Give an example of a finite ordered set \( (A, \leq) \) and a subset \( B \) of \( A \) so that \( B \) has no greatest element.

Certainly! Let's construct a finite partially ordered set \( (A, \leq) \) and a subset \( B \subseteq A \) that has no greatest element. ### Example: **Set \( A \):** \[ A = \{a, b, c\} \] **Partial Order \( \leq \):** - \( a \leq c \) - \( b \leq c \) - \( a \) and \( b \) are **incomparable** (i.e., neither \( a \leq b \) nor \( b \leq a \)) This partial order can be visualized as: ``` c / \ a b ``` **Subset \( B \):** \[ B = \{a, b\} \] ### Why \( B \) Has No Greatest Element: In the subset \( B \): - \( a \) is not greater than \( b \) (\( a \nleq b \)) - \( b \) is not greater than \( a \) (\( b \nleq a \)) There is **no element** in \( B \) that is greater than or equal to every other element in \( B \). Hence, \( B \) does **not** have a greatest element within the ordered set \( (A, \leq) \). ### Visualization: ``` A = {a, b, c} Partial Order: c / \ a b Subset B = {a, b} - No element in B is comparable to be the greatest. ``` This example clearly demonstrates a finite partially ordered set with a subset that lacks a greatest element.