Question

Show thata dense poset with at least two elements that are comparable is not well-founded.

Short answer

Hence, any dense poset with at least two elements that are comparable is not well founded.

Step-by-step solution

Given data

Poset is dense.

Concept used of poset

A partially ordered set (also poset) formalizes and generalizes the intuitive concept of an ordering, sequencing, or arrangement of the elements of aset.

Prove for dense poset

A poset\((R, \le )\) is said to be dense if for all \(x,y \in R\) with \(x < y\) there exists an element \(z \in R\) such that \(x < z < y\).Let \((S, \le )\) be a dense poset with at least two comparable elements. Let \(x,y \in S\) and assumethat \(x \le y\).

As \((S, \le )\) is dense, there exist \({z_0} \in S\) such that \(x \le {z_0} \le y\).

As \(x \le {z_0}\) and \(S\) is dense, there exist \({z_0} \in S\) such that \(x \le {z_0} \le y\).

As \(x \le {z_0}\) and \(S\) is dense, there exist \({z_1} \in S\) such that \(x \le {z_1} \le {z_0} \le y\).

Furthermore, there exist \({z_2},{z_3},{z_4},K,{z_i},K \in S\) such that \(x \le K \le {z_i} \le K \le {z_3} \le {z_2} \le {z_1} \le {z_0} \le y\) and this chain of decreasing elements \({z_i}\) of \(S\) produced to infinity.As \(S\) has infinite decreasing sequence of elements \({z_i}\) so by definition of well founded\(S\) is not well founded.Hence, any dense poset with at least two elements that are comparable is not well founded.