Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}, the set of all rational numbers. Let \tau be the usual topology of the real line \mathbb{R}. The following is a base that defines a topology on \mathbb{R}.

    \mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

The real line with the topology generated by \mathcal{B} is called the Michael line and is denoted by \mathbb{M}. In essense, in \mathbb{M}, points in \mathbb{P} are made isolated and points in \mathbb{Q} retain the usual Euclidean open sets.

The Euclidean topology \tau is coarser (weaker) than the Michael line topology (i.e. \tau being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, \mathbb{M} is submetrizable (submetrized by the Euclidean topology). It is clear that \mathbb{M} is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

  1. The space \mathbb{M} is paracompact.
  2. The space \mathbb{M} is not Lindelof.
  3. The extent of the space \mathbb{M} is c where c is the cardinality of the real line.
  4. The space \mathbb{M} is not locally compact.
  5. The space \mathbb{M} is not perfectly normal, thus not metrizable.
  6. The space \mathbb{M} is not a Moore space, but has a G_\delta-diagonal.
  7. The product \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology.
  8. The product \mathbb{M} \times \mathbb{P} is metacompact.
  9. The space \mathbb{M} has a point-countable base.
  10. For each n=1,2,3,\cdots, the product \mathbb{M}^n is paracompact.
  11. The product \mathbb{M}^\omega is not normal.
  12. There exist a Lindelof space L and a separable metric space W such that L \times W is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on \mathbb{Q} coincides with the Euclidean topology on \mathbb{Q}. A set is said to be a G_\delta-set if it is the intersection of countably many open sets. By the Baire category theorem, the set \mathbb{Q} is not a G_\delta-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set \mathbb{Q} is not a G_\delta-set in the Michael line. This fact is used in Result 5.

The fact that \mathbb{Q} is not a G_\delta-set in the Euclidean real line implies that \mathbb{P} is not an F_\sigma-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let \mathcal{U} be an open cover of \mathbb{M}. We proceed to derive a locally finite open refinement \mathcal{V} of \mathcal{U}. Recall that \tau is the usual topology on \mathbb{R}. Assume that \mathcal{U} consists of open sets in the base \mathcal{B}. Let \mathcal{U}_\tau=\mathcal{U} \cap \tau. Let Y=\cup \mathcal{U}_\tau. Note that Y is a Euclidean open subspace of the real line (hence it is paracompact). Then there is \mathcal{V}_\tau \subset \tau such that \mathcal{V}_\tau is a locally finite open refinement \mathcal{V}_\tau of \mathcal{U}_\tau and such that \mathcal{V}_\tau covers Y (locally finite in the Euclidean sense). Then add to \mathcal{V}_\tau all singleton sets \left\{ x \right\} where x \in \mathbb{M}-Y and let \mathcal{V} denote the resulting open collection.

The resulting \mathcal{V} is a locally finite open collection in the Michael line \mathbb{M}. Furthermore, \mathcal{V} is also a refinement of the original open cover \mathcal{U}. \blacksquare

A similar argument shows that \mathbb{M} is hereditarily paracompact.

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Result 2

To see that \mathbb{M} is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in \mathbb{P}). For example, it is possible to construct a Cantor set entirely within \mathbb{P}.

Let C be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set C is an uncountable closed and discrete set in \mathbb{M}. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line \mathbb{M} cannot be Lindelof. \blacksquare

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space X is the smallest infinite cardinal number \mathcal{K} such that every closed and discrete set in X has cardinality \le \mathcal{K}. The extent of the space X is denoted by e(X). When the cardinal number e(X) is e(X)=\aleph_0 (the first infinite cardinal number), the space X is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When e(X)>\aleph_0, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space X is Lindelof, the extent is e(X)=\aleph_0. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of \mathbb{M} of cardinality c. Thus we have e(\mathbb{M})=c. \blacksquare

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Result 4

The Michael line \mathbb{M} is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in \mathbb{M}. \blacksquare

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Result 5

A set is said to be a G_\delta-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a G_\delta-set. In the Michael line \mathbb{M}, the set \mathbb{Q} of rational numbers is a closed set. Yet, \mathbb{Q} is not a G_\delta-set in the Michael line (see the discussion above on the Baire category theorem). Thus \mathbb{M} is not perfectly normal and hence not a metrizable space. \blacksquare

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Result 6

The diagonal of a space X is the subset of its square X \times X that is defined by \Delta=\left\{(x,x): x \in X \right\}. If the space is Hausdorff, the diagonal is always a closed set in the square. If \Delta is a G_\delta-set in X \times X, the space X is said to have a G_\delta-diagonal. It is well known that any metric space has G_\delta-diagonal. Since \mathbb{M} is submetrizable (submetrized by the usual topology of the real line), it has a G_\delta-diagonal too.

