Adding up to a non-meager set

The preceding post gives a topological characterization of bounded subsets of \omega^\omega. From it, we know what it means topologically for a set to be unbounded. In this post we prove a theorem that ties unbounded sets to Baire category.

A set is nowhere dense if its closure has empty interior. A set is a meager set if it is the union of countably many nowhere dense sets. By definition, the union of countably many meager sets is always a meager set. In order for meager sets to add up to a non-meager set (though taking union), the number of meager sets must be uncountable. What is this uncountable cardinal number? We give an indication of how big this number is. In this post we give a constructive proof to the following fact:

Theorem 1 …. Given an unbounded set F \subset \omega^\omega, there exist \kappa=\lvert F \lvert many meager subsets of the real line whose union is not meager.

We will discuss the implications of this theorem after giving background information.

We use \omega to denote the set of all non-negative integers \{ 0,1,2,\cdots \}. The set \omega^\omega is the set of all functions from \omega into \omega. It is called the Baire space when it is topologized with the product space topology. It is well known that the Baire space is homeomorphic to the space of irrational numbers \mathbb{P} (see here).

The notion of boundedness or unboundedness used in Theorem 1 refers to the eventual domination order (\le^*) for functions in the product space. For f,g \in \omega^\omega, by f \le^* g, we mean f(n) \le g(n) for all but finitely many n. A set F \subset \omega^\omega is bounded if it has an upper bound with respect to the partial order \le^*, i.e. there is some f \in \omega^\omega such that g \le^* f for all g \in F. The set F is unbounded if it is not bounded. To spell it out, F is unbounded if for each f \in \omega^\omega, there exists g \in F such that g \not \le^* f, i.e. f(n)<g(n) for infinitely many n.

All countable subsets of the Baire space are bounded (using a diagonal argument). Thus unbounded sets must be uncountable. It does not take extra set theory to obtain an unbounded set. The Baire space \omega^\omega is unbounded. More interesting unbounded sets are those of a certain cardinality, say unbounded sets of cardinality \omega_1 or unbounded sets with cardinality less than continuum. Another interesting unbounded set is one that is of the least cardinality. In the literature, the least cardinality of an unbounded subset of \omega^\omega is called \mathfrak{b}, the bounding number.

Another notion that is part of Theorem 1 is the topological notion of small sets – meager sets. This is a topological notion and is defined in topological spaces. For the purpose at hand, we consider this notion in the context of the real line. As mentioned at the beginning of the post, a set is nowhere dense set if its closure has empty interior (i.e. the closure contains no open subset). Let A \subset \mathbb{R}. The set A is nowhere dense if no open set is a subset of the closure \overline{A}. An equivalent definition: the set A is nowhere dense if for every nonempty open subset U of the real line, there is a nonempty subset V of U such that V contains no points of A. Such a set is “thin” since it is dense no where. In any open set, we can also find an open subset that has no points of the nowhere dense set in question. A subset A of the real line is a meager set if it is the union of countably many nowhere dense sets. Another name of meager set is a set of first category. Any set that is not of first category is called a set of second category, or simply a non-meager set.

Corollaries

Subsets of the real line are either of first category (small sets) or of second category (large sets). Countably many meager sets cannot fill up the real line. This is a consequence of the Baire category theorem (see here). By definition, caountably many meager sets cannot fill up any non-meager subset of the real line. How many meager sets does it take to add up to a non-meager set?

Theorem 1 gives an answer to the above question. It can take as many meager sets as the size of an unbounded subset of the Baire space. If \kappa is a cardinal number for which there exists an unbounded subset of \omega^\omega whose cardinality is \kappa, then there exists a non-meager subset of the real line that is the union of \kappa many meager sets. The bounding number \mathfrak{b} is the least cardinality of an unbounded set. Thus there is always a non-meager subset of the real line that is the union of \mathfrak{b} many meager sets.

Let \kappa_A be the least cardinal number \kappa such that there exist \kappa many meager subsets of the real line whose union is not meager. Based on Theorem 1, the bounding number \mathfrak{b} is an upper bound of \kappa_A. These two corollaries just discussed are:

  • There always exists a non-meager subset of the real line that is the union of \mathfrak{b} many meager sets.
  • \kappa_A \le \mathfrak{b}.

The bounding number \mathfrak{b} points to a non-meager set that is the union of \mathfrak{b} many meager sets. However, the cardinal \kappa_A is the least number of meager sets whose union is a non-meager set and this number is no more than the bounding number. The cardinal \kappa_A is called the additivity number.

There are other corollaries to Theorem 1. Let A(c) be the statement that the union of fewer than continuum many meager subsets of the real line is a meager set. For any cardinal number \kappa, let A(\kappa) be the statement that the union of fewer than \kappa many meager subsets of the real line is a meager set. We have the following corollaries.

  • The statement A(c) implies that there are no unbounded subsets of \omega^\omega that have cardinalities less than continuum. In other words, A(c) implies that the bounding number \mathfrak{b} is continuum.
  • Let \kappa \le continuum. The statement A(\kappa) implies that there are no unbounded subsets of \omega^\omega that have cardinalities less than \kappa. In other words, A(\kappa) implies that the bounding number \mathfrak{b} is at least \kappa, i.e. \mathfrak{b} \ge \kappa.

Let B(c) be the statement that the real line is not the union of less than continuum many meager sets. Clearly, the statement A(c) implies the statement B(c). Thus, it follows from Theorem 1 that A(c) \Longrightarrow B(c) + \mathfrak{b}=2^{\aleph_0}. This is a result proven in Miller [1]. Theorem 1.2 in [1] essentially states that A(c) is equivalent to B(c) + \mathfrak{b}=2^{\aleph_0}. The proof of Theorem 1 given here is essentially the proof of one direction of Theorem 1.2 in [1]. Our proof has various omitted details added. As a result it should be easier to follow. We also realize that the proof of Theorem 1.2 in [1] proves more than that theorem. Therefore we put the main part of the constructive in a separate theorem. For example, Theorem 1 also proves that the additivity number \kappa_A is no more than \mathfrak{b}. This is one implication in the Cichon’s diagram.

