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

Compact metrizable scattered spaces

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space X is a scattered space if every non-empty subspace Y of X has a point y \in Y such that y is an isolated point in Y, i.e. the singleton set \left\{y \right\} is open in the subspace Y. A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space X, the following conditions are equivalent:

  • The function space C_p(X) is a Frechet-Urysohn space.
  • The function space C_p(X) is a k space.
  • X is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space X, let C(X) be the set of all continuous real-valued functions defined on the space X. Since C(X) is a subspace of the product space \mathbb{R}^X, a natural topology that can be given to C(X) is the subspace topology inherited from the product space \mathbb{R}^X. Then C_p(X) is simply the set C(X) with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space X is countable and infinite. Then the function space C_p(X) is metrizable since it is a subspace of \mathbb{R}^X, a product of countably many lines. Thus the function space C_p(X) has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space X is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let X be a space. Let A \subset X. Let p \in X. We say that p is a limit point of A if every open subset of X containing p contains a point of A distinct from p. So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let X be a hereditarily Lindelof space (i.e. every subspace of X is Lindelof). Then for any uncountable subset A of X, all but countably many points of A are limit points of A.

We now discuss the main result.

Theorem 3
Let X be a compact metrizable space such that every point of X is a limit point of X. Then there exists an uncountable closed subset C of X such that every point of C is a limit point of C.

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric \rho on the space X. One fact that we will use is that if there is a sequence of closed sets X \supset H_1 \supset H_2 \supset H_3 \supset \cdots such that the diameters of the sets H (based on the complete metric \rho) decrease to zero, then the sets H_n collapse to one point.

The uncountable closed set C we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points p_0,p_1 \in X such that p_0 \ne p_1. By assumption, both points are limit points of the space X. Choose open sets U_0,U_1 \subset X such that

  • p_0 \in U_0,
  • p_1 \in U_1,
  • K_0=\overline{U_0} and K_1=\overline{U_1},
  • K_0 \cap K_1 = \varnothing,
  • the diameters for K_0 and K_1 with respect to \rho are less than 0.5.

Note that each of these open sets contains infinitely many points of X. Then we can pick two points in each of U_0 and U_1 in the same manner. Before continuing, we set some notation. If \sigma is an ordered string of 0’s and 1’s of length n (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus \sigma is extended as \sigma 0 and \sigma 1 (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the nth stage where n \ge 1 is completed. This means that the points p_\sigma and the open sets U_\sigma have been chosen such that p_\sigma \in U_\sigma for each length n string of 0’s and 1’s \sigma. Now we continue the picking for the (n+1)st stage. For each \sigma, an n-length string of 0’s and 1’s, choose two points p_{\sigma 0} and p_{\sigma 1} and choose two open sets U_{\sigma 0} and U_{\sigma 1} such that

  • p_{\sigma 0} \in U_{\sigma 0},
  • p_{\sigma 1} \in U_{\sigma 1},
  • K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma} and K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma},
  • K_{\sigma 0} \cap K_{\sigma 1} = \varnothing,
  • the diameters for K_{\sigma 0} and K_{\sigma 1} with respect to \rho are less than 0.5^{n+1}.

For each positive integer m, let C_m be the union of all K_\sigma over all \sigma that are m-length strings of 0’s and 1’s. Each C_m is a union of finitely many compact sets and is thus compact. Furthermore, C_1 \supset C_2 \supset C_3 \supset \cdots. Thus C=\bigcap \limits_{m=1}^\infty C_m is non-empty. To complete the proof, we need to show that

  • C is uncountable (in fact of cardinality continuum),
  • every point of C is a limit point of C.