Any Moore space has a G_\delta-diagonal. However, the Michael line is an example of a space with G_\delta-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus \mathbb{M} is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. \blacksquare

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Result 7

We now show that \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology. In this proof, the following two facts are crucial:

  • The set \mathbb{P} is not an F_\sigma-set in the real line.
  • The set \mathbb{P} is dense in the real line.

Let H and K be defined by the following:

    H=\left\{(x,x): x \in \mathbb{P} \right\}
    K=\mathbb{Q} \times \mathbb{P}.

The sets H and K are disjoint closed sets in \mathbb{M} \times \mathbb{P}. We show that they cannot be separated by disjoint open sets. To this end, let H \subset U and K \subset V where U and V are open sets in \mathbb{M} \times \mathbb{P}.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval (a,b), we mean the set of all real numbers t with a<t<b. By (a,b)^*, we mean (a,b) \cap \mathbb{P}. For each x \in \mathbb{P}, choose an open interval U_x=(a,b)^* such that \left\{x \right\} \times U_x \subset U. We also assume that x is the midpoint of the open interval U_x. For each positive integer k, let P_k be defined by:

    P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}

Note that \mathbb{P}=\bigcup \limits_{k=1}^\infty P_k. For each k, let T_k=\overline{P_k} (Euclidean closure in the real line). It is clear that \bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k. On the other hand, \bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P} (otherwise \mathbb{P} would be an F_\sigma-set in the real line). So there exists T_n=\overline{P_n} such that \overline{P_n} \not\subset \mathbb{P}. So choose a rational number r such that r \in \overline{P_n}.

Choose a positive integer j such that \frac{2}{j}<\frac{1}{n}. Since \mathbb{P} is dense in the real line, choose y \in \mathbb{P} such that r-\frac{1}{j}<y<r+\frac{1}{j}. Now we have (r,y) \in K \subset V. Choose another integer m such that \frac{1}{m}<\frac{1}{j} and (r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V.

Since r \in \overline{P_n}, choose x \in \mathbb{P} such that r-\frac{1}{m}<x<r+\frac{1}{m}. Now it is clear that (x,y) \in V. The following inequalities show that (x,y) \in U.

    \lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}

The open interval U_x is chosen to have length > \frac{1}{n}. Since \lvert x-y \lvert < \frac{1}{n}, y \in U_x. Thus (x,y) \in \left\{ x \right\} \times U_x \subset U. We have shown that U \cap V \ne \varnothing. Thus \mathbb{M} \times \mathbb{P} is not normal. \blacksquare

Remark
As indicated above, the proof of Result 7 hinges on two facts about \mathbb{P}, namely that it is not an F_\sigma-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set D that is isolated is not an F_\sigma-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space D (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product \mathbb{M} \times \mathbb{P} is not paracompact since it is not normal. However, \mathbb{M} \times \mathbb{P} is metacompact.

A collection of subsets of a space X is said to be point-finite if every point of X belongs to only finitely many sets in the collection. A space X is said to be metacompact if each open cover of X has an open refinement that is a point-finite collection.

Note that \mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P}). The first \mathbb{P} in \mathbb{P} \times \mathbb{P} is discrete (a subspace of the Michael line) and the second \mathbb{P} has the Euclidean topology.

Let \mathcal{U} be an open cover of \mathbb{M} \times \mathbb{P}. For each a=(x,y) \in \mathbb{Q} \times \mathbb{P}, choose U_a \in \mathcal{U} such that a \in U_a. We can assume that U_a=A \times B where A is a usual open interval in \mathbb{R} and B is a usual open interval in \mathbb{P}. Let \mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace.

Fix x \in \mathbb{P}. For each b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}, choose some U_b \in \mathcal{U} such that b \in U_b. We can assume that U_b=\lbrace{x}\rbrace \times B where B is a usual open interval in \mathbb{P}. Let \mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace.

As a subspace of the Euclidean plane, \bigcup \mathcal{G} is metacompact. So there is a point-finite open refinement \mathcal{W} of \mathcal{G}. For each x \in \mathbb{P}, \mathcal{H}_x has a point-finite open refinement \mathcal{I}_x. Let \mathcal{V} be the union of \mathcal{W} and all the \mathcal{I}_x where x \in \mathbb{P}. Then \mathcal{V} is a point-finite open refinement of \mathcal{U}.

Note that the point-finite open refinement \mathcal{V} may not be locally finite. The vertical open intervals in \lbrace{x}\rbrace \times \mathbb{P}, x \in \mathbb{P} can “converge” to a point in \mathbb{Q} \times \mathbb{P}. Thus, metacompactness is the best we can hope for. \blacksquare

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base \mathcal{G} for a space X is said to be a point-countable base if \mathcal{G}, in addition to being a base for the space X, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for \mathbb{M}:

    \mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

where \mathcal{H} is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. 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.

