# The Sorgenfrey plane is subnormal

The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form $[a,b)$. 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 $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. 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 $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The examples of the Sorgenfrey plane and $\omega_1 \times (\omega_1+1)$ show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal but not subnormal (see this previous post for the non-subnormality).

A space $Y$ is said to be a perfect space if every closed subset of $Y$ is a $G_\delta$ subset of $Y$ (equivalently, every open subset of $Y$ is an $F_\sigma$-subset of $Y$). 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 $S$ denote the Sorgenfrey line, i.e., the real line $\mathbb{R}$ topologized using the base of half-open intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x . The Sorgenfrey plane is the product space $S \times S$. We show the following:

Proposition 1
The Sorgenfrey line $S$ is perfect.

Proof of Proposition 1
Let $U$ be a non-empty subset of $S$. We show that $U$ is a $F_\sigma$-set. Let $U_0$ be the interior of $U$ in the usual topology. In other words, $U_0$ is the following set:

$U_0=\left\{x \in U: \exists \ (a,b) \text{ such that } x \in (a,b) \text{ and } (a,b) \subset U \right\}$

The real line with the usual topology is perfect. Thus $U_0=\bigcup_{n=1}^\infty F_n$ where each $F_n$ is a closed subset of the real line $\mathbb{R}$. Since the Sorgenfrey topology is finer than the usual topology, each $F_n$ is also closed in the Sorgenfrey line.

Consider $Y=U-U_0$. We claim that $Y$ is countable. Suppose $Y$ is uncountable. Since the Sorgenfrey line is hereditarily Lindelof, there exists $y \in Y$ such that $y$ is a limit point of $Y$ (see Corollary 2 in this previous post). Since $y \in Y \subset U$, $[y,t) \subset U$ for some $t$. Note that $(y,t) \subset U_0$, which means that no point of the open interval $(y,t)$ can belong to $Y$. On the other hand, since $y$ is a limit point of $Y$, $y for some $w \in Y$, a contradiction. Thus $Y$ must be countable. It follows that $U$ is the union of countably many closed subsets of $S$. $\blacksquare$

Proposition 2
If $X$ is perfect and $Y$ is metrizable, then $X \times Y$ is perfect.

Proof of Proposition 2
Let $X$ be perfect. Let $Y$ be a space with a base $\mathcal{B}=\bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$, 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 $U$ be a non-empty open subset of $X \times Y$. We show that it is an $F_\sigma$-set in $X \times Y$. For each $x \in U$, there is some open subset $V$ of $X$ and there is some $W \in \mathcal{B}$ such that $x \in V \times W$ and $V \times \overline{W} \subset U$. Thus $U$ is the union of a collection of sets of the form $V \times \overline{W}$. Thus we have:

$U=\bigcup \mathcal{O} \text{ where } \mathcal{O}=\left\{ V_\alpha \times \overline{W_\alpha}: \alpha \in A \right\}$

for some index set $A$. For each positive integer $m$, let $\mathcal{O}_m$ be defined by

$\mathcal{O}_m=\left\{V_\alpha \times \overline{W_\alpha} \in \mathcal{O}: W_\alpha \in \mathcal{B}_m \right\}$

For each $\alpha \in A$, let $V_\alpha=\bigcup_{n=1}^\infty V_{\alpha,n}$ where each $V_{\alpha,n}$ is a closed subset of $X$. For each pair of positive integers $n$ and $m$, define $\mathcal{O}_{n,m}$ by

$\mathcal{O}_{n,m}=\left\{V_{\alpha,n} \times \overline{W_\alpha}: V_\alpha \times \overline{W_\alpha} \in \mathcal{O}_m \right\}$

We claim that each $\mathcal{O}_{n,m}$ is a discrete collection of sets in the space $X \times Y$. Let $(a,b) \in X \times Y$. Since $\mathcal{B}_m$ is discrete, there exists some open subset $H_b$ of $Y$ with $b \in H_b$ such that $H_b$ can intersect at most one $\overline{W}$ where $W \in \mathcal{B}_m$. Then $X \times H_b$ is an open subset of $X \times Y$ with $(a,b) \in X \times H_b$ such that $X \times H_b$ can intersect at most one set of the form $V_{\alpha,n} \times \overline{W_\alpha}$. Then $C_{n,m}=\bigcup \mathcal{O}_{n,m}$ is a closed subset of $X \times Y$. It is clear that $U$ is the union of $C_{n,m}$ over all countably many possible pairs $n,m$. Thus $U$ is an $F_\sigma$-set in $X \times Y$. $\blacksquare$

Proposition 3
The Sorgenfrey plane $S \times S$ is perfect.

Proof of Proposition 3
To get ready for the proof, consider the product spaces $X_1=\mathbb{R} \times S$ and $X_2=S \times \mathbb{R}$ where $\mathbb{R}$ has the usual topology. By both Proposition 1 and Proposition 2, both $X_1$ and $X_2$ are perfect. Also note that the Sorgenfrey plane topology is finer than the topologies for both $X_1$ and $X_2$. Thus a closed set in $X_1$ (in $X_2$) is also a closed set in $S \times S$. It follows that any $F_\sigma$-set in $X_1$ (in $X_2$) is also an $F_\sigma$-set in $S \times S$.

Let $U$ be a non-empty subset of $S \times S$. We show that $U$ is a $F_\sigma$-set. We assume that $U$ is the union of basic open sets of the form $[a,b) \times [a,b)$. Consider the sets $U_1$ and $U_2$ defined by:

$U_1=\left\{x \in U: \exists \ (a,b) \times [a,b) \text{ such that } x \in (a,b) \times [a,b) \text{ and } (a,b) \times [a,b) \subset U \right\}$

$U_2=\left\{x \in U: \exists \ [a,b) \times (a,b) \text{ such that } x \in [a,b) \times (a,b) \text{ and } [a,b) \times (a,b) \subset U \right\}$

Note that $U_1$ is the interior of $U$ when $U$ is considered as a subspace of $X_1$. Likewise, $U_2$ is the interior of $U$ when $U$ is considered as a subspace of $X_2$. Since both $X_1$ and $X_2$ are perfect, $U_1$ and $U_2$ are $F_\sigma$ in $X_1$ and $X_2$, respectively. Hence both $U_1$ and $U_2$ are $F_\sigma$-sets in $S \times S$.

