19.7: Activities
Activity \(\PageIndex{1}\)
Let \(F \subseteq \mathbb{N}\) represent the set of all divisors of \(30\text{.}\) Let \(A = \{a,b,c\}\text{.}\)
Note: In Task c you will compare your work from Task a and Task a, so keep your work!
- Draw the Hasse diagram for the subset partial order \(\mathord{\subseteq}\) on \(\mathscr{P}(A)\text{.}\)
- Draw the Hasse diagram for the “divides” partial order \(\mathord{\mid}\) on \(F\text{.}\)
- Compare your two Hasse diagrams. Can you devise a function \(f: F \rightarrow \mathscr{P}(A)\) that would deserve to be called an order-preserving correspondence between \(F\) and \(\mathscr{P}(A)\text{?}\)
Activity \(\PageIndex{2}\)
Suppose \(\mathord{\preceq}\) is a partial order on a set \(A\text{.}\) Verify that the inverse relation \(\preceq^{-1}\) is also a partial order on \(A\) by verifying that it is reflexive , antisymmetric , and transitive .
Activity \(\PageIndex{3}\)
Let \(A = \{a,b,c,d,e\}\text{.}\) Carry out the following steps for each of the scenarios below.
- Draw the Hasse diagram for a partial order on \(A\) with the requested features.
- In your diagram, identify all maximal/minimal elements.
- Identify all pairs of incomparable elements.
- \(A\) has both a maximum and a minimum.
- \(A\) has a maximum but no minimum.
- \(A\) has a minimum but no maximum.
- \(A\) has neither a maximum nor a minimum.
Activity \(\PageIndex{4}\)
Suppose \(\mathord{\preceq}\) is a partial order on the set \(A = \{0,1,2\}\) such that \(1\) is a maximal element. What are the possibilities for the Hasse diagram of \(\mathord{\preceq}\text{?}\)
Activity \(\PageIndex{5}\)
Using the proper strategy for proving uniqueness (see Procedure 6.10.1 ), prove that if a partially ordered set \(A\) has a maximum element, then that element is the unique maximum element.
How can your proof be modified to show that a minimum element is also unique?
Activity \(\PageIndex{6}\)
Recall that \((a,b)\subseteq \mathbb{R}\) means an open interval on the real number line:
\begin{equation*} (a,b) = \{x \in \mathbb{R} \vert a \lt x \lt b\} \text{.} \end{equation*}
Let \(\mathord{\le}\) be the usual “less than or equal to” total order on the set
\begin{equation*} A = (-2,0)\cup(0,2) \text{.} \end{equation*}
Consider the subset
\begin{equation*} B = \{-\dfrac{1}{n} \vert n \in \mathbb{N}, \, n \ge 1\} \subseteq A \text{.} \end{equation*}
Determine an upper bound for \(B\) in \(A\text{.}\) Then formally prove that \(B\) has no
least
upper bound in \(A\) by arguing that every element of \(A\) fails the criteria in the definition of
least upper bound
.