In between G-delta diagonal and submetrizable

Posted on March 21, 2018 by Dan Ma

This post discusses the property of having a $G_\delta$ -diagonal and related diagonal properties. The focus is on the diagonal properties in between $G_\delta$ -diagonal and submetrizability. The discussion is followed by a diagram displaying the relative strengths of these properties. Some examples and questions are discussed.

G-delta Diagonal

In any space $Y$ , a subset $A$ is said to be a $G_\delta$ -set in the space $Y$ (or $A$ is a $G_\delta$ -subset of $Y$ ) if $A$ is the intersection of countably many open subsets of $Y$ . A subset $A$ of $Y$ is an $F_\sigma$ -set in $Y$ (or $A$ is an $F_\sigma$ -subset of $Y$ ) if $A$ is the union of countably closed subsets of the space $Y$ . Of course, the set $A$ is a $G_\delta$ -set if and only if $Y-A$ , the complement of $A$ , is an $F_\sigma$ -set.

The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$ , which is a subset of the square $X \times X$ . When the set $\Delta$ is a $G_\delta$ -set in the space $X \times X$ , we say that the space $X$ has a $G_\delta$ -diagonal.

It is straightforward to verify that the space $X$ is a Hausdorff space if and only if the diagonal $\Delta$ is a closed subset of $X \times X$ . As a result, if $X$ is a Hausdorff space such that $X \times X$ is perfectly normal, then the diagonal would be a closed set and thus a $G_\delta$ -set. Such spaces, including metric spaces, would have a $G_\delta$ -diagonal. Thus any metric space has a $G_\delta$ -diagonal.

A space $X$ is submetrizable if there is a metrizable topology that is weaker than the topology for $X$ . Then the diagonal $\Delta$ would be a $G_\delta$ -set with respect to the weaker metrizable topology of $X \times X$ and thus with respect to the orginal topology of $X$ . This means that the class of spaces having $G_\delta$ -diagonals also include the submetrizable spaces. As a result, Sorgenfrey line and Michael line have $G_\delta$ -diagonals since the Euclidean topology are weaker than both topologies.

A space having a $G_\delta$ -diagonal is a simple topological property. Such spaces form a wide class of spaces containing many familiar spaces. According to the authors in [2], the property of having a $G_\delta$ -diagonal is an important ingredient of submetrizability and metrizability. For example, any compact space with a $G_\delta$ -diagonal is metrizable (see this blog post). Any paracompact or Lindelof space with a $G_\delta$ -diagonal is submetrizable. Spaces with $G_\delta$ -diagonals are also interesting in their own right. It is a property that had been research extensively. It is also a current research topic; see [7].

A Closer Look

To make the discussion more interesting, let’s point out a few essential definitions and notations. Let $X$ be a space. Let $\mathcal{U}$ be a collection of subsets of $X$ . Let $A \subset X$ . The notation $St(A, \mathcal{U})$ refers to the set $St(A, \mathcal{U})=\cup \{U \in \mathcal{U}: A \cap U \ne \varnothing \}$ . In other words, $St(A, \mathcal{U})$ is the union of all the sets in $\mathcal{U}$ that intersect the set $A$ . The set $St(A, \mathcal{U})$ is also called the star of the set $A$ with respect to the collection $\mathcal{U}$ .

If $A=\{ x \}$ , we write $St(x, \mathcal{U})$ instead of $St(\{ x \}, \mathcal{U})$ . Then $St(x, \mathcal{U})$ refers to the union of all sets in $\mathcal{U}$ that contain the point $x$ . The set $St(x, \mathcal{U})$ is then called the star of the point $x$ with respect to the collection $\mathcal{U}$ .

Note that the statement of $X$ having a $G_\delta$ -diagonal is defined by a statement about the product $X \times X$ . It is desirable to have a translation that is a statement about the space $X$ .

Theorem 1
Let $X$ be a space. Then the following statements are equivalent.

The space $X$ has a $G_\delta$ -diagonal.
There exists a sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ such that for each $x \in X$ , $\{ x \}=\bigcap \{ St(x, \mathcal{U}_n): n=0,1,2,\cdots \}$ .

The sequence of open covers in condition 2 is called a $G_\delta$ -diagonal sequence for the space $X$ . According to condition 2, at any given point, the stars of the point with respect to the open covers in the sequence collapse to the given point.

One advantage of a $G_\delta$ -diagonal sequence is that it is entirely about points of the space $X$ . Thus we can work with such sequences of open covers of $X$ instead of the $G_\delta$ -set $\Delta$ in $X \times X$ . Theorem 1 is not a word for word translation. However, the proof is quote natural.

Suppose that $\Delta=\cap \{U_n: n=0,1,2,\cdots \}$ where each $U_n$ is an open subset of $X \times X$ . Then let $\mathcal{U}_n=\{U \subset X: U \text{ open and } U \times U \subset U_n \}$ . It can be verify that $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a $G_\delta$ -diagonal sequence for $X$ .

Suppose that $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a $G_\delta$ -diagonal sequence for $X$ . For each $n$ , let $U_n=\cup \{ U \times U: U \in \mathcal{U}_n \}$ . It follows that $\Delta=\bigcap_{n=0}^\infty U_n$ . $\square$

It is informative to compare the property of $G_\delta$ -diagonal with the definition of Moore spaces. A development for the space $X$ is a sequence $\mathcal{D}_0,\mathcal{D}_1,\mathcal{D}_2,\cdots$ of open covers of $X$ such that for each $x \in X$ , $\{ St(x, \mathcal{D}_n): n=0,1,2,\cdots \}$ is a local base at the point $x$ . A space is said to be developable if it has a development. The space $X$ is said to be a Moore space if $X$ is a Hausdorff and regular space that has a development.

The stars of a given point with respect to the open covers of a development form a local base at the given point, and thus collapse to the given point. Thus a development is also a $G_\delta$ -diagonal sequence. It then follows that any Moore space has a $G_\delta$ -diagonal.

A point in a space is a $G_\delta$ -point if the point is the intersection of countably many open sets. Then having a $G_\delta$ -diagonal sequence implies that that every point of the space is a $G_\delta$ -point since every point is the intersection of the stars of that point with respect to a $G_\delta$ -diagonal sequence. In contrast, any Moore space is necessarily a first countable space since the stars of any given point with respect to the development is a countable local base at the given point. The parallel suggests that spaces with $G_\delta$ -diagonals can be thought of as a weak form of Moore spaces (at least a weak form of developable spaces).

Regular G-delta Diagonal

We discuss other diagonal properties. The space $X$ is said to have a regular $G_\delta$ -diagonal if $\Delta=\cap \{\overline{U_n}:n=0,1,2,\cdots \}$ where each $U_n$ is an open subset of $X \times X$ such that $\Delta \subset U_n$ . This diagonal property also has an equivalent condition in terms of a diagonal sequence.

Theorem 2
Let $X$ be a space. Then the following statements are equivalent.

The space $X$ has a regular $G_\delta$ -diagonal.
There exists a sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ such that for every two distinct points $x,y \in X$ , there exist open sets $U$ and $V$ with $x \in U$ and $y \in V$ and there also exists an $n$ such that no member of $\mathcal{U}_n$ intersects both $U$ and $V$ .

For convenience, we call the sequence described in Theorem 2 a regular $G_\delta$ -diagonal sequence. It is clear that if the diagonal of a space is a regular $G_\delta$ -diagonal, then it is a $G_\delta$ -diagonal. It can also be verified that a regular $G_\delta$ -diagonal sequence is also a $G_\delta$ -diagonal sequence. To see this, let $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ be a regular $G_\delta$ -diagonal sequence for $X$ . Suppose that $y \ne x$ and $y \in \bigcap_k St(x, \mathcal{U}_k)$ . Choose open sets $U$ and $V$ and an integer $n$ guaranteed by the regular $G_\delta$ -diagonal sequence. Since $y \in St(x, \mathcal{U}_n)$ , choose $B \in \mathcal{U}_n$ such that $x,y \in B$ . Then $B$ would be an element of $\mathcal{U}_n$ that meets both $U$ and $V$ , a contradiction. Then $\{ x \}= \bigcap_k St(x, \mathcal{U}_k)$ for all $x \in X$ .

To proof Theorem 2, suppose that $X$ has a regular $G_\delta$ -diagonal. Let $\Delta=\bigcap_{k=0}^\infty \overline{U_k}$ where each $U_k$ is open in $X \times X$ and $\Delta \subset U_k$ . For each $k$ , let $\mathcal{U}_k$ be the collection of all open subsets $U$ of $X$ such that $U \times U \subset U_k$ . It can be verified that $\{ \mathcal{U}_k \}$ is a regular $G_\delta$ -diagonal sequence for $X$ .

On the other hand, suppose that $\{ \mathcal{U}_k \}$ is a regular $G_\delta$ -diagonal sequence for $X$ . For each $k$ , let $U_k=\cup \{U \times U: U \in \mathcal{U}_k \}$ . It can be verified that $\Delta=\bigcap_{k=0}^\infty \overline{U_k}$ . $\square$

Rank-k Diagonals

Metric spaces and submetrizable spaces have regular $G_\delta$ -diagonals. We discuss this fact after introducing another set of diagonal properties. First some notations. For any family $\mathcal{U}$ of subsets of the space $X$ and for any $x \in X$ , define $St^1(x, \mathcal{U})=St(x, \mathcal{U})$ . For any integer $k \ge 2$ , let $St^k(x, \mathcal{U})=St^{k-1}(St(x, \mathcal{U}))$ . Thus $St^{2}(x, \mathcal{U})$ is the star of the star $St(x, \mathcal{U})$ with respect to $\mathcal{U}$ and $St^{3}(x, \mathcal{U})$ is the star of $St^{2}(x, \mathcal{U})$ and so on.

Let $X$ be a space. A sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ is said to be a rank- $k$ diagonal sequence of $X$ if for each $x \in X$ , we have $\{ x \}=\bigcap_{j=0}^\infty St^k(x,\mathcal{U}_j)$ . When the space $X$ has a rank- $k$ diagonal sequence, the space is said to have a rank- $k$ diagonal. Clearly a rank-1 diagonal sequence is simply a $G_\delta$ -diagonal sequence as defined in Theorem 1. Thus having a rank-1 diagonal is the same as having a $G_\delta$ -diagonal.

It is also clear that having a higher rank diagonal implies having a lower rank diagonal. This follows from the fact that a rank $k+1$ diagonal sequence is also a rank $k$ diagonal sequence.