Let $Y=U-(U_1 \cup U_2)$. We claim that $Y$ is an $F_\sigma$-set in $S \times S$. Proposition 3 is established when this claim is proved. To get ready to prove this claim, for each $x=(x_1,x_2) \in S \times S$, and for each positive integer $k$, let $B_k(x)$ be the half-open square $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$. Then $\mathcal{B}(x)=\left\{B_k(x): k=1,2,3,\cdots \right\}$ is a local base at the point $x$. For each positive integer $k$, define $Y_k$ by

$Y_k=\left\{y=(y_1,y_2) \in Y: B_k(y) \subset U \right\}$

Clearly $Y=\bigcup_{k=1}^\infty Y_k$. We claim that each $Y_k$ is closed in $S \times S$. Suppose $x=(x_1,x_2) \in S \times S-Y_k$. In relation to the point $x$, $Y_k$ can be broken into several subsets as follows:

$Y_{k,1}=\left\{y=(y_1,y_2) \in Y_k: y_1=x_1 \text{ and } y_2 \ne x_2 \right\}$

$Y_{k,2}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 = x_2 \right\}$

$Y_{k,\varnothing}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 \ne x_2 \right\}$

Since $x \notin Y_k$, it follows that $Y_k=Y_{k,1} \cup Y_{k,2} \cup Y_{k,\varnothing}$. We show that for each of these three sets, there is an open set containing the point $x$ that is disjoint from the set.

Consider $Y_{k,1}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,1}$, then we are done. So assume $B_k(x) \cap Y_{k,1} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,1}$. Note that $t_1=x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_2

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,1}=\varnothing$. Suppose $g \in G \cap Y_{k,1}$. Then $g_1=x_1$ and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

Note that $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. Thus $H$ is a subset of the interior of $U$ (as a subspace of $X_2$). We have $H \subset U_2$. It follows that $t \in H$ since

$x_1=g_1=t_1$

$x_2

On the other hand, $t \in Y_{k,1} \subset Y_k \subset Y$. Hence $t \notin U_2$, a contradiction. Thus the claim that $G \cap Y_{k,1}=\varnothing$ must be true.

The case $Y_{k,2}$ is symmetrical to the case $Y_{k,1}$. Thus by applying a similar argument, there is an open set containing the point $x$ that is disjoint from the set $Y_{k,2}$.

Now consider the case $Y_{k,\varnothing}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,\varnothing}$, then we are done. So assume $B_k(x) \cap Y_{k,\varnothing} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,\varnothing}$. Note that $t_1>x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,\varnothing}=\varnothing$. Suppose $g \in G \cap Y_{k,\varnothing}$. Then $x_1 and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

As in the previous case, $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. As before, $H \subset U_2$. We also have a contradiction in that $t \in H$ (based on the following)

$x_1

$x_2

and on the one hand and $t \in Y_{k,\varnothing} \subset Y=U-(U_1 \cup U_2)$. Thus the claim that $G \cap Y_{k,\varnothing}=\varnothing$ is true. Take the intersection of the three open sets from the three cases, we have an open set containing $x$ that is disjoint from $Y_k$. Thus $Y_k$ is closed in $S \times S$ and $Y=\bigcup_{k=1}^\infty Y_k$ is $F_\sigma$ in $S \times S$ . $\blacksquare$

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 $S^n$ is perfect, then $S^{n+1}$ 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 $X_1,X_2,X_3,\cdots$ such that the product of any finite number of these spaces is perfect, the product $\prod_{n=1}^\infty X_n$ is perfect. Then $S^\omega$ 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 $X$ be a subnormal and non-normal space. Let $Y$ be a normal space that is not perfectly normal. There are many possible choices for $Y$. If a specific example is needed, one can take $Y=\omega_1$ with the order topology. Let $X \bigoplus Y$ be the disjoint sum (union) of $X$ and $Y$. The presence of $Y$ destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint $G_\delta$-sets.

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Reference

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

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$\copyright \ 2014 \text{ by Dan Ma}$

# Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line $S=\mathbb{R}$ where the topology is generated by a base consisting the half open intervals of the form $[a,b)$. The Sorgenfrey plane is the square $S \times S$.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let $X$ be a regular space. A development for $X$ is a sequence $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, and for each open subset $U$ of $X$ with $x \in U$, there exists one cover $\mathcal{G}_n$ satisfying the condition that for any open set $V \in \mathcal{G}_n$, $x \in V \Rightarrow V \subset U$. When $X$ has a development, $X$ is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$, the open sets in $\mathcal{G}_n$ are considered “smaller” as the index $n$ increases. In fact, this is how a development is defined for a metric space, where $\mathcal{G}_n$ consists of all open balls with diameters less than $\frac{1}{n}$. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a $G_\delta$-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a $G_\delta$-diagonal (the diagonal $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$). It is a well known theorem that every compact space with a $G_\delta$-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that $X$ is compact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a finite subcover $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a countable base for $X$. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that $X$ is paracompact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a locally finite open refinement $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a $\sigma$-locally finite base for $X$. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a $\sigma$-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a $\sigma$-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

Bing’s Metrization Theorem
A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space $X$ is said to be collectionwise normal if $X$ is a $T_1$-space and for every discrete collection $\left\{W_\alpha: \alpha \in A \right\}$ of closed sets in $X$, there exists a discrete collection $\left\{V_\alpha: \alpha \in A \right\}$ of open subsets of $X$ such that $W_\alpha \subset V_\alpha$. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# The Euclidean topology of the real line (1)

In this post we discuss the Euclidean topology of the real line as a way of motivating the notion of topological space. We also contrast the Euclidean topology with the Sorgenfrey Line, another topology that can be defined on the real number line. In the comparison of the two topologies, we focus our attention on continuous functions. For some (the author of this post included), exposure to the Euclidean topology (especially the real line) is a gateway to the study of general topology.

