Helly Space

This is a discussion on a compact space called Helly space. The discussion here builds on the facts presented in Counterexample in Topology [2]. Helly space is Example 107 in [2]. The space is named after Eduard Helly.

Let I=[0,1] be the closed unit interval with the usual topology. Let C be the set of all functions f:I \rightarrow I. The set C is endowed with the product space topology. The usual product space notation is I^I or \prod_{t \in I} W_t where each W_t=I. As a product of compact spaces, C=I^I is compact.

Any function f:I \rightarrow I is said to be increasing if f(x) \le f(y) for all x<y (such a function is usually referred to as non-decreasing). Helly space is the subspace X consisting of all increasing functions. This space is Example 107 in Counterexample in Topology [2]. The following facts are discussed in [2].

  • The space X is compact.
  • The space X is first countable (having a countable base at each point).
  • The space X is separable.
  • The space X has an uncountable discrete subspace.

From the last two facts, Helly space is a compact non-metrizable space. Any separable metric space would have countable spread (all discrete subspaces must be countable).

The compactness of X stems from the fact that X is a closed subspace of the compact space C.

Further Discussion

Additional facts of concerning Helly space are discussed.

  1. The product space \omega_1 \times X is normal.
  2. Helly space X contains a copy of the Sorgenfrey line.
  3. Helly space X is not hereditarily normal.

The space \omega_1 is the space of all countable ordinals with the order topology. Recall C is the product space I^I. The product space \omega_1 \times C is Example 106 in [2]. This product is not normal. The non-normality of \omega_1 \times C is based on this theorem: for any compact space Y, the product \omega_1 \times Y is normal if and only if the compact space Y is countably tight. The compact product space C is not countably tight (discussed here). Thus \omega_1 \times C is not normal. However, the product \omega_1 \times X is normal since Helly space X is first countable.

To see that X contains a copy of the Sorgenfrey line, consider the functions h_t:I \rightarrow I defined as follows:

    \displaystyle  h_t(x) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ 0 \le x \le t \\            \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ t<x \le 1 \\                                 \end{array} \right.

for all 0<t<1. Let S=\{ h_t: 0<t<1 \}. Consider the mapping \gamma: (0,1) \rightarrow S defined by \gamma(t)=h_t. With the domain (0,1) having the Sorgenfrey topology and with the range S being a subspace of Helly space, it can be shown that \gamma is a homeomorphism.

With the Sorgenfrey line S embedded in X, the square X \times X contains a copy of the Sorgenfrey plane S \times S, which is non-normal (discussed here). Thus the square of Helly space is not hereditarily normal. A more interesting fact is that Helly space is not hereditarily normal. This is discussed in the next section.

Finding a Non-Normal Subspace of Helly Space

As before, C is the product space I^I where I=[0,1] and X is Helly space consisting of all increasing functions in C. Consider the following two subspaces of X.

    Y_{0,1}=\{ f \in X: f(I) \subset \{0, 1 \} \}

    Y=X - Y_{0,1}

The subspace Y_{0,1} is a closed subset of X, hence compact. We claim that subspace Y is separable and has a closed and discrete subset of cardinality continuum. This means that the subspace Y is not a normal space.

First, we define a discrete subspace. For each x with 0<x<1, define f_x: I \rightarrow I as follows:

    \displaystyle  f_x(y) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ 0 \le y < x \\           \text{ } & \text{ } \\          \displaystyle  \frac{1}{2} &\ \ \ \ \ y=x \\            \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ x<y \le 1 \\                                 \end{array} \right.

Let H=\{ f_x: 0<x<1 \}. The set H as a subspace of X is discrete. Of course it is not discrete in X since X is compact. In fact, for any f \in Y_{0,1}, f \in \overline{H} (closure taken in X). However, it can be shown that H is closed and discrete as a subset of Y.

We now construct a countable dense subset of Y. To this end, let \mathcal{B} be a countable base for the usual topology on the unit interval I=[0,1]. For example, we can let \mathcal{B} be the set of all open intervals with rational endpoints. Furthermore, let A be a countable dense subset of the open interval (0,1) (in the usual topology). For convenience, we enumerate the elements of A and \mathcal{B}.

    A=\{ a_1,a_2,a_3,\cdots \}

    \mathcal{B}=\{B_1,B_2,B_3,\cdots \}

We also need the following collections.

