In mathematics, the order topology is a topology defined on a totally ordered set X (equipped with a strict order relation <) by generating open sets from intervals determined by the order.[1][2]The basis for the order topology consists of all open intervals (a, b) = \{x \in X \mid a < x < b\} for a, b \in X with a < b, along with half-open intervals of the form [a_0, b) = \{x \in X \mid a_0 \leq x < b\} if X has a least element a_0, and (a, b_0] = \{x \in X \mid a < x \leq b_0\} if X has a greatest element b_0.[1][2] This collection forms a basis because it covers X and satisfies the intersection condition for basis elements.[2]Notable examples include the standard topology on the real numbers \mathbb{R}, which coincides with the order topology induced by the usual less-than order.[1][2] Another example is the order topology on \mathbb{R} \times \mathbb{R} induced by the dictionary (lexicographic) order.[1] Additionally, the order topology on the positive integers \mathbb{Z}^+ yields the discrete topology, as singletons are open.[1]Key properties of the order topology include the openness of unbounded rays such as (a, +\infty) = \{x \in X \mid a < x\} and (-\infty, b) = \{x \in X \mid x < b\}, which form a subbasis for the topology.[2] This structure allows the order topology to capture continuity and compactness in ordered spaces, making it foundational for studying ordered topological spaces like ordinals and the long line.[1]
Fundamentals
Definition
A totally ordered set, also known as a linearly ordered set, consists of a set X equipped with a binary relation \leq that is reflexive, antisymmetric, transitive, and total: for any x, y \in X, either x \leq y or y \leq x. The strict order < is defined by x < y if x \leq y and x \neq y. Common intervals in such a set include the open interval (a, b) = \{x \in X \mid a < x < b\}, the half-open intervals [a, b) = \{x \in X \mid a \leq x < b\} and (a, b] = \{x \in X \mid a < x \leq b\}, and the closed interval [a, b] = \{x \in X \mid a \leq x \leq b\}, where a, b \in X.[2]The order topology \tau on a totally ordered set (X, \leq) is the topology generated by taking all open intervals as a basis. More precisely, \tau is the unique topology having as a basis the collection of all open intervals (a, b) together with, if X has a least element a_0, the intervals [a_0, b), and if X has a greatest element b_0, the intervals (a, b_0].[1][2]Formally, for a totally ordered set (X, \leq), a subset U \subseteq X is open in the order topology if and only if it is a union of open intervals of the form (a, b) = \{x \in X \mid a < x < b\} for a, b \in X with a < b, along with rays (-\infty, b) = \{x \in X \mid x < b\} and (a, +\infty) = \{x \in X \mid x > a\} for a, b \in X. This construction ensures that the basis elements cover X and satisfy the intersection property required for a basis.[1]This concept was introduced by Felix Hausdorff in his 1914 monograph Grundzüge der Mengenlehre, where he developed foundational ideas in set theory and topology based on ordered structures.[3]
Basis Elements
In the order topology on a totally ordered set X, a standard basis \mathcal{B} consists of all open intervals of the form (a, b) = \{ x \in X \mid a < x < b \}, where a, b \in X \cup \{-\infty, +\infty\} with a < b. This collection includes unbounded rays such as (-\infty, b) = \{ x \in X \mid x < b \} for b \in X and (a, +\infty) = \{ x \in X \mid x > a \} for a \in X, accommodating sets without minimal or maximal elements. These basis elements generate all open sets as arbitrary unions, so every open set U \subseteq X can be expressed as U = \bigcup_{i \in I} (a_i, b_i) for some index set I and suitable a_i, b_i.