The following lemma builds intuition of the rank- $k$ diagonal sequence. For any two distinct points $x$ and $y$ of a space $X$ , and for any integer $d \ge 2$ , a $d$ -link path from $x$ to $y$ is a set of open sets $W_1,W_2,\cdots,W_d$ such that $x \in W_1$ , $y \in W_d$ and $W_t \cap W_{t+1} \ne \varnothing$ for all $t=1,2,\cdots,d-1$ . By default, a single open set $W$ containing both $x$ and $y$ is a d-link path from $x$ to $y$ for any integer $d \ge 1$ .

Lemma 3
Let $X$ be a space. Let $k$ be a positive integer. Let $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ be a sequence of open covers of $X$ . Then the following statements are equivalent.

The sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a rank- $k$ diagonal sequence for the space $X$ .
For any two distinct points $x$ and $y$ of $X$ , there is an integer $n$ such that $y \notin St^k(x,\mathcal{U}_n)$ .
For any two distinct points $x$ and $y$ of $X$ , there is an integer $n$ such that there is no $k$ -link path from $x$ to $y$ consisting of elements of $\mathcal{U}_n$ .

It can be seen directly from definition that Condition 1 and Condition 2 are equivalent. For Condition 3, observe that the set $St^k(x,\mathcal{U}_n)$ is the union of $k$ types of open sets – open sets in $\mathcal{U}_n$ containing $x$ , open sets in $\mathcal{U}_n$ that intersect the first type, open sets in $\mathcal{U}_n$ that intersect the second type and so on down to the open sets in $\mathcal{U}_n$ that intersect $St^{k-1}(x,\mathcal{U}_n)$ . A path is formed by taking one open set from each type.

We now show a few basic results that provide further insight on the rank- $k$ diagonal.

Theorem 4
Let $X$ be a space. If the space $X$ has a rank-2 diagonal, then $X$ is a Hausdorff space.

Theorem 5
Let $X$ be a Moore space. Then $X$ has a rank-2 diagonal.

Theorem 6
Let $X$ be a space. If $X$ has a rank-3 diagonal, then $X$ has a regular $G_\delta$ -diagonal.

Once Lemma 3 is understood, Theorem 4 is also easily understood. If a space $X$ has a rank-2 diagonal sequence $\{ \mathcal{U}_n \}$ , then for any two distinct points $x$ and $y$ , we can always find an $n$ where there is no 2-link path from $x$ to $y$ . Then $x$ and $y$ can be separated by open sets in $\mathcal{U}_n$ . Thus these diagonal ranking properties confer separation axioms. We usually start off a topology discussion by assuming a reasonable separation axiom (usually implicitly). The fact that the diagonal ranking gives a bonus makes it even more interesting. Apparently many authors agree since $G_\delta$ -diagonal and related topics had been researched extensively over decades.

To prove Theorem 5, let $\{ \mathcal{U}_n \}$ be a development for the space $X$ . Let $x$ and $y$ be two distinct points of $X$ . We claim that there exists some $n$ such that $y \notin St^2(x,\mathcal{U}_n)$ . Suppose not. This means that for each $n$ , $y \in St^2(x,\mathcal{U}_n)$ . This also means that $St(x,\mathcal{U}_n) \cap St(y,\mathcal{U}_n) \ne \varnothing$ for each $n$ . Choose $x_n \in St(x,\mathcal{U}_n) \cap St(y,\mathcal{U}_n)$ for each $n$ . Since $X$ is a Moore space, $\{ St(x,\mathcal{U}_n) \}$ is a local base at $x$ . Then $\{ x_n \}$ converges to $x$ . Since $\{ St(y,\mathcal{U}_n) \}$ is a local base at $y$ , $\{ x_n \}$ converges to $y$ , a contradiction. Thus the claim that there exists some $n$ such that $y \notin St^2(x,\mathcal{U}_n)$ is true. By Lemma 3, a development for a Moore space is a rank-2 diagonal sequence.

To prove Theorem 6, let $\{ \mathcal{U}_n \}$ be a rank-3 diagonal sequence for the space $X$ . We show that $\{ \mathcal{U}_n \}$ is also a regular $G_\delta$ -diagonal sequence for $X$ . Suppose $x$ and $y$ are two distinct points of $X$ . By Lemma 3, there exists an $n$ such that there is no 3-link path consisting of open sets in $\mathcal{U}_n$ that goes from $x$ to $y$ . Choose $U \in \mathcal{U}_n$ with $x \in U$ . Choose $V \in \mathcal{U}_n$ with $y \in V$ . Then it follows that no member of $\mathcal{U}_n$ can intersect both $U$ and $V$ (otherwise there would be a 3-link path from $x$ to $y$ ). Thus $\{ \mathcal{U}_n \}$ is also a regular $G_\delta$ -diagonal sequence for $X$ .

We now show that metric spaces have rank- $k$ diagonal for all integer $k \ge 1$ .

Theorem 7
Let $X$ be a metrizable space. Then $X$ has rank- $k$ diagonal for all integers $k \ge 1$ .

If $d$ is a metric that generates the topology of $X$ , and if $\mathcal{U}_n$ is the collection of all open subsets with diameters $\le 2^{-n}$ with respect to the metrix $d$ then $\{ \mathcal{U}_n \}$ is a rank- $k$ diagonal sequence for $X$ for any integer $k \ge 1$ .

We instead prove Theorem 7 topologically. To this end, we use an appropriate metrization theorem. The following theorem is a good candidate.

Alexandrov-Urysohn Metrization Theorem. A space $X$ is metrizable if and only if the space $X$ has a development $\{ \mathcal{U}_n \}$ such that for any $U_1,U_2 \in \mathcal{U}_{n+1}$ with $U_1 \cap U_2 \ne \varnothing$ , the set $U_1 \cup U_2$ is contained in some element of $\mathcal{U}_n$ . See Theorem 1.5 in p. 427 of [5].

Let $\{ \mathcal{U}_n \}$ be the development from Alexandrov-Urysohn Metrization Theorem. It is a development with a strong property. Each open cover in the development refines the preceding open cover in a special way. This refinement property allows us to show that it is a rank- $k$ diagonal sequence for $X$ for any integer $k \ge 1$ .

First, we make a few observations about $\{ \mathcal{U}_n \}$ . From the statement of the theorem, each $\mathcal{U}_{n+1}$ is a refinement of $\mathcal{U}_n$ . As a result of this observation, $\mathcal{U}_{m}$ is a refinement of $\mathcal{U}_n$ for any $m>n$ . Furthermore, for each $x \in X$ , $\text{St}(x,\mathcal{U}_m) \subset \text{St}(x,\mathcal{U}_n)$ for any $m>n$ .

Let $x, y \in X$ with $x \ne y$ . Based on the preceding observations, it follows that there exists some $m$ such that $\text{St}(x,\mathcal{U}_m) \cap \text{St}(y,\mathcal{U}_m)=\varnothing$ . We claim that there exists some integer $h>m$ such that there are no $k$ -link path from $x$ to $y$ consisting of open sets from $\mathcal{U}_h$ . Then $\{ \mathcal{U}_n \}$ is a rank- $k$ diagonal sequence for $X$ according to Lemma 3.

We show this claim is true for $k=2$ . Observe that there cannot exist $U_1, U_2 \in \mathcal{U}_{m+1}$ such that $x \in U_1$ , $y \in U_2$ and $U_1 \cap U_2 \ne \varnothing$ . If there exists such a pair, then $U_1 \cup U_2$ would be contained in $\text{St}(x,\mathcal{U}_m)$ and $\text{St}(y,\mathcal{U}_m)$ , a contradiction. Putting it in another way, there cannot be any 2-link path $U_1,U_2$ from $x$ to $y$ such that the open sets in the path are from $\mathcal{U}_{m+1}$ . According to Lemma 3, the sequence $\{ \mathcal{U}_n \}$ is a rank-2 diagonal sequence for the space $X$ .

In general for any $k \ge 2$ , there cannot exist any $k$ -link path $U_1,\cdots,U_k$ from $x$ to $y$ such that the open sets in the path are from $\mathcal{U}_{m+k-1}$ . The argument goes just like the one for the case for $k=2$ . Suppose the path $U_1,\cdots,U_k$ exists. Using the special property of $\{ \mathcal{U}_n \}$ , the 2-link path $U_1,U_2$ is contained in some open set in $\mathcal{U}_{m+k-2}$ . The path $U_1,\cdots,U_k$ is now contained in a $(k-1)$ -link path consisting of elements from the open cover $\mathcal{U}_{m+k-2}$ . Continuing the refinement process, the path $U_1,\cdots,U_k$ is contained in a 2-link path from $x$ to $y$ consisting of elements from $\mathcal{U}_{m+1}$ . Like before this would lead to a contradiction. According to Lemma 3, $\{ \mathcal{U}_n \}$ is a rank- $k$ diagonal sequence for the space $X$ for any integer $k \ge 2$ .

Of course, any metric space already has a $G_\delta$ -diagonal. We conclude that any metrizable space has a rank- $k$ diagonal for any integer $k \ge 1$ . $\square$

We have the following corollary.

Corollary 8
Let $X$ be a submetrizable space. Then $X$ has rank- $k$ diagonal for all integer $k \ge 1$ .

In a submetrizable space, the weaker metrizable topology has a rank- $k$ diagonal sequence, which in turn is a rank- $k$ diagonal sequence in the original topology.

Examples and Questions

The preceding discussion focuses on properties that are in between $G_\delta$ -diagonal and submetrizability. In fact, one of the properties has infinitely many levels (rank- $k$ diagonal for integers $k \ge 1$ ). We would like to have a diagram showing the relative strengths of these properties. Before we do so, consider one more diagonal property.

Let $X$ be a space. The set $A \subset X$ is said to be a zero-set in $X$ if there is a continuous $f:X \rightarrow [0,1]$ such that $A=f^{-1}(0)$ . In other words, a zero-set is a set that is the inverse image of zero for some continuous real-valued function defined on the space in question.

A space $X$ has a zero-set diagonal if the diagonal $\Delta=\{ (x,x): x \in X \}$ is a zero-set in $X \times X$ . The space $X$ having a zero-set diagonal implies that $X$ has a regular $G_\delta$ -diagonal, and thus a $G_\delta$ -diagonal. To see this, suppose that $\Delta=f^{-1}(0)$ where $f:X \times X \rightarrow [0,1]$ is continuous. Then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where $U_n=f^{-1}([0,1/n))$ . Thus having a zero-set diagonal is a strong property.

