

\(x < y\) if and only if \(|x| < |y|\).

Concept used totally ordered set

A total order (or "totally ordered set," or "linearly ordered set") is asetplus a relation on the set (called atotal order) that satisfies the conditions for apartial orderplus an additional condition known as the comparability condition.

Prove for totally ordered set

If \(x\) is an integer in a decreasing sequence of elements of this poset, then at

most \(|x|\)elements can follow \(x\) in the sequence, namely integers whose absolute values are \(|x| - 1,|x| - 2, \ldots \ldots . \ldots .1,0\). Therefore there can be no infinite decreasing sequence. Hence given poset is well founded. This is not a totally ordered set, since 5 and \( - 5\) for example are incomparable; from the definition given here it is neither true that \(5 < - 5\) nor that \( - 5 < 5\), because neither one of \(|5|\) or \(| - 5|\) is less than the other. Hence, the poset \((Z, < )\) is well defined but is not a totally ordered set.