Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space X=\omega_1 \times I^I, which is the product of \omega_1 with the interval topology and the product space I^I=\prod_{t \in I} I where I is of course the unit interval [0,1]. The notation of \omega_1, the first uncountable ordinal, in Steen and Seebach is [0,\Omega).

Another way to notate the example X is the product space \prod_{t \in I} X_t where X_0 is \omega_1 and X_t is the unit interval I for all t>0. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of \omega_1 makes this product space an interesting example.

The following lists out the basic topological properties of the space that X=\omega_1 \times I^I are covered in [2].

  • The space X is Hausdorff and completely regular.
  • The space X is countably compact.
  • The space X is neither compact nor sequentially compact.
  • The space X is neither separable, Lindelof nor \sigma-compact.
  • The space X is not first countable.
  • The space X is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

  • The space X is not normal.
  • The space X is a dense subspace that is normal.

It follows from these bullet points that the space X is an example of a completely regular space that is not normal. Not being a normal space, X is then not metrizable. Of course there are other ways to show that X is not metrizable. One is that neither of the two factors \omega_1 or I^I is metrizable. Another is that X is not first countable.

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The space X is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example X=\omega_1 \times I^I is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example X=\omega_1 \times I^I is to show whether the second factor I^I is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space Y, the product \omega_1 \times Y is normal if and only if Y is countably tight. Thus the normality of the space X (or the lack of) hinges on whether the compact factor I^I=\prod_{t \in I} I is countably tight.

A space Y is countably tight (or has countable tightness) if for each S \subset Y and for each x \in \overline{S}, there exists some countable B \subset S such that x \in \overline{B}. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space I^I=\prod_{t \in I} I is not countably tight, we let S be the subspace of I^I consisting of points, each of which is non-zero on at most countably many coordinates. Specifically S is defined as follows:

    S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}

The set S just defined is also called the \Sigma-product of copies of unit interval I. Let g \in I^I be defined by g(t)=1 for all t \in I. It follows that g \in \overline{S}. It can also be verified that g \notin \overline{B} for any countable B \subset S. This shows that the product space I^I=\prod_{t \in I} I is not countably tight.

By Theorem 1 found in this link, the space X=\omega_1 \times I^I is not normal.

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The space X has a dense subspace that is normal

Now that we know X=\omega_1 \times I^I is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace \omega_1 \times S where S is the \Sigma-product S=\Sigma_{t \in I} I defined in the preceding section. The subspace S is dense in the product space I^I. Thus \omega_1 \times S is dense in X=\omega_1 \times I^I. The space S is normal since the \Sigma-product of separable metric spaces is normal. Furthermore, \omega_1 can be embedded as a closed subspace of S=\Sigma_{t \in I} I. Then \omega_1 \times S is homeomorphic to a closed subspace of S \times S. Since S \times S \cong S, the space \omega_1 \times S is normal.

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Reference

  1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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\copyright \ 2015 \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}

Comparing two function spaces

Let \omega_1 be the first uncountable ordinal, and let \omega_1+1 be the successor ordinal to \omega_1. Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function f defined on either \omega_1 or \omega_1+1 is eventually constant, i.e., there exists some \alpha<\omega_1 such that the function f is constant on the ordinals beyond \alpha. Now consider the function spaces C_p(\omega_1) and C_p(\omega_1+1). Thus individually, elements of these two function spaces appear identical. Any f \in C_p(\omega_1) matches a function f^* \in C_p(\omega_1+1) where f^* is the result of adding the point (\omega_1,a) to f where a is the eventual constant real value of f. This fact may give the impression that the function spaces C_p(\omega_1) and C_p(\omega_1+1) are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

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Tightness

One topological property that is different between C_p(\omega_1) and C_p(\omega_1+1) is that of tightness. The function space C_p(\omega_1+1) is countably tight, while C_p(\omega_1) is not countably tight.

Let X be a space. The tightness of X, denoted by t(X), is the least infinite cardinal \kappa such that for any A \subset X and for any x \in X with x \in \overline{A}, there exists B \subset A for which \lvert B \lvert \le \kappa and x \in \overline{B}. When t(X)=\omega, we say that X has countable tightness or is countably tight. When t(X)>\omega, we say that X has uncountable tightness or is uncountably tight.

