Order topology
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In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets.
If X is a totally ordered set, the order topology on X is generated by the subbase of open rays
- [(a, \infty) = \]
- [(-\infty, b) = \]
- [(a,b) = \]
The order topology makes X into a completely normal Hausdorff space.
The standard topologies on R, Q, and N are the order topologies.
Induced order topology
If Y is a subset of X, then Y inherits a total order from X. Y therefore has an order topology, the induced order topology. As a subset of X, Y also has a subspace topology. The subspace topology is always finer than the induced order topology, but they are not in general the same.
For example, consider the subset Y = ∪ n∈N in the rationals. Under the subspace topology, the singleton set is open in Y, but under the induced order topology, any open set containing –1 must contain all but finitely many members of the space.
An example of a subspace of a linearly ordered space whose topology is not an order topology
Though the subspace topology of Y = ∪ n∈N in the section above is shown to be not generated by the induced order on Y, it is nonetheless an order topology on Y; indeed, in the subspace topology every point is isolated (i.e., singleton is open in Y for every y in Y), so the subspace topology is the discrete topology on Y (the topology in which every subset of Y is an open set), and the discrete topology on any set is an order topology. To define a total order on Y that generates the discrete topology on Y, simply modify the induced order on Y by defining -1 to be the greatest element of Y and otherwise keeping the same order for the other points, so that in this new order (call it say <1) we have 1/n <1 –1 for all n∈N. Then, in the order topology on Y generated by <1, every point of Y is isolated in Y.We wish to define here a subset Z of a linearly ordered topological space X such that no total order on Z generates the subspace topology on Z, so that the subspace topology will not be an order topology even though it is the subspace topology of a space whose topology is an order topology.
Let [Z =\\cup\ = \\cup (0,1) ] in the real line. The same argument as before shows that the subspace topology on Z is not equal to the induced order topology on Z, but one can show that the subspace topology on Z cannot be equal to any order topology on Z.
An argument follows. Suppose by way of contradiction that there is some total order < = <1 on Z such that the order topology generated by < is equal to the subspace topology on Z (note that we are not assuming that < is the induced order on Z, but rather an arbitrarily given total order on Z that generates the subspace topology).
Let M=Z\. Note that, as a topological space, M is just the open unit interval (0,1) (referring to the interval taken with respect to the usual order, but in what follows we write (a,b) to refer to the interval taken with respect to the order relation < = <1). Since is open in Z, there is some point p in M such that either -1 < p and the interval (-1,p) is empty or p < -1 and the interval (p,-1) is empty. Without loss of generality we can assume that -1 < p and (-1,p) is empty. Then M\ = A ∪ B, where A and B are open and disjoint connected subsets of M. By connectedness, no point of Z\B can lie between two points of B, and no point of Z\A can lie between two points of A. Therefore, either A < B or B < A (A < B means a < b for all a in A and b in B). Assume without loss of generality that A < B. Then, since is not open in Z, there is a point a in A such that p < a and (p,a)[\subseteq] A Then (-1,a)=[p,a)[\subseteq] ∪ A, so that ∪A is an open subset of M and hence M = (∪A) ∪ B is the union of two disjoint open subsets of M so M is not connected, a contradiction since, as a topological space, M is just the open unit interval.
A space whose topology is an order topology is called a Linearly Ordered Topological Space (LOTS), and a subspace of a linearly ordered topological space is called a Generalized Ordered Space (GO-space). Thus the example Z above is an example of a GO-space that is not a linearly ordered topological space.
Ordinal space
For any ordinal number λ one can consider the spaces of ordinal numbers
- [[0,\lambda) = \\,]
- [[0,lambda] = \\,]
When λ = ω (the first infinite ordinal), the space [0,ω) is just N with the usual topology, while [0,ω] is the one-point compactification of N.
Of particular interest is the case when λ = ω1, the first uncountable ordinal. The element ω1 is a limit point of the subset [0,ω1) even though no sequence of elements in [0,ω1) has the element ω1 as its limit. In particular, [0,ω1] is not first-countable. The subspace [0,ω1) is first-countable however, since the only point without a countable local base is ω1. Some further properties include
- neither [0,ω1) or [0,ω1] is separable or second-countable
- [0,ω1] is compact while [0,ω1) is sequentially compact and countably compact, but not compact or paracompact
Left and right order topologies
Several interesting variants of the order topology can be given:
- The left order topology on X is the topology whose open sets consist of intervals of the form (a, ∞).
- The right order topology on X is the topology whose open sets consist of intervals of the form (−∞, b).
The left order topology is the standard topology used for many set-theoretic purposes on a Boolean algebra.
See also
This article incorporates material from on PlanetMath, which is licensed under the [Text of the GNU Free Documentation LicenseGFDL].
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