nLab pseudo-order

Redirected from "pseudo-ordered set".

Contents

Idea

A pseudo-order or strict total order or strict linear order is the irreflexive version of a total order. This is sometimes called linear order, but linear order is also used to refer to the non-strict total order.

A pseudo-ordered set or strictly totally ordered set is a set equipped with a pseudo-order.

In classical mathematics, the distinction between strict total orders and total orders is merely a terminological technicality, which is not always observed; more precisely, there is a natural bijection between the set of total orders on a given set SS and the set of strict total orders on SS, and one distinguishes them by the notation <\lt (for the strict total orders) and \leq (for the total order). In constructive mathematics, however, they are irreducibly different.

Definitions

A pseudo-order or strict total order on a set SS is a (binary) relation <\lt with the following properties:

In classical mathematics, one may see these versions of asymmetry and connectedness:

  • xyx \nless y or yxy \nless x;
  • x<yx \lt y or y<xy \lt x or x=yx = y.

Using excluded middle, these are equivalent to asymmetry and connectedness as given above, but they need not hold for all pseudo-orders in constructive mathematics.

Actually, these axioms are overkill; to begin with, irreflexivity is simply a special case of asymmetry and so can be dropped. Additionally, one can either drop transitivity or drop asymmetry (which then requires keeping irreflexivity); they will still follow from the other axioms. Dropping transitivity shows manifestly the duality (see below) between pseudo-orders and total orders (even in constructive mathematics), and dropping asymmetry shows that a pseudo-order is a weakly linear strict preorder. Keeping transitivity and irreflexivity (thus allowing one to drop asymmetry) shows manifestly that a pseudo-order is a special kind of strict preorder.

Also, because the relation is asymmetric, xyyxx \nless y \vee y \nless x holds, which means that the inequality relation x#yx \# y, defined by

x#yx<yy<xx \# y \coloneqq x \lt y \vee y \lt x

can equivalently be defined using the exclusive disjunction:

x#yx<y̲y<xx \# y \coloneqq x \lt y \underline{\vee} y \lt x

Thus, the connectedness axiom can be expressed using exclusive disjunction:

  • if not (x<yx \lt y xor y<xy \lt x), then x=yx = y.

In classical mathematics, there are even more options. Now one can prove trichotomy: since the proposition x<y̲y<xx \lt y \underline{\vee} y \lt x is a decidable proposition, (x<yx \lt y xor y<xy \lt x) xor x=yx = y. Also, weak linearity can be dropped, as it follows from transitivity and connectedness.

Thus the most common definition uses only trichotomy and transitivity.

One can also interpret connectedness not as an axiom but as a definition of equality, getting a pseudo-order on a quotient set of SS. In other words, if <\lt is an asymmetric and weakly linear relation on SS, and we define xyx \equiv y to mean that neither x<yx \lt y nor y<xy \lt x, then \equiv is an equivalence relation and <\lt induces a pseudo-order on S/S/{\equiv}.

Examples

Classically, any total order defines an example of a pseudo-order (as explained below), and this also holds constructively in discrete mathematics. Since most pseudo-orders in these cases are usually described in terms of their total orders, the focus here is on constructive analysis. (The first item, however, is an exception.)

  • A lexicographic order, even in a classical or discrete context, is more easily described as a pseudo-order than as a total order.
  • The big example in analysis is the field of real numbers. Both the Dedekind reals and the Cauchy reals (even if weak countable choice fails so these are not equivalent) have a pseudo-order <\lt that extends the (decidable) pseudo-order on the rational numbers. Since the corresponding partial order \leq cannot be proved total (and in some classically invalid versions of constructive mathematics can even be proved not total), <\lt is more directly useful than \leq in constructive analysis. In any case, <\lt is more fundamental, since \leq can be defined in terms of <\lt but not the other way around. (The Mac Neille real numbers have both <\lt and \leq; however, in this case, neither is necessarily a pseudo-order or a total order, nor can they necessarily be defined in terms of one another.)
  • Baire space and Cantor space, being representable as subspaces of the real line, of course are strictly totally ordered. It's also interesting to see them as coming from the (decidable) pseudo-orders on N\mathbf{N} and 2\mathbf{2}, which they are N\mathbf{N}-fold products of.

Properties

Relation to total orders

Using excluded middle, one can move between strict linear orders and total orders using negation; that is, the negation of a strict linear order is a total order and vice versa. Actually one usually swaps the order too, as follows:

  • x<yx \lt y iff yxy \nleq x;
  • xyx \leq y iff yxy \nless x.