First, we show that the tightness of C_p(\omega_1) is greater than \omega. For each \alpha<\omega_1, define f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\} such that f_\alpha(\beta)=0 for all \beta \le \alpha and f_\alpha(\beta)=1 for all \beta>\alpha. Let g \in C_p(\omega_1) be the function that is identically zero. Then g \in \overline{F} where F is defined by F=\left\{f_\alpha: \alpha<\omega_1 \right\}. It is clear that for any countable B \subset F, g \notin \overline{B}. Thus C_p(\omega_1) cannot be countably tight.

The space \omega_1+1 is a compact space. The fact that C_p(\omega_1+1) is countably tight follows from the following theorem.

Theorem 1
Let X be a completely regular space. Then the function space C_p(X) is countably tight if and only if X^n is Lindelof for each n=1,2,3,\cdots.

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for C_p(\omega_1+1).

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The Frechet-Urysohn property

In fact, C_p(\omega_1+1) has a property that is stronger than countable tightness. The function space C_p(\omega_1+1) is a Frechet-Urysohn space (see this previous post). Of course, C_p(\omega_1) not being countably tight means that it is not a Frechet-Urysohn space.

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Normality

The function space C_p(\omega_1+1) is not normal. If C_p(\omega_1+1) is normal, then C_p(\omega_1+1) would have countable extent. However, there exists an uncountable closed and discrete subset of C_p(\omega_1+1) (see this previous post). On the other hand, C_p(\omega_1) is Lindelof. The fact that C_p(\omega_1) is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if X is a space consisting of ordinals such that X is first countable and countably compact, then C_p(X) is Lindelof.

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Embedding one function space into the other

The two function space C_p(\omega_1+1) and C_p(\omega_1) are very different topologically. However, one of them can be embedded into the other one. The space \omega_1+1 is the continuous image of \omega_1. Let g: \omega_1 \longrightarrow \omega_1+1 be a continuous surjection. Define a map \psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1) by letting \psi(f)=f \circ g. It is shown in this previous post that \psi is a homeomorphism. Thus C_p(\omega_1+1) is homeomorphic to the image \psi(C_p(\omega_1+1)) in C_p(\omega_1). The map g is also defined in this previous post.

The homeomposhism \psi tells us that the function space C_p(\omega_1), though Lindelof, is not hereditarily normal.

On the other hand, the function space C_p(\omega_1) cannot be embedded in C_p(\omega_1+1). Note that C_p(\omega_1+1) is countably tight, which is a hereditary property.

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Remark

There is a mapping that is alluded to at the beginning of the post. Each f \in C_p(\omega_1) is associated with f^* \in C_p(\omega_1+1) which is obtained by appending the point (\omega_1,a) to f where a is the eventual constant real value of f. It may be tempting to think of the mapping f \rightarrow f^* as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Buzyakova, R. Z., In search of Lindelof C_p‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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\copyright \ 2014 \text{ by Dan Ma}

Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

This is another post that discusses what C_p(X) is like when X is a compact space. In this post, we discuss the example C_p(\omega_1+1) where \omega_1+1 is the first compact uncountable ordinal. Note that \omega_1+1 is the successor to \omega_1, which is the first (or least) uncountable ordinal. The function space C_p(\omega_1+1) is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by C_p(X) for all compact spaces X. The second property is possessed by all compact scattered spaces. After we discuss C_p(\omega_1+1), we discuss briefly the general results for C_p(X).

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Initial discussion

The function space C_p(\omega_1+1) is a dense subspace of the product space \mathbb{R}^{\omega_1}. In fact, C_p(\omega_1+1) is homeomorphic to a subspace of the following subspace of \mathbb{R}^{\omega_1}:

    \Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}

The subspace \Sigma(\omega_1) is the \Sigma-product of \omega_1 many copies of the real line \mathbb{R}. The \Sigma-product of separable metric spaces is monolithic (see here). The \Sigma-product of first countable spaces is Frechet-Urysohn (see here). Thus \Sigma(\omega_1) has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space C_p(\omega_1+1) has both of these properties. The key to the discussion is then to show that C_p(\omega_1+1) is homeopmophic to a subspace of the \Sigma-product \Sigma(\omega_1).