The Real Line
Let $\mathbb{R}$ be the set of all real numbers. For any two real numbers $a$ and $b$ with $a, by the open interval $(a,b)$, we mean the set of all real numbers $x$ such that $a. Let $U$ be a subset of $\mathbb{R}$. The set $U$ is said to be an open set if for each $x \in U$, there is an open interval $(a,b)$ containing $x$ such that $(a,b) \subset U$. A set $C \subset \mathbb{R}$ is said to be a closed set if the complement $\mathbb{R}-C$ is an open set.

By definition, all open intervals $(a,b)$ are open sets. The set $(0,1) \cup (2,3)$ is an open set. Note that it is the union of two open intervals. The set of all positive numbers $(0,\infty)$ is open. Note that $(0,\infty)=\bigcup \limits_{x>0}(0,x)$.

Let $\tau$ be the set of all subsets of $\mathbb{R}$ that are open sets. The set $\tau$ is called a topology. To distinguish this topology from other topologies that may be defined on the real line, we call the topology just defined the Euclidean topology or the usual topology on the real line. We have some observations about open sets in the real line (stated in Theorem $1$).

Theorem 1

1. Both $\mathbb{R}$ and the empty set $\phi$ are open sets.
2. The union of open sets is an open set.
3. The intersection of finitely many open sets is an open set.

Often times we work with subsets of the real lines. A few examples are: $\mathbb{N}$, $\mathbb{Q}$, $\mathbb{P}$, $[0,1]$. These are the set of all nonnegative integers, the set of all rational numbers, the set of all irrational numbers and the unit interval, respectively. Given a subset $Y \subset \mathbb{R}$, $Y$ can also have a topology. The open sets in the subspace $Y$ are inherited from the overall real line (i.e. they are simply the open sets in the overall real line but points not in $Y$ are excluded). For example, for each nonnegative integer $x$, the singleton set $\lbrace{x}\rbrace$ is obviously not an open set in the real line. However $\lbrace{x}\rbrace$, where $x \in \mathbb{N}$, is an open set in the space $\mathbb{N}$. This is because we have:

$\mathbb{N} \cap (x-0.1,x+0.1)=\lbrace{x}\rbrace$

Since every single point in $\mathbb{N}$ is an open set, the space $\mathbb{N}$ is said to be a discrete space. An open interval in $\mathbb{Q}$ is simply $(a,b) \cap \mathbb{Q}$ (i.e. the set of all rational bumbers $x$ with $a). Note that every open interval contains infinitely many rational numbers. Thus $\mathbb{Q}$ is not a discrete space. For the unit interval $[0,1]$, an open interval containing the point $1$ is $(a,1]=[0,1] \cap (a,b)$ where $0. Likewise, an open interval containing the left endpoint $0$ is $[0,c)$. In general, for the subset $Y \subset \mathbb{R}$, the open sets are generated by the open intervals of the form $(a,b) \cap Y$. The set of all open sets just defined for the subset $Y$ is called the relative topology for $Y$ since it is inherited from the topology for the overall real line.

The notion of continuous functions is a topological one. For some students, the notion of continuous functions is introduced in courses such as calculus and elementary analysis where the $\epsilon \delta$-defintion is used. We consider continuous functions from a topological point of view.

Defintion 1
Let’s see how continuous function is defined in a typical calculus text such as $[1]$. The function $f: (s,t) \rightarrow \mathbb{R}$ is continuous at $a \in (s,t)$ if $\lim \limits_{x \rightarrow a} f(x)=f(a)$. For the definition of $\lim \limits_{x \rightarrow a} F(x)=L$, we have the following from $[1]$:

The number $L$ is the limit of $F(x)$ as $x$ approaches $a$ provided that, given any number $\epsilon>0$, there exists a number $\delta>0$ such that $\lvert F(x)-L \lvert<\epsilon$ for all $x$ such that $0<\lvert x-a \lvert <\delta$.

Defintion 2
Let $X \subset \mathbb{R}$ and let $f: X \rightarrow \mathbb{R}$ be a function. The function $f$ is said to be continuous at $x \in X$ if for each open interval $(a,b)$ containing $f(x)$, there is some open interval $(c,d)$ containing $x$ such that $f(X \cap (c,d)) \subset (a,b)$. The function $f$ is said to be a continuous function if it is continuous at every $x \in X$.

Of course, definition $2$ is equivalent to defintion $1$. In definition $2$, if $f(x)$ is the midpoint of the open interval $(a,b)$, then $\epsilon=0.5(b-a)$ corresponds to the $\epsilon$ in definition $1$. Likewise $\delta=0.5(d-c)$ is the $\delta$ in definition $1$.

Defintion $1$ is wedded to the Euclidean metric in the real line. When considering functions defined on a higher dimensional Euclidean space or another type of spaces, defintion $1$ will have to be rewired so to speak. It also feels cluttered, not to mention being confusing to the typical students in a calculus class. Defintion $2$, as we will see below, is applicable in a wide variety of settings. For example, when the topology changes, just replace with the new notion of open intervals in definition $2$. We have the following characterizations of continuous functions:

Theorem 2
Let $X \subset \mathbb{R}$ and let $f: X \rightarrow \mathbb{R}$ be a function. Then the following conditions are equivalent:

1. The function $f$ is continuous.
2. For each open set $U \subset \mathbb{R}$, the inverse image $f^{-1}(U)$ is an open set in the domain space $X$.
3. For each closed set $C \subset \mathbb{R}$, the inverse image $f^{-1}(C)$ is a closed set in the domain space $X$.