Proof of Theorem 1

Let 2=\{ 0,1 \}. The set 2^\omega is the set of all functions from \omega into \{0, 1 \}. When 2^\omega is endowed with the product space topology, it is called the Cantor space and is homemorphic to the middle-third Cantor set in the unit interval [0,1]. We use \{ [s]: \exists \ n \in \omega \text{ such that } s \in 2^n \} as a base for the product topology where [s]=\{ t \in 2^\omega:  s \subset t \}.

Let F \subset \omega^\omega be an unbounded set. We assume that the unbounded set F satisfies two properties.

  • Each g \in F is an increasing function, i.e. g(i)<g(j) for any i<j.
  • For each g \in F, if j>g(n), then g(j)>g(n+1).

One may wonder if the two properties are satisfied by any given unbounded set. Since F is unbounded, we can increase the values of each function g \in F, the resulting set will still be an unbounded set. More specifically, for each g \in F, define g^*\in \omega^\omega as follows:

  • g^*(0)=g(0)+1,
  • for each n \ge 1, g^*(n)=g(n)+\text{max}\{ g^*(i): i<n \} + n+1.

The set F^*=\{ g^*: g \in F \} is also an unbounded set. Therefore we use F^* and rename it as F.

Fix g \in F. Define an increasing sequence of non-negative integers n_0,n_1,n_2,\cdots as follows. Let n_0 be any integer greater than 1. For each integer j \ge 1, let n_j=g(n_{j-1}). Since n_0>1, we have n_1=g(n_0)>g(1). It follows that for all integer k \ge 1, n_k>g(k).

For each g \in F, we have an associated sequence n_0,n_1,n_2,\cdots as described in the preceding paragraph. Now define C(g)=\{ q \in 2^\omega: \forall \ k, q(n_k)=1 \}. It is straightforward to verify that each C(g) is a closed and nowhere dense subset of the Cantor space 2^\omega. Let X=\bigcup \{C(g): g \in F \}. The set X is a union of meager sets. We show that it is a non-meager subset of 2^\omega. We prove the following claim.

Claim 1
For any countable family \{C_n: n \in \omega \} where each C_n is a nowhere dense subset of 2^\omega, we have X \not \subset \bigcup \{C_n: n \in \omega \}.

According to Claim 1, the set X cannot be contained in any arbitrary meager subset of 2^\omega. Thus X must be non-meager. To establish the claim, we define an increasing sequence of non-negative integers m_0,m_1,m_2,\cdots with the property that for any k \ge 1, for any i<k, and for any s \in 2^{m_k}, there exists t \in 2^{m_{k+1}} such that s \subset t and [t] \cap C_i=\varnothing.

The desired sequence is derived from the fact that the sets C_n are nowhere dense. Choose any m_0<m_1 to start. With m_1 determined, the only nowhere dense set to consider is C_0. For each s \in 2^{m_1}, choose some integer y>m_1 such that there exists t \in 2^{y+1} such that s \subset t and [t] \cap C_0=\varnothing. Let m_2 be an integer greater than all the possible y‘s that have been chosen. The integer m_2 can be chosen since there are only finitely many s \in 2^{m_1}.

Suppose m_0<\cdots<m_{k-1}<m_k have been chosen. Then the only nowhere dense sets to consider are C_0,\cdots,C_{k-1}. Then for each i \le k-1, for each s \in 2^{m_k}, choose some integer y>m_k such that there exists t \in 2^{y+1} such that s \subset t and [t] \cap C_i=\varnothing. As before let m_{k+1} be an integer greater than all the possible y‘s that have been chosen. Again m_{k+1} is possible since there are only finitely many i \le k-1 and only finitely many s \in 2^{m_k}.

Let Z=\{ m_k: k \in \omega \}. We make the following claim.

Claim 2
There exists h \in F such that the associated sequence n_0, n_1,n_2,\cdots satisfies the condition: \lvert [n_k,n_{k+1}) \cap Z \lvert \ge 2 for infinitely many k where [n_k,n_{k+1}) is the set \{ m \in \omega: n_k \le m < m_{k+1} \}.

Suppose Claim 2 is not true. For each g \in F and its associated sequence n_0, n_1,n_2,\cdots,

    (*) there exists some integer b such that for all k>b, \lvert [n_k,n_{k+1}) \cap Z \lvert \le 1.

Let f \in \omega^\omega be defined by f(k)=m_k for all k. Choose \overline{f} \in \omega^\omega in the following manner. For each k \in \omega, define d_k \in \omega^\omega by d_k(n)=f(n+k) for all n. Then choose \overline{f} \in \omega^\omega such that d_k \le^* \overline{f} for all k.

Fix g \in F. Let m_j be the least element of [n_b, \infty) \cap Z. Then for each k>b, we have g(k) \le n_k \le m_{j+k}=f(j+k)=d_j(k). Note that the inequality n_k \le m_{j+k} holds because of the assumption (*). It follows that g \le^* d_j \le^* \overline{f}. This says that \overline{f} is an upper bound of F contradicting that F is an unbounded set. Thus Claim 2 must be true.

Let h \in F be as described in Claim 2. We now prove another claim.

Claim 3
For each n, C_n is a nowhere dense subset of C(h).

