\problemdata      
        {functions-relations-6} % latex-friendly label for the prob.
        {partial order} % The topic(s) of the problem content.
        {Velleman 4.4 Exercise 1} % Source (if known)
        {S02 cp3F; F00 pset3} % Usage (list ones you are aware of).
        {Radhika Nagpal, S02} % Last revision info (author, date).

\begin{problem}

In each case, say whether or not $R$ is a partial order on $A$.  If
so, is it a total order?

\begin{problemparts}

\problempart $A=\{a,b,c\}, R=\{(a,a),(b,a),(b,b),(b,c),(c,c)\}.$

\problempart $A=$\{the set of all words in English\} 

$R = \{(x,y) \in A \times A$ and the word $x$ comes before the word $y$
alphabetically\}.


\problempart $A=\reals, R=\{(x,y) \in \reals \times \reals \mbox{ and
} |x| < |y|$ or $x=y\}$.

\problempart $A=\reals, R=\{(x,y) \in \reals \times \reals \mbox{ and }
|x|\leq |y|\}$.

\end{problemparts}

\solution{
a. $R$ is reflexive, transitive, and antisymmetric.  Therefore, $R$ is
a partial order on $A$.  However, it is not a total order on $A$ since
neither the pair $(a,c)$ nor $(c,a)$ is in $R$.\\


b. $R$ is not reflexive (although it is transitive and antisymmetric),
and thus it is not a partial order on $A$.\\

c. $R$ is reflexive, transitive, and antisymmetric.  Therefore, $R$ is
a partial order on $A$. However, $R$ is not a total order on $A$. As a
counterexample, if x = 5, y = -5, then none of the following holds: $x
= y, |x| < |y|, |y| < |x|$.

d. $R$ is reflexive and transitive. However, it is not
antisymmetric. As a counterexample, $|-3| \leq |3|, |3| \leq |-3|$,
but $3 \neq -3$.  
}

\end{problem}
