# Jones’ Lemma

Jones’ lemma is a great tool in working with normal spaces. It is useful, under cerntain conditions, to show the non-normality of a space. The lemma establishes an upper bound for the cardinality of closed and discrete sets in any separable and normal space. Thus, whenever you have a separable space with a closed and discrete set whose cardinality exceeds the upper bound, you have a non-normal space (see examples discussed below). One way to prove the Jones’ lemma is to explore the set-theoretic relationship between the density (the minimum cardinality of a dense set) and the cardinality of closed and discrete sets in normal spaces. We sketch a proof of this lemma and give some examples. We also state an extension of Jones’ lemma. All spaces under consideration are at least Hausdorff.

Let $X$ be a space. A subset $D$ of $X$ is said to be a closed and discrete set in $X$ if $D$ is a closed set in $X$ and $D$, in the relative topology, is a discrete space. Good basic references are [1] and [3]. For more detailed information about cardinal functions, see [2].
____________________________________________________________________________

Jones’ Lemma
Let $X$ be a separable and normal space. Then for any set $D$ that is a closed and discrete set in $X$, we have $\displaystyle 2^{\lvert D \lvert} \le 2^{\omega}$.

Corollary 1
Let $X$ be a separable and normal space. Then for any set $D$ that is a closed and discrete set in $X$, we have $\displaystyle \lvert D \lvert < 2^{\omega}$.

Jones' lemma, as stated above, is essentially saying that the cardinality of continuum is an upper bound of the cardinalities of the power sets of closed and discrete sets in any separable and normal space. The corollary says that the cardinality of continuum is an upper bound of the cardinalities of closed and discrete sets in any separable and normal space. As indicated at the beginning, the corollary is a great way for checking the non-normality of a separable space.

We now give a sketch of the proof Jones' lemma. Let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$. Suppose $X$ is a separable space. A key point is that the cardinality of $C(X)$ is sandwiched between the two cardinalities in the lemma:

$\displaystyle (1) \ \ \ \ \ 2^{\lvert D \lvert} \le \lvert C(X) \lvert \le 2^{\omega}$

The first inequality in $(1)$ says that there are at least as many continuous real-valued functions as there are subsets of the closed and discrete set $D$. To see this, we appeal for some help from Urysohn's lemma. For each $E \subset D$, $E$ and $D-E$ are disjoint closed sets in $X$. By Urysohn's lemma, there is a continuous function $f_E:X \rightarrow [0,1]$ such that $f_E$ maps $E$ to $1$ and $f_E$ maps $D-E$ to $0$. Note that the mapping $\Psi: \mathcal{P}(D) \rightarrow C(X)$ defined by $\Psi(E)=f_E$ is a one-to-one map, where $\mathcal{P}(D)$ is the collection of all subsets of $D$.

The second inequality in $(1)$ says that there are at most continuum many continuous real-valued functions defined on the space $X$. In other words, the number of continuous functions is capped by the cardinality continuum (actually equals continuum in this case, but one inequality is all we need here). Let $G$ be a countable dense subset of $X$. Consider the map $\rho:C(X) \rightarrow \mathbb{R}^{G}$ defined by $\rho(f)=f \upharpoonright G$, which is the function $f$ restricted to the set $G$. The notation $\mathbb{R}^{G}$ refers to the set of all functions from the set $G$ into $\mathbb{R}$.

The key point here is that any continuous functions $f:X \rightarrow \mathbb{R}$ and $g:X \rightarrow \mathbb{R}$, if they agree on the countable dense set $G$ ($f \upharpoonright G=g \upharpoonright G$), then $f=g$ on the whole space $X$.So $\rho$ is a one-to-one map. Some elementary cardinal arithmetic shows that $\lvert \mathbb{R}^G \lvert=2^\omega$. Thus the second inequality in $(1)$ is established.

