Normality in Cp(X)

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

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

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

When Function Spaces are Normal

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

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

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

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

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

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

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

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

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

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

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

Remarks

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

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

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

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

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

Reference

  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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\copyright 2017 – Dan Ma

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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\copyright \ 2016 \text{ by Dan Ma}

Normality in the powers of countably compact spaces

Let \omega_1 be the first uncountable ordinal. The topology on \omega_1 we are interested in is the ordered topology, the topology induced by the well ordering. The space \omega_1 is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of \omega_1. We also make comments on normality in the powers of a countably compact non-compact space.

Let \omega be the first infinite ordinal. It is well known that \omega^{\omega_1}, the product space of \omega_1 many copies of \omega, is not normal (a proof can be found in this earlier post). This means that any product space \prod_{\alpha<\kappa} X_\alpha, with uncountably many factors, is not normal as long as each factor X_\alpha contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space \prod_{\alpha<\kappa} X_\alpha, the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of \omega). In other words, the interesting case is that each factor X_\alpha is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in X^{\kappa} where X is a countably non-compact space. In this post we start with the space X=\omega_1 of the countable ordinals. We examine \omega_1 power \omega_1^{\omega_1} as well as the countable power \omega_1^{\omega}. The former is not normal while the latter is normal. The proof that \omega_1^{\omega} is normal is an application of the normality of \Sigma-product of the real line.

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The uncountable product

Theorem 1
The product space \prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1} is not normal.

Theorem 1 follows from Theorem 2 below. For any space X, a collection \mathcal{C} of subsets of X is said to have the finite intersection property if for any finite \mathcal{F} \subset \mathcal{C}, the intersection \cap \mathcal{F} \ne \varnothing. Such a collection \mathcal{C} is called an f.i.p collection for short. It is well known that a space X is compact if and only collection \mathcal{C} of closed subsets of X satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space X=\omega_1, there is an f.i.p. collection of cardinality \omega_1 using its linear order. For each \alpha<\omega_1, let C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}. Let \mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}. It is a collection of closed subsets of X=\omega_1. It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space X with an f.i.p. collection of cardinality \omega_1 that has empty intersection, the product space X^{\omega_1} is not normal.

Theorem 2
Let X be a countably compact space. Suppose that there exists a collection \mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\} of closed subsets of X such that \mathcal{C} has the finite intersection property and that \mathcal{C} has empty intersection. Then the product space X^{\omega_1} is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space X^{\omega_1}=\prod_{\alpha<\omega_1} X, we mean a set of the form O=\prod_{\alpha<\omega_1} O_\alpha such that each O_\alpha is an open subset of X and that O_\alpha=X for all but finitely many \alpha<\omega_1. Given a standard basic open set O=\prod_{\alpha<\omega_1} O_\alpha, the notation \text{Supp}(O) refers to the finite set of \alpha for which O_\alpha \ne X. For any set M \subset \omega_1, the notation \pi_M refers to the projection map from \prod_{\alpha<\omega_1} X to the subproduct \prod_{\alpha \in M} X. Each element d \in X^{\omega_1} can be considered a function d: \omega_1 \rightarrow X. By (d)_\alpha, we mean (d)_\alpha=d(\alpha).

For each t \in X, let f_t: \omega_1 \rightarrow X be the constant function whose constant value is t. Consider the following subspaces of X^{\omega_1}.

    H=\prod_{\alpha<\omega_1} C_\alpha

    \displaystyle K=\left\{f_t: t \in X  \right\}

Both H and K are closed subsets of the product space X^{\omega_1}. Because the collection \mathcal{C} has empty intersection, H \cap K=\varnothing. We show that H and K cannot be separated by disjoint open sets. To this end, let U and V be open subsets of X^{\omega_1} such that H \subset U and K \subset V.

Let d_1 \in H. Choose a standard basic open set O_1 such that d_1 \in O_1 \subset U. Let S_1=\text{Supp}(O_1). Since S_1 is the support of O_1, it follows that \pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U. Since \mathcal{C} has the finite intersection property, there exists a_1 \in \bigcap_{\alpha \in S_1} C_\alpha.

Define d_2 \in H such that (d_2)_\alpha=a_1 for all \alpha \in S_1 and (d_2)_\alpha=(d_1)_\alpha for all \alpha \in \omega_1-S_1. Choose a standard basic open set O_2 such that d_2 \in O_2 \subset U. Let S_2=\text{Supp}(O_2). It is possible to ensure that S_1 \subset S_2 by making more factors of O_2 different from X. We have \pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U. Since \mathcal{C} has the finite intersection property, there exists a_2 \in \bigcap_{\alpha \in S_2} C_\alpha.

Now choose a point d_3 \in H such that (d_3)_\alpha=a_2 for all \alpha \in S_2 and (d_3)_\alpha=(d_2)_\alpha for all \alpha \in \omega_1-S_2. Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

  • a sequence d_1,d_2,d_3,\cdots of point of H=\prod_{\alpha<\omega_1} C_\alpha,
  • a sequence S_1 \subset S_2 \subset S_3 \subset \cdots of finite subsets of \omega_1,
  • a sequence a_1,a_2,a_3,\cdots of points of X

such that for all n \ge 2, (d_n)_\alpha=a_{n-1} for all \alpha \in S_{n-1} and \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U. Let A=\left\{a_1,a_2,a_3,\cdots \right\}. Either A is finite or A is infinite. Let’s examine the two cases.