    \mathcal{G}=\{G \subset \mathcal{B}: G \text{ is finite and is pairwise disjoint} \}

    \mathcal{A}=\{F \subset A: F \text{ is finite} \}

For each G \in \mathcal{G} and for each F \in \mathcal{A} with \lvert G \lvert=\lvert F \lvert=n, we would like to arrange the elements in increasing order, notated as follow:

    F=\{t_1,t_2,\cdots,t_n \}

    G=\{E_1,E_2,\cdots,E_n \}

For the set F, we have 0<t_1<t_2< \cdots <t_n<1. For the set G, E_i is to the left of E_j for i<j. Note that elements of G are pairwise disjoint. Furthermore, write E_i=(p_i,q_i). If 0 \in E_1, then E_1=[p_1,q_1)=[0,q_1). If 1 \in E_n, then E_n=(p_n,q_n]=(p_n,1].

For each F and G as detailed above, we define a function L(F,G):I \rightarrow I as follows:

    \displaystyle  L(F,G)(x) = \left\{ \begin{array}{ll}                     \displaystyle  t_1 &\ \ \ \ \ 0 \le x < q_1 \\           \text{ } & \text{ } \\          \displaystyle  t_2 &\ \ \ \ \ q_1 \le x < q_2 \\           \text{ } & \text{ } \\          \displaystyle  \vdots &\ \ \ \ \ \vdots \\           \text{ } & \text{ } \\          \displaystyle  t_{n-1} &\ \ \ \ \ q_{n-2} \le x < q_{n-1} \\           \text{ } & \text{ } \\          \displaystyle  t_n &\ \ \ \ \ q_{n-1} \le x \le 1 \\                                             \end{array} \right.

The following diagram illustrates the definition of L(F,G) when both F and G have 4 elements.

Figure 1 – Member of a countable dense set

Let D be the set of L(F,G) over all F \in \mathcal{A} and G \in \mathcal{G}. The set D is a countable set. It can be shown that D is dense in the subspace Y. In fact D is dense in the entire Helly space X.

To summarize, the subspace Y is separable and has a closed and discrete subset of cardinality continuum. This means that Y is not normal. Hence Helly space X is not hereditarily normal. According to Jones’ lemma, in any normal separable space, the cardinality of any closed and discrete subspace must be less than continuum (discussed here).

Remarks

The preceding discussion shows that both Helly space and the square of Helly space are not hereditarily normal. This is actually not surprising. According to a theorem of Katetov, for any compact non-metrizable space V, the cube V^3 is not hereditarily normal (see Theorem 3 in this post). Thus a non-normal subspace is found in V, V \times V or V \times V \times V. In fact, for any compact non-metric space V, an excellent exercise is to find where a non-normal subspace can be found. Is it in V, the square of V or the cube of V? In the case of Helly space X, a non-normal subspace can be found in X.

A natural question is: is there a compact non-metric space V such that both V and V \times V are hereditarily normal and V \times V \times V is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyond ZFC, any compact non-metric space V is likely to fail hereditarily normality in either V or V \times V. See here for a discussion of this set-theoretic question.

Reference

  1. Kelly, J. L., General Topology, Springer-Verlag, New York, 1955.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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Drawing more Sorgenfrey continuous functions

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

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

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A stroll in Bing’s Example G

In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, F \times [0,1], is a normal space.

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The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let P be any uncountable set. Let Q be the set of all subsets of P. In other words, Q is the power set of P. Consider the product of \lvert Q \lvert many copies of the two element set \left\{0,1 \right\}. The usual notation of this product space is 2^Q. The elements of 2^Q are simply the functions from Q into \left\{0,1 \right\}. An arbitrary element of 2^Q is a function f that maps every subset of P to either 0 or 1.

Though the base set P can be any uncountable set, it is a good idea to visualize clearly what P is. In the remainder of this section, think of P as the real line \mathbb{R}. Then Q is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are 2^c many subsets of the real line where c is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of 2^Q. For each real number p, define the function f_p such that f_p always maps any set of real numbers that contains p to 1 and maps any set of real numbers that does not contain p to 0. For example, the following are several values of the function f_0.

    f_0([0,1])=1

    f_0([1,2])=0

    f_0(\left\{0 \right\})=1

    f_0(\mathbb{R}-\left\{0 \right\})=0

    f_0(\mathbb{R})=1

    f_0(\varnothing)=0

    f_0(\mathbb{P})=0

where \mathbb{P} is the set of all irrational numbers. Consider the subspace F_P=\left\{f_p: p \in P \right\}. Members of F_P are easy to describe. Each function in F_P maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that F_P is important is that Bing’s Example is defined by declaring all points not in F_P isolated points and by allowing all points in F_P retaining the open sets in the product topology.