Fix C_n. Let p be an integer such that [n_p,n_{p+1}) \cap Z has at least two points, say m_k and m_{k+1}. We can choose p large enough such that n<k. Choose s \in 2^{m_k}. Since n_p is arbitrary, [s] is an arbitrary open set in 2^\omega. Since m_k is in between n_p and n_{p+1}, [s] contains a point of C(h). Thus [s] \cap C(h) is an arbitrary open set in C(h). By the way m_k and m_{k+1} are chosen originally, there exists t \in 2^{m_{k+1}} such that s \subset t and [t] \cap C_n=\varnothing. Because m_k and m_{k+1} are in between n_p and n_{p+1}, [t] \cap C(h) \ne \varnothing. This establishes the claim that C_n is nowhere dense subset of C(h).

Note that C(h) is a closed subset of the Cantor space 2^\omega and hence is also compact. Thus C(h) is a Baire space and cannot be the union of countably many nowhere dense sets. Thus C(h) \not \subset \cup \{C_n: n \in \omega \}. Otherwise, C(h) would be the union of countably many nowhere dense sets. This means that X=\bigcup \{C(g): g \in F \} \not \subset \cup \{C_n: n \in \omega \}. This establishes Claim 1.

Considering the Cantor space 2^\omega as a subspace of the real line, each C(g) is also a closed nowhere dense subset of the real line. The set X=\bigcup \{C(g): g \in F \} is also not a meager subset of the real line. This establishes Theorem 1. \square

Reference

  1. Miller A. W., Some properties of measure and category, Trans. Amer. Math. Soc., 266, 93-114, 1981.

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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}

Bernstein Sets and the Michael Line

Let \mathbb{M} be the Michael line and let \mathbb{P} be the set of all irrational numbers with the Euclidean topology. In the post called “Michael Line Basics”, we show that the product \mathbb{M} \times \mathbb{P} is not normal. This is a classic counterexample showing that the product of two paracompact spaces need not be normal even when one of the factors is a complete metric space. The Michael line \mathbb{M} is not Lindelof. A natural question is: can the first factor be made a Lindelof space? In this post, as an application of Bernstein sets, we present a non-normal product space where one factor is Lindelof and the other factor is a separable metric space. It is interesting to note that while one factor is upgraded (from paracompact to Lindelof), the other factor is downgraded (from a complete metric space to just a separable metric space).

Bernstein sets have been discussed previously in this blog. They are special subsets of the real line and with the Euclidean subspace topology, they are spaces in which the Banach-Mazur game is undecidable (see the post “Bernstein Sets Are Baire Spaces”). A Bernstein set is a subset B of the real line such that every uncountable closed subset of the real line has non-empty intersection with both B and the complement of B.

Bernstein sets are constructed by transfinite induction. The procedure starts by ordering all uncountable closed subsets of the real line in a sequence of length that is as long as the cardinality of continuum. To see how Bernstein sets are constructed, see the post “Bernstein Sets Are Baire Spaces”.

After we discuss a generalization of the definition of the Michael line, we discuss the non-normal product space based on Bernstein sets.
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Generalizing the Michael Line

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers and let \mathbb{Q}=\mathbb{R}-\mathbb{P}. Recall that the Michael line is the real line \mathbb{R} topologized by letting points in \mathbb{P} discrete and letting points in \mathbb{Q} retain their usual open neighborhoods. We can carry out the same process on any partition of the real number line.

Let D and E be disjoint sets such that \mathbb{R}=D \cup E where the set E is dense in the real line. The intention is to make D the discrete part and E the Euclidean part. In other words, we topologize \mathbb{R} be letting points in D discrete and letting points in E retain their Euclidean open sets. Let X_D denote the resulting topological space. For the lack of a better term, we call the space X_D the modified Michael line. An open set in the space X_D is of the form U \cup V where U is a Euclidean open subset of the real line and V \subset D. We have the following result:

    Proposition
    Suppose that D is not an F_\sigma-set in the Euclidean real line and that D is dense in the Euclidean real line. Then the product space X_D \times D is not normal (the second factor D is considered a subspace of the Euclidean real line).

In the post “Michael Line Basics”, we give a proof that \mathbb{M} \times \mathbb{P} is not normal. This proof hinges on the same two facts about the set D in the hypothesis in the above proposition. Thus the proof for the above proposition is just like the one for \mathbb{M} \times \mathbb{P}. Whenever we topologize the modified Michael line by using a non-F_\sigma-set as the discrete part, we can always be certain that we have a non-normal product as indicated here.

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Non-Normal Product Space

Let B be any Bernstein set. The set B is clearly not an F_\sigma-set in the real line and is clearly dense in the real line. Then X_B \times B is not normal. Note that in X_B, the set B is discrete and its complement \mathbb{R}-B has the usual topology. To see that X_B is Lindelof, note that any open cover of X_B has a countable subcollection that covers \mathbb{R}-B. This countable subcollection consists of Euclidean open sets. Furthermore, the complement of the union of these countably many Euclidean open sets must contain all but countably many points of the Bernstein set B (otherwise there would be an uncountable Euclidean closed set that misses B).

As commented at the beginning, in obtaining this non-normal product space, one factor is enhanced at the expense of the other factor (one is made Lindelof while the other is no longer a complete metric space). Even though any Bernstein set (with the Euclidean topology) is a separable metric space, it cannot be completely metrizable. Any completely metrizable subset of the real line must be a G_\delta-set in the real line. Furthermore any uncountable G_\delta subset of the real line must contain a Cantor set and thus cannot be a Bernstein set.

A similar example to X_B \times B is presented in E. Michael’s paper (see [3]). It is hinted in footnote 4 of that paper that with the additional assumption of continuum hypothesis (CH), one can have a non-normal product space where one factor is a Lindelof space and the second factor is the space of irrationals. So with an additional set-theoretic assumption, we can keep one factor from losing complete metrizability. For this construction, see point (d) in Example 3.2 of [2].