Case 1
Suppose that A is infinite. Since X is countably compact, A has a limit point a. That means that every open set containing a contains some a_n \ne a. For each n \ge 2, define y_n \in \prod_{\alpha< \omega_1} X such that

  • (y_n)_\alpha=(d_n)_\alpha=a_{n-1} for all \alpha \in S_n,
  • (y_n)_\alpha=a for all \alpha \in \omega_1-S_n

From the induction step, we have y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U for all n. Let t=f_a \in K, the constant function whose constant value is a. It follows that t is a limit of \left\{y_1,y_2,y_3,\cdots \right\}. This means that t \in \overline{U}. Since t \in K \subset V, U \cap V \ne \varnothing.

Case 2
Suppose that A is finite. Then there is some m such that a_m=a_j for all j \ge m. For each n \ge 2, define y_n \in \prod_{\alpha< \omega_1} X such that

  • (y_n)_\alpha=(d_n)_\alpha=a_{n-1} for all \alpha \in S_n,
  • (y_n)_\alpha=a_m for all \alpha \in \omega_1-S_n

As in Case 1, we have y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U for all n. Let t=f_{a_m} \in K, the constant function whose constant value is a_m. It follows that t=y_n for all n \ge m+1. Thus U \cap V \ne \varnothing.

Both cases show that U \cap V \ne \varnothing. This completes the proof the product space X^{\omega_1} is not normal. \blacksquare

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The countable product

Theorem 3
The product space \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is collectionwise normal, which is stronger than normality. The proof makes use of the \Sigma-product of \kappa many copies of \mathbb{R}, which is the following subspace of the product space \mathbb{R}^{\kappa}.

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

It is well known that \Sigma(\kappa) is collectionwise normal (see this earlier post). We show that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is a closed subspace of \Sigma(\kappa) where \kappa=\omega_1. Thus \omega_1^{\omega} is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space \omega_1 is embedded as a closed subspace of \Sigma(\omega_1).

For each \beta<\omega_1, define f_\beta:\omega_1 \rightarrow \mathbb{R} such that f_\beta(\gamma)=1 for all \gamma<\beta and f_\beta(\gamma)=0 for all \beta \le \gamma <\omega_1. Let W=\left\{f_\beta: \beta<\omega_1 \right\}. We show that W is a closed subset of \Sigma(\omega_1) and W is homeomorphic to \omega_1 according to the mapping f_\beta \rightarrow W.

First, we show W is closed by showing that \Sigma(\omega_1)-W is open. Let y \in \Sigma(\omega_1)-W. We show that there is an open set containing y that contains no points of W.

Suppose that for some \gamma<\omega_1, y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}. Consider the open set Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1) where Q_\alpha=\mathbb{R} except that Q_\gamma=O. Then y \in Q and Q \cap W=\varnothing.

So we can assume that for all \gamma<\omega_1, y_\gamma \in \left\{0, 1 \right\}. There must be some \theta such that y_\theta=1. Otherwise, y=f_0 \in W. Since y \ne f_\theta, there must be some \delta<\gamma such that y_\delta=0. Now choose the open interval T_\theta=(0.9,1.1) and the open interval T_\delta=(-0.1,0.1). Consider the open set M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1) such that M_\alpha=\mathbb{R} except for M_\theta=T_\theta and M_\delta=T_\delta. Then y \in M and M \cap W=\varnothing. We have just established that W is closed in \Sigma(\omega_1).

Consider the mapping f_\beta \rightarrow W. Based on how it is defined, it is straightforward to show that it is a homeomorphism between \omega_1 and W.

Claim 2
The \Sigma-product \Sigma(\omega_1) has the interesting property it is homeomorphic to its countable power, i.e.

    \Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}.

Because each element of \Sigma(\omega_1) is nonzero only at countably many coordinates, concatenating countably many elements of \Sigma(\omega_1) produces an element of \Sigma(\omega_1). Thus Claim 2 can be easily verified. With above claims, we can see that

    \displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)

Thus \omega_1^{\omega} is a closed subspace of \Sigma(\omega_1). Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that \omega_1^{\omega} is normal. \blacksquare

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The normality in the powers of X

We have established that \prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1} is not normal. Hence any higher uncountable power of \omega_1 is not normal. We have also established that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}, the countable power of \omega_1 is normal (in fact collectionwise normal). Hence any finite power of \omega_1 is normal. However \omega_1^{\omega} is not hereditarily normal. One of the exercises below is to show that \omega_1 \times \omega_1 is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let X be a countably compact space has an f.i.p. collection \mathcal{C} of closed sets such that \bigcap \mathcal{C}=\varnothing. Then X^{\kappa} is not normal where \kappa=\lvert \mathcal{C} \lvert.

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let X be a countably compact space that is not compact. Then there exists a cardinal number \kappa such that X^{\kappa} is not normal and X^{\tau} is normal for all cardinal number \tau<\kappa.

By the non-compactness of X, there exists an f.i.p. collection \mathcal{C} of closed subsets of X such that \bigcap \mathcal{C}=\varnothing. Let \kappa be the least cardinality of such an f.i.p. collection. By Theorem 4, that X^{\kappa} is not normal. Because \kappa is least, any smaller power of X must be normal.