The Banach-Mazur Game

A topological space X is said to be a Baire space if for every countable family \left\{U_0,U_1,U_2,\cdots \right\} of open and dense subsets of X, the intersection \bigcap \limits_{n=0}^\infty U_n is dense in X (equivalently if every nonempty open subset of X is of second category in X). By the Baire category theorem, every complete metric space is a Baire space. The Baire property (i.e. being a Baire space) can be characterized using the Banach-Mazur game, which is the focus of this post.

Baire category theorem and Baire spaces are discussed in this previous post. We define the Banach-Mazur game and show how this game is related to the Baire property. We also define some completeness properties stronger than the Baire property using this game. For a survey on Baire spaces, see [4]. For more information about the Banach-Mazur game, see [1]. Good references for basic topological terms are [3] and [5]. All topological spaces are assumed to be at least Hausdorff.

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The Banach-Mazur Game

The Banach-Mazur game is a two-person game played on a topological space. Let X be a space. There are two players, \alpha and \beta. They take turn choosing nested decreasing nonempty open subsets of X as follows. The player \beta goes first by choosing a nonempty open subset U_0 of X. The player \alpha then chooses a nonempty open subset V_0 \subset U_0. At the nth play where n \ge 1, \beta chooses an open set U_n \subset V_{n-1} and \alpha chooses an open set V_n \subset U_n. The player \alpha wins if \bigcap \limits_{n=0}^\infty V_n \ne \varnothing. Otherwise the player \beta wins.

If the players in the game described above make the moves U_0,V_0,U_1,V_1,U_2,V_2,\cdots, then this sequence of open sets is said to be a play of the game.

The Banach-Mazur game, as described above, is denoted by BM(X,\beta). In this game, the player \beta makes the first move. If we modify the game by letting \alpha making the first move, we denote this new game by BM(X,\alpha). In either version, the goal of player \beta is to reach an empty intersection of the chosen open sets while player \alpha wants the chosen open sets to have nonempty intersection.

A Remark About Topological Games

Before relating the Banach-Mazur game to Baire spaces, we give a remark about topological games. For any two-person game played on a topological space, we are interested in the following question.

  • Can a player, by making his/her moves judiciously, insure that he/she will always win no matter what moves the other player makes?

If the answer to this question is yes, then the player in question is said to have a winning strategy. For an illustration, consider a space X that is of first category in itself, so that X=\bigcup \limits_{n=0}^\infty X_n where each X_n is nowhere dense in X. Then player \beta has a winning strategy in the Banach-Mazur game BM(X,\beta). The player \beta always wins the game by making his/her nth play U_n \subset V_{n-1} - \overline{X_n}.

In general, a strategy for a player in a game is a rule that specifies what moves he/she will make in every possible situation. In other words, a strategy for a player is a function whose domain is the set of all partial plays of the game, and this function tells the player what the next move should be. A winning strategy for a player is a strategy such that this player always wins if that player makes his/her moves using this strategy. A strategy for a player in a game is not a winning strategy if of all the plays of the game resulting from using this strategy, there is at least one specific play of the game resulting in a win for the other player.

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Strategies in the Banach-Mazur Game

With the above discussion in mind, let us discuss the strategies in the Banach-Mazur game. We show that the strategies in this game code a great amount of information about the topological space in which the game is played.

First we discuss strategies for player \beta in the game BM(X,\beta). A strategy for player \beta is a function \sigma such that U_0=\sigma(\varnothing) (the first move) and for each partial play of the game (n \ge 1)

\displaystyle (*) \ \ \ \ \ \ U_0,V_0,U_1,V_1,\cdots,U_{n-1},V_{n-1},

U_n=\sigma(U_0,V_0,U_1,V_1,\cdots,U_{n-1},V_{n-1}) is a nonempty open set such that U_n \subset V_{n-1}. If player \beta makes all his/her moves using the strategy \sigma, then the strategy \sigma for player \beta contains information on all moves of \beta. We adopt the convention that a strategy for a player in a game depends only on the moves of the other player. Thus for the partial play of the Banach-Mazur game denoted by (*) above, U_n=\sigma(V_0,V_1,\cdots,V_{n-1}).

If \sigma is a winning strategy for player \beta in the game BM(X,\beta), then using this strategy will always result in a win for \beta. On the other hand, if \sigma is a not a winning strategy for player \beta in the game BM(X,\beta), then there exists a specific play of the Banach-Mazur game

\displaystyle . \ \ \ \ \ \ U_0,V_0,U_1,V_1,\cdots,U_{n-1},V_{n-1},\cdots

such that U_0=\sigma(\varnothing), and for each n \ge 1, U_n=\sigma(V_0,\cdots,V_{n-1}) and player \alpha wins in this play of the game, that is, \bigcap \limits_{n=0}^\infty V_n \ne \varnothing.

In the game BM(X,\alpha) (player \alpha making the first move), a strategy for player \beta is a function \gamma such that for each partial play of the game

\displaystyle (**) \ \ \ \ \ V_0,U_1,V_1,\cdots,U_{n-1},V_{n-1},

U_n=\gamma(V_0,V_1,\cdots,V_{n-1}) is a nonempty open set such that U_n \subset V_{n-1}.