We have the following diagram.

The diagram summarizes the preceding discussion. From top to bottom, the stronger properties are at the top. From left to right, the stronger properties are on the left. The diagram shows several properties in between $G_\delta$ -diagonal at the bottom and submetrizability at the top.

Note that the statement at the very bottom is not explicitly a diagonal property. It is placed at the bottom because of the classic result that any compact space with a $G_\delta$ -diagonal is metrizable.

In the diagram, “rank-k diagonal” means that the space has a rank- $k$ diagonal where $k \ge 1$ is an integer, which in terms means that the space has a rank- $k$ diagonal sequence as defined above. Thus rank- $k$ diagonal is not to be confused with the rank of a diagonal. The rank of the diagonal of a given space is the largest integer $k$ such that the space has a rank- $k$ diagonal. For example, for a space that has a rank-2 diagonal but has no rank-3 diagonal, the rank of the diagonal is 2.

To further make sense of the diagram, let’s examine examples.

The Mrowka space is a classic example of a space with a $G_\delta$ -diagonal that is not submetrizable (introduced here). Where is this space located in the diagram? The Mrowka space, also called Psi-space, is defined using a maximal almost disjoint family of subsets of $\omega$ . We denote such a space by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ . It is a pseudocompact Moore space that is not submetrizable. As a Moore space, it has a rank-2 diagonal sequence. A well known result states that any pseudocompact space with a regular $G_\delta$ -diagonal is metrizable (see here). As a non-submetrizable space, the Mrowka space cannot have a regular $G_\delta$ -diagonal. Thus $\Psi(\mathcal{A})$ is an example of a space with a rank-2 diagonal but not a rank-3 diagonal sequence.

Examples of non-submetrizable spaces with stronger diagonal properties are harder to come by. We discuss examples that are found in the literature.

Example 2.9 in [2] is a Tychonoff separable Moore space $Z$ that has a rank-3 diagonal but not of higher diagonal rank. As a result of not having a rank-4 diagonal, $Z$ is not submetrizable. Thus $Z$ is an example of a space with rank-3 diagonal (hence with a regular $G_\delta$ -diagonal) that is not submetrizable. According to a result in [6], any separable space with a zero-set diagonal is submetrizable. Then the space $Z$ is an example of a space with a regular $G_\delta$ -diagonal that does not have a zero-set diagonal. In fact, the authors of [2] indicated that this is the first such example.

Example 2.9 of [2] shows that having a rank-3 diagonal does not imply having a zero-set diagonal. If a space is strengthened to have a rank-4 diagonal, does it imply having a zero-set diagonal? This is essentially Problem 2.13 in [2].

On the other hand, having a rank-3 diagonal implies a rank-2 diagonal. If we weaken the hypothesis to just having a regular regular $G_\delta$ -diagonal, does it imply having a rank-2 diagonal? This is essentially Problem 2.14 in [2].

The authors of [2] conjectured that for each $n$ , there exists a space $X_n$ with a rank- $n$ diagonal but not having a rank- $(n+1)$ diagonal. This conjecture was answered affirmatively in [8] by constructing, for each integer $k \ge 4$ , a Tychonoff space with a rank- $k$ diagonal but not having a rank- $(k+1)$ diagonal. Thus even for high $k$ , a non-submetrizable space can be found with rank- $k$ diagonal.

One natural question is this. Is there a non-submetrizable space that has rank- $k$ diagonal for all $k \ge 1$ ? We have not seen this question stated in the literature. But it is clearly a natural question.

Example 2.17 in [2] is a non-submetrizable Moore space that has a zero-set diagonal and has rank-3 diagonal exactly (i.e. it does not have a higher rank diagonal). This example shows that having a zero-set diagonal does not imply having a rank-4 diagonal. A natural question is then this. Does having a zero-set diagonal imply having a rank-3 diagonal? This appears to be an open question. This is hinted by Problem 2.19 in [2]. It asks, if $X$ is a normal space with a zero-set diagonal, does $X$ have at least a rank-2 diagonal?

The property of having a $G_\delta$ -diagonal and related properties is a topic that had been researched extensively over the decades. It is still an active topic of research. The discussion in this post only touches on the surface. There are many other diagonal properties not covered here. To further investigate, check with the papers listed below and also consult with information available in the literature.

Reference

Arhangelskii A. V., Burke D. K., Spaces with a regular $G_\delta$ -diagonal, Topology and its Applications, Vol. 153, No. 11, 1917–1929, 2006.
Arhangelskii A. V., Buzyakova R. Z., The rank of the diagonal and submetrizability, Comment. Math. Univ. Carolinae, Vol. 47, No. 4, 585-597, 2006.
Buzyakova R. Z., Cardinalities of ccc-spaces with regular $G_\delta$ -diagonals, Topology and its Applications, Vol. 153, 1696–1698, 2006.
Buzyakova R. Z., Observations on spaces with zeroset or regular $G_\delta$ -diagonals, Comment. Math. Univ. Carolinae, Vol. 46, No. 3, 469-473, 2005.
Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
Martin H. W., Contractibility of topological spaces onto metric spaces, Pacific J. Math., Vol. 61, No. 1, 209-217, 1975.
Xuan Wei-Feng, Shi Wei-Xue, On spaces with rank k-diagonals or zeroset diagonals, Topology Proceddings, Vol. 51, 245{251, 2018.
Yu Zuoming, Yun Ziqiu, A note on the rank of diagonals, Topology and its Applications, Vol. 157, 1011–1014, 2010.

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Pseudocompact spaces with regular G-delta diagonals

Posted on February 19, 2018 by Dan Ma

This post complements two results discussed in two previous blog posts concerning $G_\delta$ -diagonal. One result is that any compact space with a $G_\delta$ -diagonal is metrizable (see here). The other result is that the compactness in the first result can be relaxed to countably compactness. Thus any countably compact space with a $G_\delta$ -diagonal is metrizable (see here). The countably compactness in the second result cannot be relaxed to pseudocompactness. The Mrowka space is a pseudocompact space with a $G_\delta$ -diagonal that is not submetrizable, hence not metrizable (see here). However, if we strengthen the $G_\delta$ -diagonal to a regular $G_\delta$ -diagonal while keeping pseudocompactness fixed, then we have a theorem. We prove the following theorem.

Theorem 1
If the space $X$ is pseudocompact and has a regular $G_\delta$ -diagonal, then $X$ is metrizable.

All spaces are assumed to be Hausdorff and completely regular. The assumption of completely regular is crucial. The proof of Theorem 1 relies on two lemmas concerning pseudocompact spaces (one proved in a previous post and one proved here). These two lemmas work only for completely regular spaces.

The proof of Theorem 1 uses a metrization theorem. The best metrization to use in this case is Moore metrization theorem (stated below). The result in Theorem 1 is found in [2].

First some basics. Let $X$ be a space. The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$ . When the diagonal $\Delta$ , as a subset of $X \times X$ , is a $G_\delta$ -set, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$ , the space $X$ is said to have a $G_\delta$ -diagonal.

The space $X$ is said to have a regular $G_\delta$ -diagonal if the diagonal $\Delta$ is a regular $G_\delta$ -set in $X \times X$ , i.e. $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$ . If $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ , then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}=\bigcap_{n=1}^\infty U_n$ . Thus if a space has a regular $G_\delta$ -diagonal, it has a $G_\delta$ -diagonal. We will see that there exists a space with a $G_\delta$ -diagonal that fails to be a regular $G_\delta$ -diagonal.

The space $X$ is a pseudocompact space if for every continuous function $f:X \rightarrow \mathbb{R}$ , the image $f(X)$ is a bounded set in the real line $\mathbb{R}$ . Pseudocompact spaces are discussed in considerable details in this previous post. We will rely on results from this previous post to prove Theorem 1.

The following lemma is used in proving Theorem 1.

Lemma 2
Let $X$ be a pseudocompact space. Suppose that $O_1,O_2,O_2,\cdots$ is a decreasing sequence of non-empty open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ for some point $x \in X$ . Then $\{ O_n \}$ is a local base at the point $x$ .

Proof of Lemma 2
Let $O_1,O_2,O_2,\cdots$ be a decreasing sequence of open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ . Let $U$ be open in $X$ with $x \in U$ . If $O_n \subset U$ for some $n$ , then we are done. Suppose that $O_n \not \subset U$ for each $n$ .

Choose open $V$ with $x \in V \subset \overline{V} \subset U$ . Consider the sequence $\{ O_n \cap (X-\overline{V}) \}$ . This is a decreasing sequence of non-empty open subsets of $X$ . By Theorem 2 in this previous post, $\bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing$ . Let $y$ be a point in this non-empty set. Note that $y \in \bigcap_{n=1}^\infty \overline{O_n}$ . This means that $y=x$ . Since $x \in \overline{O_n \cap (X-\overline{V})}$ for each $n$ , any open set containing $x$ would contain a point not in $\overline{V}$ . This is a contradiction since $x \in V$ . Thus it must be the case that $x \in O_n \subset U$ for some $n$ . $\square$

The following metrization theorem is useful in proving Theorem 1.

Theorem 3 (Moore Metrization Theorem)
Let $X$ be a space. Then $X$ is metrizable if and only if the following condition holds.

There exists a decreasing sequence $\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\cdots$ of open covers of $X$ such that for each $x \in X$ , the sequence $\{ St(St(x,\mathcal{B}_n),\mathcal{B}_n):n=1,2,3,\cdots \}$ is a local base at the point $x$ .

For any family $\mathcal{U}$ of subsets of $X$ , and for any $A \subset X$ , the notation $St(A,\mathcal{U})$ refers to the set $\cup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$ . In other words, it is the union of all sets in $\mathcal{U}$ that contain points of $A$ . The set $St(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the family $\mathcal{U}$ . If $A=\{ x \}$ , we write $St(x,\mathcal{U})$ instead of $St(\{ x \},\mathcal{U})$ . The set $St(St(x,\mathcal{B}_n),\mathcal{B}_n)$ indicated in Theorem 3 is the star of the set $St(x,\mathcal{B}_n)$ with respect to the open cover $\mathcal{B}_n$ .

Theorem 3 follows from Theorem 1.4 in [1], which states that for any $T_0$ -space $X$ , $X$ is metrizable if and only if there exists a sequence $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots$ of open covers of $X$ such that for each open $U \subset X$ and for each $x \in U$ , there exist an open $V \subset X$ and an integer $n$ such that $x \in V$ and $St(V,\mathcal{G}_n) \subset U$ .