Theorem 6
Let X be a space that is not countably compact. Then X^{\kappa} is not normal for any cardinal number \kappa \ge \omega_1.

Since the space X in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, \omega^{\omega_1} is not normal. Then \omega^{\omega_1} is a closed subspace of X^{\omega_1}.

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space X, X^{\kappa} is not normal for some cardinal number \kappa. The \kappa from either Theorem 5 or Theorem 6 is at least \omega_1. Interestingly for some spaces, the \kappa can be much smaller. For example, for the Sorgenfrey line, \kappa=2. For some spaces (e.g. the Michael line), \kappa=\omega.

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space X is normal, then X is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of X is just another way of discussing Theorem 7. According to Theorem 7, if X is not compact, some power of X is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in X^{\kappa} for more examples of countably compact non-compact X. One particular interesting example would be a countably compact non-compact X such that the least power \kappa for non-normality in X^{\kappa} is more than \omega_1. A possible candidate could be the second uncountable ordinal \omega_2. By Theorem 2, \omega_2^{\omega_2} is not normal. The issue is whether the \omega_1 power \omega_2^{\omega_1} and countable power \omega_2^{\omega} are normal.

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Exercises

Exercise 1
Show that \omega_1 \times \omega_1 is not hereditarily normal.

Exercise 2
Show that the mapping f_\beta \rightarrow W in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space \omega_1 is a closed subspace of the \Sigma-product of the real line. Show that \omega_1 can be embedded in the \Sigma-product of arbitrary spaces.

For each \alpha<\omega_1, let X_\alpha be a space with at least two points. Let p \in \prod_{\alpha<\omega_1} X_\alpha. The \Sigma-product of the spaces X_\alpha is the following subspace of the product space \prod_{\alpha<\omega_1} X_\alpha.

    \Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}

The point p is the center of the \Sigma-product. Show that the space \Sigma(X_\alpha) contains \omega_1 as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that \omega_1^{\omega} is normal.

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\copyright \ 2015 \text{ by Dan Ma}

An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space C_p(X) that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space X such that

  • C_p(X) is collectionwise normal and not paracompact,
  • C_p(X) is not Lindelof but contains a dense Lindelof subspace,
  • C_p(X) is not first countable but is a Frechet space,
  • As a corollary of the previous point, C_p(X) cannot contain a copy of the compact space \omega_1+1,
  • C_p(X) is homeomorphic to C_p(X)^\omega,
  • C_p(X) is not hereditarily normal,
  • C_p(X) is not metacompact.

A short and quick description of the space X is that X is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space C_p(X) is intimately related to a \Sigma-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the \Sigma-products of copies of real lines. Deep results about \Sigma-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space X=L_\tau. In the discussion that follows, the Greek letter \tau is always an uncountable cardinal number. Let D_\tau be a set with cardinality \tau. Let p be a point not in D_\tau. Let L_\tau=D_\tau \cup \left\{p \right\}. Consider the following topology on L_\tau:

  • Each point in D_\tau an isolated point, and
  • open neighborhoods at the point p are of the form L_\tau-K where K \subset D_\tau is countable.

It is clear that L_\tau is a Lindelof space. The Lindelof space L_\tau is sometimes called the one-point Lindelofication of the discrete space D_\tau since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space C_p(L_\tau). See this post for general information on the pointwise convergence topology of C_p(Y) for any completely regular space Y.

All the facts about C_p(X)=C_p(L_\tau) mentioned at the beginning follow from the fact that C_p(L_\tau) is homeomorphic to the \Sigma-product of \tau many copies of the real lines. Specifically, C_p(L_\tau) is homeomorphic to the following subspace of the product space \mathbb{R}^\tau.

    \Sigma_{\alpha<\tau}\mathbb{R}=\left\{ x \in \mathbb{R}^\tau: x_\alpha \ne 0 \text{ for at most countably many } \alpha<\tau \right\}

Thus understanding the function space C_p(L_\tau) is a matter of understanding a \Sigma-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of C_p(L_\tau) indicated above.

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The Homeomorphism

For each f \in C_p(L_\tau), it is easily seen that there is a countable set C \subset D_\tau such that f(p)=f(y) for all y \in D_\tau-C. Let W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}. Then each f \in W_0 has non-zero values only on a countable subset of D_\tau. Naturally, W_0 and \Sigma_{\alpha<\tau}\mathbb{R} are homeomorphic.

We claim that C_p(L_\tau) is homeomorphic to W_0 \times \mathbb{R}. For each f \in C_p(L_\tau), define h(f)=(f-f(p),f(p)). Here, f-f(p) is the function g \in C_p(L_\tau) such that g(x)=f(x)-f(p) for all x \in L_\tau. Clearly h(f) is well-defined and h(f) \in W_0 \times \mathbb{R}. It can be readily verified that h is a one-to-one map from C_p(L_\tau) onto W_0 \times \mathbb{R}. It is not difficult to verify that both h and h^{-1} are continuous.

We use the notation X_1 \cong X_2 to mean that the spaces X_1 and X_2 are homeomorphic. Then we have:

    C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R})  \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}

Thus C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}. This completes the proof that C_p(L_\tau) is topologically the \Sigma-product of \tau many copies of the real lines.

