Regarding the Sorgenfrey Line, we have a couple of points to add in addition to the contents in the previous post on the Sorgenfrey line. We show the following:

- The Sorgenfrey Line does not have a countable network.
- An alternative proof that is not normal.

In point G in the previous post, we prove that the Sorgenfrey line has no countable base. So the result in this post improves on the previous post. In point E in the previous post, we prove the Sorgenfrey plane is not normal using the Jones’ Lemma. The alternative method is to use the Baire Category Theorem.

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**The first bullet point**

Given a space , given a collection of subsets of , we say is a network of if for each open set and for each , there is some such that . The network weight of , denoted by , is the least cardinallity of a network of .

Of interest here are the spaces with countable network. Note that spaces with countable network are Lindelof. Note that the product of two spaces (each with a countable network) also has a countable network. If has a countable network, then would have a countable network and thus Lindelof. So the Sorgenfrey Line has no countable network.

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**The second bullet point**

To prove that is not normal using the Baire Category Theorem, define and as follows. It can be shown that these two closed subsets of cannot be separated by disjoint open sets.

Suppose and are open subsets of such that and . It is shown below that .

Let be the set of all irrational numbers and let be the set of all rational numbers. For each , choose some real number such that

Let . Obviously . Since is not an subset of , there exists and there exists an such that is in the closure of in the usual topology of . It is shown below that the point is in the closure of in . Since , the open set would have to contain points of . Thus .

To see that the point is in the closure of in , let be an open set containing the point . To make it easier to work with, assume is of the form

for some positive real number . Since is in the Euclidean closure of , there is a such that . It does not matter whether the point is to the left or right of , we have the following two observations:

- The interval must overlap with the interval . Then pick in the intersection.
- The interval must overlap with the interval . Then pick in the intersection.

Immediately, the point belongs to the open set . Consider the following derivations:

The above derivations show that the point belongs to the open set as defined above. The open set is chosen to be a subset of . Thus , establishing that the point is in the closure of in . The proof that the Sorgenfrey plane is not normal is now completed.

Note that in using the Baire Category Theorem, a pair of disjoint closed sets is produced. The proof using the Jones Lemma only implies that such a pair exists.

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