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Connection to \Sigma-product

We show that the function space C_p(\omega_1+1) is homeomorphic to a subspace of the \Sigma-product of \omega_1 many copies of the real lines. Let Y_0 be the following subspace of C_p(\omega_1+1):

    Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}

Every function in Y_0 has non-zero values at only countably points of \omega_1+1. Thus Y_0 can be regarded as a subspace of the \Sigma-product \Sigma(\omega_1).

By Theorem 1 in this previous post, C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}, i.e, the function space C_p(\omega_1+1) is homeomorphic to the product space Y_0 \times \mathbb{R}. On the other hand, the product Y_0 \times \mathbb{R} can also be regarded as a subspace of the \Sigma-product \Sigma(\omega_1). Basically adding one additional factor of the real line to Y_0 still results in a subspace of the \Sigma-product. Thus we have:

    C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

Thus C_p(\omega_1+1) possesses all the hereditary properties of \Sigma(\omega_1). Another observation we can make is that \Sigma(\omega_1) is not hereditarily normal. The function space C_p(\omega_1+1) is not normal (see here). The \Sigma-product \Sigma(\omega_1) is normal (see here). Thus \Sigma(\omega_1) is not hereditarily normal.

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A closer look at C_p(\omega_1+1)

In fact C_p(\omega_1+1) has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let h be a homeomorphism from C_p(\omega_1+1) onto Y_0 \times \mathbb{R}. For each \alpha<\omega_1, let H_\alpha be defined as follows:

    H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}

Clearly H_\alpha \subset Y_0. Furthermore H_\alpha can be considered as a subspace of \mathbb{R}^\omega and is thus metrizable. Let A be a countable subset of C_p(\omega_1+1). Then h(A) \subset H_\alpha \times \mathbb{R} for some \alpha<\omega_1. The set H_\alpha \times \mathbb{R} is metrizable. The set H_\alpha \times \mathbb{R} is also a closed subset of Y_0 \times \mathbb{R}. Then \overline{A} is contained in H_\alpha \times \mathbb{R} and is therefore metrizable. We have shown that the closure of every countable subspace of C_p(\omega_1+1) is metrizable. In other words, every separable subspace of C_p(\omega_1+1) is metrizable. This property follows from the fact that C_p(\omega_1+1) is strongly monolithic.

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Monolithicity and Frechet-Urysohn property

As indicated at the beginning, the \Sigma-product \Sigma(\omega_1) is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space C_p(\omega_1+1) is both strongly monolithic and Frechet-Urysohn.

Let \tau be an infinite cardinal. A space X is \tau-monolithic if for any A \subset X with \lvert A \lvert \le \tau, we have nw(\overline{A}) \le \tau. A space X is monolithic if it is \tau-monolithic for all infinite cardinal \tau. It is straightforward to show that X is monolithic if and only of for every subspace Y of X, the density of Y equals to the network weight of Y, i.e., d(Y)=nw(Y). A longer discussion of the definition of monolithicity is found here.

A space X is strongly \tau-monolithic if for any A \subset X with \lvert A \lvert \le \tau, we have w(\overline{A}) \le \tau. A space X is strongly monolithic if it is strongly \tau-monolithic for all infinite cardinal \tau. It is straightforward to show that X is strongly monolithic if and only if for every subspace Y of X, the density of Y equals to the weight of Y, i.e., d(Y)=w(Y).

In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.

In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space C_p([0,1]). It is clear that all metrizable spaces are strongly monolithic.

The function space C_p(\omega_1+1) is not separable. Since it is strongly monolithic, every separable subspace of C_p(\omega_1+1) is metrizable. We can see this by knowing that C_p(\omega_1+1) is a subspace of the \Sigma-product \Sigma(\omega_1), or by using the homeomorphism h as in the previous section.