Proof of Theorem 1

Suppose $X$ is pseudocompact such that its diagonal $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$ . We can assume that $U_1 \supset U_2 \supset \cdots$ . For each $n \ge 1$ , define the following:

$\mathcal{U}_n=\{ U \subset X: U \text{ open in } X \text{ and } U \times U \subset U_n \}$

Note that each $\mathcal{U}_n$ is an open cover of $X$ . Also note that $\{ \mathcal{U}_n \}$ is a decreasing sequence since $\{ U_n \}$ is a decreasing sequence of open sets. We show that $\{ \mathcal{U}_n \}$ is a sequence of open covers of $X$ that satisfies Theorem 3. We establish this by proving the following claims.

Claim 1. For each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ .

To prove the claim, let $x \ne y$ . There is an integer $n$ such that $(x,y) \notin \overline{U_n}$ . Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$ . Note that $(x,y) \notin U_k$ and $(U \times V) \cap U_n=\varnothing$ .

We want to show that $V \cap St(x,\mathcal{U}_n)=\varnothing$ , which implies that $y \notin \overline{St(x,\mathcal{U}_n)}$ . Suppose $V \cap St(x,\mathcal{U}_n) \ne \varnothing$ . This means that $V \cap W \ne \varnothing$ for some $W \in \mathcal{U}_n$ with $x \in W$ . Then $(U \times V) \cap (W \times W) \ne \varnothing$ . Note that $W \times W \subset U_n$ . This implies that $(U \times V) \cap U_n \ne \varnothing$ , a contradiction. Thus $V \cap St(x,\mathcal{U}_n)=\varnothing$ . Since $y \in V$ , $y \notin \overline{St(x,\mathcal{U}_n)}$ . We have established that for each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ .

Claim 2. For each $x \in X$ , $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$ .

Note that $\{ St(x,\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ . By Lemma 2, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$ .

Claim 3. For each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}=\{ x \}$ .

Let $x \ne y$ . There is an integer $n$ such that $(x,y) \notin \overline{U_n}$ . Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$ . It follows that $(U \times V) \cap \overline{U_t}=\varnothing$ for all $t \ge n$ . Furthermore, $(U \times V) \cap U_t=\varnothing$ for all $t \ge n$ . By Claim 2, choose integers $i$ and $j$ such that $St(x,\mathcal{U}_i) \subset U$ and $St(y,\mathcal{U}_j) \subset V$ . Choose an integer $k \ge \text{max}(n,i,j)$ . It follows that $(St(x,\mathcal{U}_i) \times St(y,\mathcal{U}_j)) \cap U_k=\varnothing$ . Since $\mathcal{U}_k \subset \mathcal{U}_i$ and $\mathcal{U}_k \subset \mathcal{U}_j$ , it follows that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k=\varnothing$ .

We claim that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$ . Suppose not. Choose $w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)$ . It follows that $w \in B$ for some $B \in \mathcal{U}_k$ such that $B \cap St(x,\mathcal{U}_k) \ne \varnothing$ and $B \cap St(y,\mathcal{U}_k) \ne \varnothing$ . Furthermore $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap (B \times B)=\varnothing$ . Note that $B \times B \subset U_k$ . This means that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k \ne \varnothing$ , contradicting the fact observed in the preceding paragraph. It must be the case that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$ .

Because there is an open set containing $y$ , namely $St(y,\mathcal{U}_k)$ , that contains no points of $St(St(x,\mathcal{U}_k), \mathcal{U}_k)$ , $y \notin \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}$ . Thus Claim 3 is established.

Claim 4. For each $x \in X$ , $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n)) \}$ is a local base at the point $x$ .

Note that $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n))}=\{ x \}$ . By Lemma 2, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a local base at the point $x$ .

In conclusion, the sequence $\mathcal{U}_1,\mathcal{U}_2,\mathcal{U}_3,\cdots$ of open covers satisfies the properties in Theorem 3. Thus any pseudocompact space with a regular $G_\delta$ -diagonal is metrizable. $\square$

Example

Any submetrizable space has a $G_\delta$ -diagonal. The converse is not true. A classic example of a non-submetrizable space with a $G_\delta$ -diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal family of almost disjoint subsets of $\omega$ . Actually $\Psi(\mathcal{A})$ would be a family of spaces since $\mathcal{A}$ is any maximal almost disjoint family. For any maximal $\mathcal{A}$ , $\Psi(\mathcal{A})$ is a pseudocompact non-submetrizable space that has a $G_\delta$ -diagonal. This example shows that the requirement of a regular $G_\delta$ -diagonal in Theorem 1 cannot be weakened to a $G_\delta$ -diagonal. See here for a more detailed discussion of this example.

Reference

Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
McArthur W. G., $G_\delta$ -Diagonals and Metrization Theorems, Pacific Journal of Mathematics, Vol. 44, No. 2, 613-317, 1973.

$\text{ }$

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

Drawing more Sorgenfrey continuous functions

Posted on January 1, 2018 by Dan Ma

In this previous post, we draw continuous functions on the Sorgenfrey line $S$ to gain insight about the function $C_p(S)$ . In this post, we draw more continuous functions with the goal of connecting $C_p(S)$ and $C_p(D)$ where $D$ is the double arrow space. For example, $C_p(D)$ can be embedded as a subspace of $C_p(S)$ . More interestingly, both function spaces $C_p(D)$ and $C_p(S)$ share the same closed and discrete subspace of cardinality continuum. As a result, the function space $C_p(D)$ is not normal.

Double Arrow Space

The underlying set for the double arrow space is $D=[0,1] \times \{ 0,1 \}$ , which is a subset in the Euclidean plane.

Figure 1 – The Double Arrow Space

The name of double arrow comes from the fact that an open neighborhood of a point in the upper line segment points to the right while an open neighborhood of a point in the lower line segment points to the left. This is demonstrated in the following diagram.

Figure 2 – Open Neighborhoods in the Double Arrow Space

More specifically, for any $a$ with $0 \le a < 1$ , a basic open set containing the point $(a,1)$ is of the form $\displaystyle \biggl[ [a,b) \times \{ 1 \} \biggr] \cup \biggl[ (a,b) \times \{ 0 \} \biggr]$ , painted red in Figure 2. One the other hand, for any $a$ with $0<a \le 1$ , a basic open set containing the point $(a,0)$ is of the form $\biggl[ (c,a) \times \{ 1 \} \biggr] \cup \biggl[ (c,a] \times \{ 0 \} \biggr]$ , painted blue in Figure 2. The upper right point $(1,1)$ and the lower left point $(0,0)$ are made isolated points.

The double arrow space is a compact space that is perfectly normal and not metrizable. Basic properties of this space, along with those of the lexicographical ordered space, are discussed in this previous post.

The drawing of continuous functions in this post aims to show the following results.

The function space $C_p(D)$ can be embedded as a subspace in the function $C_p(S)$ .
Both function spaces $C_p(D)$ and $C_p(S)$ share the same closed and discrete subspace of cardinality continuum.
The function space $C_p(D)$ is not normal.

Drawing a Map from Sorgenfrey Line onto Double Arrow Space

In order to show that $C_p(D)$ can be embedded into $C_p(S)$ , we draw a continuous map from the Sorgenfrey line $S$ onto the double arrow space $D$ . The following diagram gives the essential idea of the mapping we need.

Figure 3 – Mapping Sorgenfrey Line onto Double Arrow Space

The mapping shown in Figure 3 is to map the interval $[0,1]$ onto the upper line segment of the double arrow space, as demonstrated by the red arrow. Thus $x \mapsto (x,1)$ for any $x$ with $0 \le x \le 1$ . Essentially on the interval $[0,1]$ , the mapping is the identity map.

On the other hand, the mapping is to map the interval $[-1,0)$ onto the lower line segment of the double arrow space less the point $(0, 0)$ , as demonstrated by the blue arrow in Figure 3. Thus $-x \mapsto (x,0)$ for any $-x$ with $0<x \le 1$ . Essentially on the interval $[-1,0)$ , the mapping is the identity map times -1.

The mapping described by Figure 3 only covers the interval $[-1,1]$ in the domain. To complete the mapping, let $x \mapsto (1,1)$ for any $x \in (1, \infty)$ and $x \mapsto (0,0)$ for any $x \in (-\infty, -1)$ .

Let $h$ be the mapping that has been described. It maps the Sorgenfrey line onto the double arrow space. It is straightforward to verify that the map $h: S \rightarrow D$ is continuous.

Embedding

We use the following fact to show that $C_p(D)$ can be embedded into $C_p(S)$ .

Suppose that the space $Y$ is a continuous image of the space $X$ . Then $C_p(Y)$ can be embedded into $C_p(X)$ .

Based on this result, $C_p(D)$ can be embedded into $C_p(S)$ . The embedding that makes this true is $E(f)=f \circ h$ for each $f \in C_p(D)$ . Thus each function $f$ in $C_p(D)$ is identified with the composition $f \circ h$ where $h$ is the map defined in Figure 3. The fact that $E(f)$ is an embedding is shown in this previous post (see Theorem 1).

Same Closed and Discrete Subspace in Both Function Spaces

The following diagram describes a closed and discrete subspace of $C_p(S)$ .

Figure 4 – a family of Sorgenfrey continuous functions

For each $0<a<1$ , let $f_a: S \rightarrow \{0,1 \}$ be the continuous function described in Figure 4. The previous post shows that the set $F=\{ f_a: 0<a<1 \}$ is a closed and discrete subspace of $C_p(S)$ . We claim that $F \subset C_p(D) \subset C_p(S)$ .

To see that $F \subset C_p(D)$ , we define continuous functions $U_a: D \rightarrow \{0,1 \}$ such that $f_a=U_a \circ h$ . We can actually back out the map $U_a$ from $f_a$ in Figure 4 and the mapping $h$ . Here’s how. The function $f_a$ is piecewise constant (0 or 1). Let’s focus on the interval $[-1,1]$ in the domain of $f_a$ .

Consider where the function $f_a$ maps to the value 1. There are two intervals, $[a,1)$ and $[-1,-a)$ , where $f_a$ maps to 1. The mapping $h$ maps $[a,1)$ to the set $[a,1) \times \{ 1 \}$ . So the function $U_a$ must map $[a,1) \times \{ 1 \}$ to the value 1. The mapping $h$ maps $[-1,-a)$ to the set $(a,1] \times \{ 0 \}$ . So $U_a$ must map $(a,1] \times \{ 0 \}$ to the value 1.