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A Brief Remark

Note that the Lindelof space X_B presented here is not hereditarily Lindelof, since it has uncountably many isolated points. Can a hereditarily Lindelof example be constructed such that its product with a particular separable metric space is not normal? The answer is no. The product of a hereditarily Lindelof space and any separable metric space is hereditarily Lindelof (see Result 4 in the post Cartesian Products of Two Paracompact Spaces – Continued).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
  3. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
  4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Bernstein Sets Are Baire Spaces

A topological space X is a Baire space if the intersection of any countable family of open and dense sets in X is dense in X (or equivalently, every nonempty open subset of X is of second category in X). One version of the Baire category theorem implies that every complete metric space is a Baire space. The real line \mathbb{R} with the usual Euclidean metric \lvert x-y \lvert is a complete metric space, and hence is a Baire space. The space of irrational numbers \mathbb{P} is also a complete metric space (not with the usual metric \lvert x-y \lvert but with another suitable metric that generates the Euclidean topology on \mathbb{P}) and hence is also a Baire space. In this post, we show that there are subsets of the real line that are Baire space but not complete metric spaces. These sets are called Bernstein sets.

A Bernstein set, as discussed here, is a subset B of the real line such that both B and \mathbb{R}-B intersect with every uncountable closed subset of the real line. We present an algorithm on how to generate such a set. Bernstein sets are not Lebesgue measurable. Our goal here is to show that Bernstein sets are Baire spaces but not weakly \alpha-favorable, and hence are spaces in which the Banach-Mazur game is undecidable.

Baire spaces are defined and discussed in this post. The Banach-Mazur game is discussed in this post. The algorithm of constructing Bernstein set is found in [2] (Theorem 5.3 in p. 23). Good references for basic terms are [1] and [3].
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In constructing Bernstein sets, we need the following lemmas.

Lemma 1
In the real line \mathbb{R}, any uncountable closed set has cardinality continuum.

Proof
In the real line, every uncountable subset of the real line has a limit point. In fact every uncountable subset of the real line contains at least one of its limit points (see The Lindelof property of the real line). Let A \subset \mathbb{R} be an uncountable closed set. The set A has to contain at least one of its limit point. As a result, at most countably many points of A are not limit points of A. Take away these countably many points of A that are not limit points of A and call the remainder A^*. The set A^* is still an uncountable closed set but with an additional property that every point of A^* is a limit point of A^*. Such a set is called a perfect set. In Perfect sets and Cantor sets, II, we demonstrate a procedure for constructing a Cantor set out of any nonempty perfect set. Thus A^* (and hence A) contains a Cantor set and has cardinality continuum. \blacksquare

Lemma 2
In the real line \mathbb{R}, there are continuum many uncountable closed subsets.

Proof
Let \mathcal{B} be the set of all open intervals with rational endpoints, which is a countable set. The set \mathcal{B} is a base for the usual topology on \mathbb{R}. Thus every nonempty open subset of the real line is the union of some subcollection of \mathcal{B}. So there are at most continuum many open sets in \mathbb{R}. Thus there are at most continuum many closed sets in \mathbb{R}. On the other hand, there are at least continuum many uncountable closed sets (e.g. [-b,b] for b \in \mathbb{R}). Thus we can say that there are exactly continuum many uncountable closed subsets of the real line. \blacksquare

Constructing Bernstein Sets

Let c denote the cardinality of the real line \mathbb{R}. By Lemma 2, there are only c many uncountable closed subsets of the real line. So we can well order all uncountable closed subsets of \mathbb{R} in a collection indexed by the ordinals less than c, say \left\{F_\alpha: \alpha < c \right\}. By Lemma 1, each F_\alpha has cardinality c. Well order the real line \mathbb{R}. Let \prec be this well ordering.

Based on the well ordering \prec, let x_0 and y_0 be the first two elements of F_0. Let x_1 and y_1 be the first two elements of F_1 (based on \prec) that are different from x_0 and y_0. Suppose that \alpha < c and that for each \beta < \alpha, points x_\beta and y_\beta have been selected. Then F_\alpha-\bigcup_{\beta<\alpha} \left\{x_\beta,y_\beta \right\} is nonempty since F_\alpha has cardinality c and only less than c many points have been selected. Then let x_\alpha and y_\alpha be the first two points of F_\alpha-\bigcup_{\beta<\alpha} \left\{x_\beta,y_\beta \right\} (according to \prec). Thus x_\alpha and y_\alpha can be chosen for each \alpha<c.

Let B=\left\{ x_\alpha: \alpha<c \right\}. Then B is a Bernstein set. Note that B meets every uncountable closed set F_\alpha with the point x_\alpha and the complement of B meets every uncountable closed set F_\alpha with the point y_\alpha.

The algorithm described here produces a unique Bernstein set that depends on the ordering of the uncountable closed sets F_\alpha and the well ordering \prec of \mathbb{R}.

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Key Lemmas

Baire spaces are defined and discussed in this previous post. Baire spaces can also be characterized using the Banach-Mazur game. The following lemmas establish that any Bernstein is a Baire space that is not weakly \alpha-favorable. Lemma 3 is applicable to all topological spaces. Lemmas 4, 5, 6, and 7 are specific to the real line.

Lemma 3
Let Y be a topological space. Let F \subset Y be a set of first category in Y. Then Y-F contains a dense G_\delta subset.

Proof
Let F \subset Y be a set of first category in Y. Then F=\bigcup \limits_{n=0}^\infty F_n where each F_n is nowhere dense in Y. The set X-\bigcup \limits_{n=0}^\infty \overline{F_n} is a dense G_\delta set in the space X and it is contained in the complement of F. We have:

\displaystyle . \ \ \ \ \ X-\bigcup \limits_{n=0}^\infty \overline{F_n} \subset X-F \blacksquare

We now set up some notaions in preparation of proving Lemma 4 and Lemma 7. For any set A \subset \mathbb{R}, let \text{int}(A) be the interior of the set A. Denote each positive integer n by n=\left\{0,1,\cdots,n-1 \right\}. In particular, 2=\left\{0,1\right\}. Let 2^{n} denote the collection of all functions f: n \rightarrow 2. Identify each f \in 2^n by the sequence f(0),f(1),\cdots,f(n-1). This identification makes notations in the proofs of Lemma 4 and Lemma 7 easier to follow. For example, for f \in 2^n, I_f denotes a closed interval I_{f(0),f(1),\cdots,f(n-1)}. When we choose two disjoint subintervals of this interval, they are denoted by I_{f,0} and I_{f,1}. For f \in 2^n, f \upharpoonright 1 refers to f(0), f \upharpoonright 2 refers to the sequence f(0),f(1), and f \upharpoonright 3 refers to the sequence f(0),f(1),f(2) and so on.