Any point f in F_P determines f(q)=0 \text{ or } 1 based on membership (whether the reference point belongs to the set q). Points not in F_P have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any f in F_P maps equally to 0 or 1. So the constant functions f(q)=0 and f(q)=1 are not in F_P. Furthermore, any f such that f(q)=1 for at most countably many q would not be in F_P.

Let’s continue focusing on the product space for the time being. When F_P is considered as a subspace of the product space 2^Q, F_P is a discrete space. For each p \in P, there is an open set W_p containing f_p such that W_p contains no other points of F_P. So F_P is relatively discrete in the product space 2^Q. Of course F_P cannot be closed in 2^Q since 2^Q is a compact space. The open set W_p is defined as follows:

    W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}

It is clear that f_p \in W_p and that f_t \notin W_p for any real number t \ne p.

Two properties of the product space 2^Q would be very relevant for the discussion. By the well known Tychonoff theorem, the product space 2^Q is compact. Since P is uncountable, 2^Q always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets W_p cannot be all pairwise disjoint. So there exist at least a pair of W_p, say W_{a} and W_{b}, with nonempty intersection.

The last observation can be generalized. For each p \in P, let V_p be any open set containing f_p (open in the product topology). We observe that there are at least two a and b from P such that V_a \cap V_b \ne \varnothing. If there are only countably many distinct sets V_p, then there are uncountably many V_p that are identical and the observation is valid. So assume that there are uncountably many distinct V_p. By the CCC in the product space, there are at least two a and b with V_a \cap V_b \ne \varnothing. This observation shows that the discrete points in F_P cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint A_1, A_2 \subset F_P can be separated by disjoint open sets. To see this, define the following two open sets E_1 and E_2 in the product topology.

    q_1=\left\{p \in P: f_p \in A_1 \right\}

    q_2=\left\{p \in P: f_p \in A_2 \right\}

    E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}

    E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}

It is clear that A_1 \subset E_1 and A_2 \subset E_2. Furthermore, E_1 \cap E_2=\varnothing. This observation will be the basis for showing that Bing’s Example G is normal.

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The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on 2^Q. Let P be any uncountable set. Let Q be the set of all subsets of P. The set F_P is defined as above. Bing’s Example G is F=2^Q with points in F_P retaining the open sets in the product topology and with points not in F_P declared isolated. For some reason, in Bing’s original paper, the notation F is used even though the example is identified by G. We will follow Bing’s notation.

The subspace F_P is discrete but not closed in the product topology. However, F_P is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space F is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set F_P cannot be separated by disjoint open sets. By declaring points not in F_P isolated, the countable chain condition in the original product space 2^Q is destroyed. However, there is still a strong trace of CCC around the points in the set F_P, which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that F is normal, let H and K be disjoint closed subsets of F. To make it easy to follow, let H=A_1 \cup B_1 and K=A_2 \cup B_2 where

    A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)

    A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)

In other words, A is the non-isolated part and B is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets E_1 and E_2 where A_1 \subset E_1 and A_2 \subset E_2. Set the following open sets.

    O_1=(E_1 \cup B_1) - K

    O_2=(E_2 \cup B_2) - H

It follows that O_1 and O_2 are disjoint open sets and that A_1 \subset O_1 and A_2 \subset O_2. Thus Bing’s Example G is a normal space.

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Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space X is countably paracompact if every countable countable open cover of X has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every countable open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent (see Corollary 2 here). Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let \mathcal{U} be a countable open cover of F. Let \mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\} be the set of all open sets in \mathcal{U} that contain points in F_P. For each i, let A_i=U_i \cap F_P. From the perspective of Bing’s Example G, the sets A_i are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let O_1,O_2,O_3,\cdots be disjoint open sets such that A_i \subset O_i for each i.

We now build a point-finite open refinement of \mathcal{U}. For each i, let V_i=U_i \cap O_i. Let V=\cup_{i=1}^\infty V_i. Consider the following.