We now present a lemma that helps translate game information into topological information.

Lemma 1
Let X be a space. Let O \subset X be a nonempty open set. Let \tau be the set of all nonempty open subsets of O. Let f: \tau \longrightarrow \tau be a function such that for each V \in \tau, f(V) \subset V. Then there exists a disjoint collection \mathcal{U} consisting of elements of f(\tau) such that \bigcup \mathcal{U} is dense in O.

Proof
This is an argument using Zorn’s lemma. If the open set O in the hypothesis has only one point, then the conclusion of the lemma holds. So assume that O has at least two points.

Let \mathcal{P} be the set consisting of all collections \mathcal{F} such that each \mathcal{F} is a disjoint collection consisting of elements of f(\tau). First \mathcal{P} \ne \varnothing. To see this, let V and W be two disjoint open sets such that V \subset O and W \subset O. This is possible since O has at least two points. Let \mathcal{F^*}=\left\{ f(V),f(W)\right\}. Then we have \mathcal{F^*} \in \mathcal{P}. Order \mathcal{P} by set inclusion. It is straightforward to show that (\mathcal{P}, \subset) is a partially ordered set.

Let \mathcal{T} \subset \mathcal{P} be a chain (a totally ordered set). We wish to show that \mathcal{T} has an upper bound in \mathcal{P}. The candidate for an upper bound is \bigcup \mathcal{T} since it is clear that for each \mathcal{F} \in \mathcal{T}, \mathcal{F} \subset \bigcup \mathcal{T}. We only need to show \bigcup \mathcal{T} \in \mathcal{P}. To this end, we need to show that any two elements of \bigcup \mathcal{T} are disjoint open sets.

Note that elements of \bigcup \mathcal{T} are elements of f(\tau). Let T_1,T_2 \in \bigcup \mathcal{T}. Then T_1 \in \mathcal{F}_1 and T_2 \in \mathcal{F}_2 for some \mathcal{F}_1 \in \mathcal{T} and \mathcal{F}_2 \in \mathcal{T}. Since \mathcal{T} is a chain, either \mathcal{F}_1 \subset \mathcal{F}_2 or \mathcal{F}_2 \subset \mathcal{F}_1. This means that T_1 and T_2 belong to the same disjoing collection in \mathcal{T}. So they are disjoint open sets that are members of f(\tau).

By Zorn’s lemma, (\mathcal{P}, \subset) has a maximal element \mathcal{U}, which is a desired disjoint collection of sets in f(\tau). Since \mathcal{U} is maximal with respect to \subset, \bigcup \mathcal{U} is dense in O. \blacksquare

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Characterizing Baire Spaces using the Banach-Mazur Game

Lemma 1 is the linkage between the Baire property and the strategies in the Banach-Mazur game. The thickness in Baire spaces and spaces of second category allow us to extract a losing play in any strategy for player \beta. The proofs for both Theorem 1 and Theorem 2 are very similar (after adjusting for differences in who makes the first move). Thus we only present the proof for Theorem 1.

Theorem 1
The space X is a Baire space if and only if player \beta has no winning strategy in the game BM(X,\beta).

Proof
\Longleftarrow Suppose that X is not a Baire space. We define a winning strategy in the game BM(X,\beta) for player \beta. The space X not being a Baire space implies that there is some nonempty open set U_0 \subset X such that U_0 is of first category in X. Thus U_0=\bigcup \limits_{n=1}^\infty F_n where each F_n is nowhere dense in X.

We now define a winning strategy for \beta. Let U_0 be the first move of \beta. For each n \ge 1, let player \beta make his/her move by letting U_n \subset V_{n-1} - \overline{F_n} if V_{n-1} is the last move by \alpha. It is clear that whenever \beta chooses his/her moves in this way, the intersection of the open sets has to be empty.

\Longrightarrow Suppose that X is a Baire space. Let \sigma be a strategy for the player \beta. We show that \sigma cannot be a winning strategy for \beta.

Let U_0=\sigma(\varnothing) be the first move for \beta. For each open V_0 \subset U_0, \sigma(V_0) \subset V_0. Apply Lemma 1 to obtain a disjoint collection \mathcal{U}_0 consisting of open sets of the form \sigma(V_0) such that \bigcup \mathcal{U}_0 is dense in U_0.

For each W=\sigma(V_0) \in \mathcal{U}_0, we have \sigma(V_0,V_1) \subset V_1 for all open V_1 \subset W. So the function \sigma(V_0,\cdot) is like the function f in Lemma 1. We can then apply Lemma 1 to obtain a disjoint collection \mathcal{U}_1(W) consisting of open sets of the form \sigma(V_0,V_1) such that \bigcup \mathcal{U}_1(W) is dense in W. Then let \mathcal{U}_1=\bigcup_{W \in \mathcal{U}_0} \mathcal{U}_1(W). Based on how \mathcal{U}_1(W) are obtained, it follows that \bigcup \mathcal{U}_1 is dense in U_0.