To show the first point, we define a one-to-one function f: \left\{0,1 \right\}^N \rightarrow C where N=\left\{1,2,3,\cdots \right\}. Note that each element of \left\{0,1 \right\}^N is a countably infinite string of 0’s and 1’s. For each \tau \in \left\{0,1 \right\}^N, let \tau \upharpoonright  n denote the string of the first n digits of \tau. For each \tau \in \left\{0,1 \right\}^N, let f(\tau) be the unique point in the following intersection:

    \displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright  n} = \left\{f(\tau) \right\}

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if \tau are 01011010…., then consider K_0 \supset K_{01} \supset K_{010} \supset \cdots. At each next step, always pick the K_{\tau \upharpoonright  n} that matches the next digit of \tau. Since the sets K_{\tau \upharpoonright  n} are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map f is one-to-one. If \tau and \gamma are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map f is reversible. Pick any point x \in C. Then the point x must belong to a nested sequence of sets K‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set C has the same cardinality as the set \left\{0,1 \right\}^N, which has cardinality continuum.

To see the second point, pick x \in C. Suppose x=f(\tau) where \tau \in \left\{0,1 \right\}^N. Consider the open sets U_{\tau \upharpoonright n} for all positive integers n. Note that x \in U_{\tau \upharpoonright n} for each n. Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set U_{\tau \upharpoonright n} contains infinitely many other points of C (this is because of all the open sets U_{\tau \upharpoonright k} that are subsets of U_{\tau \upharpoonright n} where k \ge n). Because the diameters are decreasing to zero, the sequence of U_{\tau \upharpoonright n} is a local base at the point x. Thus, the point x is a limit point of C. This completes the proof. \blacksquare

Theorem 4
Let X be a compact metrizable space. It follows that X is scattered if and only if X is countable.

Proof of Theorem 4
\Longleftarrow
In this direction, we show that if X is countable, then X is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if X is not scattered, then X is uncountable. Suppose X is not scattered. Then every point of X is a limit point of X. By Theorem 3, X would contain a Cantor set C of cardinality continuum.

\Longrightarrow
In this direction, we show that if X is scattered, then X is countable. We also show the contrapositive: if X is uncountable, then X is not scattered. Suppose X is uncountable. By Theorem 2, all but countably many points of X are limit points of X. After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of X is a limit point of X. Then by Theorem 3, X contains an uncountable closed set C such that every point of C is a limit point of C. This means that X is not scattered. \blacksquare

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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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 \ 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}

A useful representation of Cp(X)

Let X be a completely regular space. The space C_p(X) is the space of all real-valued continuous functions defined on X endowed with the pointwise convergence topology. In this post, we show that C_p(X) can be represented as the product of a subspace of C_p(X) with the real line \mathbb{R}. We prove the following theorem. See here for an application of this theorem.

Theorem 1
Let X be a completely regular space. Let x \in X. Let Y be defined by:

    Y=\left\{f \in C_p(X): f(x)=0 \right\}

Then C_p(X) is homeomorphic to Y \times \mathbb{R}.

The above theorem can be found in [1] (see Theorem I.5.4 on p. 37). In [1], the homeomorphism is stated without proof. For the sake of completeness, we provide a detailed proof of Theorem 1.

Proof of Theorem 1
Define h: C_p(X) \rightarrow Y \times \mathbb{R} by h(f)=(f-f(x),f(x)) for any f \in C_p(X). The map h is a homeomorphism.

The map is one-to-one

First, we show that it is a one-to-one map. Let f,g \in C_p(X) where f \ne g. Assume that f(x) \ne g(x). Then h(f) \ne h(g). So assume that f(x)=g(x). Then the functions f-f(x) and g-g(x) are different, which means h(f) \ne h(g).

The map is onto

Now we show h maps C_p(X) onto Y \times \mathbb{R}. Let (g,t) \in Y \times \mathbb{R}. Let f=g+t. Note that f(x)=g(x)+t=t. Then f-f(x)=g. We have h(f)=(g,t).

Note. Showing the continuity of h and h^{-1} is a matter of working with the basic open sets in the function space carefully (e.g. making the necessary shifting). Some authors just skip the details and declare them continuous, e.g. [1]. Readers are welcome to work out enough of the details to see the key idea.

