# 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

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

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.