[2]To verify that \mathcal{B} forms a basis for the order topology, two conditions must hold: first, for every x \in X, there exists some B \in \mathcal{B} containing x; second, if B_1, B_2 \in \mathcal{B} and x \in B_1 \cap B_2, then there exists B_3 \in \mathcal{B} such that x \in B_3 \subseteq B_1 \cap B_2. The first condition is satisfied because, for any x \in X, one can choose successors and predecessors in the order (or rays if at endpoints) to form an interval containing x, such as (a, b) where a is the greatest element less than x (if it exists) and b the least greater than x. For the second, suppose x \in (a_1, b_1) \cap (a_2, b_2); then set a = \max(a_1, a_2) and b = \min(b_1, b_2), yielding (a, b) \in \mathcal{B} with x \in (a, b) \subseteq (a_1, b_1) \cap (a_2, b_2), since the order ensures a < x < b. This confirms \mathcal{B} satisfies the basis axioms.[2][4]A subbasis \mathcal{S} for the order topology is given by the collection of all open rays \{ (-\infty, b) \mid b \in X \} \cup \{ (a, +\infty) \mid a \in X \}. The basis \mathcal{B} arises as all finite intersections of elements from \mathcal{S}, since the intersection of (a, +\infty) and (-\infty, b) yields (a, b) when a < b, and single rays remain when no intersection is taken. Arbitrary unions of these finite intersections then produce the full topology.[2]For example, consider the totally ordered set of rational numbers \mathbb{Q} with the standard order. The basis elements are open intervals (p, q) where p, q \in \mathbb{Q} and p < q; these are open in the order topology on \mathbb{Q}, which coincides with the subspace topology induced from the Euclidean topology on \mathbb{R}, though the order construction highlights the role of rays in generating unbounded opens like (p, +\infty) \cap \mathbb{Q}.[2]
Properties
Separation Axioms
The order topology on a totally ordered set satisfies the T0 separation axiom, as distinct points can be separated by open sets: if x < y, then x \in (-\infty, y) but y \notin (-\infty, y), and symmetrically for the other ray. It is also T1, since singletons are closed; the complement of \{x\} is the union of the open rays (-\infty, x) \cup (x, \infty), which is open. Equivalently, \{x\} = \bigcap_{y > x} (y, \infty) \cap \bigcap_{z < x} (-\infty, z), showing singletons are G_\delta sets, though T1 follows directly from the closedness. This holds for totally ordered sets, which are antisymmetric by definition.[5]The order topology is Hausdorff (T2). For distinct points x < y, disjoint open neighborhoods exist: if there is c with x < c < y, take (-\infty, c) containing x and (c, \infty) containing y; if no such c exists (i.e., x, y are consecutive), take (-\infty, y) containing x and (x, \infty) containing y, which are disjoint since nothing lies between x and y. Thus, the property holds regardless of density; for example, the order topology on the integers \mathbb{Z} is the discrete topology, which is Hausdorff.[5][6]Order topologies are regular (T3). Given a point x and a closed set C with x \notin C, the basis elements allow separation: since the space is T1 and the order structure ensures closed sets are "order-closed," there exist disjoint open sets U containing x and V containing C. Specifically, basis elements like open intervals or rays can be chosen to avoid overlap, leveraging the linear order to isolate x from the "gaps" in C. This holds generally, though in dense orders like \mathbb{R}, the standard basis simplifies the separation.[6]As a consequence of being regular and Hausdorff, order topologies are Tychonoff (completely regular, T3.5). Points and closed sets can be separated by continuous functions, often via order-preserving embeddings into products of intervals. For instance, the coordinate functions in the order can be used to construct such separations.[6]Order topologies are also normal (T4) and even completely normal (hereditarily normal). Disjoint closed sets can be separated by disjoint open sets, with the property holding in subspaces. A non-metrizable example is the long line, obtained as the lexicographic order on [0, \omega_1) \times [0,1) excluding the endpoint, equipped with the order topology; it satisfies all these separation axioms but fails metrizability due to lacking a countable basis.[6]