For any compact space X, C_p(X) is countably tight (see this previous post). In the case of the compact uncountable ordinal \omega_1+1, C_p(\omega_1+1) has the stronger property of being Frechet-Urysohn. A space Y is said to be a Frechet-Urysohn space (also called a Frechet space) if for each y \in Y and for each M \subset Y, if y \in \overline{M}, then there exists a sequence \left\{y_n \in M: n=1,2,3,\cdots \right\} such that the sequence converges to y. As we shall see below, C_p(X) is rarely Frechet-Urysohn.

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General discussion

For any compact space X, C_p(X) is monolithic but does not have to be strongly monolithic. The monolithicity of C_p(X) follows from the following theorem, which is Theorem II.6.8 in [1].

Theorem 1
Then the function space C_p(X) is monolithic if and only if X is a stable space.

See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space X is stable if for any continuous image Y of X, the weak weight of Y, denoted by ww(Y), coincides with the network weight of Y, denoted by nw(Y). In [1], ww(Y) is notated by iw(Y). The cardinal function ww(Y) is the minimum cardinality of all w(T), the weight of T, for which there exists a continuous bijection from Y onto T.

All compact spaces are stable. Let X be compact. For any continuous image Y of X, Y is also compact and ww(Y)=w(Y), since any continuous bijection from Y onto any space T is a homeomorphism. Note that ww(Y) \le nw(Y) \le w(Y) always holds. Thus ww(Y)=w(Y) implies that ww(Y)=nw(Y). Thus we have:

Corollary 2
Let X be a compact space. Then the function space C_p(X) is monolithic.

However, the strong monolithicity of C_p(\omega_1+1) does not hold in general for C_p(X) for compact X. As indicated above, C_p([0,1]) is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of C_p(X).

Theorem 3
Let X be a space. Then C_p(X) is strongly monolithic if and only if X is simple.

A space X is \tau-simple if whenever Y is a continuous image of X, if the weight of Y \le \tau, then the cardinality of Y \le \tau. A space X is simple if it is \tau-simple for all infinite cardinal numbers \tau. Interestingly, any separable metric space that is uncountable is not \omega-simple. Thus [0,1] is not \omega-simple and C_p([0,1]) is not strongly monolithic, according to Theorem 3.

For compact spaces X, C_p(X) is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].

Theorem 4
Let X be a compact space. Then the following conditions are equivalent.

  1. C_p(X) is a Frechet-Urysohn space.
  2. C_p(X) is a k-space.
  3. The compact space X is a scattered space.

A space X is a scattered space if for every non-empty subspace Y of X, there exists an isolated point of Y (relative to the topology of Y). Any space of ordinals is scattered since every non-empty subset has a least element. Thus \omega_1+1 is a scattered space. On the other hand, the unit interval [0,1] with the Euclidean topology is not scattered. According to this theorem, C_p([0,1]) cannot be a Frechet-Urysohn space.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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\copyright \ 2014 \text{ by Dan Ma}

Cp(omega 1 + 1) is not normal

In this and subsequent posts, we consider C_p(X) where X is a compact space. Recall that C_p(X) is the space of all continuous real-valued functions defined on X and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is \omega_1+1, the first compact uncountable ordinal. There are many interesting results about the function space C_p(\omega_1+1). In this post we show that C_p(\omega_1+1) is not normal. An even more interesting fact about C_p(\omega_1+1) is that C_p(\omega_1+1) does not have any dense normal subspace [1].

Let \omega_1 be the first uncountable ordinal, and let \omega_1+1 be the successor ordinal to \omega_1. The set \omega_1 is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space \omega_1+1 is the first compact uncountable ordinal. In proving that C_p(\omega_1+1) is not normal, a theorem that is due to D. P. Baturov is utilized [2]. This theorem is also proved in this previous post.

For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).

The fact that C_p(\omega_1+1) is not normal is established by the following two points.

  • If C_p(\omega_1+1) is normal, then C_p(\omega_1+1) has countable extent, i.e. every closed and discrete subspace of C_p(\omega_1+1) is countable.
  • There exists an uncountable closed and discrete subspace of C_p(\omega_1 +1).

We discuss each of the bullet points separately.

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The function space C_p(\omega_1+1) is a dense subspace of \mathbb{R}^{\omega_1}, the product of \omega_1 many copies of \mathbb{R}. According to a result of D. P. Baturov [2], any dense normal subspace of the product of \omega_1 many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus C_p(\omega_1+1) cannot be normal if the second bullet point above is established.