The Greek letter \omega denotes the first infinite ordinal. We equate it as the set of all nonnegative integers \left\{0,1,2,\cdots \right\}. Let 2^\omega denote the set of all functions from \omega to 2=\left\{0,1 \right\}.

Lemma 4
Let W \subset \mathbb{R} be a dense G_\delta set. Let U be a nonempty open subset of \mathbb{R}. Then W \cap U contains a Cantor set (hence an uncountable closed subset of the real line).

Proof
Let W=\bigcap \limits_{n=0}^\infty O_n where each O_n is an open and dense subset of \mathbb{R}. We describe how a Cantor set can be obtained from the open sets O_n. Take a closed interval I_\varnothing=[a,b] \subset O_0 \cap U. Let C_0=I_\varnothing. Then pick two disjoint closed intervals I_{0} \subset O_1 and I_{1} \subset O_1 such that they are subsets of the interior of I_\varnothing and such that the lengths of both intervals are less than 2^{-1}. Let C_1=I_0 \cup I_1.

At the n^{th} step, suppose that all closed intervals I_{f(0),f(1),\cdots,f(n-1)} (for all f \in 2^n) are chosen. For each such interval, we pick two disjoint closed intervals I_{f,0}=I_{f(0),f(1),\cdots,f(n-1),0} and I_{f,1}=I_{f(0),f(1),\cdots,f(n-1),1} such that each one is subset of O_n and each one is subset of the interior of the previous closed interval I_{f(0),f(1),\cdots,f(n-1)} and such that the lenght of each one is less than 2^{-n}. Let C_n be the union of I_{f,0} \cup I_{f,1} over all f \in 2^n.

Then C=\bigcap \limits_{j=0}^\infty C_j is a Cantor set that is contained in W \cap U. \blacksquare

Lemma 5
Let X \subset \mathbb{R}. If X is not of second category in \mathbb{R}, then \mathbb{R}-X contains an uncountable closed subset of \mathbb{R}.

Proof
Suppose X is of first category in \mathbb{R}. By Lemma 3, the complement of X contains a dense G_\delta subset. By Lemma 4, the complement contains a Cantor set (hence an uncountable closed set). \blacksquare

Lemma 6
Let X \subset \mathbb{R}. If X is not a Baire space, then \mathbb{R}-X contains an uncountable closed subset of \mathbb{R}.

Proof
Suppose X \subset \mathbb{R} is not a Baire space. Then there exists some open set U \subset X such that U is of first category in X. Let U^* be an open subset of \mathbb{R} such that U^* \cap X=U. We have U=\bigcup \limits_{n=0}^\infty F_n where each F_n is nowhere dense in X. It follows that each F_n is nowhere dense in \mathbb{R} too.

By Lemma 3, \mathbb{R}-U contains W, a dense G_\delta subset of \mathbb{R}. By Lemma 4, there is a Cantor set C contained in W \cap U^*. This uncountable closed set C is contained in \mathbb{R}-X. \blacksquare

Lemma 7
Let X \subset \mathbb{R}. Suppose that X is a weakly \alpha-favorable space. If X is dense in the open interval (a,b), then there is an uncountable closed subset C of \mathbb{R} such that C \subset X \cap (a,b).

Proof
Suppose X is a weakly \alpha-favorable space. Let \gamma be a winning strategy for player \alpha in the Banach-Mazur game BM(X,\beta). Let (a,b) be an open interval in which X is dense. We show that a Cantor set can be found inside X \cap (a,b) by using the winning strategy \gamma.

Let I_{-1}=[a,b]. Let t=b-a. Let U_{-1}^*=(a,b) and U_{-1}=U^* \cap X. We take U_{-1} as the first move by the player \beta. Then the response made by \alpha is V_{-1}=\gamma(U_{-1}). Let C_{-1}=I_{-1}.

Choose two disjoint closed intervals I_0 and I_1 that are subsets of the interior of I_{-1} such that the lengths of these two intervals are less than 2^{-t} and such that U_0^*=\text{int}(I_0) and U_1^*=\text{int}(I_1) satisfy further properties, which are that U_0=U_0^* \cap X \subset V_{-1} and U_1=U_1^* \cap X \subset V_{-1} are open in X. Let U_0 and U_1 be two possible moves by player \beta at the next stage. Then the two possible responses by \alpha are V_0=\gamma(U_{-1},U_0) and V_1=\gamma(U_{-1},U_1). Let C_1=I_0 \cup I_1.

At the n^{th} step, suppose that for each f \in 2^n, disjoint closed interval I_f=I_{f(0),\cdots,f(n-1)} have been chosen. Then for each f \in 2^n, we choose two disjoint closed intervals I_{f,0} and I_{f,1}, both subsets of the interior of I_f, such that the lengths are less than 2^{-(n+1) t}, and:

  • U_{f,0}^*=\text{int}(I_{f,0}) and U_{f,1}^*=\text{int}(I_{f,1}),
  • U_{f,0}=U_{f,0}^* \cap X and U_{f,1}=U_{f,1}^* \cap X are open in X,
  • U_{f,0} \subset V_f and U_{f,1} \subset V_f

We take U_{f,0} and U_{f,1} as two possible new moves by player \beta from the path f \in 2^n. Then let the following be the responses by player \alpha:

  • V_{f,0}=\gamma(U_{-1},U_{f \upharpoonright 1}, U_{f \upharpoonright 2}, \cdots,U_{f \upharpoonright (n-1)},U_f, U_{f,0})
  • V_{f,1}=\gamma(U_{-1},U_{f \upharpoonright 1}, U_{f \upharpoonright 2}, \cdots,U_{f \upharpoonright (n-1)},U_f, U_{f,1})

The remaining task in the n^{th} induction step is to set C_n=\bigcup \limits_{f \in 2^n} I_{f,0} \cup I_{f,1}.

