# In between G-delta diagonal and submetrizable

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.

1. The space $X$ has a $G_\delta$-diagonal.
2. 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 $X=\cap \{U_n: n=0,1,2,\cdots \}$ 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.

1. The space $X$ has a regular $G_\delta$-diagonal.
2. 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.

1. The sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a rank-$k$ diagonal sequence for the space $X$.
2. For any two distinct points $x$ and $y$ of $X$, there is an integer $n$ such that $y \notin St^k(x,\mathcal{U}_n)$.
3. 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

1. Arhangelskii A. V., Burke D. K., Spaces with a regular $G_\delta$-diagonal, Topology and its Applications, Vol. 153, No. 11, 1917–1929, 2006.
2. Arhangelskii A. V., Buzyakova R. Z., The rank of the diagonal and submetrizability, Comment. Math. Univ. Carolinae, Vol. 47, No. 4, 585-597, 2006.
3. Buzyakova R. Z., Cardinalities of ccc-spaces with regular $G_\delta$-diagonals, Topology and its Applications, Vol. 153, 1696–1698, 2006.
4. Buzyakova R. Z., Observations on spaces with zeroset or regular $G_\delta$-diagonals, Comment. Math. Univ. Carolinae, Vol. 46, No. 3, 469-473, 2005.
5. 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.
6. Martin H. W., Contractibility of topological spaces onto metric spaces, Pacific J. Math., Vol. 61, No. 1, 209-217, 1975.
7. Xuan Wei-Feng, Shi Wei-Xue, On spaces with rank k-diagonals or zeroset diagonals, Topology Proceddings, Vol. 51, 245{251, 2018.
8. Yu Zuoming, Yun Ziqiu, A note on the rank of diagonals, Topology and its Applications, Vol. 157, 1011–1014, 2010.

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# Every Corson compact space has a dense first countable subspace

In any topological space $X$, a point $x \in X$ is a $G_\delta$ point if the one-point set $\left\{ x \right\}$ is the intersection of countably many open subsets of $X$. It is well known that any compact Hausdorff space is first countable at every $G_\delta$ point, i.e., if a point of a compact space is a $G_\delta$ point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space $[0,1]^{\omega_1}$ can be a $G_\delta$ point. In this post, we show that any Corson compact space has a dense set of $G_\delta$ point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.

The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.

For any infinite cardinal number $\kappa$, the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$ is the following subspace of $\mathbb{R}^\kappa$:

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^\kappa: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \kappa \right\}$

A compact space is said to be a Corson compact space if it can be embedded in $\Sigma(\kappa)$ for some infinite cardinal $\kappa$.

For each $x \in \Sigma(\kappa)$, let $S(x)$ denote the support of the point $x$, i.e., $S(x)$ is the set of all $\alpha<\kappa$ such that $x_\alpha \ne 0$.

Proposition 1
Let $Y$ be a Corson compact space. Then $Y$ has a $G_\delta$ point.

Proof of Proposition 1
If $Y$ is finite, then every point is isolated and is thus a $G_\delta$ point. Assume $Y$ is infinite. Let $\kappa$ be an infinite cardinal number such that $Y \subset \Sigma(\kappa)$. For $f,g \in Y$, define $f \le g$ if the following holds:

$\forall \ \alpha \in S(f)$, $f(\alpha)=g(\alpha)$

It is relatively straightforward to verify that the following three properties are satisfied:

• $f \le f$ for all $f \in Y$. (reflexivity)
• For all $f,g \in Y$, if $f \le g$ and $g \le f$, then $f=g$. (antisymmetry)
• For all $f,g,h \in Y$, if $f \le g$ and $g \le h$, then $f \le h$. (transitivity)

Thus $\le$ as defined here is a partial order on the compact space $Y$. Let $C \subset Y$ such that $C$ is a chain with respect to $\le$, i.e., for all $f,g \in C$, $f \le g$ or $g \le f$. We show that $C$ has an upper bound (in $Y$) with respect to the partial order $\le$. We need this for an argument using Zorn’s lemma.