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Now we show that there exists an uncountable closed and discrete subspace of C_p(\omega_1 +1). For each \alpha with 0<\alpha<\omega_1, define h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\} by:

    h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1  \end{cases}

Clearly, h_\alpha \in C_p(\omega_1 +1) for each \alpha. Let H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}. We show that H is a closed and discrete subspace of C_p(\omega_1 +1). The fact that H is closed in C_p(\omega_1 +1) is establish by the following claim.

Claim 1
Let h \in C_p(\omega_1 +1) \backslash H. There exists an open subset U of C_p(\omega_1 +1) such that h \in U and U \cap H=\varnothing.

First we get some easy cases out of the way. Suppose that there exists some \alpha<\omega_1 such that h(\alpha) \notin \left\{0,1 \right\}. Then let U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}. Clearly h \in U and U \cap H=\varnothing.

Another easy case: If h(\alpha)=0 for all \alpha \le \omega_1, then consider the open set U where U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}. Clearly h \in U and U \cap H=\varnothing.

From now on we can assume that h(\omega_1+1) \subset \left\{0,1 \right\} and that h is not identically the zero function. Suppose Claim 1 is not true. Then h \in \overline{H}. Next observe the following:

    Observation.
    If h(\beta)=1 for some \beta \le \omega_1, then h(\alpha)=1 for all \alpha \le \beta.

To see this, if h(\alpha)=0, h(\beta)=1 and \alpha<\beta, then define the open set V by V=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (-0.1,0.1) \text{ and } f(\beta) \in (0.9,1.1) \right\}. Note that h \in V and V \cap H=\varnothing, contradicting that h \in \overline{H}. So the above observation is valid.

Now either h(\omega_1)=1 or h(\omega_1)=0. We claim that h(\omega_1)=1 is not possible. Suppose that h(\omega_1)=1. Let V=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in (0.9,1.1) \right\}. Then h \in V and V \cap H=\varnothing, contradicting that h \in \overline{H}. It must be the case that h(\omega_1)=0.

Because of the continuity of h at the point x=\omega_1, of all the \gamma<\omega_1 for which h(\gamma)=1, there is the largest one, say \beta. Now h(\beta)=1. According to the observation made above, h(\alpha)=1 for all \alpha \le \beta. This means that h=h_\beta. This is a contradiction since h \notin H. Thus Claim 1 must be true and the fact that H is closed is established.

Next we show that H is discrete in C_p(\omega_1 +1). Fix h_\alpha where 0<\alpha<\omega_1. Let W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}. It is clear that h_\alpha \in W. Furthermore, h_\gamma \notin W for all \alpha < \gamma and h_\gamma \notin W for all \gamma <\alpha. Thus W is open such that \left\{h_\alpha \right\}=W \cap H. This completes the proof that H is discrete.

We have established that H is an uncountable closed and discrete subspace of C_p(\omega_1 +1). This implies that C_p(\omega_1 +1) is not normal.

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Remarks

The set H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\} as defined above is closed and discrete in C_p(\omega_1 +1). However, the set H is not discrete in a larger subspace of the product space. The set H is also a subset of the following \Sigma-product:

    \Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}

Because \Sigma(\omega_1) is the \Sigma-product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, \Sigma(\omega_1) would have countable extent. Thus the set H cannot be closed and discrete in \Sigma(\omega_1). We can actually see this directly. Let \alpha<\omega_1 be a limit ordinal. Define t:\omega_1 + 1 \rightarrow \left\{0,1 \right\} by t(\beta)=1 for all \beta<\alpha and t(\beta)=0 for all \beta \ge \alpha. Clearly t \notin C_p(\omega_1 +1) and t \in \Sigma(\omega_1). Furthermore, t \in \overline{H} (the closure is taken in \Sigma(\omega_1)).

