In mathematics, an order topology is a specific 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"
for all a, b in X. Provided X has at least two elements, this is equivalent to saying that the open intervals
together with the above rays form a base for the order topology. The open sets in X are the sets that are a union of (possibly infinitely many) such open intervals and rays.
A topological space X is called orderable or linearly orderable[1] if there exists a total order on its elements such that the order topology induced by that order and the given topology on X coincide. The order topology makes X into a completely normal Hausdorff space.
The standard topologies on R, Q, Z, and N are the order topologies.
Induced order topology
editIf Y is a subset of X, X a totally ordered set, then Y inherits a total order from X. The set 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 at least as fine as the induced order topology, but they are not in general the same.
For example, consider the subset Y = {−1} ∪ {1/n }n∈N of the rationals. Under the subspace topology, the singleton set {−1} 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.
Example of a subspace of a linearly ordered space whose topology is not an order topology
editThough the subspace topology of Y = {−1} ∪ {1/n }n∈N in the section above is shown not to be 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 {y} 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 open), 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 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 strict total order < 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 \ {−1} = (0,1), then M is connected, so M is dense on itself and has no gaps, in regards to <. If −1 is not the smallest or the largest element of Z, then and separate M, a contradiction. Assume without loss of generality that −1 is the smallest element of Z. Since {−1} is open in Z, there is some point p in M such that the interval (−1,p) is empty, so p is the minimum of M. Then M \ {p} = (0,p) ∪ (p,1) is not connected with respect to the subspace topology inherited from R. On the other hand, the subspace topology of M \ {p} inherited from the order topology of Z coincides with the order topology of M \ {p} induced by <, which is connected since there are no gaps in M \ {p} and it is dense. This is a contradiction.
Left and right order topologies
editSeveral variants of the order topology can be given:
- The right order topology[2] on X is the topology having as a base all intervals of the form , together with the set X.
- The left order topology on X is the topology having as a base all intervals of the form , together with the set X.
The left and right order topologies can be used to give counterexamples in general topology. For example, the left or right order topology on a bounded set provides an example of a compact space that is not Hausdorff.
The left order topology is the standard topology used for many set-theoretic purposes on a Boolean algebra.[clarification needed]
Ordinal space
editFor any ordinal number
together with the natural order topology. These spaces are called ordinal spaces. (Note that in the usual set-theoretic construction of ordinal numbers we have
When
Of particular interest is the case when
- neither [0,
ω 1) or [0,ω 1] is separable or second-countable - [0,
ω 1] is compact, while [0,ω 1) is sequentially compact and countably compact, but not compact or paracompact
Topology and ordinals
editOrdinals as topological spaces
editAny ordinal number can be viewed as a topological space by endowing it with the order topology (indeed, ordinals are well-ordered, so in particular totally ordered). Unless otherwise specified, this is the usual topology given to ordinals. Moreover, if we are willing to accept a proper class as a topological space, then we may similarly view the class of all ordinals as a topological space with the order topology.
The set of limit points of an ordinal
The closed sets of a limit ordinal
Any ordinal is, of course, an open subset of any larger ordinal. We can also define the topology on the ordinals in the following inductive way: 0 is the empty topological space,
As topological spaces, all the ordinals are Hausdorff and even normal. They are also totally disconnected (connected components are points), scattered (every non-empty subspace has an isolated point; in this case, just take the smallest element), zero-dimensional (the topology has a clopen basis: here, write an open interval (
The topological spaces
The space
Ordinal-indexed sequences
editIf
If X is a topological space, we say that an
Ordinal-indexed sequences are more powerful than ordinary (
However, ordinal-indexed sequences are not powerful enough to replace nets (or filters) in general: for example, on the Tychonoff plank (the product space ), the corner point is a limit point (it is in the closure) of the open subset , but it is not the limit of an ordinal-indexed sequence.
See also
editNotes
edit- ^ Lynn, I. L. (1962). "Linearly orderable spaces". Proceedings of the American Mathematical Society. 13 (3): 454–456. doi:10.1090/S0002-9939-1962-0138089-6.
- ^ Steen & Seebach, p. 74
References
edit- Steen, Lynn A. and Seebach, J. Arthur Jr.; Counterexamples in Topology, Holt, Rinehart and Winston (1970). ISBN 0-03-079485-4.
- Stephen Willard, General Topology, (1970) Addison-Wesley Publishing Company, Reading Massachusetts.
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