____________________________________________________________________________

Examples

Let $S$ be the Sorgenfrey line. This is the real line with the topology generated by half open intervals of the form $[a,b)$. The Sorgenfrey line is a classic example of a normal space whose square is not normal. The Sorgenfrey plane $S \times S$ is separable with a closed and discrete set of cardinality continuum. By the corollary of Jones’ lemma, the Sorgenfrey plane is not normal. The Sorgenfrey line is discussed in greater details in this post.

The tangent disc space is another example of a separable space with a closed and discrete set of size continuum, hence ensuring that it is a non-normal space.

____________________________________________________________________________

Generalization

The generalization involves the notion of density and the notion of extent. The density of an infinite Hausdorff space $X$ is the smallest cardinal number of the form $\lvert G \lvert$ where $G$ is a dense subset of $X$. This cardinal number is denoted by $d(X)$. For any separable space $X$, we have $d(X)=\omega$. If the space $X$ is finite, then the convention adopted by many authors is that $d(X)=\omega$.

The extent of a space $X$ is the smallest infinite cardinal $\mathcal{K}$ such that every closed and discrete set in $X$ has cardinality $\le \mathcal{K}$. The extent of the space $X$ is denoted by $e(X)$. For more detailed information about cardinal functions, see [2].

The following are the generalization of Jones’ lemma.

Jones’ Lemma
Let $X$ be a normal space. Then for any set $D$ that is a closed and discrete set in $X$, we have $\displaystyle 2^{\lvert D \lvert} \le 2^{d(X)}$.

Corollary 2
Let $X$ be a normal space. Then for any set $D$ that is a closed and discrete set in $X$, we have $\displaystyle \lvert D \lvert < 2^{d(X)}$, which implies $\displaystyle e(X) \le 2^{d(X)}$.

Corollary 2 suggests that the cardinal number $2^{d(X)}$ is an upper bound of the cardinalities of closed and discrete sets in any normal space $X$. As a result of this, the cardinal number $2^{d(X)}$ dominates the extent $e(X)$ in a normal space.
____________________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Hodel, R., Cardinal Functions I, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 1-61.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# Examples of Lindelof Spaces that are not Hereditarily Lindelof

We observe from the following statement two examples of Lindelof spaces that are not hereditarily Lindelof.

• Any product space contains a discrete subspace having the same cardinality as the number of factor spaces.

Using the above observation, by choosing the factor spaces judiciously, the product of uncountably many spaces is a handy way of obtaining Lindelof spaces (in some cases $\sigma$-compact spaces) that are not hereditarily Lindelof. For definition and basic information about product spaces, see this previous post.

________________________________________________________________________

All spaces under consideration are at least Hausdorff. For each $\alpha \in A$, let $X_\alpha$ be a space with at least two points. For each $\alpha \in A$, fix two points $p_\alpha, q_\alpha \in X_\alpha$. Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains a discrete subspace $Y$ that has the same cardinality as the cardinality of the index set $A$.

For each $\alpha \in A$, define $y_\alpha \in \prod \limits_{\alpha \in A} X_\alpha$ by the following:

$\displaystyle (1) \ \ \ \ \ \ y_\alpha(\gamma)=\left\{\begin{matrix}p_\alpha&\ \gamma=\alpha\\{q_\alpha}&\ \gamma \ne \alpha \end{matrix}\right.$

Let $Y=\left\{ y_\alpha: \alpha \in A\right\}$. It follows that $\lvert Y \lvert = \lvert A \lvert$ and that $Y$ is a discrete space.

________________________________________________________________________

Whenever the index set $A$ is uncountable, the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains an uncountable discrete subspace. Thus even if the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof, one of its subspace $Y$ cannot be Lindelof. Taking the product of uncountably many factor spaces is a handy way to obtain Lindelof space that is not hereditarily Lindelof. Some examples are shown below.

________________________________________________________________________
Examples

Let the index set $A$ be uncountable. To make the product space Lindelof, we can make every one of its factor $X_\alpha$ compact. Thus the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is compact and not hereditarily Lindelof.

