The order topology of a linearly ordered set, with the open rays as a subbasis; order-convex sets, order-density, the least upper bound property, and linear continua

Let $(L, \le)$ be a linearly ordered set (Partial order and partially ordered set): a poset in which any two elements are comparable. Write $<$ for the associated strict order.

Rays, intervals, and the order topology

For $a \in L$ the open rays at $a$ are

$L_{<a} \;:=\; \{\, t \in L : t < a \,\}, \qquad L_{>a} \;:=\; \{\, t \in L : a < t \,\},$

and $\mathcal{S}_L := \{\, L_{<a} : a \in L \,\} \cup \{\, L_{>a} : a \in L \,\}$ is the family of all of them. The order topology on $L$ is

$\mathcal{T}_{<} \;:=\; \langle \mathcal{S}_L \rangle,$

the topology generated by $\mathcal{S}_L$ (Basis and subbasis for a topology, and the topology generated by a family of sets, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). A linearly ordered topological space is a linearly ordered set carrying its order topology. For $a, b \in L$ write

$(a,b) := \{\, t \in L : a < t < b \,\}, \quad [a,b] := \{\, t \in L : a \le t \le b \,\},$ $[a,b) := \{\, t \in L : a \le t < b \,\}, \quad (a,b] := \{\, t \in L : a < t \le b \,\},$

so that $(a,b) = L_{>a} \cap L_{<b}$.

A basis, and the obligation is discharged here. By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis claim 2 the intersections of finitely many members of $\mathcal{S}_L$ form a basis for $\mathcal{T}_{<}$. This library takes the empty intersection to be $L$, so $L$ itself is among them. An intersection of finitely many rays is computed by collecting the lower cuts and the upper cuts separately: since $\le$ is linear, a finite nonempty set of elements of $L$ has a greatest and a least member, so $L_{<a_1} \cap \dots \cap L_{<a_m} = L_{<a}$ with $a$ the least of the $a_i$, and $L_{>b_1} \cap \dots \cap L_{>b_k} = L_{>b}$ with $b$ the greatest of the $b_j$. Hence every finite intersection is $L$, an open ray, or an open interval $(b,a) = L_{>b} \cap L_{<a}$, and

$\mathcal{B}_L \;:=\; \{L\} \;\cup\; \mathcal{S}_L \;\cup\; \{\, (a,b) : a, b \in L \,\}$

is a basis for $\mathcal{T}_{<}$ (Basis and subbasis for a topology, and the topology generated by a family of sets).

What a basic neighbourhood of a point looks like. Let $x \in L$. If $x$ is neither the least nor the greatest element of $L$ (Maximal element and greatest element), then some $a < x$ and some $b > x$ exist and $x \in (a,b)$; if $x$ is least, the sets $L_{<b}$ with $b > x$ are the basic sets containing $x$ apart from $L$ itself; if $x$ is greatest, they are the sets $L_{>a}$ with $a < x$. These three cases are the only ones, and every proof below that argues at a point splits along them.

Order-convex sets. A subset $C \subseteq L$ is order-convex when

$x, z \in C \text{ and } x \le w \le z \;\Longrightarrow\; w \in C .$

Every ray and every one of the four interval forms above is order-convex, by transitivity of $\le$; so are $\varnothing$, every singleton, and $L$.

A convention that is fixed once here. A subset $C \subseteq L$ inherits two topologies that need not agree: the subspace topology from $(L, \mathcal{T}_{<})$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) and the order topology of the restricted order on $C$. In this library "a subspace of a linearly ordered topological space" always means the subspace topology, and the phrase "the order topology of $C$" is written in full whenever the second is meant. The two do agree when $C$ is order-convex, which is the only case used here: for order-convex $C$ the trace $L_{<a} \cap C$ is $C$ if $a$ is above every element of $C$, is $\varnothing$ if $a$ is below or equal to every element of $C$, and is otherwise the ray $C_{<a}$ when $a \in C$, and $C_{<c}$ for any $c \in C$ above $a$ has the same trace description; in every case the trace of a subbasic set of $L$ is a subbasic set of $C$ or is $\varnothing$ or $C$, and conversely every ray of $C$ is such a trace. The general statement, for a subset that is not order-convex, is not asserted here.

Order-density and the least upper bound property

Let $(L, \le)$ be linearly ordered.