The map is continuous

Show that h is continuous. Let f \in C_p(X). Let U \times V be an open set in Y \times \mathbb{R} such that h(f) \in U \times V and,

    U=\left\{g \in Y: \forall \ i=1,\cdots,n, g(x_i) \in U_i \right\}

    \forall \ i=1,\cdots,n, \  U_i=(f(x_i)-f(x)-\frac{1}{k},f(x_i)-f(x)+\frac{1}{k})

    V=(f(x)-\frac{1}{k},f(x)+\frac{1}{k})

where x_1,\cdots,x_n are arbitrary points in X and k is some large positive integer. Define the following:

    \forall \ i=1,\cdots,n, \ W_i=(f(x_i)-\frac{1}{2k},f(x_i)+\frac{1}{2k})

    W_{n+1}=(f(x)-\frac{1}{2k},f(x)+\frac{1}{2k})

    x_{n+1}=x

Then define the open set W as follows:

    W=\left\{q \in C_p(X): \forall \ i=1,\cdots,n,n+1, q(x_i) \in W_i \right\}

Clearly f \in W. We need to show h(W) \subset U \times V. Let q \in W. Then h(q)=(q-q(x),q(x)). We need to show that q-q(x) \in U and q(x) \in V. Note that q(x_{n+1})=q(x) \in W_{n+1}. For each i=1,\cdots,n, q(x_i) \in W_i. So we have the following:

    f(x_i)-\frac{1}{2k}<q(x_i)<f(x_i)+\frac{1}{2k}

    f(x)-\frac{1}{2k}<q(x)<f(x)+\frac{1}{2k}

Subtracting the above two inequalities, we have the following:

    f(x_i)-f(x)-\frac{1}{k}<q(x_i)-q(x)<f(x_i)-f(x)+\frac{1}{k}

The above inequality shows that for each i=1,\cdots,n, q(x_i) -q(x) \in U_i. Hence q-q(x) \in U. It is clear that q(x) \in V. This completes the proof that the map h is continuous.

The inverse is continuous

We now show that h^{-1} is continuous. Let (g,t) \in Y \times \mathbb{R}. Note that h^{-1}(g,t)=g+t. Let M be an open set in C_p(X) such that g+t \in M and

    M=\left\{f \in C_p(X): \forall \ i=1,\cdots,n+1, f(x_i) \in M_i \right\}

    \forall \ i=1,\cdots,n, \  M_i=(g(x_i)+t-\frac{1}{m},g(x_i)+t+\frac{1}{m})

    x_{n+1}=x

    M_{n+1}=(t-\frac{1}{m},t+\frac{1}{m})

where x_1,\cdots,x_n are arbitrary points of X and m is some large positive integer. Now define an open subset G \times T of Y \times \mathbb{R} such that (g,t) \in G \times T and

    G=\left\{q \in Y: \forall \ i=1,\cdots,n+1, q(x_i) \in G_i \right\}

    \forall \ i=1,\cdots,n, \  G_i=(g(x_i)-\frac{1}{2m},g(x_i)+\frac{1}{2m})

    T=(t-\frac{1}{2m},t+\frac{1}{2m})

We need to show that h^{-1}(G \times T) \subset M. Let (q,a) \in G \times T. We then have the following inequalities.

    \forall \ i=1,\cdots,n, \ g(x_i)-\frac{1}{2m}<q(x_i)<g(x_i)+\frac{1}{2m}

    t-\frac{1}{2m}<a<t+\frac{1}{2m}

Adding the above two inequalities, we obtain:

    \forall \ i=1,\cdots,n, \ g(x_i)+t-\frac{1}{m}<q(x_i)+a<g(x_i)+t+\frac{1}{m}

The above implies that \forall \ i=1,\cdots,n, q(x_i)+a \in M_i. It is clear that q(x_{n+1})+a=q(x)+a=a \in M_{n+1}. Thus q+a \in M. This completes the proof that h^{-1} is continuous.

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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}

A useful embedding for Cp(X)

Let X be a Tychonoff space (also called completely regular space). By C_p(X) we mean the space of all continuous real-valued functions defined on X endowed with the pointwise convergence topology. In this post we discuss a scenario in which a function space can be embedded into another function space. We prove the following theorem. An example follows the proof.