Now consider where the function $f_a$ maps to the value 0. There are two intervals, $[0,a)$ and $[-a,0)$ , where $f_a$ maps to 0. The mapping $h$ maps $[0,a)$ to the set $[0,a) \times \{ 1 \}$ . So the function $U_a$ must map $[0,a) \times \{ 1 \}$ to the value 0. The mapping $h$ maps $[-a,0)$ to the set $(0,a] \times \{ 0 \}$ . So $U_a$ must map $(0,a] \times \{ 0 \}$ to the value 0.

To take care of the two isolated points $(1,1)$ and $(0,0)$ of the double arrow space, make sure that $U_a$ maps these two points to the value 0. The following is a precise definition of the function $U_a$ .

$\displaystyle U_a(y) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ y \in [a,1) \times \{ 1 \} \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ y \in (a,1] \times \{ 0 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y \in (0,a] \times \{ 0 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y \in [0,a) \times \{ 1 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y=(0,0) \text{ or } y = (1,1) \end{array} \right.$

The resulting $U_a$ is a translation of $f_a$ . Under the embedding $E$ defined earlier, we see that $E(U_a)=f_a$ . Let $U=\{ U_a: 0<a<1 \}$ . The set $U$ in $C_p(D)$ is homeomorphic to the set $F$ in $C_p(S)$ . Thus $U$ is a closed and discrete subspace of $C_p(D)$ since $F$ is a closed and discrete subspace of $C_p(S)$ .

Remarks

The drawings and the embedding discussed here and in the previous post establish that $C_p(D)$ , the space of continuous functions on the double arrow space, contains a closed and discrete subspace of cardinality continuum. It follows that $C_p(D)$ is not normal. This is due to the fact that if $C_p(X)$ is normal, then $C_p(X)$ must have countable extent (i.e. all closed and discrete subspaces must be countable).

While $C_p(D)$ is embedded in $C_p(S)$ , the function space $C_p(S)$ is not embedded in $C_p(D)$ . Because the double arrow space is compact, $C_p(D)$ has countable tightness. If $C_p(S)$ were to be embedded in $C_p(D)$ , then $C_p(S)$ would be countably tight too. However, $C_p(S)$ is not countably tight due to the fact that $S \times S$ is not Lindelof (see Theorem 1 in this previous post).

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

$\text{ }$

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

Drawing Sorgenfrey continuous functions

Posted on September 5, 2017 by Dan Ma

The Sorgenfrey line is a well known topological space. It is the real number line with open intervals defined as sets of the form $[a,b)$ . Though this is a seemingly small tweak, it generates a vastly different space than the usual real number line. In this post, we look at the Sorgenfrey line from the continuous function perspective, in particular, the continuous functions that map the Sorgenfrey line into the real number line. In the process, we obtain insight into the space of continuous functions on the Sorgenfrey line.

The next post is a continuation on the theme of drawing Sorgenfrey continuous functions.

The Sorgenfrey Line

Let $\mathbb{R}$ denote the real number line. The usual open intervals are of the form $(a,b)=\left\{x \in \mathbb{R}: a<x<b \right\}$ . The union of such open intervals is called an open set. If more than one topologies are considered on the real line, these open sets are referred to as the usual open sets or Euclidean open sets (on the real line). The open intervals $(a,b)$ form a base for the usual topology on the real line. One important fact abut the usual open sets is that the usual open sets can be generated by the intervals $(a,b)$ where both end points are rational numbers. Thus the usual topology on the real line is said to have a countable base.

Now tweak the usual topology by calling sets of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x<b \right\}$ open intervals. Then form open sets by taking unions of all such open intervals. The collection of such open sets is called the Sorgenfrey topology (on the real line). The real number line $\mathbb{R}$ with the Sorgenfry topology is called the Sorgenfrey line, denoted by $\mathbb{S}$ . The Sorgenfrey line has been discussed in this blog, starting with this post. This post examines continuous functions from $\mathbb{S}$ into the real line. In the process, we gain insight on the space of continuous functions defined on $\mathbb{S}$ .

Note that any usual open interval $(a,b)$ is the union of intervals of the form $[c,d)$ . Thus any usual (Euclidean) open set is an open set in the Sorgenfrey line. Thus the usual topology (on the real line) is contained in the Sorgenfrey topology, i.e. the usual topology is a weaker (coarser) topology.

Let $C(\mathbb{R})$ be the set of all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ where the domain is the real number line with the usual topology. Let $C(\mathbb{S})$ be the set of all continuous functions $f:\mathbb{S} \rightarrow \mathbb{R}$ where the domain is the Sorgenfrey line. In both cases, the range is always the number line with the usual topology. Based on the preceding paragraph, any continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is also continuous with respect to the Sorgenfrey line, i.e. $C(\mathbb{R}) \subset C(\mathbb{S})$ .

Pictures of Continuous Functions

Consider the following two continuous functions.

Figure 1 – CDF of the standard normal distribution

Figure 2 – CDF of the uniform distribution

The first one (Figure 1) is the cumulative distribution function (CDF) of the standard normal distribution. The second one (Figure 2) is the CDF of the uniform distribution on the interval $(0,a)$ where $a>0$ . Both of these are continuous in the usual Euclidean topology (in the domain). Such graphs would make regular appearance in a course on probability and statistics. They also show up in a calculus course as an everywhere differentiable curve (Figure 1) and as a differentiable curve except at finitely many points (Figure 2). Both of these functions can also be regarded as continuous functions on the Sorgenfrey line.

Consider a function that is continuous in the Sorgenfrey line but not continuous in the usual topology.

Figure 3 – Right continuous function

Figure 3 is a function that maps the interval $(-\infty,0)$ to -1 and maps the interval $[0,\infty)$ to 1. It is not continuous in the usual topology because of the jump at $x=0$ . But it is a continuous function when the domain is considered to be the Sorgenfrey line. Because of the open intervals being $[a,b)$ , continuous functions defined on the Sorgenfrey line are right continuous.

The cumulative distribution function of a discrete probability distribution is always right continuous, hence continuous in the Sorgenfrey line. Here’s an example.

Figure 4 – CDF of a discrete uniform distribution

Figure 4 is the CDF of the uniform distribution on the finite set $\left\{0,1,2,3,4 \right\}$ , where each point has probability 0.2. There is a jump of height 0.2 at each of the points from 0 to 4. Figure 3 and Figure 4 are step functions. As long as the left point of a step is solid and the right point is hollow, the step functions are continuous on the Sorgenfrey line.

The take away from the last four figures is that the real-valued continuous functions defined on the Sorgenfrey line are right continuous and that step functions (with the left point solid and the right point hollow) are Sorgenfrey continuous.

A Family of Sorgenfrey Continuous Functions

The four examples of continuous functions shown above are excellent examples to illustrate the Sorgenfrey topology. We now introduce a family of continuous functions $f_a:\mathbb{S} \rightarrow \mathbb{R}$ for $0<a<1$ . These continuous functions will lead to additional insight on the function space whose domain space is the Sorgenfrey line.

For any $0<a<1$ , the following gives the definition and the graph of the function $f_a$ .

$\displaystyle f_a(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ -\infty<x<-1 \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ -1 \le x<-a \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ -a \le x <a \\ \text{ } & \text{ } \\ 1 &\ \ \ \ \ \ a \le x <1 \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ 1 \le x <\infty \end{array} \right.$

Figure 5 – a family of Sorgenfrey continuous functions

Function Space on the Sorgenfrey Line

This is the place where we switch the focus to function space. The set $C(\mathbb{S})$ is a subset of the product space $\mathbb{R}^\mathbb{R}$ . So we can consider $C(\mathbb{S})$ as a topological space endowed with the topology inherited as a subspace of $\mathbb{R}^\mathbb{R}$ . This topology on $C(\mathbb{S})$ is called the pointwise convergence topology and $C(\mathbb{S})$ with the product subspace topology is denoted by $C_p(\mathbb{S})$ . See here for comments on how to work with the pointwise convergence topology.

For the present discussion, all we need is some notation on a base for $C_p(\mathbb{S})$ . For $x \in \mathbb{S}$ , and for any open interval $(a,b)$ (open in the usual topology of the real number line), let $[x,(a,b)]=\left\{h \in C_p(\mathbb{S}): h(x) \in (a, b) \right\}$ . Then the collection of intersections of finitely many $[x,(a,b)]$ would form a base for $C_p(\mathbb{S})$ .

The following is the main fact we wish to establish.

The function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum. In particular, the set $F=\left\{f_a: 0<a<1 \right\}$ is a closed and discrete subspace of $C_p(\mathbb{S})$ .

The above result will derive several facts on the function space $C_p(\mathbb{S})$ , which are discussed in a section below. More interestingly, the proof of the fact that $F=\left\{f_a: 0<a<1 \right\}$ is a closed and discrete subspace of $C_p(\mathbb{S})$ is based purely on the definition of the functions $f_a$ and the Sorgenfrey topology. The proof given below does not use any deep or high powered results from function space theory. So it should be a nice exercise on the Sorgenfrey topology.

I invite readers to either verify the fact independently of the proof given here or follow the proof closely. Lots of drawing of the functions $f_a$ on paper will be helpful in going over the proof. In this one instance at least, drawing continuous functions can help gain insight on function spaces.

Working out the Proof

The following diagram was helpful to me as I worked out the different cases in showing the discreteness of the family $F=\left\{f_a: 0<a<1 \right\}$ . The diagram is a valuable aid in convincing myself that a given case is correct.

Figure 6 – A comparison of three Sorgenfrey continuous functions

Now the proof. First, $F$ is relatively discrete in $C_p(\mathbb{S})$ . We show that for each $a$ , there is an open set $O$ containing $f_a$ such that $O$ does not contain $f_w$ for any $w \ne a$ . To this end, let $O=[a,V_1] \cap [-a,V_2]$ where $V_1$ and $V_2$ are the open intervals $V_1=(0.9,1.1)$ and $V_2=(-0.1,0.1)$ . With Figure 6 as an aid, it follows that for $0<b<a$ , $f_b \notin O$ and for $a<c<1$ , $f_c \notin O$ .

The open set $O=[a,V_1] \cap [-a,V_2]$ contains $f_a$ , the function in the middle of Figure 6. Note that for $0<b<a$ , $f_b(-a)=1$ and $f_b(-a) \notin V_2$ . Thus $f_b \notin O$ . On the other hand, for $a<c<1$ , $f_c(a)=0$ and $f_c(a) \notin V_1$ . Thus $f_c \notin O$ . This proves that the set $F$ is a discrete subspace of $C_p(\mathbb{S})$ relative to $F$ itself.