Example 1
Some familiar examples of continuous functions include polynomial functions with real coefficients, trigonometric functions such as $sin(x)$ and $cos(x)$ as well as exponential functions such as $e^x$ and logarithmic functions such as $ln(x)$.

Example 2
The following function $F(x)$ is not continuous at $x=0$.

$\displaystyle F(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace x\ge0\end{matrix}\right.$

Example 3
The following function $G(x)$ is not continuous at every $x \in \mathbb{R}$.

$\displaystyle G(x)=\left\{\begin{matrix}1&\thinspace \text{x is a rational number}\\{0}&\thinspace \text{x is an irrational number}\end{matrix}\right.$

Example 4
Let $A=\lbrace{x \in \mathbb{R}: x \ne 0}\rbrace$. Define $H(x):A \rightarrow \mathbb{R}$ by $H(x)=\frac{1}{x}$. The function $H(x)$ is continuous at every $x \in A$. Thus it is a continuous function (as a function defined over $A$). However, it cannot be extended to a continuous function over the entire real line.

The Sorgenfrey Line
We now modify the definition of open intervals to be of the form $[a,b)$ where $a. In other words, the open intervals include the left endpoint. As before, a set $U \subset \mathbb{R}$ is open if for each point $x \in U$, there is an open interval $[a,b)$ containing $x$ such that $[a,b) \subset U$. Theorem $1$ still hold true with this definition of open sets defined in the real line. Let $\tau_s$ be the set of all open sets generated by the open intervals $[a,b)$. The set $\tau_s$ is called the Sorgenfrey topology of the real line. Thus we have two topologies defined on $\mathbb{R}$, $\tau$ and $\tau_s$. To avoid confusion, when we discuss the both types of open sets at the same time, we refer to the open sets in $\tau$ (defined by the open intervals $(a,b)$) as the Euclidean open sets or the usual open sets. We refer to the open sets in $\tau_s$ (defined by the open intervals $[a,b)$) as the Sorgenfrey open sets.

How are the Sorgenfrey open sets different from the usual open sets? First off, every usual open set is a Sorgenfrey open set (i.e. $\tau \subset \tau_s$). To see this, for each $x \in (a,b)$, we have $[x,t) \subset (a,b)$ where $x. Thus $(a,b)$ is a Sorgenfrey open set. As a result, the union of however many $(a,b)$ is also a Sorgenfrey open set.

Another dramatic difference is the notion of continuous functions. Consider the function $F(x)$ in Example $2$ above. Even though $F(x)$ is not continuous at $x=0$ with respect to the usual open sets, it is continuous with respect to the Sorgenfrey open sets. Note that for the Sorgenfrey open intervals $[0,c)$ of $x=0$, only points on the right side of $x=0$ are relevant. Thus, continuous functions with repsect to the Sorgenfrey open sets are called right continuous functions (or upper continuous). A related observation is that with respect to the usual open sets, we can have sequences of points converge to $x \in \mathbb{R}$ from either side. With respect to Sorgenfrey open sets, we can only have points converge to $x \in \mathbb{R}$ from the right.

Another difference is that the usual open sets are generated by a metric (a function indicating the distance between any two points). The metric in question is the Euclidean metric ($\lvert x-y \lvert$ being the distance between $x$ and $y$). On the other hand, the Sorgenfrey open sets cannot be generated by the Euclidean metric or any other metric. This is a deeper result (see another post A Note On The Sorgenfrey Line in this blog).

Topological Spaces
We have just presented two examples of topological spaces, both defined on the space of the real numbers. In general, a topological space is a pair $(X, \tau)$ where $X$ is the underlying space and $\tau_0$ is the collection of all open sets, which is called a topology. In our first example, $X=\mathbb{R}$ and $\tau_0=\tau$ is the set of all open sets generated by the open intervals $(a,b)$, which is called the usual topology (or the Euclidean topology of the real line). In the second example, $X=\mathbb{R}$ and $\tau_0=\tau_s$ is the set of all open sets generated by the open intervals $[a,b)$, which is called the Sorgenfrey topology. The real line with the Sorgenfrey topology is called the Sorgenfrey line.

Any topological space $(X, \tau_0)$ satisfies the same three conditions stated in Theorem $1$. These three conditions are usually part of the definition of a topological space. So we restate these in the following definition.

Definition 3
A topological space is a pair $(X, \tau_0)$ such that $X$ is a set of points and $\tau_0$ is a collection of subsets of $X$ satisfying the following three conditions:

1. Both $X$ and the empty set $\phi$ belong to $\tau_0$,
2. Any union of elements of $\tau$ is also an element of $\tau_0$,
3. The intersection of finitely many elements of $\tau_0$ is also an element of $\tau_0$.

In many instances, topological spaces are defined by decalring what the “open intervals” are. Then the open sets are simply the unions of open intervals. The set of all “open intervals” is called a base for a topology. Let $X$ be a set. Let $\mathcal{B}$ be a set of subsets of $X$. If $\mathcal{B}$ satisfies the following two conditions, $\mathcal{B}$ is a base for a topology on $X$.

• The set $\mathcal{B}$ is a cover of $X$ (i.e. every point of $X$ belongs to some element in $\mathcal{B}$).
• If $B_1,B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there is some $B_3 \in \mathcal{B}$ such that $x \in B_3$ and $B_3 \subset B_1 \cap B_2$.