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Reference

  1. Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
  2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

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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}

The normality of the product of the first uncountable ordinal with a compact factor

The product of a normal space with a compact space needs not be normal. For example, the product space \omega_1 \times (\omega_1+1) is not normal where \omega_1 is the first uncountable ordinal with the order topology and \omega_1+1 is the immediate successor of \omega_1 (see this post). However, \omega_1 \times I is normal where I=[0,1] is the unit interval with the usual topology. The topological story here is that I has countable tightness while the compact space \omega_1+1 does not. In this post, we prove the following theorem:

Theorem 1

    Let Y be an infinite compact space. Then the following conditions are equivalent:

    1. The product space \omega_1 \times Y is normal.
    2. Y has countable tightness, i.e., t(Y)=\omega.

Theorem 1 is a special case of the theorem found in [4]. The proof for the direction of countable tightness of Y implies \omega_1 \times Y is normal given in [4] relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction 2 \Longrightarrow 1, we give a complete proof. For the direction 1 \Longrightarrow 2, we essentially give the same proof as in [4], proving it by using a series of lemmas (stated below).

The authors in [2] studied the normality of X \times \omega_1 where X is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.

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Definitions and Lemmas

Let X be a topological space. The tightness of X, denoted by t(X), is the least infinite cardinal number \kappa such that for any A \subset X and for any x \in \overline{A}, there exists a B \subset A such that x \in \overline{B} and \lvert B \lvert \le \kappa. When t(X)=\omega, we say X has countable tightness or is countably tight. When t(X)>\omega, we say X has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is \omega_1+1=\omega_1 \cup \left\{\omega_1 \right\}. This space has uncountable tightness at the point \omega_1. All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.

A sequence \left\{x_\alpha: \alpha<\tau \right\} of points of a space X is said to be a free sequence if for each \alpha<\tau, \overline{\left\{x_\beta: \beta<\alpha \right\}} \cap \overline{\left\{x_\beta: \beta \ge \alpha \right\}}=\varnothing. When a free sequence is indexed by the cardinal number \tau, the free sequence is said to have length \tau. The cardinal function F(X) is the least infinite cardinal \kappa such that if \left\{x_\alpha \in X: \alpha<\tau \right\} is a free sequence of length \tau, then \tau \le \kappa. The concept of tightness was introduced by Arkhangelskii and he proved that t(X)=F(X) (see p. 15 of [3]). This fact implies the following lemma.

Lemma 2

    Let X be compact. If t(X) \ge \tau, then there exists a free sequence \left\{x_\alpha \in X: \alpha<\tau \right\} of length \tau.

A proof of Lemma 2 can be found here.

The proof of the direction 1 \Longrightarrow 2 also uses the following lemmas.

Lemma 3

    For any compact space Y, \beta (\omega_1 \times Y)=(\omega_1+1) \times Y.

Lemma 4

    Let X be a normal space. For every pair H and K of disjoint closed subsets of X, H and K have disjoint closures in \beta X.

For Lemma 3, see 3.12.20(c) on p. 237 of [1]. For Lemma 4, see 3.6.4 on p. 173 of [1].

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Proof of Theorem 1

1 \Longrightarrow 2
Let X=\omega_1 \times Y. Suppose that X is normal. Suppose that Y has uncountable tightness, i.e., t(Y) \ge \omega_1. By Lemma 2, there exists a free sequence \left\{y_\alpha \in Y: \alpha<\omega_1 \right\}. For each \beta<\omega_1, let C_\beta=\left\{y_\alpha: \alpha>\beta \right\}. Then the collection \left\{\overline{C_\beta}: \beta<\omega_1 \right\} has the finite intersection property. Since Y is compact, \bigcap_{\beta<\omega_1} \overline{C_\beta} \ne \varnothing. Let p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}. Consider the following closed subsets of X=\omega_1 \times Y.

    H=\overline{\left\{(\alpha,y_\alpha): \alpha<\omega_1 \right\}}
    K=\left\{(\alpha,p): \alpha<\omega_1 \right\}

We claim that H \cap K=\varnothing. Suppose that (\alpha,p) \in H \cap K. Either p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}} or p \in \overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}. The latter case is not possible. Note that [0,\alpha] \times Y is an open set containing (\alpha,p). This open set cannot contain points of the form (\delta,p) where \delta \ge \alpha+1. So the first case p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}} must hold. Since p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}, p \in \overline{C_\alpha}=\overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}, a contradiction. So H and K are disjoint closed subsets of X=\omega_1 \times Y.