Continue the inductive process in the same manner, we can obtain, for each n \ge 1, a disjoint collection \mathcal{U}_n consisting of open sets of the form \sigma(V_0,\dots,V_{n-1}) (these are moves made by \beta using the strategy \sigma) such that \bigcup \mathcal{U}_n is dense in U_0.

For each n, let O_n=\bigcup \mathcal{U}_n. Each O_n is dense open in U_0. Since X is a Baire space, every nonempty open subset of X is of second category in X (including U_0). Thus \bigcap \limits_{n=0}^\infty O_n \ne \varnothing. From this nonempty intersection, we can extract a play of the game such that the open sets in this play of the game have one point in common (i.e. player \alpha wins). We can extract the play of the game because the collection \mathcal{U}_n are disjoint. Thus the strategy \sigma is not a winning strategy for \beta. This completes the proof of Theorem 1. \blacksquare

Theorem 2
The space X is of second category in itself if and only if player \beta has no winning strategy in the game BM(X,\alpha).

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Some Completeness Properties

Theorem 1 shows that a Baire space is one in which the player \beta has no winning strategy in the Banach-Mazur game (the version in which \beta makes the first move). In such a space, no matter what strategy player \beta wants to use, it can be foiled by player \alpha by producing one specific play in which \beta loses. We now consider spaces in which player \alpha has a winning strategy. A space X is said to be a weakly \alpha-favorable if player \alpha has a winning strategy in the game BM(X,\beta). If \alpha always wins, then \beta has no winning strategy. Thus the property of being a weakly \alpha-favorable space is stronger than the Baire property.

In any complete metric space, the player \alpha always has a winning strategy. The same idea used in proving the Baire category theorem can be used to establish this fact. By playing the game in a complete metric space, player \alpha can ensure a win by making sure that the closure of his/her moves have diameters converge to zero (and the closure of his/her moves are subsets of the previous moves).

Based on Theorem 1, any Baire space is a space in which player \beta of the Banach-Mazur game has no winning strategy. Any Baire space that is not weakly \alpha-favorable is a space in which both players of the Banach-Mazur game have no winning strategy (i.e. the game is undecidable). Any subset of the real line \mathbb{R} that is a Bernstein set is such a space. A subset B of the real line is said to be a Bernstein set if B and its complement intersect every uncountable closed subset of the real line. Bernstein sets are discussed here.

Suppose \theta is a strategy for \alpha in the game BM(X,\beta). If at each step, the strategy \theta can provide a move based only on the other player’s last move, it is said to be a stationary strategy. For example, in the partial play U_0,V_0,\cdots,U_{n-1},V_{n-1},U_n, the strategy \theta can determine the next move for \alpha by only knowing the last move of \beta, i.e., V_n=\theta(U_n). A space X is said to be \alpha-favorable if player \alpha has a stationary winning strategy in the game BM(X,\beta). Clearly, any \alpha-favorable spaces are weakly \alpha-favorable spaces. However, there are spaces in which player \alpha has a winning strategy in the Banach-Mazur game and yet has no stationary winning strategy (see [2]). Stationary winning strategy for \alpha is also called \alpha-winning tactic (see [1]).

Reference

  1. Choquet, G., Lectures on analysis, Vol I, Benjamin, New York and Amsterdam, 1969.
  2. Deb, G., Stategies gagnantes dans certains jeux topologiques, Fund. Math. 126 (1985), 93-105.
  3. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  4. Haworth, R. C., McCoy, R. A., Baire Spaces, Dissertations Math., 141 (1977), 1 – 73.
  5. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

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Revised 4/4/2014. \copyright \ 2014 \text{ by Dan Ma}

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.

Soppuse 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 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.

Thinking about the Space of Irrationals Topologically

Let \mathbb{R} denote the real number line and \mathbb{P} denote the set of all irrational numbers. The irrational numbers and the set \mathbb{P} occupy an important place in mathematics. The set \mathbb{P} with the Euclidean topology inherited from the full real line is a topological space in its own right. Thus the space \mathbb{P} has some of the same properties inherited from the Euclidean real line, e.g., just to name a few, it is hereditarily separable, hereditarily Lindelof, paracompact and metrizable. The space of the irrational numbers \mathbb{P} has many properties apart from the full real line (e.g. \mathbb{P} is totally disconnected). In this post, we look at a topological description of the space \mathbb{P}.

Let \omega be the set of all nonnegative integers. Then the space \mathbb{P} of irrational numbers is topologically equivalent (i.e. homeomorphic to) the product space \prod \limits_{i=0}^\infty X_i where each X_i=\omega has the discrete topology. We will also denote the product space \prod \limits_{i=0}^\infty X_i by \omega^\omega. We have the following theorem.