The set of all usual open intervals $(a,b)$ forms a base for the Euclidean topology of the real line. The set of all open intervals $[a,b)$ forms a base for the Sorgenfrey topology on the real line. The topology text of Willard is an excellent reference (see $[2]$).

Reference

1. Edwards, C. H. and Penny, D. E., Calculus with Analytic Geometry, Fifth Edition, 1998, Prentice Hall, Upper Saddle River, New Jersey.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# Spaces With Countable Network

The concept of network is a useful tool in working with generalized metric spaces. A network is like a base for a topology, but the members of a network do not have to be open. After a brief discussion on network, the focus here is on the spaces with networks that are countably infinite in size. The following facts are presented:

1. Any space with a countable network is separable and Lindelof.
2. The property of having a countable network is hereditary. Thus any space with a countable network is hereditarily separable and hereditarily Lindelof.
3. The property of having a countable network is preserved by taking countable product.
4. The Sorgenfrey Line is an example of a hereditarily separable and hereditarily Lindelof space that has no countable network.
5. For any compact space $X$, $nw(X)=w(X)$. In particular, any compact space with a countable network is metrizable.
6. As a corollary to 5, $w(X) \leq \vert X \vert$ for any compact $X$.
7. A space $X$ has a countable network if and only if it is the continuous impage of a separable metric space (hence such a space is sometimes called cosmic).
8. Any continuous image of a cosmic space is cosmic.
9. Any continuous image of a compact metric space is a compact metric space.
10. As a corollary to 2, any space with countable network is perfectly normal.
11. An example is given to show that the continuous image of a separable metric space needs not be metric (i.e. an example of a cosmic space that is not metrizable).

All spaces in this discussion are at least $T_3$ (Hausdorff and regular). Let $X$ be a space. A collection $\mathcal{N}$ of subsets of $X$ is said to be a network for $X$ if for each $x \in X$ and for each open $U \subset X$ with $x \in U$, then we have $x \in N \subset U$ for some $N \in \mathcal{N}$. The network weight of a space $X$, denoted by $nw(X)$, is defined as the minimum cardinality of all the possible $\vert \mathcal{N} \vert$ where $\mathcal{N}$ is a network for $X$. The weight of a space $X$, denoted by $w(X)$, is defined as the minimum cardinality of all possible $\vert \mathcal{B} \vert$ where $\mathcal{B}$ is a base for $X$. Obviously any base is also a network. Thus $nw(X) \leq w(X)$. For any compact space $X$, $nw(X)=w(X)$. On the other hand, the set of singleton sets is a network. Thus $nw(X) \leq \vert X \vert$.

Our discussion is based on an important observation. Let $\mathcal{T}$ be the topology for the space $X$. Let $\mathcal{K}=nw(X)$. We can find a base $\mathcal{B}_0$ that generates a weaker (coarser) topology such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$. We can also find a base $\mathcal{B}_1$ that generates a finer topology such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$. These are restated as lemmas.

Lemma 1. We can define base $\mathcal{B}_0$ that generates a weaker (coarser) topology $\mathcal{S}_0$ on $X$ such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$. Thus $w(X,\mathcal{S}_0) \leq nw(X)$.

Proof. Let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$. Consider all pairs $N_0,N_1 \in \mathcal{N}$ such that there exist disjoint $O_0,O_1 \in \mathcal{T}$ with $N_0 \subset O_0$ and $N_1 \subset O_1$. Such pairs exist because we are working in a Hausdorff space. Let $\mathcal{B}_0$ be the collection of all such open sets $O_0,O_1$ and their finite interections. This is a base for a topology and let $\mathcal{S}_0$ be the topology generated by $\mathcal{B}_0$. Clearly, $\mathcal{S}_0 \subset \mathcal{T}$ and this is a Hausdorff topology. Note that $w(X,\mathcal{S}_0) \leq \vert \mathcal{B}_0 \vert =\vert \mathcal{N} \vert$.

Lemma 2. We can define base $\mathcal{B}_1$ that generates a finer topology $\mathcal{S}_1$ on $X$ such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$. Thus $w(X,\mathcal{S}_1) \leq nw(X)$.

Proof. As before, let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$. Since we are working in a regular space, we can assume that the sets in $\mathcal{N}$ are closed. If not, take closures of the elements of $\mathcal{N}$ and we still have a network. Consider $\mathcal{B}_1$ to be the set of all finite intersections of elements in $\mathcal{N}$. This is a base for a topology on $X$. Let $\mathcal{S}_1$ be the topology generated by this base. Clearly, $\mathcal{T} \subset \mathcal{S}_1$. It is also clear that $w(X,\mathcal{S}_1) \leq nw(X)$. The only thing left to show is that the finer topology is regular. Note that the network $\mathcal{N}$ consists of closed sets in the topology $\mathcal{T}$. Thus the sets in the base $\mathcal{B}_1$ also consists of closed sets with respect to $\mathcal{T}$ and the sets in $\mathcal{B}_1$ are thus closed in the finer topology. Since $\mathcal{B}_1$ is a base consisting of cloased and open sets, the topology $\mathcal{S}_1$ regular.

Discussion of 1, 2, and 3
Points 1, 2 and 3 are basic facts about countable network and they are easily verified based on definitions. They are called out for the sake of having a record.

Discussion of 4
The Sorgenfrey Line does not have a countable network for the same reason that the Sorgenfrey Plane is not Lindelof. If the Sorgenfrey Line has a countable netowrk, then the Sorgenfrey plane would have a countable network and hence Lindelof.

Discussion of 5
In general, $nw(X) \leq w(X)$. In a compact Hausdorff space, any weaker Hausdorff topology must conincide with the original topology. So the weaker topology produced in Lemma 1 must coincide with the original topology. In the countable case, any compact space with a countable network has a weaker topology with a countable base. This weaker topology must coincide with the original topology.