Let C=\bigcap \limits_{n=-1}^\infty C_n, which is a Cantor set, hence an uncountable subset of the real line. We claim that C \subset X.

Let x \in C. There there is some g \in 2^\omega such that \left\{ x \right\} = \bigcap \limits_{n=1}^\infty I_{g \upharpoonright n}. The closed intervals I_{g \upharpoonright n} are associated with a play of the Banach-Mazur game on X. Let the following sequence denote this play:

\displaystyle (1) \ \ \ \ \ U_{-1},V_{-1},U_{g \upharpoonright 1},V_{g \upharpoonright 1},U_{g \upharpoonright 2},V_{g \upharpoonright 2},U_{g \upharpoonright 3},U_{g \upharpoonright 3}, \cdots

Since the strategy \gamma is a winning strategy for player \alpha, the intersection of the open sets in (1) must be nonempty. Thus \bigcap \limits_{n=1}^\infty V_{g \upharpoonright n} \ne \varnothing.

Since the sets V_{g \upharpoonright n} \subset I_{g \upharpoonright n}, and since the lengths of I_{g \upharpoonright n} go to zero, the intersection must have only one point, i.e., \bigcap \limits_{n=1}^\infty V_{g \upharpoonright n} = \left\{ y \right\} for some y \in X. It also follows that y=x. Thus x \in X. We just completes the proof that X contains an uncountable closed subset of the real line. \blacksquare

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Conclusions about Bernstein Sets

Lemma 6 above establishes that any Bernstein set is a Baire space (if it isn’t, the complement would contain an uncountable closed set). Lemma 7 establishes that any Bernstein set is a topological space in which the player \alpha has no winning strategy in the Banach-Mazur game (if player \alpha always wins in a Bernstein set, it would contain an uncountable closed set). Thus any Bernstein set cannot be a weakly \alpha favorable space. According to this previous post about the Banach-Mazur game, Baire spaces are characterized as the spaces in which the player \beta has no winning strategy in the Banach-Mazur game. Thus any Bernstein set in a topological space in which the Banach-Mazur game is undecidable (i.e. both players in the Banach-Mazur game have no winning strategy).

One interesting observation about Lemma 6 and Lemma 7. Lemma 6 (as well as Lemma 5) indicates that the complement of a “thin” set contains a Cantor set. On the other hand, Lemma 7 indicates that a “thick” set contains a Cantor set (if it is dense in some open interval).

Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Oxtoby, J. C., Measure and Category, Graduate Texts in Mathematics, Springer-Verlag, New York, 1971.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

A Question About The Rational Numbers

Let \mathbb{R} be the real line and \mathbb{Q} be the set of all rational numbers. Consider the following question:

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Question

  • For each nonnegative integer n, let U_n be an open subset of \mathbb{R} such that that \mathbb{Q} \subset U_n. The intersection \bigcap \limits_{n=0}^\infty U_n is certainly nonempty since it contains \mathbb{Q}. Does this intersection necessarily contain some irrational numbers?

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While taking a real analysis course, the above question was posted to the author of this blog by the professor. Indeed, the question is an excellent opening of the subject of category. We first discuss the Baire category theorem and then discuss the above question. A discussion of Baire spaces follow. For any notions not defined here and for detailed discussion of any terms discussed here, see [1] and [2].

In the above question, the set \bigcap \limits_{n=0}^\infty U_n is a G_\delta set since it is the intersection of countably many open sets. It is also dense in the real line \mathbb{R} since it contains the rational numbers. So the question can be rephrased as: is the set of rational numbers \mathbb{Q} a G_\delta set? Can a dense G_\delta set in the real line \mathbb{R} be a “small” set such as \mathbb{Q}? The discussion below shows that \mathbb{Q} is too “thin” to be a dense G_\delta set. Put it another way, a dense G_\delta subset of the real line is a “thick” set. First we present the Baire category theorem.
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Baire Category Theorem

Let X be a complete metric space. For each nonnegative integer n, let O_n be an open subset of X that is also dense in X. Then \bigcap \limits_{n=0}^\infty O_n is dense in X.

Proof
Let A=\bigcap \limits_{n=0}^\infty O_n. Let V_0 be any nonempty open subset of X. We show that V_0 contains some point of A.

Since O_0 is dense in X, V_0 contains some point of O_0. Let x_0 be one such point and choose open set V_1 such that x_0 \in V_1 and \overline{V_1} \subset V_0 \cap O_0 \subset V_0 with the additional condition that the diameter of \overline{V_1} is less than \displaystyle \frac{1}{2^1}.

Since O_1 is dense in X, V_1 contains some point of O_1. Let x_1 be one such point and choose open set V_2 such that x_1 \in V_2 and \overline{V_2} \subset V_1 \cap O_1 \subset V_1 with the additional condition that the diameter of \overline{V_2} is less than \displaystyle \frac{1}{2^2}.

By continuing this inductive process, we obtain a nested sequence of open sets V_n and a sequence of points x_n such that x_n \in V_n \subset \overline{V_n} \subset V_{n-1} \cap O_{n-1} \subset V_0 for each n and that the diameters of \overline{V_n} converge to zero (according to some complete metric on X). Then the sequence of points x_n is a Cauchy sequence. Since X is a complete metric space, the sequence x_n converges to a point x \in X.