Theorem
The space \mathbb{P} of irrational numbers is homeomorphic to the product space \omega^\omega.

Because of this theorem, we can look at irrational numbers as sequences of nonnegative integers. Specifically each irrational number can be identified by a unique sequence of nonnegative integers. We can think of each such unique sequence as an address to locate an irrational number. In the remainder of the post, we describe at a high level how to define the unique addresses, which will also give us a homeomorphic map between the space \mathbb{P} and the product space \omega^\omega.

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Step 0
Put the rational numbers in a sequence r_0,r_1,r_2,r_3,\cdots such that r_0=0. We divide the real line into countably many non-overlapping intervals. Specifically, let A_0=[0,1], A_1=[-1,0], A_2=[1,2], etc (see the following figure).

To facilitate the remaining construction, we denote the left endpoint of the interval A_i by L_i and denote the right endpoint by R_i.

Step 1
In each of the interval A_i, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval A_i are L_i and R_i, respectively. We choose a sequence x_{i,0}, x_{i,1}, x_{i,2},\cdots of rational numbers converging to the right endpoint R_i. Then let A_{i,0}=[L_i,x_{i,0}], A_{i,1}=[x_{i,0},x_{i,1}], A_{i,2}=[x_{i,1},x_{i,2}], etc (see the following figure).

Two important points to consider in Step 1. One is that we make sure the rational number r_1 is chosen as an endpoint of some interval in Step 1. The second is that the length of each A_{i,j} is less than \displaystyle \frac{1}{2^1}.

Step 2
In each of the interval A_{i,j}, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval A_{i,j} are L_{i,j} and R_{i,j}, respectively. We choose a sequence x_{i,j,0}, x_{i,j,1}, x_{i,j,2},\cdots of rational numbers converging to the left endpoint L_{i,j}. Then let A_{i,j,0}=[x_{i,j,0},R_{i,j}], A_{i,j,1}=[x_{i,j,1},x_{i,j,0}], A_{i,j,2}=[x_{i,j,2},x_{i,j,1}], etc (see the following figure).

As in the previous step, two important points to consider in Step 2. One is that we make sure the rational number r_2 is chosen as an endpoint of some interval in Step 2. The second is that the length of each A_{i,j,k} is less than \displaystyle \frac{1}{2^2}.

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Remark

In the process described above, the endpoints of the intervals A_{f(0),\cdots,f(n)} are rational numbers and we make sure that all the rational numbers are used as endpoints. We also make sure that the intervals from the successive steps are nested closed intervals with lengths \rightarrow 0. The consequence of this point is that the nested decreasing closed intervals will collapse to one single point (since the real line is a complete metric space) and this single point must be an irrational number (since all the rational numbers are used up as endpoints of the nested closed intervals).

In Step i where i is an odd integer, we make the endpoints of the new intervals converge to the right. In Step i where i>1 is an even integer, we make the endpoints of the new intervals converge to the left. This manipulation is to ensure that the nested closed intervals will never share the same endpoint from one step all the way to the end of the process.

Thus if we have f \in \omega^\omega, then \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)} \ne \varnothing and in fact has only one point that is an irrational number.

On the other hand, for each point x \in \mathbb{P}, we can locate inductively a sequence of intervals, A_{f(0)}, A_{f(0),f(1)}, A_{f(0),f(1),f(2)}, \cdots, containing the point x. This sequence of nested closed intervals must collapse to a single point and the single point must be the irrational number x.

The process described above gives us a one-to-one mapping from \mathbb{P} onto the product space \omega^\omega. This mapping is also continuous in both directions, making it a homeomorphism. the nested intervals defined above form a base for the Euclidean topology on \mathbb{P}. These basic open sets have a natural correspondance with basic open sets in the product space \omega^\omega.

For example, for f \in \omega^\omega, \left\{ A_{f(0),\cdots,f(n)} \cap \mathbb{P}: n \in \omega \right\} is a local base at the point x \in \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)}. One the other hand, each A_{f(0),\cdots,f(n)} \cap \mathbb{P} has a natural counterpart in a basic open set in the product space, namely the following set:

\displaystyle . \ \ \ \ \left\{g \in \omega^\omega: g \restriction n = f \restriction n \right\}

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The above process establishes that the countable product of the integers, \omega^\omega, is equivalent topologically to the Euclidean space \mathbb{P}.

An observation about sequential spaces

This post is about an observation about sequential spaces. In a sequential space, non-trivial convergent sequences abound. Thus in the extreme case of there being no trivial convergent convergent sequences, the space in question must not be sequential. Specifically we observe that if X is a Hausdorff sequential space and if p \in X is a non-isolated point (i.e. the singleton set \left\{p\right\} is not open), there is a convergent sequence p_n of points of X-\left\{p\right\} such that p_n \mapsto p. Thus it is necessary condition that in a sequential space, there exist non-trivial convergent sequences at every non-isolated point. We present examples showing that this condition is not a sufficient condition for a space being sequential. As the following examples show, the property that there are non-trivial convergent sequences at every non-isolated is a rather weak property.

