# Linear gradings of partial orders

Order relations are often fruitfully conceived as being stratified into “levels” in a natural way. The level structure is meant to be compatible with the order, in the sense that as one moves strictly up in the order, one also ascends successively from lower levels to strictly higher levels. With the simple order relation pictured here, for example, we might naturally imagine the least element $0$ on a bottom level, the three middle nodes $a$, $b$, and $c$ on an intermediate level, and node $1$ on a level at the top. With this level structure, as we move up strictly in the order, then we also move up strictly in the hierarchy of levels.

What exactly is a level? Let us be a little more precise. We can depict the intended level structure of our example order more definitely as in the figure here. At the left appears the original order relation, while the yellow highlighted bands organize the nodes into three levels, with node $0$ on the bottom level, nodes $a$, $b$, and $c$ on the middle level, and node $1$ on the top level. This level structure in effect describes a linear preorder relation $\leq$ for which $0\leq a,b,c\leq 1$, with the three intermediate nodes all equivalent with respect to this preorder—they are on the same level.

$$a\prec b\quad\text{ implies }\quad a<b$$ Thus, any strictly lower point in the original order is on a lower level, and we define that objects are on the same level if they are equivalent with respect to the preorder. A linearly graded order is a relational structure $\<A,\preccurlyeq,\leq>$ with two orders on the same domain, the first $\preccurlyeq$ being an order relation on $A$ and the second $\leq$ being a linear preorder relation that grades the first order.

It turns out that there are often far more ways to stratify a given order by levels than one might have expected. For the simple order above, for example, there are thirteen distinct linear grading orders, as shown here.

The conclusion is inescapable that the level concept is not inherent in the order relation itself, for a given order relation may admit a huge variety of different level hierarchies, each of them compatible with the given order.

One should therefore not make the mistake of thinking that if one has an order relation, such as a philosophical reduction notion of some kind or a supervenience relation, then one is automatically entitled to speak of “levels” of the order. One might want to speak of “low-level” phenomena or “high-level” concepts in the order, but one must recognize that the order relation itself does not determine a specific hierarchy of levels, although it does place limitations on the possible stratifications. My point is that there is often surprising flexibility in the nature of the level structure, as the example above shows even in a very simple case, and so what counts as low or high in terms of levels may be much less determined than one expects. In some of the linear gradings above, for example, the node $a$ could be described as high-level, and in others, it is low-level. Therefore if one wants to speak of levels for one’s order, then one must provide further elucidation of the stratification one has in mind.

Meanwhile, we often can provide a natural level structure. In the power set $P(X)$ of a finite set $X$, ordered by the subset relation $A\of B$, for example, we can naturally stratify the relation by the sizes of the set, that is, in terms of the number of elements. Thus, we would place the $\emptyset$ on the bottom level, and the singleton sets $\{a\}$ on the next level, and then the doubletons $\{a,b\}$, and so on. This stratification by cardinality doesn’t quite work when $X$ is infinite, however, since there can be instances of strict inclusion $A\subsetneq B$ where $A$ and $B$ are both infinite and nevertheless equinumerous. Is there a level stratification of the infinite power set order?

Indeed there is, for every order relation admits a grading into levels.

Theorem. Every order relation $\<A,\preccurlyeq>$ can be linearly graded. Indeed, every order relation can be extended to a linear order (not merely a preorder), and so it can be graded into levels with exactly one node on each level.

Proof. Let us begin with the finite case, which we prove by induction. Assume $\preccurlyeq$ is an order relation on a finite set $A$. We seek to find a linear order $\leq$ on $A$ such that $x\preccurlyeq y\implies x\leq y$. If $A$ has at most one element, then we are done immediately, since $\preccurlyeq$ would itself already be linear.

Let us proceed by induction. Assume that every order of size $n$ has a linear grading, and that we have a partial order $\preccurlyeq$ on a set $A$ of size $n+1$. Every finite order has at least one maximal element, so let $a\in A$ be a $\preccurlyeq$-maximal element. If we consider the relation $\preccurlyeq$ on the remaining elements $A\setminus\{a\}$, it is a partial order of size $n$, and thus admits a linear grading order $\leq$. We can now simply place $a$ atop that order, and this will be a linear grading of $\<A,\preccurlyeq>$, because $a$ was maximal, and so making it also greatest in the grading order will cohere with the grading condition.

So by induction, every finite partial order relation can be extended to a linear order.

Now, we consider the general case. Suppose that $\preccurlyeq$ is a partial order relation on a (possibly infinite) set $A$. We construct a theory $T$ in a language with a new relation symbol $\leq$ and constant symbols $\dot a$ for every element $a\in A$. The theory $T$ should assert that $\leq$ is a linear order, that $\dot a\leq \dot b$, whenever it is actually true that $a\preccurlyeq b$ in the original order, and that $\dot a\neq\dot b$ whenever $a\neq b$. So the theory $T$ describes the situation that we want, namely, a linear order that conforms with the original partial order.

The key observation is that every finite subset of the theory will be satisfiable, since such a finite subtheory in effect reduces to the case of finite orders, which we handled above. That is, if we take only finitely many of the axioms of $T$, then it involves a finite partial order on the nodes that are mentioned, and by the finite case of the theorem, this order is refined by a linear order, which provides a model of the subtheory. So every finite subtheory of $T$ is satisfiable, and so by the compactness theorem, $T$ itself also is satisfiable.

Any model of $T$ provides a linear order $\leq$ on the constant symbols $\dot a$, which will be a linear order extending the original order relation, as desired. $\Box$

This material is adapted from my book-in-progress, Topics in Logic, which includes a chapter on relational logic, included an extended section on orders.