Now consider \beta X, the Stone-Cech compactification of X=\omega_1 \times Y. By Lemma 3, \beta X=\beta (\omega_1 \times Y)=(\omega_1+1) \times Y. Let H^*=\overline{H} and K^*=\overline{K} (closures in \beta X). We claim that (\omega_1,p) \in H^* \cap K^*. Let O=(\theta,\omega_1] \times V be an open set in \beta X with (\omega_1,p) \in O. Note that p \in \overline{C_\theta}=\left\{y_\delta: \delta>\theta \right\}. Thus V \cap \overline{C_\theta} \ne \varnothing. Choose \delta>\theta such that y_\delta \in V. We have (\delta,y_\delta) \in (\theta,\omega_1] \times V and (\delta,y_\delta) \in H^*. On the other hand, (\delta,p) \in K^*. Thus (\omega_1,p) \in H^* \cap K^*, a contradiction. Since X=\omega_1 \times Y is normal, Lemma 4 indicates that H and K should have disjoint closures in \beta X=(\omega_1+1) \times Y. Thus Y has countable tightness.

2 \Longrightarrow 1
Suppose t(Y)=\omega. Let H and K be disjoint closed subsets of \omega_1 \times Y. The following series of claims will complete the proof:

Claim 1
For each y \in Y, there exists an \alpha<\omega_1 such that either W_{H,y} \subset \alpha+1 or W_{K,y} \subset \alpha+1 where

    W_{H,y}=\left\{\delta<\omega_1: (\delta,y) \in H \right\}
    W_{K,y}=\left\{\delta<\omega_1: (\delta,y) \in K \right\}

Proof of Claim 1
Let y \in Y. The set V=\omega_1 \times \left\{y \right\} is a copy of \omega_1. It is a known fact that in \omega_1, there cannot be two disjoint closed and unbounded sets. Let V_H=V \cap H and V_K=V \cap K. If V_H \ne \varnothing and V_K \ne \varnothing, they cannot be both unbounded in V. Thus the claim follows if both V_H \ne \varnothing and V_K \ne \varnothing. Now suppose only one of V_H and V_K is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say V_K. Then W_{H,y}=\varnothing and the claim follows.

Claim 2
For each y \in Y, there exists an \alpha<\omega_1 and there exists an open set O_y \subset Y with y \in O_y such that one and only one of the following holds:

    H \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (1)
    K \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (2)

Proof of Claim 2
Let y \in Y. Let \alpha<\omega_1 be as in Claim 1. Assume that W_{H,y} \subset \alpha+1. We want to show that there exists an open set O_y \subset Y with y \in O_y such that (1) holds. Suppose that for each open O \subset Y with y \in O, there is a q \in \overline{O} and there exists \delta_q>\alpha such that (\delta_q,q) \in H. Let S be the set of all such points q. Then y \in \overline{S}. Since Y has countable tightness, there exists countable T \subset S such that y \in \overline{T}. Since T is countable, choose \gamma >\omega_1 such that \alpha<\delta_q<\gamma for all q \in T. Note that [\alpha,\gamma] \times \left\{y \right\} does not contain points of H since W_{H,y} \subset \alpha+1. For each \theta \in [\alpha,\gamma], the point (\theta,y) has an open neighborhood that contains no point of H. Since [\alpha,\gamma] \times \left\{y \right\} is compact, finitely many of these neighborhoods cover [\alpha,\gamma] \times \left\{y \right\}. Let these finitely many open neighborhoods be M_i \times N_i where i=1,\cdots,m. Let N=\bigcap_{i=1}^m N_i. Then y \in N and N would contain a point of T, say q. Then (\delta_q,q) \in M_i \times N_i for some i, a contradiction. Note that (\delta_q,q) is a point of H. Thus there exists an open O_y \subset Y with y \in O_y such that (1) holds. This completes the proof of Claim 2.

