Question

To find the dual of the poset \(\left( {{z^ + },1} \right)\).

Short answer

The dual of poset \((P(z), \supseteq )\) is \((P(z), \subseteq )\).

Step-by-step solution

Given data

The poset \(\left( {{z^ + },1} \right)\).

Concept used of partially ordered set

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).

Find dual of poset

Let us consider \(\left( {{z^ + },1} \right)\) is a poset.

Then \(S\) can be defined as \(S = \left\{ {{z^ + }} \right\}\).

and, \({\rm{R}} \in \{ (a,b),a/b\} \Rightarrow {R^{ - 1}} \in \{ (a,b)(b,a) \in R\} \)

\({1^{ - 1}} = \{ (a,b),b/a\} \)

Then, The dual of poset \(\left( {{z^ + },1} \right)\) is \(\left( {{z^ + },{1^{ - 1}}} \right)\).