We claim that x \in V_0 \cap A. To see this, note that for each n, x_j \in \overline{V_n} for each j \ge n. Since x is the sequential limit of x_j, x \in \overline{V_n} for each n. It follows that x \in O_n for each n (x \in A) and x \in V_0. This completes the proof of Baire category theorem. \blacksquare

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Discussion of the Above Question

For each nonnegative integer n, let U_n be an open subset of \mathbb{R} such that that \mathbb{Q} \subset U_n. We claim that the intersection \bigcap \limits_{n=0}^\infty U_n contain some irrational numbers.

Suppose the intersection contains no irrational numbers, that is, \mathbb{Q}=\bigcap \limits_{n=0}^\infty U_n.

Let \mathbb{Q} be enumerated by \left\{r_0,r_1,r_2,\cdots \right\}. For each n, let G_n=\mathbb{R}-\left\{ r_n \right\}. Then each G_n is an open and dense set in \mathbb{R}. Note that the set of irrational numbers \mathbb{P}=\bigcap \limits_{n=0}^\infty G_n.

We then have countably many open and dense sets U_0,U_1,U_2,\cdots,G_0,G_1,G_2,\cdots whose intersection is empty. Note that any point that belongs to all U_n has to be a rational number and any point that belongs to all G_n has to be an irrational number. On the other hand, the real line \mathbb{R} with the usual metric is a complete metric space. By the Baire category theorem, the intersection of all U_n and G_n must be nonempty. Thus the intersection \bigcap \limits_{n=0}^\infty U_n must contain more than rational numbers.

It follows that the set of rational numbers \mathbb{Q} cannot be a G_\delta set in \mathbb{R}. In fact, the discussion below will show that the in a complete metric space such as the real line, any dense G_\delta set must be a “thick” set (see Theorem 3 below).
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Baire Spaces

The version of the Baire category theorem discussed above involves complete metric spaces. However, the ideas behind the Baire category theorem are topological in nature. The following is the conclusion of the Baire category theorem:

(*) \ \ \ \ X is a topological space such that for each countable family \left\{U_0,U_1,U_2,\cdots \right\} of open and dense sets in X, the intersection \bigcap \limits_{n=0}^\infty U_n is dense in X.

A Baire space is a topological space in which the condition (*) holds. The Baire category theorem as stated above gives a sufficient condition for a topological space to be a Baire space. There are plenty of Baire spaces that are not complete metric spaces, in fact, not even metric spaces. The condition (*) is a topological property. In order to delve deeper into this property, let’s look at some related notions.

Let X be a topological space. A set A \subset X is dense in X if every open subset of X contains a point of A (i.e. \overline{A}=X). A set A \subset X is nowhere dense in X if for every open subset U of X, there is some open set V \subset U such that V contains no point of A (another way to describe this: \overline{A} contains no interior point of X).

A set is dense if its points can be found in every nonempty open set. A set is nowhere dense if every nonempty open set has an open subset that misses it. For example, the set of integers \mathbb{N} is nowhere dense in \mathbb{R}.

A set A \subset X is of first category in X if A is the union of countably many nowhere dense sets in X. A set A \subset X is of second category in X if it is not of first category in X.

To make sense of these notions, the following observation is key:

(**) \ \ \ \ F \subset X is nowhere dense in X if and only if X-\overline{F} is an open and dense set in X.

So in a Baire space, if you take away any countably many closed and nowhere dense sets (in other words, taking away a set of first category in X), there is a remainder (there are still points remaining) and the remainder is still dense in X. In thinking of sets of first category as “thin”, a Baire space is one that is considered “thick” or “fat” in that taking away a “thin” set still leaves a dense set.

A space X is of second category in X means that if you take away any countably many closed and nowhere dense sets in X, there are always points remaining. For a set Y \subset X, Y is of second category in X means that if you take away from Y any countably many closed and nowhere dense sets in X, there are still points remaining in Y. A set of second category is “thick” in the sense that after taking away a “thin” set there are still points remaining.

For example, \mathbb{N} is nowhere dense in \mathbb{R} and thus of first category in \mathbb{R}. However, \mathbb{N} is of second category in itself. In fact, \mathbb{N} is a Baire space since it is a complete metric space (with the usual metric).

For example, \mathbb{Q} is of first category in \mathbb{R} since it is the union of countably many singleton sets (\mathbb{Q} is also of first category in itself).

For example, let T=[0,1] \cup (\mathbb{Q} \cap [2,3]). The space T is not a Baire space since after taking away the rational numbers in [2,3], the remainder is no longer dense in T. However, T is of second category in itself.

For example, any Cantor set defined in the real line is nowhere dense in \mathbb{R}. However, any Cantor set is of second category in itself (in fact a Baire space).

The following theorems summarize these concepts.

Theorem 1a
Let X be a topological space. The following conditions are equivalent:

  1. X is of second category in itself.
  2. The intersection of countably many dense open sets is nonempty.

Theorem 1b
Let X be a topological space. Let A \subset X. The following conditions are equivalent:

  1. The set A is of second category in X.
  2. The intersection of countably many dense open sets in X must intersect A.

Theorem 2
Let X be a topological space. The following conditions are equivalent:

  1. X is a Baire space, i.e., the intersection of countably many dense open sets is dense in X.
  2. Every nonempty open subset of X is of second category in X.

The above theorems can be verified by appealing to the relevant definitions, especially the observation (**). Theorems 2 and 1a indicate that any Baire space is of second category in itself. The converse is not true (see the space T=[0,1] \cup (\mathbb{Q} \cap [2,3]) discussed above).