Discussion of 6
Note that $nw(X) \leq \lvert X \lvert$ always holds. For compact spaces, we have $w(X)=nw(X) \leq \lvert X \lvert$.

Discussion of 7
Let $X$ be a space with a countable network. By Lemma 2, $X$ has a finer topology that has a countable base. Let $Y$ denote $X$ with this finer second countable topology. Then the identity map from $Y$ onto $X$ is continuous.

For the other direction, let $f:Y \rightarrow X$ be a continuous function mapping a separable metric space $Y$ onto $X$. Let $\mathcal{B}$ be a countable base for $Y$. Then $\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a network for $X$.

Discussion of 8
This is easily verified. Let $X$ is the continuous image of a cosmic space $Y$. Then $Y$ is the continuous image of some separable metric space $Z$. It follows that $X$ is the continuous image of $Z$.

Discussion of 9
Let $X$ be compact metrizable and let $Y$ be a continuous image of $X$. Then $Y$ is compact. By point 7, $Y$ has a countable network. By point 5, $Y$ is metrizable.

Discussion of 10
A space is perfectly normal if it is normal and that every closed subset is a $G_\delta-$set. Let $X$ be a space with a countable network. The normality of $X$ comes from the fact that it is regular and Lindelof. Note that $X$ is also hereditarily Lindelof. In a hereditarily Lindelof and regular space, every open subspace is an $F_\sigma-$set (thus every closed set is a $G_\delta-$set.

Discussion of 11 (Example of cosmic but not separable metrizable space)
This is the “Butterfly” space or “Bow-tie” space due to L. F. McAuley. I found this example in [Michael]. Let $Y=T \cup S$ where
$T=\lbrace{(x,y) \in \mathbb{R}^2:y>0}\rbrace$ and
$S=\lbrace{(x,y) \in \mathbb{R}^2:y=0}\rbrace$.

Points in $T$ have the usual plane open neighborhoods. A basic open set at $p \in S$ is of the form $B_c(p)$ where $B_c(p)$ consists of $p$ and all points $q \in Y$ having distance $ from $p$ and lying underneath either one of the two straight lines in $Y$ which emanate from $p$ and have slopes $+c$ and $-c$, respectively.

It is clear that $Y$ is a Hausdorff and regular space. The relative “Bow-tie” topologies on $T$ and $S$ coincide with the usual topology on $T$ and $S$, respectively. Thus the union of the usual countable bases on $T$ and $S$ would be a countable network for $Y$. On the other hand, $Y$ is separable but cannot have a countable base (hence not metrizable).

Reference
[Michael]
Michael, E., $\aleph_0-$spaces, J. Math. Mech. 15, 983-1002.

# The Lexicographic Order and The Double Arrow Space

This post is a basic discussion on the unit square with the lexicographic order and one of its subspace called the double arrow space. The goal is to establish several basic facts of these two spaces. Both of these spaces are compact non-metrizable spaces. The double arrow space consists of the top and bottom edges of the unit square. The double arrow has a strong connection to the Sorgenfrey Line and its square is an example of a space that is normal that is not completely normal. The unit square with the lexicographic order is an example of a completely normal space that is not perfectly normal.

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Defining the spaces

Let $I$ be the unit interval $[0,1]$. Let $\mathcal{X}$ be $I\times I$. Consider the lexicographic order on $\mathcal{X}$. This is the linear order $<$ defined by letting $(a,b)<(c,d)$ whenever $a or $(a=c$ and $b. The goal is to consider the square with the topology induced by this linear order. For $p=(a,b)$ and $q=(c,d)$, $$ is the notation in this post for open intervals in the unit square.

To facilitate further discussion, let’s look at the following subsets of $\mathcal{X}$.

$\mathcal{A}_0=I \times \lbrace{0}\rbrace$
$\mathcal{A}_1=I \times \lbrace{1}\rbrace$
$\mathcal{B}_0=\lbrace{0}\rbrace \times I$
$\mathcal{B}_1=\lbrace{1}\rbrace \times I$
$\mathcal{S}_0=(0,1) \times \lbrace{0}\rbrace$
$\mathcal{S}_1=(0,1) \times \lbrace{1}\rbrace$
$\mathcal{A}=\mathcal{A}_0 \cup \mathcal{A}_1$

The subspace $\mathcal{A}$ is called the double arrow space. In this post, my aim is to establish some basic facts about the double arrow space $\mathcal{A}$ and then the unit square $\mathcal{X}$.

Double arrow space

A. $\mathcal{A}$ is compact.
B. $\mathcal{A}$ is hereditarily Lindelof and hereditarily separable.
C. $\mathcal{A}$ is perfectly normal.
D. $\mathcal{A}^2$ is not hereditarily normal.

Unit square with lexicographic order

E. $\mathcal{X}$ is compact.
F. $\mathcal{X}$ is first countable.
G. $\mathcal{X}$ is not separable.
H. $\mathcal{X}$ does not have the countable chain condition (ccc).
I. $\mathcal{X}$ is not perfectly normal.

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Open intervals in these spaces

Before proving the results stated above, let’s make sense of the open intervals in $\mathcal{X}$ and then in $\mathcal{A}$. The open intervals for points in $\mathcal{X}-\mathcal{A}$ are essentially the usual open intervals. For example, for the point $(0.5,0.5)$, the following is an open interval in $\mathcal{X}$.

$\lbrace{0.5}\rbrace \times (0.25,0.75)$

Because of this, $\mathcal{X}$ has an uncountable collection of pairwise disjoint open intervals (thus does not have ccc and is not separable). Also, a vertical line in $\mathcal{X}$ of the form $\lbrace{t}\rbrace \times I$ is a homeomorphic copy of the unit interval $[0,1]$ with the usual topology.

