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 subset of a space is a subset of (or a -set in ) if is the intersection of countably many open subsets of . A subset of a space is a subset of (or a -set in ) if is a -set in (equivalently if is the union of countably many closed subsets of ).

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).

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**The Sorgenfrey plane is perfect**

Let denote the Sorgenfrey line, i.e., the real line topologized using the base of half-open intervals of the form . The Sorgenfrey plane is the product space . We show the following:

*Proposition 1*

The Sorgenfrey line is perfect.

**Proof of Proposition 1**

Let be a non-empty subset of . We show that is a -set. Let be the interior of in the usual topology. In other words, is the following set:

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 .

*Proposition 2*

If is perfect and is metrizable, then is perfect.

*Proof of Proposition 2*

Let be perfect. Let be a space with a base such that each , in addition to being a collection of basic open sets, is a discrete collection. The existence of such a base is equivalent to metrizability, a well known result called Bing’s metrization theorem (see Theorem 4.4.8 in [1]). Let be a non-empty open subset of . We show that it is an -set in . For each , there is some open subset of and there is some such that and . Thus is the union of a collection of sets of the form . Thus we have:

for some index set . For each positive integer , let be defined by

For each , let where each is a closed subset of . For each pair of positive integers and , define by

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 .

*Proposition 3*

The Sorgenfrey plane is perfect.

*Proof of Proposition 3*

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 .

Let be a non-empty subset of . We show that is a -set. We assume that is the union of basic open sets of the form . Consider the sets and defined by:

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 .

Let . We claim that is an -set in . Proposition 3 is established when this claim is proved. To get ready to prove this claim, for each , and for each positive integer , let be the half-open square . Then is a local base at the point . For each positive integer , define by

Clearly . We claim that each is closed in . Suppose . In relation to the point , can be broken into several subsets as follows:

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.

Consider . If is disjoint from , then we are done. So assume . Let . Note that and . Now consider the following open set:

The set is an open set containing the point . We claim that . Suppose . Then and . Consider the following set:

Note that is an open subset of . Since , it follows that . Thus is a subset of the interior of (as a subspace of ). We have . It follows that since

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 .

Now consider the case . If is disjoint from , then we are done. So assume . Let . Note that and . Now consider the following open set:

The set is an open set containing the point . We claim that . Suppose . Then and . Consider the following set:

As in the previous case, is an open subset of . Since , it follows that . As before, . We also have a contradiction in that (based on the following)

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 .

**Remarks**

The authors of [2] showed that any finite power of the Sorgenfrey line is perfect. The proof in [2] is an inductive proof: if is perfect, then is perfect. We take the inductive proof in [2] and adapt it for the Sorgenfrey plane. The authors in [2] 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.

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**A non-perfect example**

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.

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**Reference**

- Engelking, R.,
*General Topology, Revised and Completed edition*, Heldermann Verlag, Berlin, 1989. - Heath, R. W., Michael, E.,
*A property of the Sorgenfrey line*, Compositio Math., 23, 185-188, 1971.

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