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Looking at the \Sigma-Product

Understanding the function space C_p(L_\tau) is now reduced to the problem of understanding a \Sigma-product of copies of the real lines. Most of the facts about \Sigma-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the \Sigma-product of separable metric spaces is collectionwise normal. Thus C_p(L_\tau) is collectionwise normal. The \Sigma-product of spaces, each of which has at least two points, always contains a closed copy of \omega_1 with the ordered topology (see the lemma in this previous post). Thus C_p(L_\tau) contains a closed copy of \omega_1 and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the \Sigma-product \Sigma_{\alpha<\tau}\mathbb{R}:

    \sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}

In this previous post, it is shown that \sigma_\tau is a Lindelof space. Though C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is not Lindelof, it has a dense Lindelof subspace, namely \sigma_\tau.

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A space Y is first countable if there exists a countable local base at each point y \in Y. A space Y is a Frechet space (or is Frechet-Urysohn) if for each y \in Y, if y \in \overline{A} where A \subset Y, then there exists a sequence \left\{y_n: n=1,2,3,\cdots \right\} of points of A such that the sequence converges to y. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number \tau, the product \mathbb{R}^\tau is not first countable. In fact, any dense subspace of \mathbb{R}^\tau is not first countable. In particular, the \Sigma-product \Sigma_{\alpha<\tau}\mathbb{R} is not first countable. In this previous post, it is shown that the \Sigma-product of first countable spaces is a Frechet space. Thus C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is a Frechet space.

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As a corollary of the previous point, C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space \omega_1+1 is not embeddable in C_p(L_\tau). The interest in compact subspaces of C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is that any compact space that is topologically embeddable in a \Sigma-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

    \Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}

Thus C_p(L_\tau) \cong C_p(L_\tau)^\omega. In particular, C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau) due to the following observation:

    C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)

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As a result of the peculiar fact that C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau), it can be concluded that C_p(L_\tau), though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if Y_1 \times Y_2 is hereditarily normal, then either Y_1 is perfectly normal or every countably infinite subset of Y_2 is closed and discrete (see this previous post). The function space C_p(L _\tau) is not perfectly normal since it contains a closed copy of \omega_1. On the other hand, there are plenty of countably infinite subsets of C_p(L _\tau) that are not closed and discrete. As a Frechet space, C_p(L _\tau) has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let \left\{x_1,x_2,x_3,\cdots \right\} be an infinite subset of D_\tau and consider the following:

    C=\left\{f_n: n=1,2,3,\cdots \right\}

where f_n is such that f_n(x_n)=n and f_n(x)=0 for each x \in L_\tau with x \ne x_n. Note that C is not closed and not discrete since the points in C converge to g \in \overline{C} where g is the zero-function. Thus C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau) is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since C_p(L_\tau) is collectionwise normal and not paracompact, C_p(L_\tau) can never be metacompact.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
  3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

(Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Fix a point b \in \prod_{\alpha \in A} X_\alpha, called the base point. The \Sigma-product of the spaces \left\{X_\alpha: \alpha \in A \right\} is the following subspace of the product space X:

    \Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}

In other words, the space \Sigma_{\alpha \in A} X_\alpha is the subspace of the product space X=\prod_{\alpha \in A} X_\alpha consisting of all points that deviate from the base point on at most countably many coordinates \alpha \in A. We also consider the following subspace of \Sigma_{\alpha \in A} X_\alpha.

    \sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}

For convenience , we call \Sigma_{\alpha \in A} X_\alpha the (upper case) Sigma-product (or \Sigma-product) of the spaces X_\alpha and we call the space \sigma the (lower case) sigma-product (or \sigma-product). Clearly, the space \sigma is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor X_\alpha is a separable metric space with at least two points, the \Sigma-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) \Sigma-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer n=0,1,2,3,\cdots, we define \sigma_n as follows:

    \sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}

Clearly, \sigma=\bigcup_{n=0}^\infty \sigma_n. We prove the following theorem. The fact that \sigma is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space X=\prod_{\alpha \in A} X_\alpha. We say V is a standard basic open subset of the product space X if V is of the form V=\prod_{\alpha \in A} V_\alpha such that each V_\alpha is an open subset of the factor space X_\alpha and V_\alpha=X_\alpha for all but finitely many \alpha \in A. The finite set F of all \alpha \in A such that V_\alpha \ne X_\alpha is called the support of the open set V.

Theorem 1
Let \sigma be the \sigma-product of the separable metrizable spaces \left\{X_\alpha: \alpha \in A \right\}. For each n, let \sigma_n be defined as above. The product space \sigma_n \times Y is Lindelof for each non-negative integer n and for all separable metric space Y.

Proof of Theorem 1
We prove by induction on n. Note that \sigma_0=\left\{b \right\}, the base point. Clearly \sigma_0 \times Y is Lindelof for all separable metric space Y. Suppose the theorem hold for the integer n. We show that \sigma_{n+1} \times Y for all separable metric space Y. To this end, let \mathcal{U} be an open cover of \sigma_{n+1} \times Y where Y is a separable metric space. Without loss of generality, we assume that each element of \mathcal{U} is of the form V \times W where V=\prod_{\alpha \in A} V_\alpha is a standard basic open subset of the product space X=\prod_{\alpha \in A} X_\alpha and W is an open subset of Y.