Compactness and Connectedness
In the order topology induced by a total order on a set X, the space is connected if and only if X is densely ordered (for any a < b in X, there exists c \in X with a < c < b) and Dedekind-complete (every nonempty subset of X that is bounded above has a least upper bound in X).[7] Such an ordered set is called a linear continuum, and connectedness follows from the fact that any disconnection would correspond to a Dedekind cut without a supremum or a gap in the ordering.[7] For instance, the real numbers \mathbb{R} under the standard ordering form a linear continuum, so the order topology on \mathbb{R} (which coincides with the standard topology) is connected.[7]Order topologies on linear continua are locally connected, as the basis consists of open intervals and rays, each of which is itself a connected subspace (a linear continuum without endpoints or one-sided).[8] This local connectedness arises because small neighborhoods around any point are homeomorphic to open intervals in \mathbb{R}, which are connected.[8]The order topology on a totally ordered set X is compact if and only if X has a minimum and maximum element and satisfies the least upper bound property (Dedekind-completeness).[9] Under these conditions, every open cover has a finite subcover, as the completeness ensures that suprema exist to "cap" chains of basis elements in any cover.[9] For example, the closed interval [0,1] with the order topology is compact, and this topology agrees with the subspace topology from \mathbb{R}, satisfying the Heine-Borel theorem.[9] In contrast, \mathbb{R} itself is not compact in the order topology, as the open cover \{(n, n+2) \mid n \in \mathbb{Z}\} has no finite subcover, mirroring the failure of Heine-Borel for unbounded sets.[10]When X is well-ordered, the order topology is compact if and only if X has a maximum element (i.e., its order type is a successor ordinal). Examples include finite ordinals and [0, \omega] (order type \omega + 1), as well as larger successor ordinals like \omega_1 + 1.[10] In this case, the space is sequentially compact, and since it is Hausdorff, compactness follows; larger well-orderings like [0, \omega_1] (order type \omega_1 + 1? Wait, no: [0, ω₁] has type ω₁ +1, compact; but [0, ω₁) type ω₁, not compact due to uncountable open covers without finite subcovers.[10]The Lindelöf property—that every open cover has a countable subcover—holds in the order topology on spaces of countable ordinals (e.g., [0, \omega] or finite ordinals) but fails for uncountable ordinals like \omega_1, the first uncountable ordinal.[11] For \omega_1, the open cover \{(0, \alpha) \mid \alpha < \omega_1\} requires uncountably many sets to cover the space, as any countable subcollection covers only up to a countable supremum.[11]
One-Sided Variants
Left Ray Topology
The left ray topology on a totally ordered set (X, <) is the topology generated by the subbasis consisting of all left rays (−∞, b) = {x ∈ X | x < b} for b ∈ X. The open sets are arbitrary unions of these left rays, which, owing to their nested structure, are either the empty set, the whole space X, or individual left rays (−∞, b) for some b ∈ X. This construction yields a topology coarser than the standard order topology, as the latter's subbasis includes both left and right rays, producing a larger collection of open sets; consequently, the identity map from (X with the order topology) to (X with the left ray topology) is continuous, but the reverse map is not, since bounded open intervals (a, b) from the order topology are not open in the left ray topology.