May 14th
Today I learned a little more about doing algebra over posets, from Combinatorial Reciprocity Theorems . To review, we are thinking about posets as "sets with structure,'' where the "structure'' is the partial order we provided. This motivates us to define order-preserving maps, which are maps $\varphi:X\to Y$ between posets $(X,\le)$ and $(Y,\le)$ such that\[x_1\le x_2\implies\varphi(x_1)\le\varphi(x_2).\]We say that $\varphi$ is strictly order-preserving if $x_1 \lt x_2$ implies $\varphi(x_1) \lt \varphi(x_2).$
These order-preserving maps behave more or less like homomorphisms. Trying to move in a more categorical framework, we can study their "kernels.'' There is no obvious notion of an identity element in a poset, but a minimum element can do. To be explicit, given a poset $(X,\le)$ and order-preserving $\varphi:X\to[n],$ we can ask what the possible\[\varphi^{-1}(\{1\})\]are. The main property here is that $x\in\varphi^{-1}(\{1\})$ and $x'\le x,$ then we had better have\[\varphi(x')\le\varphi(x)=1,\]implying $x'\in\varphi^{-1}(\{1\}).$ This motivates us to take the following definition.
Definition. Given a poset $(X,\le),$ we say that $I\subseteq X$ is an order ideal if and only if, given $x\in I,$ we have \[x'\le x\implies x'\in I.\] At a high level, this means that we are taking some elements of $X$ and everything smaller.
To show that these actually make kernels, we note that for any order ideal $I\subseteq X,$ we can define the order-preserving map $\varphi:X\to[n]$ by taking $\varphi(I)=\{1\}$ and $\varphi(X\setminus I)=\{n\}.$ To show this is order-preserving, we need to know that\[x_1\le x_2\implies\varphi(x_1)\le\varphi(x_2).\]Well, if $x_2\in I,$ then we see $x_1\in I$ because $I$ is an order ideal, so $\varphi(x_1)=\varphi(x_2)=1.$ Alternatively, if $x_2\notin I,$ then $\varphi(x_2)=n,$ which is the largest element, so it doesn't matter what $\varphi(x_1)$ is.
As a quick remark about language, I think we're associating posets with rings, which is why we are calling these ideals. So our "extra-structured'' posets, namely $n$-chains, have the nice property that all of their ideals are "principal.'' Namely, an order ideal $I$ of $[n]$ is a sub-poset and therefore is itself a chain.
But $I$ is a finite chain, so it has a maximum element, say $m\in[n].$ To finish, we note\[x\le m\implies x\in I,\]so we basically have $I=[m]\subseteq[n].$ So indeed, $I$ is basically "generated by $m.$'' I am obligated to say that things are a bit funnier in the infinite case: $(-\infty,0)$ and $(-\infty,0]$ are both order ideals of $(\RR,\le),$ but it's not clear which should be "generated by $0.$''
Very briefly, we note that any set of subsets of $X$ can be endowed with poset structure by $\subseteq.$
Definition. Given a poset $X,$ we define the lattice of order ideals as the set of order ideals ordered by $\subseteq.$
Anyways, the reason we care about order ideals is that they're going to let us talk intelligently about order-preserving maps only using data internal to the poset $X.$ We have the following.
Proposition. Fix $(X,\le)$ a poset and $n$ a positive integer. We can biject order-preserving maps $\varphi:X\to[n]$ with a sequence of ascending order ideals: \[\emp=I_0\subseteq I_1\subseteq\cdots\subseteq I_{n-1}\subseteq I_n=X.\]
Our correspondence is to take $\varphi:X\to[n]$ to the sequence\[I_k=\varphi^{-1}(\{1,2,\ldots,k\}),\]for $k\in\{1,2,\ldots,n-1\}.$ Note this gives $I_0=\emp$ and $I_n=X$ for free. As an abuse of notation, we will write $[k]=\{1,2,\ldots,k\}\subseteq[n].$ We begin by showing this correspondence is well-defined. All of the $I_\bullet$ are order ideals because, given $x\in I,$ we see $x'\le x$ implies\[\varphi(x')\le\varphi(x)\in[k],\]so $\varphi(x')\in[k]$ as well, meaning $x'\in I.$ And this sequence is ascending because $k\le\ell$ implies $[k]\subseteq[\ell]$ so that\[I_k=\varphi^{-1}([k])\subseteq\varphi^{-1}([\ell])=I_\ell.\]This finishes our check that our correspondence is well-defined.
