With this, and all future assignments, you should expect to have all necessary material to complete the assignment by the end of the week when it is posted.
Suppose that $(S, \preceq)$ is a partially ordered set; recall that for $a, b \in S$ we write $a \prec b$ if $a\preceq b$ and $a\neq b$. Let $\not\prec$ be the relation on $S$ defined by setting $a \not\prec b$ if and only if $a \prec b$ is false.
Suppose $(S, \preceq)$ is an ordered set.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment. They are coloured by approximate difficulty: easy/medium/hard.
Suppose that $\mathcal{N}$ is a set with the following properties:
The set $\mathcal{N}$ is a model of the natural numbers proposed by von Neumann (cf. Wikipedia). That is to say, $\mathcal{N}$ satisfies the Peano axioms for arithmetic, but can be defined directly from the axioms of set theory without needing to introduce any new objects.
Let $\mathcal{P}$ be the set of polynomial functions from $\Q \to \Q$. Define a relation on $\mathcal{P}$ by $p \sqsubseteq q$ if and only if $p(10) \leq q(10)$. Is $\sqsubseteq$ a partial order? Is it a total order?
A (total) order $\preceq$ on a set $S$ is said to be a well-order if every $T \subseteq S$ contains a minimum element.
Recall that for $a, b \in \N$ we write $a \mid b$ if $a$ divides $b$, i.e., if there is some $k \in \N$ so that $b = ak$.
Prove that any partial order may be extended to a total order, regardless of whether or not the underlying set is finite. To prove this, you may assume the following:
Zorn's Lemma. Suppose that $Z$ is a set and $\preceq$ is a partial order on $Z$ with the property that every totally-ordered subset of $Z$ is bounded above in $Z$. Then $Z$ contains a maximal element. (A subset $W \subseteq Z$ is totally ordered if $\preceq$ restricted to $W$ is an order, i.e., if whenever $x, y \in W$ we have either $x \preceq y$ or $y \preceq x$. An element $x \in Z$ is maximal if whenever $y \in Z$ with $x \preceq y$, we have $y = x$.)