Let $W=\bigcup_{f \in C} S(f)$. For each $\alpha \in W$, choose some $f \in C$ such that $\alpha \in S(f)$ and define $u_\alpha=f_\alpha$. For all $\alpha \notin W$, define $u_\alpha=0$. Because $C$ is a chain, the point $u$ is well-defined. It is also clear that $f \le u$ for all $f \in C$. If $u \in Y$, then $u$ is a desired upper bound of $C$. So assume $u \notin Y$. It follows that $u$ is a limit point of $C$, i.e., every open set containing $u$ contains a point of $C$ different from $u$. Hence $u$ is a limit point of $Y$ too. Since $Y$ is compact, $u \in Y$, a contradiction. Thus it must be that $u \in Y$. Thus every chain in the partially ordered set $(Y,\le)$ has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order $\le$, i.e., there exists $t \in Y$ such that $f \le t$ for all $f \in Y$.

We now show that $t$ is a $G_\delta$ point in $Y$. Let $S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}$. For each $p \in \mathbb{R}$ and for each positive integer $n$, let $B_{p,n}$ be the open interval $B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n})$. For each positive integer $n$, define the open set $O_n$ as follows:

$O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y$

Note that $t \in \bigcap_{n=1}^\infty O_n$. Because $t$ is a maximal element, note that if $g \in Y$ such that $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$, then it must be the case that $g=t$. Thus if $g \in \bigcap_{n=1}^\infty O_n$, then $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$. We have $\left\{t \right\}= \bigcap_{n=1}^\infty O_n$. $\blacksquare$

Lemma 2
Let $Y$ be a compact space such that for every non-empty compact subspace $K$ of $Y$, there exists a $G_\delta$ point in $K$. Then every non-empty open subset of $Y$ contains a $G_\delta$ point.

Proof of Lemma 2
Let $U_1$ be a non-empty open subset of the compact space $Y$. If there exists $y \in U_1$ such that $\left\{y \right\}$ is open in $Y$, then $y$ is a $G_\delta$ point. So assume that every point of $U_1$ is a non-isolated point of $Y$. By regularity, choose an open subset $U_2$ of $Y$ such that $\overline{U_2} \subset U_1$. Continue in the same manner and obtain a decreasing sequence $U_1,U_2,U_3,\cdots$ of open subsets of $Y$ such that $\overline{U_{n+1}} \subset U_n$ for each positive integer $n$. Then $K=\bigcap_{n=1}^\infty \overline{U_n}$ is a non-empty closed subset of $Y$ and thus compact. By assumption, $K$ has a $G_\delta$ point, say $p \in K$.

Then $\left\{p \right\}=\bigcap_{n=1}^\infty W_n$ where each $W_n$ is open in $K$. For each $n$, let $V_n$ be open in $Y$ such that $W_n=V_n \cap K$. For each $n$, let $V_n^*=V_n \cap U_n$, which is open in $Y$. Then $\left\{p \right\}=\bigcap_{n=1}^\infty V_n^*$. This means that $p$ is a $G_\delta$ point in the compact space $Y$. Note that $p \in U_1$, the open set we start with. This completes the proof that every non-empty open subset of $Y$ contains a $G_\delta$ point. $\blacksquare$

Proposition 3
Let $Y$ be a Corson compact space. Then $Y$ has a dense set of $G_\delta$ points.

Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of $Y$ is also Corson compact. By Proposition 1, every compact subspace of $Y$ has a $G_\delta$ point. By Lemma 2, $Y$ has a dense set of $G_\delta$ points. $\blacksquare$

Corollary 4
Every Corson compact space has a dense first countable subspace.

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Blog posts on Corson compact spaces

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Reference

1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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

# The Sorgenfrey plane is subnormal

The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form $[a,b)$. The Sorgenfrey line is one of the most important counterexamples in general topology. One of the often recited facts about this counterexample is that the Sorgenfrey plane (the square of the Sorgengfrey line) is not normal. We show that, though far from normal, the Sorgenfrey plane is subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal. The Sorgenfrey plane is an example of a subnormal space that is not normal.

