An exercise gleaned from the proof of a theorem on pseudocompact space

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space X is a pseudocompact space if every continuous real-valued function defined on X is bounded, i.e., if f:X \rightarrow \mathbb{R} is a continuous function, then f(X) is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

    \text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space \Psi(\mathcal{A}) where \mathcal{A} is a maximal almost disjoint family of subsets of \omega (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let Y be a compact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto a space Z. Then g is a homeomorphism.

Theorem 2
Let Y be a pseudocompact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto Z where Z is a separable and metrizable space. Then g is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map g as described in Theorem 2, the domain Y must be compact. Then Theorem 1 will finish the job.

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Reference

  1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

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\copyright \ 2015 \text{ by Dan Ma}

Looking for non-normal subspaces of the square of a compact X

A theorem of Katetov states that if X is compact with a hereditarily normal cube X^3, then X is metrizable (discussed in this previous post). This means that for any non-metrizable compact space X, Katetov’s theorem guarantees that some subspace of the cube X^3 is not normal. Where can a non-normal subspace of X^3 be found? Is it in X, in X^2 or in X^3? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable X (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.

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Katetov’s theorems

First we state the results of Katetov for reference. These results are proved in this previous post.

Theorem 1
If X \times Y is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

  • The factor X is perfectly normal.
  • Every countable and infinite subset of the factor Y is closed.

Theorem 2
If X and Y are compact and X \times Y is hereditarily normal, then both X and Y are perfectly normal.

Theorem 3
Let X be a compact space. If X^3=X \times X \times X is hereditarily normal, then X is metrizable.

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Examples of compact non-metrizable spaces

The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can X^3 in Theorem 3 be replaced by X^2? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.

  1. One-point compactification of the Tychonoff plank.
  2. One-point compactification of \psi(\mathcal{A}) where \mathcal{A} is a maximal almost disjoint family of subsets of \omega.
  3. The first compact uncountable ordinal, i.e. \omega_1+1.
  4. The one-point compactification of an uncountable discrete space.
  5. Alexandroff double circle.
  6. Double arrow space.
  7. Unit square with the lexicographic order.

Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, X is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.

For #3 through #7, X is hereditarily normal. For #3 through #5, each X has a closed subset that is not a G_\delta set (hence not perfectly normal). In #3 and #4, the non-G_\delta-set is a single point. In #5, the the non-G_\delta-set is the inner circle. Thus the compact space X in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.

For #6 and #7, each X^2 contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.

In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:

    Question. Is there an example of a compact non-metrizable space X such that the failure of hereditary nornmality occurs in “dimension” 3? Specifically, is there a compact non-metrizable X such that X^2 is hereditarily normal but X^3 is not hereditarily normal?

Such a space X would be an example to show that the condition “X^3 is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.

The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable X such that X^2 is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable X such that X^2 is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.

In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact X with hereditarily normal X^2 is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming X^2 is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable X is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.

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Reference

  1. Gruenhage G., Nyikos P. J., Normality in X^2 for Compact X, Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
  2. Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
  3. Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791

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\copyright \ 2015 \text{ by Dan Ma}

When a product space is hereditarily normal

When the spaces X and Y are normal spaces, the product space X \times Y is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors X and Y can have no influence on the normality of the product X \times Y. The dynamics in the other direction are totally different. When the product X \times Y is hereditarily normal, the two factors X and Y are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset W of a space X is said to be a G_\delta-set in X if W is the intersection of countably many open subsets of X. A space X is perfectly normal if it is normal and that every closed subset of X is a G_\delta-set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product X \times Y will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product X \times Y is a very strong property.

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Katetov’s theorems

Theorem 1
If X \times Y is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

  • The factor X is perfectly normal.
  • Every countable and infinite subset of the factor Y is closed.

Proof of Theorem 1
The strategy we use is to define a subspace of X \times Y that is not normal after assuming that none of the two conditions is true. So assume that X has a closed subspace W that is not a G_\delta-set and assume that T=\left\{t_n: n=1,2,3,\cdots \right\} is an infinite subset of Y that is not closed. Let p \in Y be a limit point of T such that p \notin T. The candidate for a non-normal subspace of X \times Y is:

    M=X \times Y-W \times \left\{p \right\}

Note that M is an open subspace of X \times Y since it is the result of subtracting a closed set from X \times Y. The following are the two closed sets that demonstrate that M is not normal.