    \mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}

It follows that \mathcal{V} is an open cover of F. All points of F_P belong to the open sets V_i. Any point that is not in one of the V_i belongs to a singleton open set. It is also clear that \mathcal{V} is a refinement of \mathcal{U}. For each i, V_i \subset U_i and each singleton set is contained in some member of \mathcal{U}. It follows that each point in F belongs to at most finitely many sets in \mathcal{V}. In fact, each point belongs to exactly one set in \mathcal{V}. Each point in F_P belongs to exactly one V_i since the open sets O_i are disjoint. Any point in V belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, F \times [0,1], is a normal space.

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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 <b \right\}. 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<w<t 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<t_2 \right\}

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<g_2<t_2. Consider the following set:

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

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<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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<t_1 \text{ and }y_2<t_2 \right\}

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<g_1<t_1 and x_2<g_2<t_2. Consider the following set:

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

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<g_1<t_1<x_1+\frac{1}{k}<g_1+\frac{1}{k}

    x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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}

An exercise involving non-normal spaces

A space is normal if any two disjoint closed subsets of the space can be separated by disjoint open sets. A space is pseudonormal if any two disjoint closed subsets of the space, one of which is countable, can be separated by disjoint open sets. In this post, we present an interesting exercise that deals with non-normal spaces:

    Take a space that is not normal. Then determine whether it is pseudonormal. You can supply your own examples or you can start with several non-normal spaces listed below. Once you have a list, determine which ones are psuedonormal and which ones are not.

To make the exercise more interesting, we propose that the focus is on spaces that are T_1 (i.e. singleton sets are closed) and regular. Since regular Lindelof spaces are normal, we will be certain that any non-normal (and regular) space is not Lindelof.

In the previous post called Pseudonormal spaces, we identify four spaces that are known to be non-normal. Three of these spaces are not normal because one countable closed set and another closed set cannot be separated, hence not pseudonormal (one is the Sorgenfrey plane and one is the Niemmytzkis’ plane). The fourth non-normal space is pseudonormal.

Here’s a list of several other non-normal spaces previously discussed in this blog.

  • The Tychonoff Plank.
  • The sigma-product of \omega_1 many copies of \omega_1+1.
  • The product space \omega^{\omega_1}.
  • The product of the Michael line and the space of irrationals.
  • The product of countably many copies of the Michael line.
  • The product of a Lindelof space and a Bernstein set.
  • The Pixley-Roy space \mathcal{F}[\mathbb{R}].
  • Mrowka space, defined on a maximal almost disjoint family of subsets of \omega.

Readers are welcome to submit other examples of non-normal spaces. Submit examples by entering a comment below. Submitted examples that are different from the ones listed above will be appended to this post.

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\copyright \ 2014 \text{ by Dan Ma}

Pseudonormal spaces

When two disjoint closed sets in a topological space cannot be separated by disjoint open sets, the space fails to be a normal space. When one of the two closed sets is countable, the space fails to satisfy a weaker property than normality. A space X is said to be a pseudonormal space if H and K can always be separated by two disjoint open sets whenever H and K are disjoint closed subsets of X and one of them is countable. In this post, we discuss several non-normal spaces that actually fail to be pseudonormal. We also give an example of a pseudonormal space that is not normal.

We work with spaces that are at minimum T_1 spaces, i.e., spaces in which singleton sets are closed. Then any pseudonormal space is regular. To see this, let X be T_1 and pseudonormal. For any closed subset C of X and for any point x \in X-C, we can always separate the disjoint closed sets \left\{ x \right\} and C by disjoint open sets. This is one reason why we insist on having T_1 separation axiom as a starting point. We now show some examples of spaces that fail to be pseudonormal.

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Some Non-Pseudonormal Examples

All three examples in this section are spaces where the failure of normality is exhibited by the inability of separating a countable closed set and another disjoint closed set.

Example 1
This example of a non-normal space that fails to be pseudonormal is defined in the previous post called An Example of a Completely Regular Space that is not Normal. This is an example of a Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular space that is not normal. We state the definition of the space and present a proof that it is not pseudonormal.

Let E be the set of all points (x,y) \in \mathbb{R} \times \mathbb{R} such that y \ge 0. For each real number x, define the following sets:

    V_x=\left\{(x,y) \in E: 0 \le y \le 2 \right\}

    D_x=\left\{(s,s-x) \in E: x \le s \le x+2 \right\}

    O_x=V_x \cup D_x

The set V_x is the vertical line of height 2 at the point (x,0). The set D_x is the line originating at (x,0) and going in the Northeast direction reaching the same vertical height as V_x as shown in the following figure.