[12]The finite intersections of subbasis elements are left rays with the minimum upper bound, so the collection of all left rays {(−∞, b) | b ∈ X} ∪ {X} forms a basis for the topology. On the real line ℝ, this topology has open sets ∅, ℝ, and (−∞, b) for b ∈ ℝ, providing a model for one-sided (left) continuity in ordered spaces, where functions that are continuous with respect to left limits can be analyzed using these unbounded open sets. The order topology on X is generated by taking both left rays and right rays as a subbasis. The lower limit topology (Sorgenfrey line) on ℝ is a different refinement with basis [a, b) for a < b ∈ ℝ.[12]The left ray topology satisfies the T0 separation axiom but is not Hausdorff. For distinct points x < y in X, the open set (−∞, y) contains x but not y, satisfying T0, but every open set containing y is of the form (−∞, b) with b > y and thus contains x, preventing separation of x and y by disjoint opens. On the integers ℤ with the usual order, this manifests as points not being separable from points to their left, with any open set containing n containing all m < n. The topology is also not regular, as there exist closed sets and points outside them that cannot be separated by disjoint open sets; for example, in ℝ, the closed set [0, +∞) and the point -1 cannot be separated in this manner.[13][14]
Right Ray Topology
The right ray topology on a totally ordered set (X, <) is the topology generated by taking as a subbasis the collection of all right rays \{(a, +\infty) \mid a \in X \}, where (a, +\infty) = \{x \in X \mid x > a \}, along with the empty set and X itself; the open sets are arbitrary unions of these right rays. Owing to their nested structure, the open sets are either \emptyset, X, or individual right rays (a, +\infty) for some a \in X.[15] This topology is coarser than the order topology, and the identity map from (X with order topology) to (X with right ray topology) is continuous, but not conversely, as bounded intervals are not open in the right ray topology.The finite intersections of subbasis elements are right rays with the maximum lower bound, so the collection of all right rays \{(a, +\infty) \mid a \in X\} \cup \{X\} forms a basis for the topology. On the real line \mathbb{R}, this topology has open sets \emptyset, \mathbb{R}, and (a, +\infty) for a \in \mathbb{R}, providing a model for one-sided (right) continuity in ordered spaces. The upper limit topology on \mathbb{R} is a distinct refinement generated by the basis \{(a, b] \mid a, b \in \mathbb{R}, a < b\}.[16]The right ray topology satisfies the T_0 separation axiom but is not T_1 or Hausdorff. For distinct points x < y in X, the open set (x, +\infty) contains y but not x, satisfying T_0, but every open set containing x is of the form (a, +\infty) with a < x and thus contains y, preventing separation by disjoint opens. Any two non-empty open sets intersect, as they are cofinal to +\infty. On the integers \mathbb{Z} with the usual order, points cannot be separated from points to their right, with any open set containing n containing all m > n. The topology is also not regular; for example, in \mathbb{R}, the closed set (-\infty, 0] and the point $1$ cannot be separated by disjoint open sets.[17]There is a natural duality with the left ray topology: the order-reversing map f: X \to X, or on \mathbb{R} the homeomorphism x \mapsto -x, sends right rays (a, +\infty) to left rays (-\infty, -a), establishing a homeomorphism between the two topologies.[18] On \mathbb{R}, the right ray topology emphasizes right-unbounded opens and is coarser than the standard order topology, as standard open intervals cannot be expressed as unions of right rays without overflowing to +\infty. In contrast to the left ray topology, which emphasizes left-unbounded opens, the right ray topology focuses on right-unbounded ones, and their combination as subbasis generates the full order topology.