Theorem 1
Suppose that the space Y is a continuous image of the space X. Then C_p(Y) can be embedded into C_p(X).

Proof of Theorem 1
Let t:X \rightarrow Y be a continuous surjection, i.e., t is a continuous function from X onto Y. Define the map \psi: C_p(Y) \rightarrow C_p(X) by \psi(f)=f \circ t for all f \in C_p(Y). We show that \psi is a homeomorphism from C_p(Y) into C_p(X).

First we show \psi is a one-to-one map. Let f,g \in C_p(Y) with f \ne g. There exists some y \in Y such that f(y) \ne g(y). Choose some x \in X such that t(x)=y. Then f \circ t \ne g \circ t since (f \circ t)(x)=f(t(x))=f(y) and (g \circ t)(x)=g(t(x))=g(y).

Next we show that \psi is continuous. Let f \in C_p(Y). Let U be open in C_p(X) with \psi(f) \in U such that

    U=\left\{q \in C_p(X): \forall \ i=1,\cdots,n, \ q(x_i) \in U_i \right\}

where x_1,\cdots,x_n are arbitrary points of X and each U_i is an open interval of the real line \mathbb{R}. Note that for each i, f(t(x_i)) \in U_i. Now consider the open set V defined by:

    V=\left\{r \in C_p(Y): \forall \ i=1,\cdots,n, \ r(t(x_i)) \in U_i \right\}

Clearly f \in V. It follows that \psi(V) \subset U since for each r \in V, it is clear that \psi(r)=r \circ t \in U.

Now we show that \psi^{-1}: \psi(C_p(Y)) \rightarrow C_p(Y) is continuous. Let \psi(f)=f \circ t \in \psi(C_p(Y)) where f \in C_p(Y). Let G be open with \psi^{-1}(f \circ t)=f \in G such that

    G=\left\{r \in C_p(Y): \forall \ i=1,\cdots,m, \ r(y_i) \in G_i \right\}

where y_1,\cdots,y_m are arbitrary points of Y and each G_i is an open interval of \mathbb{R}. Choose x_1,\cdots,x_m \in X such that t(x_i)=y_i for each i. We have f(t(x_i)) \in G_i for each i. Define the open set H by:

    H=\left\{q \in \psi(C_p(Y)) \subset C_p(X): \forall \ i=1,\cdots,m, \ q(x_i) \in G_i \right\}

Clearly f \circ t \in H. Note that \psi^{-1}(H) \subset G. To see this, let r \circ t \in H where r \in C_p(Y). Now r(t(x_i))=r(y_i) \in G_i for each i. Thus \psi^{-1}(r \circ t)=r \in G. It follows that \psi^{-1} is continuous. The proof of the theorem is now complete. \blacksquare

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Example

The proof of Theorem 1 is not difficult. It is a matter of notating carefully the open sets in both function spaces. However, the embedding makes it easy in some cases to understand certain function spaces and in some cases to relate certain 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. As an application of Theorem 1, we show that C_p(\omega_1+1) can be embedded as a subspace of C_p(\omega_1). Define a continuous surjection g:\omega_1 \rightarrow \omega_1+1 as follows:

    g(\gamma) = \begin{cases} \omega_1 & \mbox{if } \ \gamma =0 \\ \gamma-1 & \mbox{if } \ 1 \le \gamma < \omega \\ \gamma & \mbox{if } \ \omega \le \gamma < \omega_1  \end{cases}

The map g is continuous from \omega_1 onto \omega_1+1. By Theorem 1, C_p(\omega_1+1) can be embedded as a subspace of C_p(\omega_1). On the other hand, C_p(\omega_1) cannot be embedded in C_p(\omega_1+1). The function space C_p(\omega_1+1) is a Frechet-Urysohn space, which is a property that is carried over to any subspace. The function C_p(\omega_1) is not Frechet-Urysohn. Thus C_p(\omega_1) cannot be embedded in C_p(\omega_1+1). A further comparison of these two function spaces is found in this subsequent post.