Now we show that $F$ is closed in $C_p(\mathbb{S})$ . To this end, we show that

$g \in C_p(\mathbb{S})$

$U$

$g$

$U$

$F$

Actually, this has already been done above with points $g$ that are in $F$ . One thing to point out is that the range of $f_a$ is $\left\{0,1 \right\}$ . As we consider $g \in C_p(\mathbb{S})$ , we only need to consider $g$ that maps into $\left\{0,1 \right\}$ . Let $g \in C_p(\mathbb{S})$ . The argument is given in two cases regarding the function $g$ .

Case 1. There exists some $a \in (0,1)$ such that $g(a) \ne g(-a)$ .

We assume that $g(a)=0$ and $g(-a)=1$ . Then for all $0<b<a$ , $f_b(a)=1$ and for all $a<c<1$ , $f_c(-a)=0$ . Let $U=[a,(-0.1,0.1)] \cap [-a,(0.9,1.1)]$ . Then $g \in U$ and $U$ contains no $f_b$ for any $0<b<a$ and $f_c$ for any $a<c<1$ . To help see this argument, use Figure 6 as a guide. The case that $g(a)=1$ and $g(-a)=0$ has a similar argument.

Case 2. For every $a \in (0,1)$ , we have $g(a) = g(-a)$ .

Claim. The function $g$ is constant on the interval $(-1,1)$ . Suppose not. Let $0<b<a<1$ such that $g(a) \ne g(b)$ . Suppose that $0=g(b) < g(a)=1$ . Consider $W=\left\{w<a: g(w)=0 \right\}$ . Clearly the number $a$ is an upper bound of $W$ . Let $u \le a$ be a least upper bound of $W$ . The function $g$ has value 1 on the interval $(u,a)$ . Otherwise, $u$ would not be the least upper bound of the set $W$ . There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=0$ for all $n$ . Otherwise, $u$ would not be the least upper bound of the set $W$ .

It follows that $g(u)=1$ . Otherwise, the function $g$ is not continuous at $u$ . Now consider the 6 points $-a<-u<-b<b<u<a$ . By the assumption in Case 2, $g(u)=g(-u)=1$ and $g(b)=g(-b)=0$ . Since $g(x_n)=0$ for all $n$ , $g(-x_n)=0$ for all $n$ . Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=0$ , contradicting $g(-u)=1$ . Thus we cannot have $0=g(b) < g(a)=1$ .

Now suppose we have $1=g(b) > g(a)=0$ where $0<b<a<1$ . Consider $W=\left\{w<a: g(w)=1 \right\}$ . Clearly $W$ has an upper bound, namely the number $a$ . Let $u \le a$ be a least upper bound of $W$ . The function $g$ has value 0 on the interval $(u,a)$ . Otherwise, $u$ would not be the least upper bound of the set $W$ . There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=1$ for all $n$ . Otherwise, $u$ would not be the least upper bound of the set $W$ .

It follows that $g(u)=0$ . Otherwise, the function $g$ is not continuous at $u$ . Now consider the 6 points $-a<-u<-b<b<u<a$ . By the assumption in Case 2, $g(u)=g(-u)=0$ and $g(b)=g(-b)=1$ . Since $g(x_n)=1$ for all $n$ , $g(-x_n)=1$ for all $n$ . Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=1$ , contradicting $g(-u)=0$ . Thus we cannot have $1=g(b) > g(a)=0$ .

The claim that the function $g$ is constant on the interval $(-1,1)$ is established. To wrap up, first assume that the function $g$ is 1 on the interval $(-1,1)$ . Let $U=[0,(0.9,1.1)]$ . It is clear that $g \in U$ . It is also clear from Figure 5 that $U$ contains no $f_a$ . Now assume that the function $g$ is 0 on the interval $(-1,1)$ . Since $g$ is Sorgenfrey continuous, it follows that $g(-1)=0$ . Let $U=[-1,(-0.1,0.1)]$ . It is clear that $g \in U$ . It is also clear from Figure 5 that $U$ contains no $f_a$ .

We have established that the set $F=\left\{f_a: 0<a<1 \right\}$ is a closed and discrete subspace of $C_p(\mathbb{S})$ .

What does it Mean?

The above argument shows that the set $F$ is a closed an discrete subspace of the function space $C_p(\mathbb{S})$ . We have the following three facts.

Three Results

$C_p(\mathbb{S})$ is separable.
$C_p(\mathbb{S})$ is not hereditarily separable.
$C_p(\mathbb{S})$ is not a normal space.

To show that $C_p(\mathbb{S})$ is separable, let’s look at one basic helpful fact on $C_p(X)$ . If $X$ is a separable metric space, e.g. $X=\mathbb{R}$ , then $C_p(X)$ has quite a few nice properties (discussed here). One is that $C_p(X)$ is hereditarily separable. Thus $C_p(\mathbb{R})$ , the space of real-valued continuous functions defined on the number line with the pointwise convergence topology, is hereditarily separable and thus separable. Recall that continuous functions in $C_p(\mathbb{R})$ are also Soregenfrey line continuous. Thus $C_p(\mathbb{R})$ is a subspace of $C_p(\mathbb{S})$ . The space $C_p(\mathbb{R})$ is also a dense subspace of $C_p(\mathbb{S})$ . Thus the space $C_p(\mathbb{S})$ contains a dense separable subspace. It means that $C_p(\mathbb{S})$ is separable.

Secondly, $C_p(\mathbb{S})$ is not hereditarily separable since the subspace $F=\left\{f_a: 0<a<1 \right\}$ is a closed and discrete subspace.

Thirdly, $C_p(\mathbb{S})$ is not a normal space. According to Jones’ lemma, any separable space with a closed and discrete subspace of cardinality of continuum is not a normal space (see Corollary 1 here). The subspace $F=\left\{f_a: 0<a<1 \right\}$ is a closed and discrete subspace of the separable space $C_p(\mathbb{S})$ . Thus $C_p(\mathbb{S})$ is not normal.

Remarks

The topology of the Sorgenfrey line is vastly different from the usual topology on the real line even though the the Sorgenfrey topology is obtained by a seemingly small tweak from the usual topology. The real line is a metric space while the Sorgenfrey line is not metrizable. The real number line is connected while the Sorgenfrey line is not. The countable power of the real number line is a metric space and thus a normal space. On the other hand, the Sorgenfrey line is a classic example of a normal space whose square is not normal. See here for a basic discussion of the Sorgenfrey line.

The pictures of Sorgenfrey continuous functions demonstrated here show that the real number line and the Sorgenfrey line are also very different from a function space perspective. The function space $C_p(\mathbb{R})$ has a whole host of nice properties: normal, Lindelof (hence paracompact and collectionwise normal), hereditarily Lindelof (hence hereditarily normal), hereditarily separable, and perfectly normal (discussed here).

Though separable, the function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum, making it not hereditarily separable and not normal.

For more information about $C_p(X)$ in general and $C_p(\mathbb{S})$ in particular, see [1] and [2]. A different proof that $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum can be found in Problem 165 in [2].

The next post is a continuation on the theme of drawing Sorgenfrey continuous functions.

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

Normality in Cp(X)

Posted on August 15, 2017 by Dan Ma

Any collectionwise normal space is a normal space. Any perfectly normal space is a hereditarily normal space. In general these two implications are not reversible. In function spaces $C_p(X)$ , the two implications are reversible. There is a normal space that is not countably paracompact (such a space is called a Dowker space). If a function space $C_p(X)$ is normal, it is countably paracompact. Thus normality in $C_p(X)$ is a strong property. This post draws on Dowker’s theorem and other results, some of them are previously discussed in this blog, to discuss this remarkable aspect of the function spaces $C_p(X)$ .

Since we are discussing function spaces, the domain space $X$ has to have sufficient quantity of real-valued continuous functions, e.g. there should be enough continuous functions to separate the points from closed sets. The ideal setting is the class of completely regular spaces (also called Tychonoff spaces). See here for a discussion on completely regular spaces in relation to function spaces.

Let $X$ be a completely regular space. Let $C(X)$ be the set of all continuous functions from $X$ into the real line $\mathbb{R}$ . When $C(X)$ is endowed with the pointwise convergence topology, the space is denoted by $C_p(X)$ (see here for further comments on the definition of the pointwise convergence topology).

When Function Spaces are Normal

Let $X$ be a completely regular space. We discuss these four facts of $C_p(X)$ :

If the function space $C_p(X)$ is normal, then $C_p(X)$ is countably paracompact.
If the function space $C_p(X)$ is hereditarily normal, then $C_p(X)$ is perfectly normal.
If the function space $C_p(X)$ is normal, then $C_p(X)$ is collectionwise normal.
Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e. every closed and discrete subset of $X$ is countable, implying that $X$ is collectionwise normal.

Fact #1 and Fact #2 rely on a representation of $C_p(X)$ as a product space with one of the factors being the real line. For $x \in X$ , let $Y_x=\left\{f \in C_p(X): f(x)=0 \right\}$ . Then $C_p(X) \cong Y_x \times \mathbb{R}$ . This representation is discussed here.

Another useful tool is Dowker’s theorem, which essentially states that for any normal space $W$ , the space $W$ is countably paracompact if and only if $W \times C$ is normal for all compact metric space $C$ if and only if $W \times [0,1]$ is normal. For the full statement of the theorem, see Theorem 1 in this previous post, which has links to the proofs and other discussion.

To show Fact #1, suppose that $C_p(X)$ is normal. Immediately we make use of the representation $C_p(X) \cong Y_x \times \mathbb{R}$ where $x \in X$ . Since $Y_x \times \mathbb{R}$ is normal, $Y_x \times [0,1]$ is also normal. By Dowker’s theorem, $Y_x$ is countably paracompact. Note that $Y_x$ is a closed subspace of the normal $C_p(X)$ . Thus $Y_x$ is also normal.

One more helpful tool is Theorem 5 in in this previous post, which is like an extension of Dowker’s theorem, which states that a normal space $W$ is countably paracompact if and only if $W \times T$ is normal for any $\sigma$ -compact metric space $T$ . This means that $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal.

We want to show $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact. Since $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal (based on the argument in the preceding paragraph), $(Y_x \times \mathbb{R}) \times [0,1]$ is normal. Thus according to Dowker’s theorem, $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact.

