19.7: Activities

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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.