To prove this, it's enough to see that the properties of a strict linear order are dual to the properties of a total order, as follows:

strict linear ordertotal order
irreflexivityreflexivity
asymmetrytotality
transitivityweak linearity
weak linearitytransitivity
connectednessantisymmetry

Constructively, these are still the default definitions to use; that is, if one is given a strict linear order or a total order and wants to interpret the other symbol, then one does so using these definitions. However, the result will not necessarily be a total order or a strict linear order. To be specific, if one starts with a strict linear order <\lt and defines \leq as above, then totality does not follow; and if one starts with a total order \leq and defines <\lt as above, then weak linearity does not follow. Nevertheless, at least \leq will be a partial order, and least <\lt will be a strict preorder. Furthermore, the duality between the axioms is still there, even though negation no longer mediates between them; although weak linearity need not hold for a total order constructively, the duality is preserved if one defines strict linear orders without using transitivity.

One often sees x<yx \lt y defined as xyx \le y but xyx \ne y; this is equivalent, but doesn't show the de Morgan duality explicitly. Similarly, one often sees xyx \leq y defined as x<yx \lt y or x=yx = y; this is not even equivalent constructively, although it is classically.

Keep in mind, however, that the only use of excluded middle in the classical theory is the assumption that x=yx = y, x<yx \lt y, and xyx \leq y are always either true or false. There is therefore a perfect correspondence between decidable strict linear orders and decidable total orders on sets with decidable equality.

Decidable pseudo-orders

A pseudo-order on a set AA is decidable if for all xx and yy in AA, x<yx \lt y or ¬(x<y)\neg (x \lt y).

Given a proposition PP, PP can be made into a subsingleton set by considering the subset {|P}{p{}|(p=)P}\{\top \vert P\} \coloneqq \{p \in \{\top\} \vert (p = \top) \wedge P\} of the singleton {}\{\top\}. Let ABA \uplus B denote the disjoint union of sets AA and BB, and let B AB^A denote the function set with domain AA and codomain BB.

Theorem

Every decidable pseudo-order on a set AA is purely cotransitive: Given elements xx, yy, and zz in AA, one can construct an element of the function set

cotransitive(x,y,z)({|x<y}{|y<z}) {|x<z}\mathrm{cotransitive}(x, y, z) \in \left(\{\top \vert x \lt y\} \uplus \{\top \vert y \lt z\}\right)^{\{\top \vert x \lt z\}}

Proof

We prove this by case analysis.

Suppose that x<yx \lt y. Then one can construct the element {|x<y}\top \in \{\top \vert x \lt y\} and by definition of disjoint union an element (0,){|x<y}{|y<z}(0, \top) \in \{\top \vert x \lt y\} \uplus \{\top \vert y \lt z\}. cotransitive(x,y,z)\mathrm{cotransitive}(x, y, z) is given by the constant function t(0,)t \mapsto (0, \top).

Now suppose that ¬(x<y)\neg (x \lt y). This means that yx<zy \leq x \lt z, and one can construct the element {|y<z}\top \in \{\top \vert y \lt z\}. By definition of disjoint union one can construct an element (1,){|x<y}{|y<z}(1, \top) \in \{\top \vert x \lt y\} \uplus \{\top \vert y \lt z\}. cotransitive(x,y,z)\mathrm{cotransitive}(x, y, z) is given by the constant function t(1,)t \mapsto (1, \top).

Since x<yx \lt y is decidable this covers every possible case. Thus, every decidable pseudo-order is purely cotransitive.

The above proof first appeared for the pseudo-order of the real numbers in theorem 11.4.3 of the HoTT book in the context of dependent type theory.

Corollary

Suppose that the rational numbers are a subset of the decidably pseudo-ordered set AA, and the canonical injection i:Ai:\mathbb{Q} \to A is strictly monotonic. Then for every element xx of AA one can construct a locator for xx.

Proof

The locator is given by the dependent function

(q,r)cotransitive(i(q),x,i(r)) q,r({|q<x}{|x<r}) {|q<r}(q, r) \mapsto \mathrm{cotransitive}(i(q), x, i(r)) \in \prod_{q, r \in \mathbb{Q}} \left(\{\top \vert q \lt x\} \uplus \{\top \vert x \lt r\}\right)^{\{\top \vert q \lt r\}}

which always exists by the previous theorem and by the fact that the rational numbers are a subset of AA which preserves the pseudo-order.

See also

 References

For a proof that the decidable pseudo-order on the real numbers is purely cotransitive, see theorem 11.4.3 of:

Last revised on September 25, 2024 at 22:51:49. See the history of this page for a list of all contributions to it.