Subspace Considerations
Induced Topology on Subsets
In a linearly ordered set (X, \leq) equipped with the order topology \tau_X, consider a subset Y \subseteq X ordered by the induced order \leq_Y. The subspace topology \tau_Y on Y is defined as \tau_Y = \{ U \cap Y \mid U \in \tau_X \}.[19]The subspace topology \tau_Y coincides with the order topology generated by \leq_Y for certain subsets Y. A key case is when Y is convex in X (i.e., an interval, so that if a, b \in Y and a < x < b with x \in X, then x \in Y); in this situation, the open sets in the order topology on Y—such as intervals (a, b)_Y = \{ y \in Y \mid a < y < b \} with a, b \in Y—align precisely with intersections of open intervals from \tau_X with Y.[19] Convexity is sufficient but not necessary for coincidence; for example, the subspace topology on the disconnected set (0,1) \cup (2,3) \subset \mathbb{R} also matches its order topology, as both are unions of open intervals in each component.[20]Another case is when Y is order-dense in X (i.e., between any two elements of X, there lies an element of Y). A representative example is the rational numbers \mathbb{Q} as a subset of the real numbers \mathbb{R}, where \mathbb{R} carries its standard order topology (equivalent to the Euclidean topology). Although \mathbb{Q} is not convex in \mathbb{R}, it is order-dense, and the subspace topology on \mathbb{Q} coincides with the order topology on \mathbb{Q}. Basis elements in the subspace topology, such as (c, d) \cap \mathbb{Q} for c, d \in \mathbb{R}, can be expressed as unions of open intervals (p, q) \cap \mathbb{Q} with p, q \in \mathbb{Q}, which form the basis for the order topology on \mathbb{Q}.[6]To see this more generally, note that the basis for the order topology on Y consists of sets of the form (a, b) \cap Y with a, b \in Y. Each such set is open in \tau_Y because (a, b) \cap Y = ((a, b) \cap X) \cap Y and (a, b) \cap X is open in \tau_X. Conversely, for convex or order-dense Y, every basis element of \tau_Y—namely, (c, d) \cap Y with c, d \in X—contains, for each y \in (c, d) \cap Y, an order basis element (a', b') \cap Y \subseteq (c, d) \cap Y with a', b' \in Y, ensuring the topologies match.[19][6]
Non-Order Subspace Examples
A classic example illustrating the discrepancy between the subspace topology and the induced order topology arises in the real line \mathbb{R} equipped with its standard order topology. Consider the subset Y = [0, 1) \cup \{2\} \subseteq \mathbb{R}. In the subspace topology on Y, the singleton \{2\} is open, as it equals the intersection of Y with the open interval (1.5, 2.5) in \mathbb{R}. However, in the order topology induced on Y by its natural linear order, \{2\} is not open. Any basis element containing 2 takes the form (a, 2] where a \in Y and a < 2; since all such a lie in [0, 1), this interval includes points from [0, 1) arbitrarily close to 1, making it impossible to isolate 2 without including elements from the left accumulation.This failure occurs because the gap between 1 and 2 in Y is recognized by the ambient topology of \mathbb{R}, allowing open sets to separate the isolated point 2 from the accumulating interval [0, 1), whereas the induced order on Y treats the space as a single linear order without "filling" the gap, forcing neighborhoods of 2 to extend leftward into the accumulation at 1.In the context of ordinal spaces, an analogous example is found in the ordered set \omega + 2 = \omega \cup \{\omega, \omega + 1\} with its order topology, considering the subspace Y = \omega \cup \{\omega + 1\} (omitting \omega). In the subspace topology, \{\omega + 1\} is open, as it equals (\omega, \omega + 1] \cap Y, leveraging the ambient basis element that isolates the successor point after the limit ordinal \omega. In contrast, the induced order topology on Y renders \{\omega + 1\} not open, since basis neighborhoods of \omega + 1 are rays (\alpha, \omega + 1] for \alpha < \omega + 1 in Y, which always include cofinal tails of \omega (the natural numbers).The general reason for these mismatches lies in subspaces where accumulation points or gaps do not align with the order intervals of the induced linear order; the ambient topology can exploit gaps or limits external to the subset to create finer open sets, such as isolated points, that the intrinsic order topology cannot replicate without including extraneous accumulation. Consequently, the order topology is not always hereditary: while the subspace topology inherits the ambient structure faithfully, it may fail to coincide with (and can be strictly finer than) the order topology induced on the subset.