Let \mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\} be a countable subcollection of \mathcal{U} such that \mathcal{U}_0 covers \left\{b \right\} \times Y. For each j, let U_j=V_j \times W_j where V_j=\prod_{\alpha \in A} V_{j,\alpha} is a standard basic open subset of the product space X with b \in V_j and W_j is an open subset of Y. For each j, let F_j be the support of V_j. Note that \alpha \in F_j if and only if V_{j,\alpha} \ne X_\alpha. Also for each \alpha \in F_j, b_\alpha \in V_{j,\alpha}. Furthermore, for each \alpha \in F_j, let V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}. With all these notations in mind, we define the following open set for each \beta \in F_j:

    H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j

Observe that for each point y \in \sigma_{n+1} such that y \in V^c_{j,\beta} \times T_\beta, the point y already deviates from the base point b on one coordinate, namely \beta. Thus on the coordinates other than \beta, the point y can only deviates from b on at most n many coordinates. Thus \sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta) is homeomorphic to V^c_{j,\beta} \times \sigma_n. Note that V^c_{j,\beta} \times W_j is a separable metric space. By inductive hypothesis, V^c_{j,\beta} \times \sigma_n \times W_j is Lindelof. Thus there are countably many open sets in the open cover \mathcal{U} that covers points of H_{j,\beta} \cap (\sigma_{n+1} \times W_j).

Note that

    \sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)

To see that the left-side is a subset of the right-side, let t=(x,y) \in \sigma_{n+1} \times Y. If t \in U_j for some j, we are done. Suppose t \notin U_j for all j. Observe that y \in W_j for some j. Since t=(x,y) \notin U_j, x_\beta \notin V_{j,\beta} for some \beta \in F_j. Then t=(x,y) \in H_{j,\beta}. It is now clear that t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j). Thus the above set equality is established. Thus one part of \sigma_{n+1} \times Y is covered by countably many open sets in \mathcal{U} while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of \mathcal{U} covers \sigma_{n+1} \times Y. \blacksquare

Corollary 2
It follows from Theorem 1 that

  • If each factor space X_\alpha is a separable metric space, then each \sigma_n is a Lindelof space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a Lindelof space.
  • If each factor space X_\alpha is a compact separable metric space, then each \sigma_n is a compact space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that \Sigma-product of compact spaces is countably compact. Thus \Sigma_{\alpha \in A} X_\alpha is a countably compact space if each X_\alpha is compact. Note that each \sigma_n is a closed subset of \Sigma_{\alpha \in A} X_\alpha and is thus countably compact. Being a Lindelof space, each \sigma_n is compact. It follows that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space. \blacksquare

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each \alpha \in A, let X_\alpha be a separable metric space with at least two points. Then the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is collectionwise normal (see this previous post). According to the lemma in this previous post, the \Sigma-product \Sigma_{\alpha \in A} X_\alpha contains a closed copy of \omega_1. Thus the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is not Lindelof. It is clear that the \sigma-product is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. By Corollary 2, the \sigma-product is a Lindelof subspace of \Sigma_{\alpha \in A} X_\alpha.

Using specific factor spaces, if each X_\alpha=\mathbb{R} with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each X_\alpha=[0,1] with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense \sigma-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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\copyright \ 2014 \text{ by Dan Ma}

Normal dense subspaces of products of “omega 1” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii (see Problem I.5.25 in [2]). One partial positive answer is a theorem attributed to Corson: if Y is a normal dense subspace of a product of separable spaces such that Y \times Y is normal, then Y is collectionwise normal. Another partial positive answer: assuming 2^\omega<2^{\omega_1}, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). Another partial positive answer to Arkhangelskii’s question is the theorem due to Reznichenko: If C_p(X), which is a dense subspace of the product space \mathbb{R}^X, is normal, then it is collectionwise normal (see Theorem I.5.12 in [2]). In this post, we highlight another partial positive answer to the question posted in [2]. Specifically, we prove the following theorem:

Theorem 1

    Let X=\prod_{\alpha<\omega_1} X_\alpha be a product space where each factor X_\alpha is a separable metric space. Let Y be a dense subspace of X. Then if Y is normal, then Y is collectionwise normal.

Since any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post), it suffices to prove the following theorem:

Theorem 1a

    Let X=\prod_{\alpha<\omega_1} X_\alpha be a product space where each factor X_\alpha is a separable metric space. Let Y be a dense subspace of X. Then if Y is normal, then every closed and discrete subspace of Y is countable, i.e., Y has countable extent.

Arkhangelskii’s question was studied by the author of [3] and [4]. Theorem 1 as presented in this post is essentially the Theorem 1 found in [3]. The proof given in [3] is a beautiful proof. The proof in this post is modeled on the proof in [3] with the exception that all the crucial details are filled in. Theorem 1a (as stated above) is used in [1] to show that the function space C_p(\omega_1+1) contains no dense normal subspace.

It is natural to wonder if Theorem 1 can be generalized to product space of \tau many separable metric factors where \tau is an arbitrary uncountable cardinal. The work of [4] shows that the question at the beginning of this post cannot be answered positively in ZFC. Recall the above mentioned result that assuming 2^\omega<2^{\omega_1}, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). A theorem in [4] implies that assuming 2^\omega=2^{\omega_1}, for any separable metric space M with at least 2 points, the product of continuum many copies of M contains a normal dense subspace Y that is not collectionwise normal. A side note: for this normal subspace Y, Y \times Y is necessarily not normal (according to Corson’s theorem). Thus [3] and [4] collectively show that Arkhangelskii’s question stated here at the beginning of the post is answered positively (in ZFC) among product spaces of \omega_1 many separable metric factors and that outside of the \omega_1 case, it is impossible to answer the question positively in ZFC.