For Fact #2, a helpful tool is Katetov’s theorem (stated and proved here), which states that for any hereditarily normal $X \times Y$ , one of the factors is perfectly normal or every countable subset of the other factor is closed (in that factor).

To show Fact #2, suppose that $C_p(X)$ is hereditarily normal. With $C_p(X) \cong Y_x \times \mathbb{R}$ and according to Katetov’s theorem, $Y_x$ must be perfectly normal. The product of a perfectly normal space and any metric space is perfectly normal (a proof is found here). Thus $C_p(X) \cong Y_x \times \mathbb{R}$ is perfectly normal.

The proof of Fact #3 is found in Problems 294 and 295 of [2]. The key to the proof is a theorem by Reznichenko, which states that any dense convex normal subspace of $[0,1]^X$ has countable extent, hence is collectionwise normal (problem 294). See here for a proof that any normal space with countable extent is collectionwise normal (see Theorem 2). The function space $C_p(X)$ is a dense convex subspace of $[0,1]^X$ (problem 295). Thus if $C_p(X)$ is normal, then it has countable extent and hence collectionwise normal.

Fact #4 says that normality of the function space imposes countable extent on the domain. This result is discussed in this previous post (see Corollary 3 and Corollary 5).

Remarks

The facts discussed here give a flavor of what function spaces are like when they are normal spaces. For further and deeper results, see [1] and [2].

Fact #1 is essentially driven by Dowker’s theorem. It follows from the theorem that whenever the product space $X \times Y$ is normal, one of the factor must be countably paracompact if the other factor has a non-trivial convergent sequence (see Theorem 2 in this previous post). As a result, there is no Dowker space that is a $C_p(X)$ . No pathology can be found in $C_p(X)$ with respect to finding a Dowker space. In fact, not only $C_p(X) \times C$ is normal for any compact metric space $C$ , it is also true that $C_p(X) \times T$ is normal for any $\sigma$ -compact metric space $T$ when $C_p(X)$ is normal.

The driving force behind Fact #2 is Katetov’s theorem, which basically says that the hereditarily normality of $X \times Y$ is a strong statement. Coupled with the fact that $C_p(X)$ is of the form $Y_x \times \mathbb{R}$ , Katetov’s theorem implies that $Y_x \times \mathbb{R}$ is perfectly normal. The argument also uses the basic fact that perfectly normality is preserved when taking product with metric spaces.

There are examples of normal but not collectionwise normal spaces (e.g. Bing’s Example G). Resolution of the question of whether normal but not collectionwise normal Moore space exists took extensive research that spanned decades in the 20th century (the normal Moore space conjecture). The function $C_p(X)$ is outside of the scope of the normal Moore space conjecture. The function space $C_p(X)$ is usually not a Moore space. It can be a Moore space only if the domain $X$ is countable but then $C_p(X)$ would be a metric space. However, it is still a powerful fact that if $C_p(X)$ is normal, then it is collectionwise normal.

On the other hand, a more interesting point is on the normality of $X$ . Suppose that $X$ is a normal Moore space. If $C_p(X)$ happens to be normal, then Fact #4 says that $X$ would have to be collectionwise normal, which means $X$ is metrizable. If the goal is to find a normal Moore space $X$ that is not collectionwise normal, the normality of $C_p(X)$ would kill the possibility of $X$ being the example.

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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Looking for spaces in which every compact subspace is metrizable

Posted on June 11, 2017 by Dan Ma

Once it is known that a topological space is not metrizable, it is natural to ask, from a metrizability standpoint, which subspaces are metrizable, e.g. whether every compact subspace is metrizable. This post discusses several classes of spaces in which every compact subspace is metrizable. Though the goal here is not to find a complete characterization of such spaces, this post discusses several classes of spaces and various examples that have this property. The effort brings together many interesting basic and well known facts. Thus the notion “every compact subspace is metrizable” is an excellent learning opportunity.

Several Classes of Spaces

The notion “every compact subspace is metrizable” is a very broad class of spaces. It includes well known spaces such as Sorgenfrey line, Michael line and the first uncountable ordinal $\omega_1$ (with the order topology) as well as Moore spaces. Certain function spaces are in the class “every compact subspace is metrizable”. The following diagram is a good organizing framework.

$\displaystyle \begin{aligned} &1. \ \text{Metrizable} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&2. \ \text{Submetrizable} \Longleftarrow 5. \ \exists \ \text{countable network} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&3. \ \exists \ G_\delta \text{ diagonal} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&4. \ \text{Every compact subspace is metrizable} \end{aligned}$

Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$ . Thus a space $X$ is submetrizable if it has a weaker metrizable topology.

Let $\mathcal{N}$ be a set of subsets of the space $X$ . $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$ , there exists $N \in \mathcal{N}$ such that $x \in N \subset O$ . Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).

The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$ . The space $X$ has a $G_\delta$ -diagonal if $\Delta$ is a $G_\delta$ -subset of $X \times X$ , i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$ .

The implication $1 \Longrightarrow 2$ is clear. For $5 \Longrightarrow 2$ , see Lemma 1 in this previous post on countable network. The implication $2 \Longrightarrow 3$ is left as an exercise. To see $3 \Longrightarrow 4$ , let $K$ be a compact subset of $X$ . The property of having a $G_\delta$ -diagonal is hereditary. Thus $K$ has a $G_\delta$ -diagonal. According to a well known result, any compact space with a $G_\delta$ -diagonal is metrizable (see here).

None of the implications in the diagram is reversible. The first uncountable ordinal $\omega_1$ is an example for $4 \not \Longrightarrow 3$ . This follows from the well known result that any countably compact space with a $G_\delta$ -diagonal is metrizable (see here). The Mrowka space is an example for $3 \not \Longrightarrow 2$ (see here). The Sorgenfrey line is an example for both $2 \not \Longrightarrow 5$ and $2 \not \Longrightarrow 1$ .

To see where the examples mentioned earlier are placed, note that Sorgenfrey line and Michael line are submetrizable, both are submetrizable by the usual Euclidean topology on the real line. Each compact subspace of the space $\omega_1$ is countable and is thus contained in some initial segment $[0,\alpha]$ which is metrizable. Any Moore space has a $G_\delta$ -diagonal. Thus compact subspaces of a Moore space are metrizable.

Function Spaces

We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) $X$ , $C_p(X)$ is the space of all continuous functions from $X$ into $\mathbb{R}$ with the pointwise convergence topology (see here for basic information on pointwise convergence topology).

Theorem 1
Suppose that $X$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

Proof
The proof here actually shows more than is stated in the theorem. We show that $C_p(X)$ is submetrizable by a separable metric topology. Let $Y$ be a countable dense subspace of $X$ . Then $C_p(Y)$ is metrizable and separable since it is a subspace of the separable metric space $\mathbb{R}^{\omega}$ . Thus $C_p(Y)$ has a countable base. Let $\mathcal{E}$ be a countable base for $C_p(Y)$ .

Let $\pi:C_p(X) \longrightarrow C_p(Y)$ be the restriction map, i.e. for each $f \in C_p(X)$ , $\pi(f)=f \upharpoonright Y$ . Since $\pi$ is a projection map, it is continuous and one-to-one and it maps $C_p(X)$ into $C_p(Y)$ . Thus $\pi$ is a continuous bijection from $C_p(X)$ into $C_p(Y)$ . Let $\mathcal{B}=\left\{\pi^{-1}(E): E \in \mathcal{E} \right\}$ .

We claim that $\mathcal{B}$ is a base for a topology on $C_p(X)$ . Once this is established, the proof of the theorem is completed. Note that $\mathcal{B}$ is countable and elements of $\mathcal{B}$ are open subsets of $C_p(X)$ . Thus the topology generated by $\mathcal{B}$ is coarser than the original topology of $C_p(X)$ .

For $\mathcal{B}$ to be a base, two conditions must be satisfied – $\mathcal{B}$ is a cover of $C_p(X)$ and for $B_1,B_2 \in \mathcal{B}$ , and for $f \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ such that $f \in B_3 \subset B_1 \cap B_2$ . Since $\mathcal{E}$ is a base for $C_p(Y)$ and since elements of $\mathcal{B}$ are preimages of elements of $\mathcal{E}$ under the map $\pi$ , it is straightforward to verify these two points. $\square$

Theorem 1 is actually a special case of a duality result in $C_p$ function space theory. More about this point later. First, consider a corollary of Theorem 1.

Corollary 2
Let $X=\prod_{\alpha<c} X_\alpha$ where $c$ is the cardinality continuum and each $X_\alpha$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

The key fact for Corollary 2 is that the product of continuum many separable spaces is separable (this fact is discussed here). Theorem 1 is actually a special case of a deep result.

Theorem 3
Suppose that $X=\prod_{\alpha<\kappa} X_\alpha$ is a product of separable spaces where $\kappa$ is any infinite cardinal. Then every compact subspace of $C_p(X)$ is metrizable.

Theorem 3 is a much more general result. The product of any arbitrary number of separable spaces is not separable if the number of factors is greater than continuum. So the proof for Theorem 1 will not work in the general case. This result is Problem 307 in [2].

A Duality Result

Theorem 1 is stated in a way that gives the right information for the purpose at hand. A more correct statement of Theorem 1 is: $X$ is separable if and only if $C_p(X)$ is submetrizable by a separable metric topology. Of course, the result in the literature is based on density and weak weight.

The cardinal function of density is the least cardinality of a dense subspace. For any space $Y$ , the weight of $Y$ , denoted by $w(Y)$ , is the least cardinaility of a base of $Y$ . The weak weight of a space $X$ is the least $w(Y)$ over all space $Y$ for which there is a continuous bijection from $X$ onto $Y$ . Thus if the weak weight of $X$ is $\omega$ , then there is a continuous bijection from $X$ onto some separable metric space, hence $X$ has a weaker separable metric topology.

There is a duality result between density and weak weight for $X$ and $C_p(X)$ . The duality result:

The density of $X$ coincides with the weak weight of $C_p(X)$ and the weak weight of $X$ coincides with the density of $C_p(X)$ . These are elementary results in $C_p$ -theory. See Theorem I.1.4 and Theorem I.1.5 in [1].

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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Extracting more information from Dowker’s theorem

Posted on May 1, 2017 by Dan Ma

Countably paracompact spaces are discussed in a previous post. The discussion of countably paracompactness in the previous post is through discussing Dowker’s theorem. In this post, we discuss a few more facts that can be derived from Dowker’s theorem.

