A Short Note About The Sorgenfrey Line

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 S \times S 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 X, given \mathcal{A} a collection of subsets of X, we say \mathcal{A} is a network of X if for each open set U \subset{X} and for each p \in {U}, there is some A \in {\mathcal{A}} such that p \in A. The network weight of X, denoted by nw(X), is the least cardinallity of a network of X.

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 S has a countable network, then S \times S 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 S \times S is not normal using the Baire Category Theorem, define H_0 and H_1 as follows. It can be shown that these two closed subsets of S \times S cannot be separated by disjoint open sets.

    H_0=\lbrace{(x,-x): x} \text{ is rational} \rbrace
    H_1=\lbrace{(y,-y): y} \text{ is irrational} \rbrace

Suppose U_0 and U_1 are open subsets of S \times S such that H_0 \subset{U_0} and H_1 \subset{U_1}. It is shown below that U_0 \cap U_1 \ne \varnothing.

Let \mathbb{P} be the set of all irrational numbers and let \mathbb{Q} be the set of all rational numbers. For each p \in {\mathbb{P}}, choose some real number a(p)>0 such that

    W_p=[p,p+a(p)) \times [-p,-p+a(p)) \subset{U_1}

Let P_n=\lbrace{p \in {\mathbb{P}}: a(p)>\frac{1}{n}}\rbrace. Obviously \mathbb{P}=\bigcup \limits_n {P_n}. Since \mathbb{P} is not an F_\sigma subset of \mathbb{R}, there exists z \in \mathbb{Q} and there exists an n such that z is in the closure of P_n in the usual topology of \mathbb{R}. It is shown below that the point (z,-z) is in the closure of U_1 in S \times S. Since (z,-z) \in H_0 \subset U_0, the open set U_0 would have to contain points of U_1. Thus U_0 \cap U_1 \ne \varnothing.

To see that the point (z,-z) is in the closure of U_1 in S \times S, let V be an open set containing the point (z,-z). To make it easier to work with, assume V is of the form

    V=[z,t) \times [-z,-z+t)

for some positive real number t<\frac{1}{2n}. Since (z,-z) is in the Euclidean closure of P_n, there is a p \in P_n such that \lvert z-p \lvert < \frac{t}{10}. It does not matter whether the point p is to the left or right of z, we have the following two observations:

  • The interval [z,z+t) must overlap with the interval [p,p+t). Then pick x in the intersection.
  • The interval [-z,-z+t) must overlap with the interval [-p,-p+t). Then pick y in the intersection.

Immediately, the point (x,y) belongs to the open set V. Consider the following derivations:

    \displaystyle p < x<p+t<p+\frac{1}{2n}<p+\frac{1}{n}<p+a(p)

    \displaystyle -p < y<-p+t<-p+\frac{1}{2n}<-p+\frac{1}{n}<-p+a(p)

The above derivations show that the point (x,y) belongs to the open set W_p as defined above. The open set W_p is chosen to be a subset of U_1. Thus V \cap U_1 \ne \varnothing, establishing that the point (z,-z) is in the closure of U_1 in S \times S. 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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