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Proving Theorem 1a

We use the following lemma. For a proof of this lemma, see the proof for Lemma 1 in this previous post.

Lemma 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(Y), meaning that \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing.

For any B \subset \omega_1, let \pi_B be the natural projection from the product space X=\prod_{\alpha<\omega_1} X_\alpha into the subproduct space \prod_{\alpha \in B} X_\alpha.

Proof of Theorem 1a
Let Y be a dense subspace of the product space X=\prod_{\alpha<\omega_1} X_\alpha where each factor X_\alpha has a countable base. Suppose that D is an uncountable closed and discrete subset of Y. We then construct a pair of disjoint closed subsets H and K of Y such that for all countable B \subset \omega_1, \pi_B(H) and \pi_B(K) are not separated, specifically \pi_B(H) \cap \overline{\pi_B(K)}\ne \varnothing. Here the closure is taken in the space \pi_B(Y). By Lemma 2, the dense subspace Y of X is not normal.

For each \alpha<\omega_1, let \mathcal{B}_\alpha be a countable base for the space X_\alpha. The standard basic open sets in the product space X are of the form O=\prod_{\alpha<\omega_1} O_\alpha such that

  • each O_\alpha is an open subset of X_\alpha,
  • if O_\alpha \ne X_\alpha, then O_\alpha \in \mathcal{B}_\alpha,
  • O_\alpha=X_\alpha for all but finitely many \alpha<\omega_1.

We use supp(O) to denote the finite set of \alpha such that O_\alpha \ne X_\alpha. Technically we should be working with standard basic open subsets of Y, i.e., sets of the form O \cap Y where O is a standard basic open set as described above. Since Y is dense in the product space, every standard open set contains points of Y. Thus we can simply work with standard basic open sets in the product space as long as we are working with points of Y in the construction.

Let \mathcal{M} be the collection of all standard basic open sets as described above. Since there are only \omega_1 many factors in the product space, \lvert \mathcal{M} \lvert=\omega_1. Recall that D is an uncountable closed and discrete subset of Y. Let \mathcal{M}^* be the following:

    \mathcal{M}^*=\left\{U \in \mathcal{M}: U \cap D \text{ is uncountable }  \right\}

Claim 1. \lvert \mathcal{M}^* \lvert=\omega_1.

First we show that \mathcal{M}^* \ne \varnothing. Let B \subset \omega_1 be countable. Consider these two cases: Case 1. \pi_B(D) is an uncountable subset of \prod_{\alpha \in B} X_\alpha; Case 2. \pi_B(D) is countable.

Suppose Case 1 is true. Since \prod_{\alpha \in B} X_\alpha is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point y \in \pi_B(D) such that every open neighborhood of y (open in \prod_{\alpha \in B} X_\alpha) contains uncountably many points of \pi_B(D). Thus every standard basic open set U=\prod_{\alpha \in B} U_\alpha, with y \in U, contains uncountably many points of \pi_B(D). Suppose Case 2 is true. There exists one point y \in \pi_B(D) such that y=\pi_B(t) for uncountably many t \in D. Then in either case, every standard basic open set V=\prod_{\alpha<\omega_1} V_\alpha, with supp(V) \subset B and y \in \pi_B(V), contains uncountably many points of D. Any one such V is a member of \mathcal{M}^*.

We can partition the index set \omega_1 into \omega_1 many disjoint countable sets B. Then for each such B, obtain a V \in \mathcal{M}^* in either Case 1 or Case 2. Since supp(V) \subset B, all such open sets V are distinct. Thus Claim 1 is established.

Claim 2.
There exists an uncountable H \subset D such that for each U \in \mathcal{M}^*, U \cap H \ne \varnothing and U \cap (D-H) \ne \varnothing.

Enumerate \mathcal{M}^*=\left\{U_\gamma: \gamma<\omega_1 \right\}. Choose h_0,k_0 \in U_0 \cap D with h_0 \ne k_0. Suppose that for all \beta<\gamma, two points h_\beta,k_\beta are chosen such that h_\beta,k_\beta \in U_\beta \cap D, h_\beta \ne k_\beta and such that h_\beta \notin L_\beta and k_\beta \notin L_\beta where L_\beta=\left\{h_\rho: \rho<\beta \right\} \cup \left\{k_\rho: \rho<\beta \right\}. Then choose h_\gamma,k_\gamma with h_\gamma \ne k_\gamma such that h_\gamma,k_\gamma \in U_\gamma \cap D and h_\gamma \notin L_\gamma and k_\gamma \notin L_\gamma where L_\gamma=\left\{h_\rho: \rho<\gamma \right\} \cup \left\{k_\rho: \rho<\gamma \right\}.

Let H=\left\{h_\gamma: \gamma<\omega_1 \right\} and let K=D-H. Note that K_0=\left\{k_\gamma: \gamma<\omega_1 \right\} \subset K. Based on the inductive process that is used to obtain H and K_0, it is clear that H satisfies Claim 2.