In the proof of the non-normality of the Sorgenfrey plane in this previous post, one of the two disjoint closed subsets of the Sorgenfrey plane that cannot be separated by disjoint open sets is countable. Thus the Sorgenfrey plane is not only not normal; it is not pseudonormal (also discussed in this previous post). A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The examples of the Sorgenfrey plane and $\omega_1 \times (\omega_1+1)$ show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal but not subnormal (see this previous post for the non-subnormality).

A space $Y$ is said to be a perfect space if every closed subset of $Y$ is a $G_\delta$ subset of $Y$ (equivalently, every open subset of $Y$ is an $F_\sigma$-subset of $Y$). It is clear that any perfect space is subnormal. We show that the Sorgenfrey plane is perfect. There are subnormal spaces that are not perfect (see the example below).

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The Sorgenfrey plane is perfect

Let $S$ denote the Sorgenfrey line, i.e., the real line $\mathbb{R}$ topologized using the base of half-open intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x . The Sorgenfrey plane is the product space $S \times S$. We show the following:

Proposition 1
The Sorgenfrey line $S$ is perfect.

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

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

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

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

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

Proof of Proposition 2
Let $X$ be perfect. Let $Y$ be a space with a base $\mathcal{B}=\bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$, in addition to being a collection of basic open sets, is a discrete collection. The existence of such a base is equivalent to metrizability, a well known result called Bing’s metrization theorem (see Theorem 4.4.8 in [1]). Let $U$ be a non-empty open subset of $X \times Y$. We show that it is an $F_\sigma$-set in $X \times Y$. For each $x \in U$, there is some open subset $V$ of $X$ and there is some $W \in \mathcal{B}$ such that $x \in V \times W$ and $V \times \overline{W} \subset U$. Thus $U$ is the union of a collection of sets of the form $V \times \overline{W}$. Thus we have:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$x_1=g_1=t_1$

$x_2

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

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

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

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

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

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

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

$x_1

$x_2

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

Remarks
The authors of [2] showed that any finite power of the Sorgenfrey line is perfect. The proof in [2] is an inductive proof: if $S^n$ is perfect, then $S^{n+1}$ is perfect. We take the inductive proof in [2] and adapt it for the Sorgenfrey plane. The authors in [2] also proved that for a sequence of spaces $X_1,X_2,X_3,\cdots$ such that the product of any finite number of these spaces is perfect, the product $\prod_{n=1}^\infty X_n$ is perfect. Then $S^\omega$ is perfect.

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A non-perfect example

Any perfect space is subnormal. Subnormal spaces do not have to be perfect. In fact subnormal non-normal spaces do not have to be perfect. From a perfect space that is not normal (e.g. the Sorgenfrey plane), one can generate a subnormal and non-normal space that is not perfect. Let $X$ be a subnormal and non-normal space. Let $Y$ be a normal space that is not perfectly normal. There are many possible choices for $Y$. If a specific example is needed, one can take $Y=\omega_1$ with the order topology. Let $X \bigoplus Y$ be the disjoint sum (union) of $X$ and $Y$. The presence of $Y$ destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint $G_\delta$-sets.

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Reference

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

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

# Normal x compact needs not be subnormal

In this post, we revisit a counterexample that was discussed previously in this blog. A previous post called “Normal x compact needs not be normal” shows that the Tychonoff product of two normal spaces needs not be normal even when one of the factors is compact. The example is $\omega_1 \times (\omega_1+1)$. In this post, we show that $\omega_1 \times (\omega_1+1)$ fails even to be subnormal. Both $\omega_1$ and $\omega_1+1$ are spaces of ordinals. Thus they are completely normal (equivalent to hereditarily normal). The second factor is also a compact space. Yet their product is not only not normal; it is not even subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal.