The first example is from the problem section of Mathematical Monthly in 1970 (see [2]). Let \mathbb{R} be the real line and let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}. Let X=\mathbb{R} and define a new topology on X by calling a subset U \subset X open if and only if U=W-H where W is a usual open subset of the real line and H is a subset of \mathbb{P} that is at most countable. This new topology on the real line is finer than the Euclidean topology. Thus X is a Hausdorff space. Every point of X is a non-isolated point and is the the sequential limit of a sequence of rational numbers, satisfying the condition that every non-isolated point is the sequential limit of a non-trivial convergent sequence.

In the topology for X, every countably infinite subset of the set \mathbb{P} is closed in X. Thus no sequence of points of \mathbb{P} can converge to a point not in \mathbb{P}. Therefore \mathbb{P} is sequentially closed and non-closed in X, making X not a sequential space.

Not only that every countably infinite subset of \mathbb{P} is closed in X, every countably infinite subset of \mathbb{P} is relatively discrete. Then it follows that for every compact K \subset X, K \cap \mathbb{P} is finite (and is thus closed in K). Thus X is also not a k-space.

Another example is that of a product space. Any uncountable product where each factor has at least two points is not sequential. This follows from the fact that 2^{\omega_1} is not sequential (see Sequential spaces, IV). Furthermore, in any product space with infinitely many factors each of which has at least two points, every point is the sequential limit of a non-trivial convergent sequence. Thus any product space with uncountably many factors, each of which has at least two points, is another example of a non-sequential space where there are non-trivial convergent sequences at every point.

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space

Reference

  1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
  3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Sequential spaces, III

This is a continuation of the discussion on sequential spaces started with the post Sequential spaces, I and Sequential spaces, II and k-spaces, I. The topology in a sequential space is generated by the convergent sequences. The convergence we are interested in is from a topological view point and not necessarily from a metric (i.e. distance) standpoint. In our discussion, a sequence \left\{x_n\right\}_{n=1}^\infty converges to x simply means for each open set O containing x, O contains x_n for all but finitely many n. In any topological space, there are always trivial convergent sequences. These are sequences of points that are eventually constant, i.e. the sequences \left\{x_n\right\} where for some n, x_n=x_j for j \ge n. Any convergent sequence that is not eventually constant is called a non-trivial convergent sequence. We present an example of a space where there are no non-trivial convergent sequences of points. This space is derived from the Euclidean topology on the real line. This space has no isolated point in this space and yet has no non-trivial convergent sequences and has no infinite compact sets. From this example, we make some observations about sequential spaces and k-spaces.

The space we define here is obtained by modifying the Euclidean topology on the real line. Let \mathbb{R} be the real line. Let \tau_e be the Euclidean topology on the real line. Consider the following collection of subsets of the real line:

\mathcal{B}=\left\{U-C:U \in \tau_e \text{ and } \lvert C \lvert \le \omega\right\}

It can be verified that \mathcal{B} is a base for a topology \tau on \mathbb{R}. In fact this topology is finer than the Euclidean topology. Denote \mathbb{R} with this finer topology by X. Clearly X is Hausdorff since the Euclidean topology is. Any countable subset of X is closed. Thus X is not separable (no countable set can be dense). This space is a handy example of a hereditarily Lindelof space that is not separable. The following lists some properties of X:

  1. X is herditarily Lindelof.
  2. There are no non-trivial convergent sequences in X.
  3. All compact subsets of X are finite.
  4. X is not a k-space and is thus not sequential.

Discussion of 1. This follows from the fact that the real line with the Euclidean topology is hereditarily Lindelof and the fact that each open set in X is an Euclidean open set minus a countable set.

Discussion of 2 This follows from the fact that every countable subset of X is closed. If a non-trivial sequence \left\{x_n\right\} were to converge to x \in X, then \left\{x_n:n=1,2,3,\cdots\right\} would be a countable subset of X that is not closed.

Discussion of 3. Let A \subset X be an infinite set. If A is bounded in the Euclidean topology, then there would be a non-trivial convergent sequence of points of A in the Euclidean topology, say, x_n \mapsto x. Let U_0=X-\left\{x_n:n=1,2,3,\cdots\right\}, which is open in X. For n \ge 1, let U_n be Euclidean open such that x_n \in U_n. We also require that all U_n are pairwise disjoint and not contain x. Then U_0,U_1,U_2,\cdots form an open cover of A (in the topology of X) that has no finite subcover. So any bounded infinite A is not compact in X.

