I know of a couple of interesting examples of posets.
- Let $\Bbb N$ be collection of all natural numbers (including $0$). We can order $\Bbb N$ by divisibility. This poset is actually a complete lattice, and it's interesting that the top of this poset is $0$ and that the supremum of any infinite collection of numbers is also $0$.
- Suppose $X$ is a set. Let $\mathcal R$ denote all of the partial orderings for $X$. Given two partial orderings $\leq$ and $\sqsubseteq$ for $X$, we say that the partial ordering $\sqsubseteq$ extends the partial ordering $\leq$ if $x\leq y$ implies $x\sqsubseteq y$ for all $x$ and $y\in X$. Now we can actually equip $\mathcal R$ with a partial ordering. Namely, we say that $\leq$ is less than $\sqsubseteq$ if $\sqsubseteq$ extends $\leq$. (It is an interesting exercise to prove that the maximal elements in this poset are the linear orderings of $X$)
- Let $X$ and $Y$ be two sets. Given two partial functions $f$ and $g$ from $X$ to $Y$, let's say $f\leq g$ if the domain of $f$ is a subset of the domain of $g$ and $f(x)=g(x)$ for all $x$ in the domain of $f$. You can verify that $\leq$ is a partial ordering on the set of all partial functions from $X$ to $Y$.
- The previous example is particularly interesting when $X$ is a collection of non-empty sets and $Y=\bigcup X$ and we restrict our attention to those partial functions $f$ such that $f(A)\in A$ for each $A$ in its domain. In this case, we are ordering the set of all partial choice functions of $X$ by restriction. (This gives an easy proof of the Axiom of Choice from the Hausdorff Maximal Principle since this poset is chain-complete).
- (This is kind of an alternative to the previous result) Let $X$ be a collection of non-empty sets. Let $\mathcal C$ be the collection of all sets $T$ such that $T\subseteq\bigcup X$ and for each $A\in X$, $A\cap T$ consists of at most one element (such a $T$ is called a partial transversal of $X$). We can order $\mathcal C$ be inclusion. (This is chain-complete, and we get an easy proof that every collection of non-empty sets has a traversal by the HMP)
- Let $V$ be a vector space. You can order the set of all linearly-independent collections of vectors by inclusion. (This poset is also chain-complete, and you can get an easy proof that every vector space has a basis by the HMP)
- Let $(X,\leq)$ be a poset. Let $\mathcal C$ denote the collection of all subchains of $X$. You can order this collection by inclusion. (This poset is chain complete, thus it is inductive. From this collection you get an easy proof that Zorn's lemma implies the HMP)
- Let $(X,\leq)$ be a poset. Let $\mathcal A$ denote the collection of all antichains of $X$. You can order this collection by inclusion also. (This is also chain-complete, and (surprise) you get an easy proof that every poset has a maximal antichain by the HMP)
- Let $X$ be a set. Let $\mathcal W$ be the set of all elements of the form $(S,\leq)$ where $S$ is a subset of $X$, and $\leq$ is a well-ordering of $S$. Given two elements $(S,\leq)$ and $(T,\preceq)$ from $\mathcal W$, let's say that $(S,\leq)\sqsubseteq (T,\preceq)$ if $S\subseteq T$ and for each $x$ and $y\in S$ we have that $x\leq y$ implies $x\preceq y$. Succinctly, we say $(S,\leq)$ is less than $(T,\preceq)$ if $(T,\preceq)$ continues $(S,\leq)$. You can verify that $\sqsubseteq$ is a partial ordering for $\mathcal W$. (And guess what! It's chain-complete. From this you get an easy proof of the Well-Ordering theorem from the HMP)
- Let $\Bbb R[x]$ be the set of all real polynomials. Given two polynomials $f$ and $g$, let's say $f\sqsubseteq g$ if there exists an $x\in\Bbb R$ such that for all $y\geq x$ we have that $f(y)\leq g(y)$. It's an interesting exercise to show that $\sqsubseteq$ is a linear order on $\Bbb R[x]$. Overmore, $\sqsubseteq$ is compatible with the operation $+$ on polynomials. So that $(\Bbb R[x], +,\sqsubseteq)$ forms a linearly ordered group.
- The same thing applies to $\Bbb Q[x]$. However, there is a proposition which states that any countable, dense, linearly ordered set is (order-) isomorphic to a subset of $\Bbb Q$. So there is an order-preserving injection from $\Bbb Q[x]$ to $\Bbb Q$ since $\Bbb Q[x]$ is dense and countable. (Note that such a function cannot preserve addition between the two linearly-ordered groups. One is Archimedean while the other is not)
- Let $(X,\leq)$ be a poset. A subset $L\subseteq X$ is called a lower-set of $X$ if for all $x\in L$ whenever $y\in X$ such that $y\leq x$ then $y\in L$. Denote by $\mathcal O(X)$ the set of all lower-sets of $X$. We can order $\mathcal O(X)$ by inclusion. This new poset is in fact a complete lattice where infimums are given by intersections (except for the empty-set of lower-sets) and supremums by unions.