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\copyright \ 2014 \text{ by Dan Ma}

Cp(X) is countably tight when X is compact

Let X be a completely regular space (also called Tychonoff space). If X is a compact space, what can we say about the function space C_p(X), the space of all continuous real-valued functions with the pointwise convergence topology? When X is an uncountable space, C_p(X) is not first countable at every point. This follows from the fact that C_p(X) is a dense subspace of the product space \mathbb{R}^X and that no dense subspace of \mathbb{R}^X can be first countable when X is uncountable. However, when X is compact, C_p(X) does have a convergence property, namely C_p(X) is countably tight.

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Tightness

Let X be a completely regular 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 Y has countable tightness or is countably tight. When t(X)>\omega, we say that X has uncountable tightness or is uncountably tight. Clearly any first countable space is countably tight. There are other convergence properties in between first countability and countable tightness, e.g., the Frechet-Urysohn property. The notion of countable tightness and tightness in general is discussed in further details here.

The fact that C_p(X) is countably tight for any compact X 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 the countable case of Theorem I.4.1 on page 33 of [1]. We prove one direction of Theorem 1, the direction that will give us the desired result for C_p(X) where X is compact.

Proof of Theorem 1
The direction \Longleftarrow
Suppose that X^n is Lindelof for each positive integer. Let f \in C_p(X) and f \in \overline{H} where H \subset C_p(X). For each positive integer n, we define an open cover \mathcal{U}_n of X^n.

Let n be a positive integer. Let t=(x_1,\cdots,x_n) \in X^n. Since f \in \overline{H}, there is an h_t \in H such that \lvert h_t(x_j)-f(x_j) \lvert <\frac{1}{n} for all j=1,\cdots,n. Because both h_t and f are continuous, for each j=1,\cdots,n, there is an open set W(x_j) \subset X with x_j \in W(x_j) such that \lvert h_t(y)-f(y) \lvert < \frac{1}{n} for all y \in W(x_j). Let the open set U_t be defined by U_t=W(x_1) \times W(x_2) \times \cdots \times W(x_n). Let \mathcal{U}_n=\left\{U_t: t=(x_1,\cdots,x_n) \in X^n \right\}.

For each n, choose \mathcal{V}_n \subset \mathcal{U}_n be countable such that \mathcal{V}_n is a cover of X^n. Let K_n=\left\{h_t: t \in X^n \text{ such that } U_t \in \mathcal{V}_n \right\}. Let K=\bigcup_{n=1}^\infty K_n. Note that K is countable and K \subset H.

We now show that f \in \overline{K}. Choose an arbitrary positive integer n. Choose arbitrary points y_1,y_2,\cdots,y_n \in X. Consider the open set U defined by

    U=\left\{g \in C_p(X): \forall \ j=1,\cdots,n, \lvert g(y_j)-f(y_j) \lvert <\frac{1}{n} \right\}.

We wish to show that U \cap K \ne \varnothing. Choose U_t \in \mathcal{V}_n such that (y_1,\cdots,y_n) \in U_t where t=(x_1,\cdots,x_n) \in X^n. Consider the function h_t that goes with t. It is clear from the way h_t is chosen that \lvert h_t(y_j)-f(x_j) \lvert<\frac{1}{n} for all j=1,\cdots,n. Thus h_t \in K_n \cap U, leading to the conclusion that f \in \overline{K}. The proof that C_p(X) is countably tight is completed.

The direction \Longrightarrow
See Theorem I.4.1 of [1].

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Remarks

As shown above, countably tightness is one convergence property of C_p(X) that is guaranteed when X is compact. In general, it is difficult for C_p(X) to have stronger convergence properties such as the Frechet-Urysohn property. It is well known C_p(\omega_1+1) is Frechet-Urysohn. According to Theorem II.1.2 in [1], for any compact space X, C_p(X) is a Frechet-Urysohn space if and only if the compact space X is a scattered 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 - 2015 \text{ by Dan Ma}