Suppose A is unbounded in the Euclidean topology. Then A contains a closed and discrete subset \left\{x_1,x_2,x_3,\cdots\right\} in the Euclidean topology. We can find Euclidean open sets U_n that are pairwise disjoint such that x_n \in U_n for each n. Let U_0=X-\left\{x_n:n=1,2,3,\cdots\right\}, which is open in X. Then U_0,U_1,U_2,\cdots form an open cover of A (in the topology of X) that has no finite subcover. So any unbounded infinite A is not compact in X.

Discussion of 4. Note that every point of X is a non-isolated point. Just pick any x \in X. Then X-\left\{x\right\} is not closed in X. However, according to 3, K \cap (X-\left\{x\right\}) is finite and is thus closed in K for every compact K \subset X. Thus X is not a k-space.

General Discussion
Suppose \tau is the topology for the space Y. Let \tau_s be the set of all sequentially open sets with respect to \tau (see Sequential spaces, II). Let \tau_k be the set of all compactly generated open sets with respect to \tau (see k-spaces, I). The space Y is a sequential space (a k-space ) if \tau=\tau_s (\tau=\tau_k). Both \tau_s and \tau_k are finer than \tau, i.e. \tau \subset \tau_s and \tau \subset \tau_k. When are \tau_s and \tau_k discrete? We discuss sequential spaces and k-spaces separately.

Observations on Sequential Spaces
With respect to the space (Y,\tau), we discuss the following four properties:

  • A. \ No non-trivial convergent sequences.
  • B. \ \tau_s is a discrete topology.
  • C. \ \tau is a discrete topology.
  • D. \ Sequential, i.e., \tau=\tau_s.

Observation 1
The topology \tau_s is discrete if and only if Y has no non-trivial convergent sequences, i.e. A \Longleftrightarrow B.

If \tau_s is a discrete topology, then every subset of Y is sequentially open and every subset is sequentially closed. Hence there can be no non-trivial convergent sequences. If there are no non-trivial convergent sequences, every subset of the space is sequentially closed (thus every subset is sequentially open).

Observation 2
Given that Y has no non-trivial convergent sequences, Y is not discrete if and only if Y is not sequential. Equivalently, given property A, C \Longleftrightarrow D.

Given that there are no non-trivial convergent sequences in Y, \tau_s is discrete. For (Y,\tau) to be sequential, \tau=\tau_s. Thus for a space Y that has no non-trivial convergent sequences, the only way for Y to be sequential is that it is a discrete space.

Observation 3
Given Y is not discrete, Y has no non-trivial convergent sequences implies that Y is not sequential, i.e. given \text{not }C, A \Longrightarrow \text{not }D. The converse does not hold.

Observation 3 is a rewording of observation 2. To see that the converse of observation 3 does not hold, consider Y=[0,\omega_1]=\omega_1+1, the successor ordinal to the first uncountable ordinal with the order topology. It is not sequential as the singleton set \left\{\omega_1\right\} is sequentially open and not open.

Observations on k-spaces
The discussion on k-spaces mirrors the one on sequential spaces. With respect to the space (Y,\tau), we discuss the following four properties:

  • E. \ No infinite compact sets.
  • F. \ \tau_k is a discrete topology.
  • G. \ \tau is a discrete topology.
  • H. \ k-space, i.e., \tau=\tau_k.

Observation 4
The topology \tau_k is discrete if and only if Y has no infinite compact sets, i.e. E \Longleftrightarrow F.

If \tau_k is a discrete topology, then every subset of Y is a compactly generated open set. In particular, for every compact K \subset Y, every subset of K is open in K. This means K is discrete and thus must be finite. Hence there can be no infinite compact sets if \tau_k is discrete. If there are no infinite compact sets, every subset of the space is a compactly generated closed set (thus every subset is a compactly generated open set).

Observation 5
Given that Y has no infinite compact sets, Y is not discrete if and only if Y is not a k-space. Equivalently, given property E, G \Longleftrightarrow H.

Given that there are no infinite compact sets in Y, \tau_k is discrete. For (Y,\tau) to be a k-space, \tau=\tau_k. Thus for a space Y that has no infinite compact sets, the only way for Y to be a k-space is that it is a discrete space.

Observation 6
Given Y is not discrete, Y has no infinite compact sets implies that Y is not a k-space, i.e. given \text{not }G, E \Longrightarrow \text{not }H. The converse does not hold.

Observation 6 is a rewording of observation 5. To see that the converse of observation 6 does not hold, consider the topological sum of a non-k-space and an infinite compact space.

Remark
In the space X defined above by removing countable sets from Euclidean open subsets of the real line, there are no infinite compact sets and no non-trivial convergent sequences. Yet the space is not discrete. Thus it can neither be a sequential space nor a k-space. Another observation we would like to make is that no infinite compact sets implies no non-trivial convergent sequences (E \Longrightarrow A). However, the converse is not true. Consider \beta(\omega), the Stone-Cech compactification of \omega, the set of all nonnegative integers.

Reference

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