The more interesting open neighborhoods are for the points in the top and bottom lines (i.e. the double arrow space). Let’s look an example of an open interval in $\mathcal{X}$ for the point $(0.5,1)$. Consider the open interval $$ where $p=(0.5,0.9)$ and $q=(0.6,0.2)$. It follows that the open interval $=L \cup M \cup R$ where

$L=\lbrace{0.5}\rbrace \times (0.9,1]$

$M=(0.5,0.6) \times I$

$R=\lbrace{0.6}\rbrace \times [0,0.2)$

Note that $M$ is the vertical strip in the middle, $L$ is the left edge, and $R$ is the right edge. If you look at $\cap{\mathcal{A}}$, then the open interval for the point $(0.5,1)$ becomes

$\biggl((0.5,0.6] \times \lbrace{0}\rbrace \biggr) \cup \biggl( [0.5,0.6) \times \lbrace{1}\rbrace \biggr)$

The above is also an open interval in $\mathcal{A}$ containing the point $(0.6,0)$. Similarly, an example of an open interval containing the point $(0.5,0)$ is:

$\biggl( (0.4,0.5] \times \lbrace{0}\rbrace \biggr) \cup \biggl([0.4,0.5) \times \lbrace{1}\rbrace\biggr)$

Based on the above examples of open intervals, open intervals in $\mathcal{A}$ have the following form, hence the term “double arrow”.

$\biggl((a,b] \times \lbrace{0}\rbrace\biggr) \cup \biggl([a,b) \times \lbrace{1}\rbrace\biggr)$

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Proofs

The double arrow space contains two copies of the Sorgenfrey line. So knowing the basic properties of the Sorgenfrey line will help follow the proofs of A through D below. Information about the Sorgenfrey line can be found in the following blog post.

Proof of A. I now proof that the double arrow space $\mathcal{A}$ is compact. I do so by showing that it is Lindelof and countably compact. A space is countably compact if any countable open cover has a finite subcover (equivalently, any infinite subspace has a limit point). Here, limit point is in the topological sense (i.e. the point $p$ is a limit point of the set $K$ if any open set containing $p$ contains a point of $K$ different from $p$).

Note that $\mathcal{S}_0$ and $\mathcal{S}_1$ as defined above are homeomorphic copies of the Sorgenfrey Line. Thus $\mathcal{A}$ is made of of two copies of the Sorgenfrey Line plus 4 points. This shows that $\mathcal{A}$ is hereditarily Lindelof.

Pick an infinite subset $W$ of $\mathcal{A}$. Then either the top or the bottom contains infinitely many points of $W$. Assume the top does. There is a point $(q,1)$ that is a limit point of $W \cap \mathcal{A}_1$ in the usual topology. If the point $(q,1)$ is a right-sided limit point of $W \cap \mathcal{A}_1$ in the usual topology, then the point $(q,1)$ is a limit point of $W$ in the order topology. If the point $(q,1)$ is a left-sided limit point of $W \cap \mathcal{A}_1$ in the usual topology, then the point $(q,0)$ is a limit point of $W$ in the order topology.

Proof of B, C and D. As observed in the proof of A, the double arrow space is the union of 2 copies of the Sorgenfrey Line plus 4 points. Thus it is hereditarily Lindelof, hereditarily separable and perfectly normal.

Note that $\mathcal{A}^2$ contains a copy of the Sorgenfrey Plane. Thus it is not hereditarily normal. This also implies that both the double arrow space and the unit square with the lexicographic order are not metrizable.

Proof of E. Let $\mathcal{U}$ be an open cover for $\mathcal{X}$ that consists of open intervals. Since $\mathcal{A}$ is compact, there exists a finite collection of open intervals $\lbrace{U_0,U_1,...,U_n}\rbrace$ from $\mathcal{U}$ that cover $\mathcal{A}$. Based on the above observation about open intervals of points in the double arrow space, the open intervals $U_i$ cover all the points in $\mathcal{X}-\mathcal{A}$ except possibly the left and right edges of the open intervals $U_i$. With the topology inherited from the order topology, each of the edges is homeomorphic to the unit interval with the usual topology. Thus there are finitely many open intervals from $\mathcal{U}$ that cover these left and right edges. Thus $\mathcal{U}$ has a finite subcover and $\mathcal{X}$ is compact.

Proof of F, G, H. These are clear based on the above observation made about the open intervals in the unit square.

Proof of I. I show here that $\mathcal{A}$ cannot be a $G_\delta$ set in the unit square $\mathcal{X}$. For each $n<\omega$, let $U_n$ be an open set in $\mathcal{X}$ such that $\mathcal{A} \subset U_n$. Since $\mathcal{A}$ is compact, we can assume that $U_n$ is the union of finitely many open intervals. Based on the observation given above, each $U_n$ covers all of $\mathcal{X}-\mathcal{A}$ except for finitely many vertical lines (the left and right edges of these open intervals). Pick one vertical line that is not one of the vertical edges from the open intervals $U_n$. Clearly this vertical line is covered by $U_n$ for each $n<\omega$. Thus $\mathcal{A}$ is not a $G_\delta$ set in $\mathcal{X}$.

Comment. The double arrow space is made up of 2 copies of the Sorgenfrey Line and the 4 corner points. It follows that it is hereditarily Lindelof and perfectly normal. However, since the square of the double arrow space contains a copy of the Sorgenfrey Plane, the square of the double arrow space is not hereditarily normal, thus showing that normality is not a hereditary notion. The square of the double arrow space $\mathcal{A}^2$ is a handy example of a normal space that is not completely normal. This implies both the double arrow space and the unit square with the lexicographic order not metrizable. The space $\mathcal{A}^2$ also demonstrates that hereditarily normality is not preserved by Cartesian product.

The unit square with the lexicographic order topology is completely normal ([Steen & Seebach]). Thus it is an example of a space that completely normal (T4) but not perfectly normal (T5).