Ordinal Applications
Ordinal Spaces
The order topology on an ordinal α is defined on the set [0, α] of all ordinals less than or equal to α, using as a basis the collection of all open intervals (β, γ) = {δ ∈ [0, α] | β < δ < γ} for ordinals β < γ ≤ α, along with the open rays (−∞, γ) = {δ ∈ [0, α] | δ < γ} and (β, +∞) = {δ ∈ [0, α] | β < δ} as a subbasis.[6] This topology makes [0, α] a linearly ordered topological space (LOTS) where the order is preserved.[6]Ordinal spaces are Hausdorff, as the order topology on any linear order separates distinct points with disjoint open intervals.[6] They are first-countable at successor ordinals, where each such point σ = ρ + 1 is isolated via the clopen singleton {σ} = (ρ, σ + 1) ∩ [0, α] (adjusting for the endpoint if necessary), and thus admits a countable local basis consisting of that singleton. At limit ordinals with countable cofinality, points are also first-countable, as a countable cofinal sequence below the point yields a countable local basis of neighborhoods. However, at limit ordinals with uncountable cofinality, such as the endpoint α = ω₁ in [0, ω₁], the space is not first-countable, since any local basis at ω₁ requires uncountably many distinct open rays (β, ω₁] for β < ω₁ to separate the cofinal structure.[21] The space [0, α] is compact for every ordinal α. For instance, [0, ω₁] is compact.[22]Ordinal spaces are scattered, meaning no nonempty subspace is dense-in-itself (or equivalently, every nonempty subspace has an isolated point), a consequence of the well-ordering: in any nonempty subset, the least element is isolated relative to that subset via an open ray from below.[23] This property holds for all ordinals, including uncountable ones like ω₁. The long line provides an example of a non-separable ordinal-like space, constructed as the set ω₁ × [0, 1) equipped with the lexicographic order (comparing first by ordinal component, then by real), and thus the order topology; it is sequentially compact and hereditarily normal but fails separability due to the uncountable "length" without a countable dense subset.[24]Ordinal spaces are zero-dimensional, as every basic open interval (β, γ) is clopen: its complement is the union of the clopen initial segment [0, β] (open as (−∞, β + 1) and closed by well-ordering) and the terminal segment [γ, α] (symmetrically clopen). For countable α, the space is second-countable (hence metrizable and separable), with a countable basis of rational-like intervals in the countable order. Uncountable ordinal spaces exhibit uncountable cellularity: for example, in [0, ω₁), the family of singletons {σ} for all successor ordinals σ < ω₁ forms an uncountable collection of pairwise disjoint nonempty open sets.[25]
Sequences in Ordinals
In the order topology on an ordinal \alpha, the convergence of a sequence (\beta_n)_{n \in \mathbb{N}} to a point \gamma \in \alpha is determined by the basis neighborhoods of \gamma. For a successor ordinal \gamma = \delta + 1, the singleton \{\gamma\} is open, so convergence requires the sequence to be eventually constant at \gamma. For a limit ordinal \gamma, the basis neighborhoods are of the form (\delta, \gamma + 1) for \delta < \gamma, so (\beta_n) converges to \gamma if and only if for every \delta < \gamma, there exists N such that \beta_n > \delta for all n > N. This means the tail of the sequence is unbounded below \gamma in the order, and the limit superior of the sequence equals \gamma.[26]In countable ordinals like \omega, sequences behave similarly to those in the natural numbers with the discrete topology, where convergence is eventual constancy except for increasing sequences approaching limit points such as \omega itself (in \omega + 1). However, in uncountable ordinals such as the first uncountable ordinal \omega_1, countable sequences cannot converge to \omega_1 because any countable collection of ordinals less than \omega_1 has a countable supremum, which is itself a countable ordinal strictly less than \omega_1. Thus, in the space [0, \omega_1], no non-constant sequence from below converges to \omega_1, making \{\omega_1\} sequentially open but not open, illustrating that the topology is not sequential.[26]Ordinal spaces also exhibit interesting sequential compactness properties. The space [0, \omega_1) with the order topology is sequentially compact: every sequence has a convergent subsequence. This follows from the well-ordering, which allows extraction of a non-decreasing subsequence whose supremum lies in [0, \omega_1) and serves as the limit. In contrast, [0, \omega_1] is sequentially compact, as it is a compact Hausdorff space. More generally, the space [0, γ] with the order topology is always sequentially compact, since it is compact.[27][28]