A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal (see this previous post). The Sorgenfrey plane is an example of a subnormal space that is not pseudonormal (see here). Thus the two weak forms of normality (pseudonormal and subnormal) are not equivalent.

The same two disjoint closed sets that prove the non-normality of $\omega_1 \times (\omega_1+1)$ are also used for proving non-subnormality. The two closed sets are:

$H=\left\{(\alpha,\alpha): \alpha<\omega_1 \right\}$

$K=\left\{(\alpha,\omega_1): \alpha<\omega_1 \right\}$

The key tool, as in the proof for non-normality, is the Pressing Down Lemma ([1]). The lemma has been used in a few places in this blog, especially for proving facts about $\omega_1$ (e.g. this previous post on the first uncountable ordinal). Lemma 1 below is a lemma that is derived from the Pressing Down Lemma.

Pressing Down Lemma
Let $S$ be a stationary subset of $\omega_1$. Let $f:S \rightarrow \omega_1$ be a pressing down function, i.e., $f$ satisfies: $\forall \ \alpha \in S, f(\alpha)<\alpha$. Then there exists $\alpha<\omega_1$ such that $f^{-1}(\alpha)$ is a stationary set.

Lemma 1
Let $L=\left\{(\alpha,\alpha) \in \omega_1 \times \omega_1: \alpha \text{ is a limit ordinal} \right\}$. Suppose that $L \subset \bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open subset of $\omega_1 \times \omega_1$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty O_n$ for some $\gamma<\omega_1$.

Proof of Lemma 1
For each $n$ and for each $\alpha<\omega_1$ where $\alpha$ is a limit, choose $g_n(\alpha)<\alpha$ such that $[g_n(\alpha),\alpha] \times [g_n(\alpha),\alpha] \subset O_n$. The function $g_n$ can be chosen since $O_n$ is open in the product $\omega_1 \times \omega_1$. By the Pressing Down Lemma, for each $n$, there exists $\gamma_n < \omega_1$ and there exists a stationary set $S_n \subset \omega_1$ such that $g_n(\alpha)=\gamma_n$ for all $\alpha \in S_n$. It follows that $[\gamma_n,\omega_1) \times [\gamma_n,\omega_1) \subset O_n$ for each $n$. Choose $\gamma<\omega_1$ such that $\gamma_n<\gamma$ for all $n$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset O_n$ for each $n$. $\blacksquare$

Theorem 2
The product space $\omega_1 \times (\omega_1+1)$ is not subnormal.

Proof of Theorem 2
Let $H$ and $K$ be defined as above. Suppose $H \subset \bigcap_{n=1}^\infty U_n$ and $K \subset \bigcap_{n=1}^\infty V_n$ where each $U_n$ and each $V_n$ are open in $\omega_1 \times (\omega_1+1)$. Without loss of generality, we can assume that $U_n \cap (\omega_1 \times \left\{\omega_1 \right\})=\varnothing$, i.e., $U_n$ is open in $\omega_1 \times \omega_1$ for each $n$. By Lemma 1, $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$ for some $\gamma<\omega_1$.

Choose $\beta>\gamma$ such that $\beta$ is a successor ordinal. Note that $(\beta,\omega_1) \in \bigcap_{n=1}^\infty V_n$. For each $n$, there exists some $\delta_n<\omega_1$ such that $\left\{\beta \right\} \times [\delta_n,\omega_1] \subset V_n$. Choose $\delta<\omega_1$ such that $\delta >\delta_n$ for all $n$ and that $\delta >\gamma$. Note that $\left\{\beta \right\} \times [\delta,\omega_1) \subset \bigcap_{n=1}^\infty V_n$. It follows that $\left\{\beta \right\} \times [\delta,\omega_1) \subset [\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$. Thus there are no disjoint $G_\delta$ sets separating $H$ and $K$. $\blacksquare$

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Reference

1. Kunen, K., Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York, 1980.

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