# 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) 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) for any $i.
• 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.

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, 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 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 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. 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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# Lindelof Exercise 2

The preceding post is an exercise showing that the product of countably many $\sigma$-compact spaces is a Lindelof space. The result is an example of a situation where the Lindelof property is countably productive if each factor is a “nice” Lindelof space. In this case, “nice” means $\sigma$-compact. This post gives several exercises surrounding the notion of $\sigma$-compactness.

Exercise 2.A

According to the preceding exercise, the product of countably many $\sigma$-compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many $\sigma$-compact spaces. More specifically, give an example of a product of uncountably many $\sigma$-compact spaces such that the product space is not Lindelof.

Exercise 2.B

Any $\sigma$-compact space is Lindelof. Since $\mathbb{R}=\bigcup_{n=1}^\infty [-n,n]$, the real line with the usual Euclidean topology is $\sigma$-compact. This exercise is to find an example of “Lindelof does not imply $\sigma$-compact.” Find one such example among the subspaces of the real line. Note that as a subspace of the real line, the example would be a separable metric space, hence would be a Lindelof space.

Exercise 2.C

This exercise is also to look for an example of a space that is Lindelof and not $\sigma$-compact. The example sought is a non-metric one, preferably a space whose underlying set is the real line and whose topology is finer than the Euclidean topology.

Exercise 2.D

Show that the product of two Lindelof spaces is a Lindelof space whenever one of the factors is a $\sigma$-compact space.

Exercise 2.E

Prove that the product of finitely many $\sigma$-compact spaces is a $\sigma$-compact space. Give an example of a space showing that the product of countably and infinitely many $\sigma$-compact spaces does not have to be $\sigma$-compact. For example, show that $\mathbb{R}^\omega$, the product of countably many copies of the real line, is not $\sigma$-compact.

The Lindelof property and $\sigma$-compactness are basic topological notions. The above exercises are natural questions based on these two basic notions. One immediate purpose of these exercises is that they provide further interaction with the two basic notions. More importantly, working on these exercise give exposure to mathematics that is seemingly unrelated to the two basic notions. For example, finding $\sigma$-compactness on subspaces of the real line and subspaces of compact spaces naturally uses a Baire category argument, which is a deep and rich topic that finds uses in multiple areas of mathematics. For this reason, these exercises present excellent learning opportunities not only in topology but also in other useful mathematical topics.

If preferred, the exercises can be attacked head on. The exercises are also intended to be a guided tour. Hints are also provided below. Two sets of hints are given – Hints (blue dividers) and Further Hints (maroon dividers). The proofs of certain key facts are also given (orange dividers). Concluding remarks are given at the end of the post.

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Hints for Exercise 2.A

Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if $X$ is a Lindelof space, then every closed subspace of $X$ is also Lindelof.

Prove that if $X$ is a Lindelof space, then every closed and discrete subset of $X$ is countable (every space that has this property is said to have countable extent).

Show that the product of uncountably many copies of the real line does not have countable extent. Specifically, focus on either one of the following two examples.

• Show that the product space $\mathbb{R}^c$ has a closed and discrete subspace of cardinality continuum where $c$ is cardinality of continuum. Hence $\mathbb{R}^c$ is not Lindelof.
• Show that the product space $\mathbb{R}^{\omega_1}$ has a closed and discrete subspace of cardinality $\omega_1$ where $\omega_1$ is the first uncountable ordinal. Hence $\mathbb{R}^{\omega_1}$ is not Lindelof.

Hints for Exercise 2.B

Let $\mathbb{P}$ be the set of all irrational numbers. Show that $\mathbb{P}$ as a subspace of the real line is not $\sigma$-compact.

Hints for Exercise 2.C

Let $S$ be the real line with the topology generated by the half open and half closed intervals of the form $[a,b)=\{ x \in \mathbb{R}: a \le x < b \}$. The real line with this topology is called the Sorgenfrey line. Show that $S$ is Lindelof and is not $\sigma$-compact.

Hints for Exercise 2.D

It is helpful to first prove: the product of two Lindelof space is Lindelof if one of the factors is a compact space. The Tube lemma is helpful.

Tube Lemma
Let $X$ be a space. Let $Y$ be a compact space. Suppose that $U$ is an open subset of $X \times Y$ and suppose that $\{ x \} \times Y \subset U$ where $x \in X$. Then there exists an open subset $V$ of $X$ such that $\{ x \} \times Y \subset V \times Y \subset U$.

Hints for Exercise 2.E

Since the real line $\mathbb{R}$ is homeomorphic to the open interval $(0,1)$, $\mathbb{R}^\omega$ is homeomorphic to $(0,1)^\omega$. Show that $(0,1)^\omega$ is not $\sigma$-compact.

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Further Hints for Exercise 2.A

The hints here focus on the example $\mathbb{R}^c$.