• $L$ is order-dense (or densely ordered) when for all $x, y \in L$ with $x < y$ there is $z \in L$ with $x < z < y$. Equivalently, no element of $L$ has an immediate successor above it: there is no pair $x < y$ with $(x,y) = \varnothing$.
• $L$ has the least upper bound property when every nonempty $S \subseteq L$ that has an upper bound in $L$ has a least upper bound in $L$ (Upper bound, least upper bound, and strict upper bound). A least upper bound is unique when it exists, by antisymmetry: two of them bound each other, and antisymmetry of $\le$ (Partial order and partially ordered set) forces them equal. We write $\sup S$ for it.

A linear continuum is a linearly ordered set with at least two elements that is order-dense and has the least upper bound property.

The two-element requirement is not decoration. Without it the empty ordered set and every one-point ordered set would qualify vacuously, and the theorems about linear continua elsewhere in this library would have degenerate instances whose statements say nothing. A linear continuum in the sense above is automatically infinite: two elements $x < y$ produce $z_1$ strictly between them, then $z_2$ strictly between $x$ and $z_1$, and so on, and each is new because the order is strict.

$\mathbb{R}$ is a linear continuum, and its order topology is its usual topology. The order of $\mathbb{R}$ (Order on the reals, Ordered field) is linear; $\mathbb{R}$ has at least two elements, namely $0$ and $1$; it is order-dense because $x < (x+y)/2 < y$ whenever $x < y$ (Ordered field); and it has the least upper bound property, which is exactly the completeness axiom (Complete ordered field (least-upper-bound property), Suprema and infima are unique, Greatest lower bound (infimum), Lower bound, bounded below, bounded set). Its order topology is the usual topology, that is the metric topology of $d_{\mathbb{R}}(s,t) = |s-t|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not): the ball $B(x,r)$ is the interval $(x - r, x + r)$ (Open ball, closed ball and sphere in a metric space, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$), which is a basic set of $\mathcal{T}_{<}$, so every set open for the metric is open for the order; and conversely $L_{<b}$ and $L_{>a}$ are open for the metric (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), so every subbasic set of $\mathcal{T}_{<}$ is metric-open and $\mathcal{T}_{<} \subseteq \mathcal{T}_{d_{\mathbb{R}}}$ by minimality of the generated topology (Basis and subbasis for a topology, and the topology generated by a family of sets). The two topologies therefore coincide, and no second topology on $\mathbb{R}$ is being introduced.

• The dictionary with the ordinal case. The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis ↗ puts a topology on an ordinal $\gamma$ using the initial segments $[0,\beta]$ and the half-open intervals $(\alpha,\beta]$ as a basis, and says in its own body that this is the general order basis rewritten so that no case analysis is needed. The two agree: $[0,\beta] = \gamma_{<\beta^{+}}$ when $\beta^{+} \in \gamma$ and is $\gamma$ otherwise, and $(\alpha,\beta] = \gamma_{>\alpha} \cap \gamma_{<\beta^{+}}$ under the same proviso, so every basic set there is a finite intersection of rays here; conversely $\gamma_{<\beta} = [0,\beta]\setminus\{\beta\}$ is the union of the sets $[0,\xi]$ with $\xi < \beta$, and $\gamma_{>\alpha}$ is the union of the sets $(\alpha,\beta]$ with $\alpha < \beta < \gamma$, so every ray here is a union of basic sets there. The two topologies have the same open sets.

• The same dictionary for the rays presentation. The order topology on a totally ordered set, with the open rays as a subbasis, and its agreement with the usual topology of $\mathbb{R}$ presents the order topology of a totally ordered set by exactly the subbasis used above and identifies it with the usual topology of $\mathbb{R}$; the present item repeats that identification because it is the one every later proof quotes, and adds the order-convexity, order-density and least-upper-bound vocabulary that the linear-continuum theorems need.

• Why the rays and not the intervals. Taking only the open intervals $(a,b)$ as a basis fails whenever $L$ has a least or a greatest element: no interval contains the least element unless some element sits below it. The rays repair this without a case split, which is why they are the subbasis of record here.

• Order-density is not topological density. Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets calls a subset $A \subseteq X$ dense when $\overline{A} = X$. Order-density is a property of the ordered set itself, not of a subset, and the two words coincide only by historical accident. Where both are in play this library writes order-dense in full.