- Let $\Sigma^*$ denote the set of all finite binary strings. Given two binary strings $u$ and $v$, let's say that $u\leq v$ if $u$ is an initial substring of $v$. That $\leq$
is a partial order is clear. This same relation can be extended to $\Sigma^{**}$, the set of all countable binary strings. (Both of these posets are trees which are posets such that every lower set is well-ordered by $\leq$. You might take the time to prove to yourself that although $\Sigma^*$ is countable, it still has an uncountable number of maximal chains)
- The collection of all subgroups of a group forms a complete lattice. And the collection of normal subgroups forms a complete sublattice of the subgroup lattice. The set of all topologies over a set form a complete lattice. The set of all ideals in a ring also form a complete lattice.
I also know some interesting theorems.
- In every non-empty finite poset, every chain is contained in a maximal chain; every antichain is contained in a maximal antichain; every element is greater than (or equal to) some minimal element; every element is less than (or equal to) some maximal element. In particular, there are maximal chains, maximal antichains, minimal elements, and maximal elements.
- Let $(X,\leq)$ be a finite poset. Then there is another partial order $\sqsubseteq$ of $X$ which extends $\leq$ and such that $(X,\sqsubseteq)$ is a linear order. (Proof by induction)
- If you have the Axiom of Choice to work with, you can extend the last result to show that all partial orderings have a linear extension (sources say that this can be proven with just the Ultrafilter lemma or Boolean Prime Ideal theorem, but I'm having difficulty finding a proof. However, the principle that every set can be linearly ordered implies the Axiom of Choice for finite sets; you can show that).
- (Dilworth's Theorem) Let $(X,\leq)$ be a finite poset. Let $k$ denote the maximum cardinality that an antichain of $X$ can attain. Then $X$ can be partitioned into $k$ chains.
- (Other results of Dilworth) Let $(X,\leq)$ be a finite poset of cardinality $n$. For each $t>0$, either $X$ contains a chain of length greater than $t$ or an antichain of size at least $\lceil n/t\rceil$. Further, $X$ either contains a chain of length greater than $\sqrt{n}$ or an antichain of size at least $\lceil\sqrt{n}\rceil$.
- Every countable, dense, linear ordering is order isomorphic to a subset of $\Bbb Q$. This problem is actually addressed on this site here.
- There is a parallel between monotonic functions and continuous functions which isn't seen too often. Namely, given two posets $(X,\leq)$ and $(Y,\preceq)$ a function $f:X\rightarrow Y$ is monotonic if and only if for every lower-set $L\subseteq Y$ then $f^{-1}(L)$ is a lower-set of $X$. This can be explained by the fact that the lower-sets of a poset form a basis for a topology on the underlying set (they form an Alexandrov topology, in fact).
- (Tarski's Fixed-Point Theorem) Let $(X,\leq)$ be a complete poset. Then the set of fixed-points of any monotonic map $f:X\rightarrow X$ forms a complete sublattice of $X$ (in particular, there is a fixed-point). This theorem can be used to prove Schroeder-Bernstein.
- (Bourbaki-Witt Theorem) Let $(X,\leq)$ be a chain-complete poset. Let $f:X\rightarrow X$ be an inflationary map (that is $x\leq f(x)$ for all $x\in X$). Then $f$ has a fixed-point.
- (Tarski Converse) If $(X,\leq)$ is a poset such that every monotonic endomorphism has a fixed-point, then $(X,\leq)$ is a complete poset.
- Every countable, dense, linear-order without a top or bottom is order isomorphic to $\Bbb Q$. (typically uses Dependent choice; might be able to force Countable choice to give us what we want)
- A poset has the least-upper-bound property if and only if it has the greatest-lower-bound property.
- If $A$ and $B$ are two countable dense subsets of $\Bbb R$ then $\Bbb R-A$ is order-isomorphic to $\Bbb R - B$. (Sit on that for a moment. What if $A$ is the algebraic real numbers and $B$ is the rational numbers?)
- The $M_3$-$N_5$ theorem states that a lattice is not distributive if and only if it contains a sublattice isomorphic to either $M_3$ (the "diamond lattice") or $N_5$ (the "pentagon lattice"). And a lattice is not modular if and only if it contains a sublattice isomorphic to $N_5$.
This is quite a big list; I hope you found something interesting to think about and share with your students (sorry, I didn't see your question sooner). If you have any questions, let me know.