The topology on E is defined by the following:

  • Each point (x,y) \in E where y>0 is isolated.
  • For each point (x,0) \in E, a basic open set is of the form O_x - F where (x,0) \notin F and F is a finite subset of O_x.

The x-axis in this example is a closed and discrete set of cardinality continuum. Amy two disjoint subsets of the x-axis are disjoint closed sets. The two closed sets that cannot be separated are:

    H=\left\{(x,0) \in E: x \text{ is rational} \right\}

    K=\left\{(x,0) \in E: x \text{ is irrational} \right\}

For each (x,0), let W_x=O_x-F_x where F_x \subset O_x is finite and (x,0) \notin F_x. Furthermore, break up F_x by letting F_{x,d}=F_x \cap D_x and F_{x,v}=F_x \cap V_x. Let U and V be defined by:

    U_H=\bigcup \limits_{(x,0) \in H} W_x

    U_K=\bigcup \limits_{(x,0) \in K} W_x

The open sets U_H and U_K are essentially arbitrary open sets containing H and K respectively. We claims that U_H \cap U_K \ne \varnothing.

Define the projection map \tau_1:\mathbb{R}^2 \rightarrow \mathbb{R} by \tau_1(x,y)=x. Let A and B be defined by:

    A=\bigcup \left\{\tau_1(F_{x,d}): (x,0) \in H \right\}

    B=\left\{(x,0) \in K: (x,0) \notin A \right\}

The set A is countable. So the set B is uncountable. Choose (x,0) \in B. Choose (a,0) \in H on the left of (x,0) and close enough to (x,0) such that V_x \cap D_a=\left\{t \right\} and t \notin F_{x,v}. This means that

    t \in V_x \cup D_x -F_x=O_x-F_x=W_x

    t \in V_a \cup D_a -F_a=O_a-F_a=W_a.

Thus U_H \cap U_K \ne \varnothing. We have shown that the space E is not pseudonormal and thus not normal.

Example 2
The Sorgenfrey line is the real line \mathbb{R} topologized by the base consisting of half open and half closed intervals of the form [a,b)=\left\{x \in \mathbb{R}: a \le x < b \right\}. In this post, we use S to denote the real line \mathbb{R} with this topology.

The Sorgenfrey line S is a classic example of a normal space whose square S \times S is not normal. In the Sorgenfrey plane S \times S, the set \left\{(x,-x) \in S \times S: x \in \mathbb{R} \right\} is a closed and discrete set and is called the anti-diagonal. The proof presented in this previous post shows that the following two disjoint closed subsets of S \times S

    H=\left\{(x,-x) \in S \times S: x \text{ is rational} \right\}

    K=\left\{(x,-x) \in S \times S: x \text{ is irrational} \right\}

cannot be separated by disjoint open sets. The argument is based on the fact that the real line with the usual topology is of second category. The key point in the argument is that the set of the irrationals cannot be the union of countably many closed and nowhere dense sets (in the usual topology of the real line).

Thus S \times S fails to be pseudonormal. This example shows that normality can fail to be preserved by taking Cartesian product in such a way that even pseudonormality cannot be achieved in the Cartesian product!

Example 3
Another example of a non-normal space that fails to be pseudonormal is the Niemmytzkis’ plane (Example 2 in in this previous post). The underlying set is N=\left\{(x,y) \in \mathbb{R} \times \mathbb{R}: y \ge 0 \right\}. The points lying above the x-axis have the usual Euclidean open neighborhoods. A point (x,0) in the x-axis has as neighborhoods \left\{(x,0) \right\} together with the interior of a disc in the upper half plane that is tangent at the point (x,0). Consider the following the two disjoint closed sets on the x-axis:

    H=\left\{(x,0): x \text{ is rational} \right\}

    K=\left\{(x,0): x \text{ is irrational} \right\}

The disjoint closed sets H and K cannot be separated by disjoint open sets (see Niemytzki’s Tangent Disc Topology in [2], Example 82). Like Example 2 above, the argument that H and K cannot be separated is also a Baire category argument.

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An Example of Pseudonormal but not Normal

Example 4
One way to find such a space is to look for spaces that are non-normal and see which one is pseudonormal. On the other hand, in a pseudonormal space, countable closed sets are easily separated from other disjoint closed sets. One space in which “countable” is nice is the first uncountable ordinal \omega_1 with the order topology. But \omega_1 is normal. So we look at the Cartesian product \omega_1 \times (\omega_1 +1). The second factor is the successor ordinal to \omega_1 or as a space that is obtained by tagging one more point to \omega_1 that is considered greater than all the points in \omega_1. Let’s use X \times Y=\omega_1 \times (\omega_1 +1) to denote this space.

