Are linearly ordered sets algebraic structures?

Question

I'm tempted to notate a poset or chain on a set $S$ as $(S, \leq)$, but I feel more comfortable using (set, stuff we're doing on set) for algebraic structures. Is there another way to notate ordered structures? That preserves the "There's an underlying set and we're working with this relation on it" feel? I don't want to use $\leq_{S}$ because the subscripts are so often used for denoting different relations on the same underlying set. Am I the only one bothered by this?

Thanks to @Ixion and @Chris Leary for pointing out errors in my original post :)

Answer 1

It is standard to denote a set equipped with extra structure by $({\tt set},{\tt structure})$ or $\langle {\tt set};{\tt structure}\rangle$, where ${\tt set}$ is some set and ${\tt structure}$ might be some combination of operations, relations, topologies, ETC. If you feel that this is notation reserved for algebraic structures, that probably just means you have more experience with algebra.

Regarding the question of whether posets or chains, defined as relational structures, may be thought of as algebraic structures, my answer would be: it depends on what you are looking for. Two other contributors have shown that one can encode the order relation on $(S,\leq)$ into a binary operation on $S$ (or on $S\cup \{0,1\}$), but this process typically alters the notion of morphism between structures. The morphisms between posets are the order-preserving maps. It is possible for a function between posets to be bijective and order-preserving and yet not an isomorphism, while it is not possible for a function between algebraic structures to be a bijective homomorphism without being an isomorphism. Thus any attempt to algebraize the category of posets must alter the category.

The situation is different if you restrict to chains. The order-preserving maps are exactly the maps that preserve the lattice operations. The answer given by William Elliot converts the category of chains from a category of relational structures to an isomorphic category of algebraic structures.

Answer 2

This idea of a general algebraic structure is formalized in universal algebra, where it is defined as a set $A$, together with a (finite or infinite) collection of operations on $A$. An $n$-ary operation on A is a function that takes $n$ ($n$ is finite) elements of A and returns a single element of $A$.".

You can see by looking into articles such as https://en.wikipedia.org/wiki/Ordered_field that, in practice, an algebraic structure is notated as $(A, f_1, f_2, ..., f_n)$, where each $f_i$ is an $n_i$-ary function on $A$, that is, $f_i:A^{n_i}\rightarrow A$. When there is additional information about this algebraic structure, such as an ordering $\leq_A$, this is not written into the brackets like $(A, f_1, f_2, ..., f_n, \leq_A)$, but written outside the brackets, like:

"A field $(F, +, \times)$ together with a total order $\leq$ on $F$ is an ordered field..."

An ordering $\leq_A$ can be formalized as a relation on $A\times A$, that is, a subset of it.

Thus, linearly ordered sets are not algebraic structures. They are more like algebraic structures (in this case, a naked set $(A)$, with no function defined on it) with a relation defined on it.

Answer 3

Lattices are algebraic structures with sup and inf operations.
Since linear orders are lattices, yes they are algebraic structures.
Define $+:S\times S \to S$ by $ (a,b) \mapsto \max\{ a,b \}$, and you have an algebraic
structure where $a \leq b$ can be taken as $a + b = b;$ max of course,
is defined using $\leq$.

Answer 4

For every poset $\langle A,\leq\rangle $ one can define a the following (“pogroupoid”) operation $*$ on $A$ setting $$ x * y := \begin{cases} x & \text{if } x\leq y \\ y & \text{if } x\not\leq y. \end{cases} $$ It satisfies $$ x \leq y \text{ if and only if } x * y = x \tag{1}, $$ hence it encodes the partial order completely. This is a generalization of the answer by @WilliamElliot.

If $A$ has more than two elements, you can even get some $*$ satisfying $(1)$ to be commutative.

Whether you can get an associative $*$ (“posemigroup”) is not generally true, and deciding whether some poset admits a posemigroup operation is not trivial.

Answer 5

The general answer in no. If you want to define ordered algebras in the formalism of universal algebras described by @maudpietherocktorate, you need to introduce an order relation compatible with the operations. A formal definition can be found in Section 1.1. of this paper.