Claim 3.
For each countable B \subset \omega_1, the sets \pi_B(H) and \pi_B(K) are not separated in the space \pi_B(Y).

Let B \subset \omega_1 be countable. Consider the two cases: Case 1. \pi_B(H) is uncountable; Case 2. \pi_B(H) is countable. Suppose Case 1 is true. Since \prod_{\alpha \in B} X_\alpha is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point p \in \pi_B(H) such that every open neighborhood of p (open in \prod_{\alpha \in B} X_\alpha) contains uncountably many points of \pi_B(H). Choose h \in H such that p=\pi_B(h). Then the following statement holds:

  1. For every basic open set U=\prod_{\alpha<\omega_1} U_\alpha with h \in U such that supp(U) \subset B, the open set U contains uncountably many points of H.

Suppose Case 2 is true. There exists some p \in \pi_B(H) such that p=\pi_B(t) for uncountably many t \in H. Choose h \in H such that p=\pi_B(h). Then statement 1 also holds.

In either case, there exists h \in H such that statement 1 holds. The open sets U described in statement 1 are members of \mathcal{M}^*. By Claim 2, the open sets described in statement 1 also contain points of K. Since the open sets described in statement 1 have supports \subset B, the following statement holds:

  1. For every basic open set V=\prod_{\alpha \in B} V_\alpha with \pi_B(h) \in V, the open set V contains points of \pi_B(K).

Statement 2 indicates that \pi_B(h) \in \overline{\pi_B(K)}. Thus \pi_B(h) \in \pi_B(H) \cap \overline{\pi_B(K)}. The closure here can be taken in either \prod_{\alpha \in B} X_\alpha or \pi_B(Y) (to apply Lemma 2, we only need the latter). Thus Claim 3 is established.

Claim 3 is the negation of condition 3 of Lemma 2. Therefore Y is not normal. \blacksquare

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Remark

The proof of Theorem 1a, though a proof in ZFC only, clearly relies on the fact that the product space is a product of \omega_1 many factors. For example, in the inductive step in the proof of Claim 2, it is always possible to pick a pair of points not chosen previously. This is because the previously chosen points form a countable set and each open set in \mathcal{M}^* contains \omega_1 many points of the closed and discrete set D. With the “\omega versus \omega_1” situation, at each step, there are always points not previously chosen. When more than \omega_1 many factors are involved, there may be no such guarantee in the inductive process.

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Reference

  1. Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
  2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  3. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
  4. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
  5. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

Normal dense subspaces of a product of “continuum” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii in [1] (see Problem I.5.25). A partial positive answer is provided by a theorem that is usually attributed to Corson: If Y is a normal dense subspace of a product of separable metric spaces and if Y \times Y is also normal, then Y is collectionwise normal. In this post, using a simple combinatorial argument, we show that any normal dense subspace of a product of continuum many separable metric space is collectionwise normal (see Corollary 4 below), which is a corollary of the following theorem.

Theorem 1
Let X be a normal space with character \le 2^\omega. If 2^\omega<2^{\omega_1}, then the following holds:

  • If Y is a closed and discrete subspace of X with \lvert Y \lvert=\omega_1, then Y contains a separated subset of cardinality \omega_1.

Theorem 1 gives the corollary indicated at the beginning and several other interesting results. The statement 2^\omega<2^{\omega_1} means that the cardinality of the power set (the set of all subsets) of \omega is strictly less than the cardinality of the power set of \omega_1. Note that the statement 2^\omega<2^{\omega_1} follows from the continuum hypothesis (CH), the statement that 2^\omega=\omega_1. With the assumption 2^\omega<2^{\omega_1}, Theorem 1 is a theorem that goes beyond ZFC. We also present an alternative to Theorem 1 that removes the assumption 2^\omega<2^{\omega_1} (see Theorem 6 below).

A subset T of a space S is a separated set (in S) if for each t \in T, there is an open subset O_t of S with t \in O_t such that \left\{O_t: t \in T \right\} is a pairwise disjoint collection. First we prove Theorem 1 and then discuss the corollaries.

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Proof of Theorem 1

Suppose Y is a closed and discrete subset of X with \lvert Y \lvert=\omega_1 such that no subset of Y of cardinality \omega_1 can be separated. We then show that 2^{\omega_1} \le 2^{\omega}.

For each y \in Y, let \mathcal{B}_y be a local base at the point y such that \lvert \mathcal{B}_y \lvert \le 2^\omega. Let \mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y. Thus \lvert \mathcal{B} \lvert \le 2^\omega. By normality, for each W \subset Y, let U_W be an open subset of X such that W \subset U_W and \overline{U_W} \cap (Y-W)=\varnothing. For each W \subset Y, consider the following collection of open sets:

    \mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W  \right\}

For each W \subset Y, choose a maximal disjoint collection \mathcal{M}_W of open sets in \mathcal{G}_W. Because no subset of Y of cardinality \omega_1 can be separated, each \mathcal{M}_W is countable. If W_1 \ne W_2, then \mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}.