The space X \times Y is not normal (shown in this previous post). In the previous post, X \times Y is presented as an example showing that the product of a normal space with a compact space needs not be normal. However, in this case at least, the product is pseudonormal.

Let \alpha < \omega_1. Then the square \alpha \times \alpha as a subspace of X \times Y is a countable space and a first countable space. So it has a countable base (second countable) and thus metrizable, and in particular normal. Any countable subset of X \times Y is contained in one of these countable squares, making it easy to separate a countable closed set from another closed set.

Let H and K be disjoint closed sets in X \times Y such that H is countable. Then there is some successor ordinal \mu < \omega_1 (\mu=\alpha+1 for some ordinal \alpha<\omega_1) such that H \subset \mu \times \mu. Based on the discussion in the preceding paragraph, there are disjoint open sets O_H and O_K in \mu \times \mu such that H \subset O_H and (K \cap (\mu \times \mu)) \subset O_K. With \mu being a successor ordinal, the square \mu \times \mu is both closed and open in X \times Y. Then the following sets

    V_H=O_H

    V_K=O_K \cup (X \times Y-\mu \times \mu)

are disjoint open sets in X \times Y separating H and K.

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Some Comments about Examples 1 – 3

In each of Examples 1, 2 and 3 discussed above, there is a closed and discrete set of cardinality continuum (the x-axis in Examples 1 and 3 and the anti-diagonal in Example 2). So the extent of each of these three spaces is continuum. Note that the extent of a space is the maximum cardinality of a closed and discrete subset.

In each of these examples, it just so happens that it is not possible to separate the rationals from the irrationals in the x-axis or the anti-diagonal by disjoint open sets, making each example not only not normal but also not pseudonormal.

What if we consider a smaller subset of the x-axis or anti-diagonal? For example, consider an uncountable set of cardinality less than continuum. Then what can we say about the pseudonormality or normality of the resulting subspaces? For Example 1, the picture is clear cut.

In Example 1, the argument that H and K cannot be separated is a “countable vs. uncountable” argument. The argument will work as long as H is a countable dense set in the x-axis (dense in the usual topology) and K is any uncountable set.

For Example 2 and Example 3, the argument that H and K cannot be separated is not a “countable vs. uncountable” argument and instead is a Baire category argument. The fact that one of the closed sets is the irrationals is a crucial point. On the other hand, both Example 2 and Example 3 (especially Example 3) are set-theoretic sensitive examples. For Example 2 and Example 3, the normality of the resulting smaller subspaces is dependent on some extra axioms beyond ZFC. For pseudonormality, it could be set-theoretic sensitive too. We give some indication here why this is so.

Let S be the Sorgenfrey line as in Example 2 above. Assuming Martin’s Axiom and the negation of the continuum hypothesis (abbreviated by MA + not CH), for any uncountable X \subset S with \lvert X \lvert < c, X \times X is normal but not paracompact (see Example 6.3 in [1] and see [3]). Even though X \times X is not exactly a comparable example, this example shows that restricting to a smaller subset on the anti-diagonal seems to make the space normal.

Example 3 has an illustrious history with respect to the normal Moore space conjecture. There is not surprise that extra set-theory axioms are used. For any subset B of the x-axis, let N(B) be the space defined as in Example 3 above except that only points of B are used on the x-axis. Assuming MA + not CH, for any uncountable B that is of cardinality less than continuum, it can be shown that N(B) is normal non-metrizable Moore space (see Example F in [4]). So by assuming extra axiom of MA + not CH, we cannot get a non-pseudonormal example out of Example 3 by restricting to a smaller uncountable subset of the x-axis. Under other set-theoretic axioms, there exists no normal non-metrizable Moore space. Just because this is a set-theoretic sensitive example, it is conceivable that N(B) could be a space that is not pseudonormal under some other axioms.

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Reference

  1. Burke, D. K., Covering Properties, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 347-422, 1984.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc, Amsterdam, New York, 1995.
  3. Przymusinski, T. C., A Lindelof space X such that X \times X is normal but not paracompact, Fund. Math., 91, 161-165, 1973.
  4. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.

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\copyright \ 2014 \text{ by Dan Ma}