    H=W \times (Y-\left\{p \right\})

    K=(X-W) \times \left\{p \right\}

It is clear that H and K are closed subsets of M. Let U and V be open subsets of M such that H \subset U and K \subset V. We show that U \cap V \ne \varnothing. To this end, define U_j=\left\{x \in X: (x,t_j) \in U \right\} for each j. It follows that for each j, W \subset U_j. Furthermore each U_j is an open subspace of X. Thus W \subset \bigcap_j U_j. Since W is not a G_\delta-set in X, there must exist t \in \bigcap_j U_j such that t \notin W. Then (t, p) \in K and (t, p) \in V.

Since V is open in the product X \times Y, choose open sets A \subset X and B \subset Y such that (t,p) \in A \times B and A \times B \subset V. With p \in B, there exists some j such that t_j \in B. First, (t,t_j) \in V. Since t \in U_j, (t,t_j) \in U. Thus U \cap V \ne \varnothing. This completes the proof that the subspace M is not normal and that X \times Y is not hereditarily normal. \blacksquare

Let’s see what happens in Theorem 1 when both factors are compact. If both X and Y are compact and if X \times Y is hereditarily normal, then both X and Y must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

Theorem 2
If X and Y are compact and X \times Y is hereditarily normal, then both X and Y are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

Theorem 3
Let X be a compact space. If X^3=X \times X \times X is hereditarily normal, then X is metrizable.

Proof of Theorem 3
Suppose that X^3 is hereditarily normal. By Theorem 2, the compact spaces X^2 and X are perfectly normal. In particular, the following subset of X^2 is a G_\delta-set in X^2.

    \Delta=\left\{(x,x): x \in X \right\}

The set \Delta is said to be the diagonal of the space X. It is a well known result that any compact space whose diagonal is a G_\delta-set in the square is metrizable (discussed here). \blacksquare

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of X^2. This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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Reference

  1. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2015 \text{ by 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}

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(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}

Basic topological properties of Corson compact spaces

A compact space is said to be a Corson compact space if it can be embedded in a \Sigma-product of real lines. Corson compact spaces play an important role in functional analysis. Corson compact spaces are also very interesting from a topological point of view. In this post, we discuss several basic topological properties of Corson compact spaces, some of which have been discussed and proved in some previous posts.

Links to other posts on Corson compact spaces are given throughout this post (as relevant results are discussed). A listing of these previous posts is also given at the end of this post.

For any infinite cardinal number \kappa, the \Sigma-product of \kappa many copies of \mathbb{R} 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\}

A compact space is said to be a Corson compact space if it can be embedded in \Sigma(\kappa) for some infinite cardinal \kappa.

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Some examples of Corson compact spaces

When \kappa=\omega, \Sigma(\omega) is simply \mathbb{R}^\omega, the product of countably many copies of the real lines. Any compact metrizable space can be embedded in \mathbb{R}^\omega; see Theorem 4.2.10 in [3]. Thus any compact metrizable space is Corson compact.

One easily described non-metrizable Corson compact space is the one-point compactification of an uncountable discrete space. Let \tau be an uncountable cardinal number. Let D_\tau be the discrete space of cardinality \tau. Let p be a point not in D_\tau. Let A_\tau=D_\tau \cup \left\{p \right\}. Consider a topology on A_\tau such that D_\tau is discrete as before and any open neighborhood of p has the form \left\{p \right\} \cup C where C \subset D_\tau and D_\tau-C is finite. The space A_\tau is better known as the one-point compactification of a discrete space of cardinality \tau. To see that A_\tau is embedded in \Sigma(\tau), for each \alpha<\tau, define f_\alpha:\tau \rightarrow \left\{0,1 \right\} such that f_\alpha(\alpha)=1 and f_\alpha(\beta)=0 for all \beta \ne \alpha. Furthermore, let g:\tau \rightarrow \left\{0,1 \right\} be defined by g(\beta)=0 for all \beta<\tau. It is easy to clear that K=\left\{f_\alpha: \alpha<\tau \right\} \cup \left\{g \right\} is a subspace of \Sigma(\tau). Note that \left\{f_\alpha: \alpha<\tau \right\} is a discrete space of cardinality \tau and that any open neighborhood of g contains all but finitely many f_\alpha. Thus K is homeomorphic to A_\tau.

Another non-metrizable Corson compact space is defined and discussed in this previous post.