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Dense G delta Subsets of a Baire Space

In answering the question stated at the beginning, we have shown that \mathbb{Q} cannot be a G_\delta set. Being a set of first category, \mathbb{Q} cannot be a dense G_\delta set. In fact, it can be shown that in a Baire space, any dense G_\delta subset is also a Baire space.

Theorem 3
Let X be a Baire space. Then any dense G_\delta subset of X is also a Baire space.

Proof
Let Y=\bigcap \limits_{n=0}^\infty U_n where each U_n is open and dense in X. We show that Y is a Baire space. In light of Theorem 2, we show that every nonempty open set of Y is of second category in Y.

Suppose that there is a nonempty open subset U \subset Y such that U is of first category in Y. Then U=\bigcup \limits_{n=0}^\infty W_n where each W_n is nowhere dense in Y. It can be shown that each W_n is also nowhere dense in X.

Since U is open in Y, there is an open set U^* \subset X such that U^* \cap Y=U. Note that for each n, F_n=X-U_n is closed and nowhere dense in X. Then we have:

\displaystyle (1) \ \ \ \ \ U^*=\bigcup \limits_{n=0}^\infty (F_n \cap U^*) \cup \bigcup \limits_{n=0}^\infty W_n

(1) shows that U^* is the union of countably many nowhere dense sets in X, contracting that every nonempty open subset of X is of second category in X. Thus we can conclude that every nonempty open subset of Y is of second category in Y. \blacksquare

Reference

  1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Revised July 3, 2019

A note on basic set theory

This is a short note listing some basic facts on set theory and set theory notations, mostly about cardinality of sets. The discussion in this note is useful for proving theorems in topology and in many other areas. For more information on basic set theory, see [2].

Let A and B be sets. The cardinality of the set A is denoted by \lvert A \lvert. A function f:A \rightarrow B is said to be one-to-one (an injection) if for x,y \in A with x \ne y, f(x) \ne f(y). A function f:A \rightarrow B is said to map A onto B (a surjection) if B=\left\{f(x): x \in A\right\}, i.e. the range of the function f is B. If the function f:A \rightarrow B is both an injection and a surjection, then f is called a bijection, in which case, we say both sets have the same cardinality and we use the notation \lvert A \lvert = \lvert B \lvert. When the function f:A \rightarrow B, we denote this condition by \lvert A \lvert \le \lvert B \lvert. The Cantor–Bernstein–Schroeder theorem states that if \lvert A \lvert \le \lvert B \lvert and \lvert B \lvert \le \lvert A \lvert then \lvert A \lvert = \lvert B \lvert (see 1.12 in [2]).

For the functions f:X \rightarrow Y and g:Y \rightarrow Z, we define the function g \circ f by (g \circ f)(x)=g(f(x)) for each x \in X. The function g \circ f is denoted by g \circ f:X \rightarrow Z and is called the composition of g and f.

We use the notation B^A to denote the set of all functions f:A \rightarrow B. It follows that if \lvert A \lvert \le \lvert B \lvert then \lvert A^C \lvert \le \lvert B^C \lvert. To see this, suppose we have a one-to-one function f:A \rightarrow B. We define a one-to-one function H: A^C \rightarrow B^C by H(h)=f \circ h for each h \in A^C. Since f:A \rightarrow B, it follows that H is one-to-one.

By \omega, we mean the first infinite ordinal, which can be viewed as the set of all nonnegative intergers. By \omega_1 we mean the first uncountable ordinal. The notation 2^\omega has dual use. With 2=\left\{0,1\right\}, the set 2^\omega denotes all functions f:\omega \rightarrow 2. It can be shown that 2^\omega has the same cardinality as the real line \mathbb{R} and the unit interval [0,1] and the middle third Cantor set (see The Cantor set, I). Thus we also use 2^\omega to denote continuum, the cardinality of the real line.

If \lvert A \lvert=2^\omega, then the set \lvert A^{\omega} \lvert=2^\omega where A^{\omega} is the set of all functions from \omega into A. Since \omega_1 is the first uncountable ordinal, we have \omega_1 \le 2^\omega. The Continuum Hypothesis states that \omega_1 = 2^\omega, i.e. the cardinality of the real line is the first uncountable cardinal number.

The union of 2^\omega many sets, each of which has cardinality 2^\omega, has cardinality 2^\omega. Furthermore, the union of \le 2^\omega many sets, each of which has cardinality \le 2^\omega, has cardinality \le 2^\omega.

Reference

  1. Kunen, K. Set Theory, An Introduction to Independence Proofs, 1980, Elsevier Science Publishing, New York.
  2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Closed uncountable subsets of the real line

This is post #12 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the previous post Perfect sets and Cantor sets, II, we show that every subset of the real line that is a perfect set contains a Cantor set and thus has cardinality continuum. We would like to add an observation that every uncountable closed subset of the real line contains a perfect set. Thus every uncountable closed subset of the real line contains a Cantor set and thus has the same cardinality as the real line itself.

Recall that a subset A of the real line is a perfect set if it is closed and every point of A is a limit point of A. In the previous post Perfect sets and Cantor sets, II, we demonstrate a procedure for constructing a Cantor set out of any nonempty perfect set. In another post The Lindelof property of the real line, we show that every uncountable subset A of the real line contains at least one of its limit points. Thus all but countably many points of A are limit points of A. Consequently, for any uncountable closed subset A of the real line, all but countably many points of A are limit points of A. By removing the countably many non-limit points (isolated points), we have a perfect set.

Links to previous posts on the topology of the real line:
1. The Euclidean topology of the real line (1)
2. The Euclidean topology of the real line (2)
3. The Euclidean topology of the real line (3) – Completeness
4. The Euclidean topology of the real line (4) – Compactness
5. The Cantor bus tour
6. The Cantor set, I
7. The Cantor set, II
8. The Cantor set, III
9. Perfect sets and Cantor sets, I
10. The Lindelof property of the real line
11. Perfect sets and Cantor sets, II