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 \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.

  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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