Right order topology

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
:$(a,b) = \$
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 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

If ''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'' = ∪ _''n''∈N of 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.

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 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 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 $Z = \ \cup (0,1)$ 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 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'' \  = (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 $(-\infty,-1)$ and $(-1,\infty)$ separate ''M'', a contradiction. Assume without loss of generality that −1 is the smallest element of ''Z''. Since 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'' \  = (0,''p'') ∪ (''p'',1) is not connected with respect to the subspace topology inherited from . On the other hand, the subspace topology of ''M'' \  inherited from the order topology of ''Z'' coincides with the order topology of ''M'' \  induced by <, which is connected since there are no gaps in ''M'' \  and it is dense. This is a contradiction.

Left and right order topologies

Several variants of the order topology can be given:

* The right order topology on ''X'' is the topology having as a base all intervals of the form $(a,\infty)=\$, together with the set ''X''.
* The left order topology on ''X'' is the topology having as a base all intervals of the form $(-\infty,a)=\$, together with the set ''X''.

These topologies naturally arise when working with semicontinuous functions, in that a real-valued function on a topological space is lower semicontinuous if and only if it is continuous when the reals are equipped with the right order.Stromberg, p. 132, Exercise 4 The (natural) compact open topology on the resulting set of continuous functions is sometimes referred to as the ''semicontinuous topology''''.''

Additionally, these 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.

Ordinal space

For any ordinal number ''λ'' one can consider the spaces of ordinal numbers
: ,\lambda) = \
:$[0,\lambda= \$
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 ''λ'' = [0, ''λ'') and ''λ'' + 1 = [0, ''λ'']). Obviously, these spaces are mostly of interest when ''λ'' is an infinite ordinal; for finite ordinals, the order topology is simply the discrete topology.

When ''λ'' = ω (the first infinite ordinal), the space ,ω) is just N with the usual (still discrete) topology, while [0,ωis the Alexandroff_extension">one-point compactification