Thus the product space $[0,1]^{\omega_1}$, the product of $\omega_1$ many copies of the unit interval, is compact and not hereditarily Lindelof. Another example is $\left\{ 0,1 \right\}^{\omega_1}$, the product of $\omega_1$ many copies of $\left\{ 0,1 \right\}$

Another way to make the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ Lindelof is to make some of the factors compact such that the product of the remaining non-compact factors is Lindelof. Then the product space is essentially the product of a compact space and a Lindelof space, which is always Lindelof.

For example, let $X_0=\mathbb{R}$ and let $X_\alpha=[0,1]$ for all $\alpha$ with $0<\alpha<\omega_1$. Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof since it is essentially the product of a compact space and a Lindelof space. However, the product $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is not hereditarily Lindelof.

In fact, the product space in the previous paragraph is $\sigma$-compact (i.e. the union of countably many compact sets). To make the example not $\sigma$-compact, simply make the first factor space a non-locally compact Lindelof space. For example, use the Sorgenfrey line or the space of the irrational numbers.

# 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.

# A Theorem About Hereditarily Normality

Let $Y$ be either the closed unit interval $\mathbb{I}=[0,1]$ or $[0,\omega]=\omega+1$. Let $X$ be any one of the following spaces:

$X=[0,\omega_1]=\omega_1+1$,
$X=[0,\omega_1)=\omega_1$,
$X=$ Michael Line,
$X=$ Sorgenfrey Line.

The cross product $X \times Y$ is normal in all four cases. The $X$ factor, in the order listed, is compact, countably compact, paracompact and Lindelof. Thus they are all countably paracompact. According to the Dowker’s theorem, the product of a normal space $X$ and a compact metric space $Y$ is normal if and only if $X$ is countably paracompact. My goal here is to show that the four cases of $X \times Y$ here cannot be hereditarily normal. This is from a theorem due to Katetov. We prove the following theorem.

Theorem. If $X \times Y$ is hereditarily normal, then either $X$ is perfectly normal or every countably infinite subspace of $Y$ is closed and discrete.

Proof. Suppose that $X$ is not perfectly normal and $Y$ has a countably infinite subset that is not closed and discrete. Let $H \subset X$ be a closed set that is not a $G_\delta-$set. Let $C=\lbrace{y_n:n \in \omega}\rbrace \subset Y$ be an infinite set with an accumulation point $y$. We assume that $y \notin C$.

We show that the open subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal. To this end, let $A=H \times (Y-\lbrace{y}\rbrace)$ and $B=(X-H) \times \lbrace{y}\rbrace$. The sets $A$ and $B$ are disjoint closed subspaces of the open subspace $U$. Suppose we have disjoint open sets $S$ and $T$ such that $A \subset S$ and $B \subset T$.

For each $n \in \omega$, let $O_n=\lbrace{x \in X:(x,y_n) \in S}\rbrace$. Each $O_n$ is open and $H \subset O_n$. Thus $H \subset \bigcap_n O_n$. Let $x \in \bigcap_n O_n$. Then $(x,y) \in \overline{S}$. This means $x \in H$. If $x \notin H$, then $(x,y) \in B \subset T$ (which is impossible). So we have $H=\bigcap_n O_n$, indicating that $H$ is a $G_\delta-$set, and leading to a contradiction. So the subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal.

Corollary. For countably compact spaces (in particular compact spaces) $X$ and $Y$, if the Cartesian product $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are prefectly normal.

Proof. Note that both factors, being countably compact, cannot have closed and discrete infinite subsets.

Comment. Note that the converse of the corollary is not true. Let both factors be the double arrow space, which is perfectly normal. But the square of the double arrow space contains a copy of the Sorgenfrey Plane, which is not normal.

Reference
[Katetov] Katetov, M., [1948] Complete normality of Cartesian products, Fund. Math., 36, 271-274.

# 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.

____________________________________________________________________

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). $\blacksquare$