To show injectivity, suppose $\varphi_1$ and $\varphi_2$ both give the same ascending sequence of order ideals. Then for any $x\in X,$ there is a smallest $k\in[n]$ such that $x\in I_k.$ (Certainly $x\in I_n,$ so well-order guarantees a smallest $k.$) But this means that $x\in I_k\setminus I_{k-1},$ so\[\varphi_\bullet(x)\in[k]\setminus[k-1]=\{k\}.\]In particular, $\varphi_1(x)=\varphi_2(x)=k.$ So the $\varphi_\bullet$ are equal on all inputs and are therefore equal.
To show surjectivity, we do the same thing. Any ascending sequence of order ideals (with $I_0=\emp$ and $I_n=X$), we will construct a $\varphi:X\to[n]$ giving the desired sequence. For some $x\in X,$ we again note that there is a smallest $k\in[n]$ such that $x\in I_k,$ so we take $x\mapsto k.$ To be explicit,\[\varphi(x):=\min\{k\in[n]:x\in I_k\}.\]Around here it is apparent why we want things finite: this makes less sense with codomain $\RR.$ We have to show that this $\varphi$ is order-preserving. Well, for $x_1,x_2\in X$ with $x_1\le x_2,$ we note that\[x_2\in I_{\varphi(x_2)}\]by definition. But now $I_{\varphi(x_2)}$ is an order ideal, so $x_1\in I_{\varphi(x_2)}$ as well, implying $\varphi(x_2)\in\{k\in[n]:x_1\in I_k\},$ so $\varphi(x_1)\le\varphi(x_2),$ as desired. This completes the proof. $\blacksquare$
We have a slight refinement for strictly order-preserving maps.
Proposition. Fix $(X,\le)$ a poset and $n$ a positive integer. We can biject order-preserving maps $\varphi:X\to[n]$ with a sequence of ascending order ideals \[\emp=I_0\subseteq I_1\subseteq\cdots\subseteq I_{n-1}\subseteq I_n=X\] such that $I_k\setminus I_{k-1}$ is an antichain of $X$ for each $k\in[n].$
We continue with the bijection of the previous proof. The key observation is that\[\varphi(I_k\setminus I_{k-1})=\big\{\min\{k\in[n]:x\in I_k\}:x\in I_k\setminus I_{k-1}\big\}=\{k\}.\]In particular, $\varphi$ is strictly order-preserving if and only if $x_1\le x_2$ and $x_1\ne x_2$ implies $\varphi(x_1)\le\varphi(x_2)$ and $\varphi(x_1)\ne\varphi(x_2).$ We already know that $\varphi(x_1)\le\varphi(x_2),$ so the conclusion we are interested in is $\varphi(x_1)\ne\varphi(x_2).$ Namely, strictly order-preserving is equivalent to\[x_1\lneq x_2\implies\varphi(x_1)\ne\varphi(x_2).\]Taking the contrapositive, this is equivalent to\[\varphi(x_1)=\varphi(x_2)\implies x_1=x_2\text{ or }x_1\not\le x_2.\]Using the key observation, this is now\[x_1,x_2\in I_k\setminus I_{k-1}\implies x_1=x_2\text{ or }x_1\not\le x_2.\]In other words, for distinct $x_1,x_2\in I_k\setminus I_{k-1},$ we cannot have $x_1\not\le x_2.$ Switching the role of $x_1$ and $x_2,$ we cannot have $x_2\not\le x_1$ either, so $x_1$ and $x_2$ are incomparable, implying $I_k\setminus I_{k-1}$ is antichain. And conversely, if $I_k\setminus I_{k-1}$ is antichain, then the above condition still holds. This finishes. $\blacksquare$
The main point here is that we have reduced the study of order-preserving maps $\varphi:X\to[n]$ to studying chains of particular kinds/lengths inside of the lattice of order ideals. Studying chains in a poset is a general poset question, and hopefully this abstraction is helpful.