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Activities 19.7 Activities
Activity 19.1 .
Let
\(F \subseteq \N \) represent the set of all divisors of
\(30 \text{.}\) Let
\(A = \{a,b,c\} \text{.}\)
(a) Draw the Hasse diagram for the subset partial order
\(\mathord{\subseteq} \) on
\(\powset{A} \text{.}\)
(b) Draw the Hasse diagram for the “divides” partial order
\(\mathord{\mid} \) on
\(F \text{.}\)
(c) Compare your two Hasse diagrams. Can you devise a function
\(\funcdef{f}{F}{\powset{A}} \) that would deserve to be called an
order-preserving correspondence between
\(F \) and
\(\powset{A} \text{?}\)
Activity 19.2 .
Activity 19.3 .
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)
\(A \) has both a maximum and a minimum.
(b)
\(A \) has a maximum but no minimum.
(c)
\(A \) has a minimum but no maximum.
(d)
\(A \) has neither a maximum nor a minimum.
Activity 19.4 .
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 19.5 .
Using the proper strategy for proving uniqueness (see
Procedure 6.10.2 ), 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 19.6 .
Recall that \((a,b)\subseteq \R \) means an open interval on the real number line:
\begin{equation*}
(a,b) = \setdef{x \in \R}{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 = \setdef{-\frac{1}{n}}{n \in \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 .
