Question

To determine a poset is well ordered if and only if it is totally ordered and well-founded.

Short answer

Hence, a poset is well ordered if and only if it is totally ordered and well-founded.

Step-by-step solution

Given data

Poset is well ordered.

Concept used of poset

A relation\(R\)is a poset if and only if,\((x,x)\)is in\({\rm{R}}\)for all x (reflexivity)

\((x,y)\)and\((y,x)\)in R implies\(x = y\)(anti-symmetry),\((x,y)\)and\((y,z)\)in R implies\((x,z)\)is in\({\rm{R}}\)(transitivity).

Prove for dense poset

Now, we let \((S, < )\) be a partially ordered set. It is enough to show that every non empty subset \(S\) contains a least element if and only if there is no infinite decreasing sequence of elements \({a_1},{a_2},{a_3}, \ldots \ldots \) in \(S\).An infinite decreasing sequence of elements clearly has no least element. Conversely, let \({\rm{A}}\) be any non empty subset of \(S\) that has no least element. Because \({\rm{A}}\) is non empty, choose \({a_1} \in A\), Because \({a_1}\) is not the least element of \({\rm{A}}\),choose \({a_2} \in A\) with \({a_1} < {a_2}\).Because \({a_2}\) is not the least element of \(A\), choose \({a_3} \in A\) with \({a_3} \prec {a_2}\).We can continue in this manner and it will produce an infinite decreasing sequence in \(S\).Hence, proved.