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Dowker’s Theorem

Essentially, Dowker’s theorem is the statement that for a normal space $X$ , the space $X$ is countably paracompact if any only if $X \times Y$ is normal for any infinite compact metric space. The following is the full statement of Dowker’s theorem. The long list of equivalent conditions is important for applications in various scenarios.

Theorem 1 (Dowker’s Theorem)
Let $X$ be a normal space. The following conditions are equivalent.

The space $X$ is countably paracompact.
Every countable open cover of $X$ has a point-finite open refinement.
If $\left\{U_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$ , there exists an open refinement $\left\{V_n: n=1,2,3,\cdots \right\}$ such that $\overline{V_n} \subset U_n$ for each $n$ .
The product space $X \times Y$ is normal for any infinite compact metric space $Y$ .
The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
The product space $X \times S$ is normal where $S$ is a non-trivial convergent sequence with the limit point. Note that $S$ can be taken as a space homeomorphic to $\left\{1,\frac{1}{2},\frac{1}{3},\cdots \right\} \cup \left\{0 \right\}$ with the Euclidean topology.
For each sequence $\left\{A_n \subset X: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\cap_n A_n=\varnothing$ , there exist open sets $B_1,B_2,B_3,\cdots$ such that $A_n \subset B_n$ for each $n$ such that $\cap_n B_n=\varnothing$ .

A Dowker space is any normal space that is not countably paracompact. The notion of Dowker space was motivated by Dowker’s theorem since such a space would be a normal space $X$ for which $X \times [0,1]$ is not normal. The search for such a space took about 20 years from 1951 when C. H. Dowker proved the theorem to 1971 when M. E. Rudin constructed a ZFC example of a Dowker space.

Theorem 1 (Dowker’s theorem) is proved here and is further discussed in this previous post on countably paracompact space. The statement appears in Condition 6 here is not found in the previous version of the theorem. However, no extra effort is required to support it. Condition 5 trivially implies condition 6. The proof of condition 5 implying condition 7 (the proof of 4 implies 5 shown here) only requires that the product of $X$ and a convergent sequence is normal. So inserting condition 6 does not require extra proof.

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Getting More from Dowker’s Theorem

As a result of Theorem 1, normal countably paracompact spaces are productive in normality with respect to compact metric spaces (condition 4 in Dowker’s theorem as stated above). Another way to look at condition 4 is that the normality in the product $X \times Y$ is a strong property. Whenever the product $X \times Y$ is normal, we know that each factor is normal. Dowker’s theorem tells us that whenever $X \times Y$ is normal and one of the factor is a compact metric space such as the unit interval $[0,1]$ , the other factor is countably paracompact. The fact can be extended. Even if the factors are not metric spaces, as long as one of the factors has a non-discrete point with “countable” tightness, normality of the product confers countably paracompactness on one of the factors. The following two theorems make this clear.

Theorem 2
Suppose that the product $X \times Y$ is normal. If one of the factor contains a non-trivial convergent sequence, then the other factor is countably paracompact.

Proof of Theorem 2
Suppose $Y$ contains a non-trivial convergent sequence. Let this sequence be denoted by $S =\left\{ x_n:n=1,2,3,\cdots \right\} \cup \left\{x \right\}$ such that the point $x$ is the limit point. Since $X \times Y$ is normal, both $X$ and $Y$ are normal and that $X \times S$ is normal. By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 3
Suppose that the product $X \times Y$ is normal. If one of the factor contains a countable subset that is non-discrete, then the other factor is countably paracompact.

Proof of Theorem 3
To discuss this fact, we need to turn to the generalized Dowker’s theorem, which is Theorem 2 in this previous post. We will not re-state the theorem. The crucial direction is $7 \longrightarrow 4$ in that theorem. To avoid confusion, we call these two conditions A7 and A4. The following are the conditions.

$X \times Y$

$Y$

$\kappa$

$\left\{F_\alpha: \alpha<\kappa \right\}$

$X$

$\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$

$\left\{G_\alpha: \alpha<\kappa \right\}$

$X$

$\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$

$F_\alpha \subset G_\alpha$

$\alpha<\kappa$

Actually the proof in the previous post shows that A7 implies another condition that is equivalent to A4 for any infinite cardinal $\kappa$ . In particular, A7 $\longrightarrow$ A4 would hold for the countably infinite $\kappa=\omega$ . Note that under $\kappa=\omega$ , A4 would be the same as condition 7 in Theorem 1 above.

Thus by Theorem 2 in this previous post for the countably infinite case and by Theorem 1 in this post, the theorem is established. $\square$

Remarks
In Theorem 2, the second factor $Y$ does not have to be a metric space. As long as it has a non-trivial convergent sequence, the normality of the product (a big if in some situation) implies countably paracompactness in the other factor.

Theorem 3 is essentially a corollary of the proof of Theorem 2 in the previous post. One way to look at Theorem 3 is that the normality of the product $X \times Y$ is a strong statement. If the product is normal and if one factor has a countable non-discrete subspace, then the other factor is countably paracompact. Another way to look at it is through the angle of Dowker spaces. By Dowker’s theorem (Theorem 1), the product of any Dowker space with any infinite compact metric space is not normal. The pathology is actually more severe. A Dowker space is severely lacking in ability to form normal product, as the following corollary makes clear.

Corollary 4
If $X$ is a Dowker space, then $X \times Y$ is not normal for any space $Y$ containing a non-discrete countable subspace.

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More Results

Two more results are discussed. According to Dowker’s theorem, the product of a countably paracompact space $X$ and any compact metric space is normal. In particular, $X \times [0,1]$ is normal. Theorem 5 is saying that with a little extra work, it can be shown that $X \times \mathbb{R}$ is normal. What makes this works is that the metric factor is $\sigma$ -compact.

Theorem 5
Let $X$ be a normal space. The following conditions are equivalent.

The space $X$ is countably paracompact.
The product space $X \times Y$ is normal for any non-discrete $\sigma$ -compact metric space $Y$ .
The product space $X \times \mathbb{R}$ is normal where $\mathbb{R}$ is the real number line with the usual Euclidean topology.

Proof of Theorem 5
$1 \rightarrow 2$
Suppose that $X$ is countably paracompact. Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. Since $Y$ is a $\sigma$ -compact metric space, it is Lindelof. The Lindelof number and the weight agree in a metric space. Thus $Y$ has a countable base. According to Urysohn’s metrization theorem (discussed here), $Y$ can be embedded into the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$ . For convenience, we consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$ . Furthermore, $X \times Y=\bigcup_{j=1}^\infty (X \times Y_j) \subset X \times\prod_{j=1}^\infty W_j$ .

By Theorem 1, each $X \times Y_j$ is normal and that $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times Y$ is an $F_\sigma$ -subset of the normal space $X \times\prod_{j=1}^\infty W_j$ . Since normality is passed to $F_\sigma$ -subsets, $X \times Y$ is normal.

Note. For a proof that $F_\sigma$ -subsets of normal spaces are normal, see 2.7.2(b) on p. 112 of Englelking [1].

$2 \rightarrow 3$ is immediate.

$3 \rightarrow 1$
Suppose that $X \times \mathbb{R}$ is normal. Then $X \times [0,1]$ is normal since it is a closed subspace of $X \times \mathbb{R}$ . By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 6
Let $X$ be a normal space. Let $Y$ be a non-discrete $\sigma$ -compact metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

Proof of Theorem 6
Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. As in the proof of Theorem 5, we use the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$ .

Suppose that $X \times Y$ is normal. Since $Y$ is a non-discrete metric space, $Y$ contains a countable non-discrete subspace. Then by either Theorem 2 or Theorem 3, $X$ is countably paracompact.

By Theorem 1, $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times \prod_{j=1}^\infty W_j \times [0,1]$ is normal since $(\prod_{j=1}^\infty W_j) \times [0,1]$ is a compact metric space. By Theorem 1 again, $X \times\prod_{j=1}^\infty W_j$ is countably paracompact.

As in the proof of Theorem 5, we can consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$ . Furthermore, $X \times Y=\bigcup_{j=1}^\infty X \times Y_j \subset X \times\prod_{j=1}^\infty W_j$ .

Note that $X \times Y$ is $F_\sigma$ -subset of the countably paracompact space $X \times\prod_{j=1}^\infty W_j$ . Since countably paracompactness is passed to $F_\sigma$ -subsets, we conclude that $X \times Y$ is countably paracompact.

Note. For a proof that countably paracompactness is passed to $F_\sigma$ -subsets, see the proof that paracompactness is passed to $F_\sigma$ -subsets in this previous post. Just apply the same proof but start with a countable open cover.

For the other direction, suppose that $X \times Y$ is countably paracompact. Since $X \times \left\{y \right\}$ is a closed subspace of $Y$ with $y \in Y$ and is a copy of $X$ , $X$ is countably paracompact. Then by Theorem 5, $X \times Y$ is a normal space. $\square$

Remarks
Theorem 5 seems like an extension of Theorem 1. But the amount of extra work is very little. So normal countably paracompact spaces are productive with not just compact metric spaces but also with $\sigma$ -compact metric spaces. The $\sigma$ -compactness is absolutely crucial. The product of a normal countably paracompact space with a metric space does not have to be normal. For example, the Michael line $\mathbb{M}$ is paracompact and thus countably paracompact. The product of $\mathbb{M}$ and metric space is not necessarily normal (discussed here). However, the product of $\mathbb{M}$ and $\mathbb{R}$ or other $\sigma$ -compact metric space is normal.

Recall that a space is called a Dowker space if it is normal and not countably paracompact. For the type of product $X \times Y$ discussed in Theorem 6, it cannot be Dowker (if it is normal, it is countably paracompact). The two notions are the same with such product $X \times Y$ . Theorem 6 actually holds for a wider class than indicated. The following is Corollary 4.3 in [2].

Theorem 7
Let $X$ be a normal space. Let $Y$ be a non-discrete metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

So $\sigma$ -compactness is not necessary for Theorem 6. However, when the metric factor is $\sigma$ -compact, the proof is simplified considerably. For the full proof, see Corollary 4.3 in [2].

Among the products $X \times Y$ , the two notions of normality and countably paracompactness are the same as long as one factor is normal and the other factor is a non-discrete metric space. For such product, determining normality is equivalent to determining countably paracompactness, a covering property. In showing countably paracompactness, a shrinking property as well as a condition about decreasing sequence of closed sets being expanded by open sets (see Theorem 4 and Theorem 5 in this previous post) can be used.

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Reference

Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

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$\copyright$ 2017 – Dan Ma

Dan Ma's Topology Blog

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