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Some basic operations

Corson compact spaces behave nicely with respect to some natural topological operations such as the following:

  1. Corson compactness is hereditary with respect to closed subspaces.
  2. Corson compactness is preserved by taking continuous images.
  3. Corson compactness is preserved by taking countable products.

It is clear that closed subspaces of every Corson compact space are also Corson compact. The fact that any continuous image of a Corson compact space is Corson compact (bullet point #2) is established in Theorem 6.2 of [6].

We prove bullet point #3, that the Tychonoff product of countably many Corson compact spaces is Corson compact. Let K_1,K_2,K_3,\cdots be compact spaces such that each K_j is a subspace of \Sigma(\tau_j) for some infinite cardinal \tau_j. We show that the product \prod_{j=1}^\infty K_j is Corson compact. Let \tau be the cardinality of \bigcup_{j=1}^\infty \tau_j. Note that \prod_{j=1}^\infty K_j is a compact subspace of \prod_{j=1}^\infty \Sigma(\tau_j). In turn, \prod_{j=1}^\infty \Sigma(\tau_j) is identical to \Sigma(\tau). Thus \prod_{j=1}^\infty K_j is Corson compact.

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Basic properties inherited from \Sigma-products

Some properties of Corson compact spaces are inherited from the \Sigma-products of real line. For example, we have the following two facts about \Sigma-products.

Both of the above two properties are hereditary. Thus we have the following:

  1. Every Corson compact space is a Frechet-Urysohn space.
  2. Every Corson compact space is monolithic.

A space X is monolithic if for each subspace Y of X, the density of Y coincides with the network weight of Y, i.e., d(Y)=nw(Y). A space X is strongly monolithic if for each subspace Y of X, the density of Y coincides with the weight of Y, i.e., d(Y)=w(Y). Monolithic spaces are discussed in this previous post. For compact spaces, the notion of being monolithic and the notion of being strongly monolithic coincide. One obvious consequence of bullet point #5 is that being separable is an indicator of metrizability among Corson compact spaces. The following bullet point captures this observation.

  1. Let X be a Corson compact space. Then X is metrizable if and only if X is separable. See Proposition 1 in this previous post.

A space Y is said to be Frechet space (also called Frechet-Urysohn 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. Thus any compact space that is not a Frechet-Urysohn space is not Corson compact. A handy example is the compact space \omega_1+1 with the order topology. Note that \omega_1+1 is monolithic. Thus monolithic compact spaces need not be Corson compact.

An extreme example of a compact non-Frechet-Urysohn space is one that has no non-trivial convergent sequence. For example, take \beta \omega, the Stone-Cech compactification of a countable discrete space, which has no non-trivial convergent sequences at any point. Thus \beta \omega is not Corson compact.

Every Corson compact space has a G_\delta point. It then follows that every Corson compact space has a dense set of G_\delta points (see this previous post). In a compact space, there is a countable local base at every G_\delta point. Thus we have the following bullet point.

  1. Every Corson compact space has a dense first countable subspace.

However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [7] for an example of a first countable Corson compact space with no dense metrizable subspace.

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Remarks

The results indicated here and proved in the previous posts represent a small sample of results on Corson compact spaces (and just focusing on the topological aspects). Many results on Corson compact spaces and Eberlein compact spaces are very deep results. The chapter c-16 in [5] is a good introduction. Some of the results proven in this and other posts in this blog are mentioned in [5] without proof. Interesting characterizations of Corson compact spaces are presented in [4].

In closing, we mention one more property. The authors in [1] showed that for any Corson compact space X, the function space C_p(X) with the pointwise convergence topology is a Lindelof space. Thus we have the following bullet point.

  1. The function space C_p(X) is Lindelof for every Corson compact space X.

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Blog posts on Corson compact spaces

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Reference

  1. Alster, K., Pol, R., On function spaces of compact subspaces of \Sigma-products of the real line, Fund. Math., 107, 35-46, 1980.
  2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  3. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  4. Gruenhage, G., Covering properties on X^2 \backslash \Delta, W-sets, and compact subsets of \Sigma-products, Topology Appl., 17, 287-304, 1984.
  5. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  6. Michael, E., Rudin, M. E., A note on Eberlein compacts, Pacific J. Math., 128, 149-156 1987.
  7. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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\copyright \ 2014 \text{ by Dan Ma}