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

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

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

${\mathcal{T}}_{<}:=⟨{\mathcal{S}}_L⟩,$

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

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

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 $L_{<a_1}\cap \cdots \cap L_{<a_m}=L_{<a}$ with $a$ the least of the $a_i$, and $L_{>b_1}\cap \cdots \cap L_{>b_k}=L_{>b}$ 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

$x,z\in C\text{ and }x\le w\le z⟹w\in C\mathrm{.}$

Every ray and every one of the four interval forms above is order-convex, by transitivity of $\le$; so are $\mathrm{\varnothing }$, 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 $\mathrm{\varnothing }$ 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 $\mathrm{\varnothing }$ 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 $(x,y)=\mathrm{\varnothing }$.
• $L$ 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 $x<(x+y)\mathrm{/}2<y$ 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 $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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 $\epsilon$-neighbourhood and the punctured $\epsilon$-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 $\bar{A}=X$. 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.