Is the Sorgenfrey line separable?
This topological space is known as the Sorgenfrey line. The rational numbers are a dense subset of this topological space, so the Sorgenfrey line is separable.
Is Sorgenfrey a compact line?
We observe also that the Sorgenfrey line cannot be compact (since the usual topology on R is coarser and not compact). However, the Sorgenfrey line is hereditarily Lindelöf, i.e. every subspace is Lindelöf (Exercise).
Is the Sorgenfrey line Lindelof?
The Sorgenfrey line E is Lindelöf. Proof. Let c be a basic open (in E) cover of R. Since R with the usual topology is second-countable, it is Lindelöf.
Is R with lower limit topology locally compact?
with the lower limit topology is not locally compact.
Is the Sorgenfrey line a subspace of a separable space?
This has a proof that for any ordered space separable implies hereditarily separable. And the Sorgenfrey line is a subspace of a separable ordered space (e.g. the double arrow). Thanks for contributing an answer to Mathematics Stack Exchange!
Which is an example of the Sorgenfrey line?
The Sorgenfrey line serves as a counterexample to several topological properties, see, for example, [a3]. For example, it is not metrizable (cf. also Metrizable space) but it is Hausdorff and perfectly normal (cf. also Hausdorff space; Perfectly-normal space ).
Is the Sorgenfrey line a Lindelof space?
The product of two Lindelöf spaces need not be Lindelöf. For example, the Sorgenfrey line is not Lindelöf. In a Lindelöf space, every locally finite family of nonempty subsets is at most countable. A space is hereditarily Lindelöf if and only if every open subspace of it is Lindelöf.
Is the Sorgenfrey half open square topology locally compact?
The Sorgenfrey topology is neither locally compact nor locally connected (cf. also Locally compact space; Locally connected space ). Consider the Cartesian product $X:=\\mathbf R^s imes\\mathbf R^s$ equipped with the product topology , which is called the Sorgenfrey half-open square topology.