Reference

[Steen & Seebach]
Steen, L. A. and Seebach, J. A., [1995] Counterexamples In Topology, Dover Edition

# 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.

____________________________________________________________________

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.

# A Note On The Sorgenfrey Line

The Sorgenfrey Line is a topological space whose underlying space is the real line. The topology is generated by the basis of the half open intervals $[a, b)$ where $a$ and $b$ are real numbers. For students of topology, the Sorgenfrey Line is a handy example of (1) “Lindelof x Lindelof” does not have to be Lindelof, (2) “normal x normal” does not have to be normal, (3) “paracompact x paracompact” does not have to be paracompact, (4) “perfectly normal x perfectly normal” does not have to be perfectly normal (does not even have to be normal). In other words, these four properties are not preserved by taking Cartesian product. The goal of this note is to prove these and a few other results about the Sorgenfrey Line. In this note, $S$ is to denote the Sorgenfrey Line.

We will show these results:

A
$S$ is Lindelof (thus is normal and paracompact).
B
$S$ is hereditarily Lindelof.
C
Compact subsets of $S$ are countable. Thus $S$ is an example of a space that is Lindelof and not $\sigma$-compact.
D
$S \times S$ is not Lindelof.
E
$S \times S$ is not normal.
F
$S \times S$ is not paramcompact.
G
$S$ is not second countable, thus not metrizable.
H
$S$ is perfectly normal.

In proving C, we will use the following lemma (Lemma 1), which was proved in a previous post. This is a special countable extent property of the real line. Note that a space $X$ has extent of cardinality $\mathcal{K}$ if $\mathcal{K}$ is the least upper bound on the sizes of all closed and discrete subsets of $X$. In proving F, the Jones Lemma will be used. This lemma is essentially saying that the extent of a separable normal space cannot be the cardinality continuum or greater. A space $X$  is perfectly normal if $X$ is normal and every closed subset is a $G_{\delta}$ set (equivalently every open subset is an $F_{\sigma}$ set). These two lemmas are stated below.

Lemma 1
Every uncountable subset of $\mathbb{R}$ has a two-sided limit point.
Lemma 2 (Jones’ Lemma)
If $X$ is a separable normal space, then it has no closed and discrete subset of cardinality continuum.

See this post for a proof of Jones’ Lemma.

Proof of A. Let $\mathcal{A}$ be an open cover of $S$ consisting of open intervals of the form $[a, b)$.

Let $T=\cup\lbrace{(a, b): [a, b)\epsilon\mathcal{A}}\rbrace$. We claim that $U=S - T$ is a countable set. Suppose that $U$ is uncountable. By Lemma 1, there exists a real number $p$ that is a two-sided limit point of $U$. This means that for every open interval $(s,t)$ (open interval in the usual topology on the real line) with $p \in (s,t)$, the interval $(s,t)$ contains points of $U$ on the left side of $p$ as well as on the right side of $p$.

Choose some $[a,b) \in \mathcal{A}$ such that $p \in [a,b)$. Note that $(a,b)$ is a subset of $T$ and so should not contain points of $U=S-T$. Because $p$ is a two-sided limit point of $U$, $(a,b)$ will contain points of $U$, a contradiction. So $U$ must be countable.

Note that $T$ is an open set in the usual topology, which is Lindelof. So we can find countably many $[c,d) \in \mathcal{A}$ such that the union of all such $(c,d)$ covers $T$. Then find countably many $[c,d) \in \mathcal{A}$ that cover the countably many points in $U=S-T$. Combining both sets of $[c,d)$, we see that $\mathcal{A}$ has a countable subcover. $\blacksquare$

Proof of B. Take any uncountable subspace of $S$, we can apply the same proof as in A.

Proof of C. Let $A\subset{S}$ be a compact subspace that is uncoutable. By Lemma 1, $A$ has a two-sided limit point $y$ (i.e. it is both a left-sided limit point and a right-sided limit point in the usual topology). Let $\lbrace{y_n}\rbrace$ be a sequence of points in $A$ that converges to $y$ from the left. Then $\lbrace{(\infty,y_n)}\rbrace$  and $[y,\infty)$ form an open cover of $A$ that has no finite subcover. Thus all compact subspaces of the Sorgenfrey Line are countable. Furthermore $S$ is an example of a Lindelof but not $\sigma$-compact space.

Proof of D. If $S \times S$ is Lindelof, then it would be a separable normal space and cannot have closed and discrete subset of cardinality continuum or greater (according to Jones’ Lemma). Note that $D=\lbrace{(x,-x): x\epsilon{S}}\rbrace$ is a closed and discrete subset of $S \times S$, which has cardinality continuum. This shows that $S \times S$ cannot be Lindeloff. $\blacksquare$

Proof of E and F. For E, use the same argument as in the proof of D. For F, note that paracompactness implies normality. $\blacksquare$

Proof of G. If $S$ is second countable (has a countable base), then it would be a separable metrizable space and $S \times S$ would also be a separable metrizable space. But $S \times S$ is not even Lindeloff. Thus the Sorgenfrey Line cannot be second countable. $\blacksquare$

Proof of H. Let $W$ be an open subset of $S$. Let $O$ be the interor of $W$ in the usual topology. By a similar argument in the proof of A above, we can show that $W-O$ is countable. Since $O$ is an open set in the usual topology, it is an $F_{\sigma}$ set in the usual topology (and thus in the Sorgenfrey topology). It follows that $O$ plus countably many points would form an $F_{\sigma}$ set. Thus every open subset of the Sorgenfrey line $S$ is a $G_\delta$-set. Since the Sorgenfrey line is Lindelof, it is normal. Thus the Sorgenfrey line is a normal space in which every open set is a $G_\delta$-set (i.e. it is perfectly normal). $\blacksquare$