Let \mathcal{P}(Y) be the power set (i.e. the set of all subsets) of Y. Let \mathcal{P}_\omega(\mathcal{B}) be the set of all countable subsets of \mathcal{B}. Then the mapping W \mapsto \mathcal{M}_W is a one-to-one map from \mathcal{P}(Y) into \mathcal{P}_\omega(\mathcal{B}). Note that \lvert \mathcal{P}(Y) \lvert=2^{\omega_1}. Also note that since \lvert \mathcal{B} \lvert \le 2^\omega, \lvert \mathcal{P}_\omega(\mathcal{B}) \lvert \le 2^\omega. Thus 2^{\omega_1} \le 2^{\omega}. \blacksquare

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Some Corollaries of Theorem 1

Here’s some corollaries that follow easily from Theorem 1. A space X has the countable chain condition (CCC) if every pairwise disjoint collection of non-empty open subset of X is countable. For convenience, if X has the CCC, we say X is CCC. The following corollaries make use of the fact that any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post).

Corollary 2
Let X be a CCC space with character \le 2^\omega. If 2^\omega<2^{\omega_1}, then the following conditions hold:

  • If X is normal, then every closed and discrete subset of X is countable, i.e., X has countable extent.
  • If X is normal, then X is collectionwise normal.

Corollary 3
Let X be a CCC space with character \le 2^\omega. If CH holds, then the following conditions hold:

  • If X is normal, then every closed and discrete subset of X is countable, i.e., X has countable extent.
  • If X is normal, then X is collectionwise normal.

Corollary 4
Let X=\prod_{\alpha<2^\omega} X_\alpha be a product where each factor X_\alpha is a separable metric space. If 2^\omega<2^{\omega_1}, then the following conditions hold:

  • If Y is a normal dense subspace of X, then Y has countable extent.
  • If Y is a normal dense subspace of X, then Y is collectionwise normal.

Corollary 4 is the result indicated in the title of the post. The product of separable spaces has the CCC. Thus the product space X and any dense subspace of X have the CCC. Because X is a product of continuum many separable metric spaces, X and any subspace of X have characters \le 2^\omega. Then Corollary 4 follows from Corollary 2.

When dealing with the topic of normal versus collectionwise normal, it is hard to avoid the connection with the normal Moore space conjecture. Theorem 1 gives the result of F. B. Jones from 1937 (see [3]). We have the following theorem.

Theorem 5
If 2^\omega<2^{\omega_1}, then every separable normal Moore space is metrizable.

Though this was not how Jones proved it in [3], Theorem 5 is a corollary of Corollary 2. By Corollary 2, any separable normal Moore space is collectionwise normal. It is well known that collectionwise normal Moore space is metrizable (Bing’s metrization theorem, see Theorem 5.4.1 in [2]).

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A ZFC Theorem

We now prove a result that is similar to Corollary 2 but uses no set-theory beyond the Zermelo–Fraenkel set theory plus axiom of choice (abbreviated by ZFC). Of course the conclusion is not as strong. Even though the assumption 2^\omega<2^{\omega_1} is removed in Theorem 6, note the similarity between the proof of Theorem 1 and the proof of Theorem 6.

Theorem 6
Let X be a CCC space with character \le 2^\omega. Then the following conditions hold:

  • If X is normal, then every closed and discrete subset of X has cardinality less than continuum.

Proof of Theorem 6
Let X be a normal CCC space with character \le 2^\omega. Let Y be a closed and discrete subset of X. We show that \lvert Y \lvert < 2^\omega. Suppose that \lvert Y \lvert = 2^\omega.

For each y \in Y, let \mathcal{B}_y be a local base at the point y such that \lvert \mathcal{B}_y \lvert \le 2^\omega. Let \mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y. Thus \lvert \mathcal{B} \lvert = 2^\omega. By normality, for each W \subset Y, let U_W be an open subset of X such that W \subset U_W and \overline{U_W} \cap (Y-W)=\varnothing. For each W \subset Y, consider the following collection of open sets:

    \mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W  \right\}

For each W \subset Y, choose \mathcal{M}_W \subset \mathcal{G}_W such that \mathcal{M}_W is a maximal disjoint collection. Since X is CCC, \mathcal{M}_W is countable. It is clear that if W_1 \ne W_2, then \mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}.

Let \mathcal{P}(Y) be the power set (i.e. the set of all subsets) of Y. Let \mathcal{P}_\omega(\mathcal{B}) be the set of all countable subsets of \mathcal{B}. Then the mapping W \mapsto \mathcal{M}_W is a one-to-one map from \mathcal{P}(Y) into \mathcal{P}_\omega(\mathcal{B}). Note that since \lvert \mathcal{B} \lvert = 2^\omega, \lvert \mathcal{P}_\omega(\mathcal{B}) \lvert = 2^\omega. Thus \lvert \mathcal{P}(Y) \lvert \le 2^{\omega}. However, Y is assumed to be of cardinality continuum. Then \lvert \mathcal{P}(Y) \lvert>2^{\omega_1}, leading to a contradiction. Thus it must be the case that \lvert Y \lvert < 2^\omega. \blacksquare

With Theorem 6, Corollary 3 still holds. Theorem 6 removes the set-theoretic assumption of 2^\omega<2^{\omega_1}. As a result, the upper bound for cardinalities of closed and discrete sets is (at least potentially) higher.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Jones, F. B., Concerning normal and completely normal spaces, Bull. Amer. Math. Soc., 43, 671-677, 1937.

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\copyright \ 2014 \text{ by Dan Ma}