The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form . The Sorgenfrey line is one of the most important counterexamples in general topology. One of the often recited facts about this counterexample is that the Sorgenfrey plane (the square of the Sorgengfrey line) is not normal. We show that, though far from normal, the Sorgenfrey plane is subnormal.
A space is normal if for any disjoint closed subsets and of , there exist disjoint open subsets and of such that and . A space is subnormal if for any disjoint closed subsets and of , there exist disjoint subsets and of such that and . Clearly any normal space is subnormal. The Sorgenfrey plane is an example of a subnormal space that is not normal.
In the proof of the non-normality of the Sorgenfrey plane in this previous post, one of the two disjoint closed subsets of the Sorgenfrey plane that cannot be separated by disjoint open sets is countable. Thus the Sorgenfrey plane is not only not normal; it is not pseudonormal (also discussed in this previous post). A space is pseudonormal if for any disjoint closed subsets and of (one of which is countable), there exist disjoint open subsets and of such that and . The examples of the Sorgenfrey plane and show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space is pseudonormal but not subnormal (see this previous post for the non-subnormality).
A space is said to be a perfect space if every closed subset of is a subset of (equivalently, every open subset of is an -subset of ). It is clear that any perfect space is subnormal. We show that the Sorgenfrey plane is perfect. There are subnormal spaces that are not perfect (see the example below).
The Sorgenfrey line is perfect.
The real line with the usual topology is perfect. Thus where each is a closed subset of the real line . Since the Sorgenfrey topology is finer than the usual topology, each is also closed in the Sorgenfrey line.
Consider . We claim that is countable. Suppose is uncountable. Since the Sorgenfrey line is hereditarily Lindelof, there exists such that is a limit point of (see Corollary 2 in this previous post). Since , for some . Note that , which means that no point of the open interval can belong to . On the other hand, since is a limit point of , for some , a contradiction. Thus must be countable. It follows that is the union of countably many closed subsets of .
If is perfect and is metrizable, then is perfect.
We claim that each is a discrete collection of sets in the space . Let . Since is discrete, there exists some open subset of with such that can intersect at most one where . Then is an open subset of with such that can intersect at most one set of the form . Then is a closed subset of . It is clear that is the union of over all countably many possible pairs . Thus is an -set in .
The Sorgenfrey plane is perfect.
To get ready for the proof, consider the product spaces and where has the usual topology. By both Proposition 1 and Proposition 2, both and are perfect. Also note that the Sorgenfrey plane topology is finer than the topologies for both and . Thus a closed set in (in ) is also a closed set in . It follows that any -set in (in ) is also an -set in .
Note that is the interior of when is considered as a subspace of . Likewise, is the interior of when is considered as a subspace of . Since both and are perfect, and are in and , respectively. Hence both and are -sets in .
Since , it follows that . We show that for each of these three sets, there is an open set containing the point that is disjoint from the set.
On the other hand, . Hence , a contradiction. Thus the claim that must be true.
The case is symmetrical to the case . Thus by applying a similar argument, there is an open set containing the point that is disjoint from the set .
and on the one hand and . Thus the claim that is true. Take the intersection of the three open sets from the three cases, we have an open set containing that is disjoint from . Thus is closed in and is in .
The authors of  showed that any finite power of the Sorgenfrey line is perfect. The proof in  is an inductive proof: if is perfect, then is perfect. We take the inductive proof in  and adapt it for the Sorgenfrey plane. The authors in  also proved that for a sequence of spaces such that the product of any finite number of these spaces is perfect, the product is perfect. Then is perfect.
Any perfect space is subnormal. Subnormal spaces do not have to be perfect. In fact subnormal non-normal spaces do not have to be perfect. From a perfect space that is not normal (e.g. the Sorgenfrey plane), one can generate a subnormal and non-normal space that is not perfect. Let be a subnormal and non-normal space. Let be a normal space that is not perfectly normal. There are many possible choices for . If a specific example is needed, one can take with the order topology. Let be the disjoint sum (union) of and . The presence of destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint -sets.
Heath, R. W., Michael, E., A property of the Sorgenfrey line, Compositio Math., 23, 185-188, 1971.