Claim 3
For each y \in Y, there exists an \alpha<\omega_1 and there exists an open set O_y \subset Y with y \in O_y such that there are disjoint open subsets Q_H and Q_K of \omega_1 \times \overline{O_y} with H \cap (\omega_1 \times \overline{O_y}) \subset Q_H and K \cap (\omega_1 \times \overline{O_y}) \subset Q_K.

Proof of Claim 3
Let y \in Y. Let \alpha and O_y be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that (\alpha+1) \times \overline{O_y} is a product of two compact spaces and is thus compact (and normal). Let R_{H,y} and R_{K,y} be disjoint open sets in (\alpha+1) \times \overline{O_y} such that H \cap (\alpha+1) \times \overline{O_y} \subset R_{H,y} and K \cap (\alpha+1) \times \overline{O_y} \subset R_{K,y}. Note that [\alpha+1,\omega_1) \times \overline{O_y} contains no points of H. Then Q_{H,y}=R_{H,y} and Q_{K,y}=R_{K,y} \cup [\alpha+1,\omega_1) \times \overline{O_y} are the desired open sets. This completes the proof of Claim 3.

To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.

Claim 4
Let Z be a space. Let C and D be disjoint closed subsets of Z. Suppose that \left\{U_1,U_2,\cdots,U_m \right\} is a collection of open subsets of Z covering C \cup D such that for each i=1,2,\cdots,m, only one of the following holds:

    C \cap \overline{U_i} \ne \varnothing \text{ and } D \cap \overline{U_i}=\varnothing
    C \cap \overline{U_i} = \varnothing \text{ and } D \cap \overline{U_i} \ne \varnothing

Then there exist disjoint open subsets of Z separating C and D.

Proof of Claim 4
Let U_C=\cup \left\{U_i: \overline{U_i} \cap C \ne \varnothing \right\} and U_D=\cup \left\{U_i: \overline{U_i} \cap D \ne \varnothing \right\}. Note that \overline{U_C}=\cup \left\{\overline{U_i}: \overline{U_i} \cap C \ne \varnothing \right\}. Likewise, \overline{U_D}=\cup \left\{\overline{U_i}: \overline{U_i} \cap D \ne \varnothing \right\}. Let V_C=U_C-\overline{U_D} and V_D=U_D-\overline{U_C}. Then V_C and V_D are disjoint open sets. Furthermore, C \subset V_C and D \subset V_D. This completes the proof of Claim 4.

Now back to the proof of Theorem 1. For each y \in Y, let O_y, Q_{H,y} and Q_{K,y} be as in Claim 3. Since Y is compact, there exists \left\{y_1,y_2,\cdots,y_n \right\} \subset Y such that \left\{O_{y_1},O_{y_2},\cdots,O_{y_n} \right\} is a cover of Y. For each i=1,\cdots,n, let L_i=Q_{H,y_i} \cap (\omega_1 \times O_y) and M_i=Q_{K,y_i} \cap (\omega_1 \times O_y). Note that both L_i and M_i are open in \omega_1 \times Y. To apply Claim 4, rearrange the open sets L_i and M_i and re-label them as U_1,U_2,\cdots,U_m. By letting Z=\omega_1 \times Y, C=H and D=K, the open sets U_i satisfy Claim 4. Tracing the U_i to L_j or M_j and then to Q_{H,y_j} and Q_{K,y_j}, it is clear that the two conditions in Claim 4 are satisfied:

    H \cap \overline{U_i} \ne \varnothing \text{ and } K \cap \overline{U_i}=\varnothing
    H \cap \overline{U_i} = \varnothing \text{ and } K \cap \overline{U_i} \ne \varnothing

Then by Claim 4, the disjoint closed sets H and K can be separated by two disjoint open subsets of \omega_1 \times Y. \blacksquare

The theorem proved in [4] is essentially the statement that for any compact space Y, the product \kappa^+ \times Y is normal if and only t(Y) \le \kappa. Here \kappa^+ is the first ordinal of the next cardinal that is greater than \kappa.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Gruenhage, G., Nogura, T., Purisch, S., Normality of X \times \omega_1, Topology and its Appl., 39, 263-275, 1991.
  3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  4. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

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\copyright \ 2014 - 2015 \text{ by Dan Ma}