Let $I=[0,1]$. Let $\omega$ be the first infinite ordinal. For convenience, consider $\omega$ the set $\{ 0,1,2,3,\cdots \}$, the set of all non-negative integers. Since $\omega^I$ is a closed subset of $\mathbb{R}^I$, any closed and discrete subset of $\omega^I$ is a closed and discrete subset of $\mathbb{R}^I$. The task at hand is to find a closed and discrete subset of $Y=\omega^I$. To this end, we define $W=\{W_x: x \in I \}$ after setting up background information.

For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that

• $\{ t \}=\bigcap_{j=1}^\infty O_{t,j}$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$).

Note. For each $t \in I-\{0,1 \}$, the open intervals $O_{t,j}$ are of the form $(a,b)$. For $t=0$, the open intervals $O_{t,j}$ are of the form $[0,b)$. For $t=1$, the open intervals $O_{t,j}$ are of the form $(a,1]$.

For each $t \in I$, define the map $f_t: I \rightarrow \omega$ as follows:

$f_t(x) = \begin{cases} 0 & \ \ \ \mbox{if } x=t \\ 1 & \ \ \ \mbox{if } x \in I-O_{t,1} \\ 2 & \ \ \ \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \ \ \ \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \\ j & \ \ \ \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \end{cases}$

We are now ready to define $W=\{W_x: x \in I \}$. For each $x \in I$, $W_x$ is the mapping $W_x:I \rightarrow \omega$ defined by $W_x(t)=f_t(x)$ for each $t \in I$.

Show the following:

• The set $W=\{W_x: x \in I \}$ has cardinality continuum.
• The set $W$ is a discrete space.
• The set $W$ is a closed subspace of $Y$.

Further Hints for Exercise 2.B

A subset $A$ of the real line $\mathbb{R}$ is nowhere dense in $\mathbb{R}$ if for any nonempty open subset $U$ of $\mathbb{R}$, there is a nonempty open subset $V$ of $U$ such that $V \cap A=\varnothing$. If we replace open sets by open intervals, we have the same notion.

Show that the real line $\mathbb{R}$ with the usual Euclidean topology cannot be the union of countably many closed and nowhere dense sets.

Further Hints for Exercise 2.C

Prove that if $X$ and $Y$ are $\sigma$-compact, then the product $X \times Y$ is $\sigma$-compact, hence Lindelof.

Prove that $S$, the Sorgenfrey line, is Lindelof while its square $S \times S$ is not Lindelof.

Further Hints for Exercise 2.D

As suggested in the hints given earlier, prove that $X \times Y$ is Lindelof if $X$ is Lindelof and $Y$ is compact. As suggested, the Tube lemma is a useful tool.

Further Hints for Exercise 2.E

The product space $(0,1)^\omega$ is a subspace of the product space $[0,1]^\omega$. Since $[0,1]^\omega$ is compact, we can fall back on a Baire category theorem argument to show why $(0,1)^\omega$ cannot be $\sigma$-compact. To this end, we consider the notion of Baire space. A space $X$ is said to be a Baire space if for each countable family $\{ U_1,U_2,U_3,\cdots \}$ of open and dense subsets of $X$, the intersection $\bigcap_{i=1}^\infty U_i$ is a dense subset of $X$. Prove the following results.

Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that $\overline{O_{n+1}} \subset O_n$ for each $n$. Then the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Fact E.2
Any compact Hausdorff space is Baire space.

Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Then $Y$ is not a $\sigma$-compact space.

Since $X=[0,1]^\omega$ is compact, it follows from Fact E.2 that the product space $X=[0,1]^\omega$ is a Baire space.

Fact E.4
Let $X=[0,1]^\omega$ and $Y=(0,1)^\omega$. The product space $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X=[0,1]^\omega$. Furthermore, $X-Y$ is a dense subset of $X$.

It follows from the above facts that the product space $(0,1)^\omega$ cannot be a $\sigma$-compact space.

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Proofs of Key Steps for Exercise 2.A

The proof here focuses on the example $\mathbb{R}^c$.

To see that $W=\{W_x: x \in I \}$ has the same cardinality as that of $I$, show that $W_x \ne W_y$ for $x \ne y$. This follows from the definition of the mapping $W_x$.

To see that $W$ is discrete, for each $x \in I$, consider the open set $U_x=\{ b \in Y: b(x)=0 \}$. Note that $W_x \in U_x$. Further note that $W_y \notin U_x$ for all $y \ne x$.

To see that $W$ is a closed subset of $Y$, let $k: I \rightarrow \omega$ such that $k \notin W$. Consider two cases.

Case 1. $k(r) \ne 0$ for all $r \in I$.
Note that $\{ O_{t,k(t)}: t \in I \}$ is an open cover of $I$ (in the usual topology). There exists a finite $H \subset I$ such that $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. Consider the open set $G=\{ b \in Y: \forall \ h \in H, \ b(h)=k(h) \}$. Define the set $F$ as follows:

$F=\{ c \in I: W_c \in G \}$

The set $F$ can be further described as follows:

\displaystyle \begin{aligned} F&=\{ c \in I: W_c \in G \} \\&=\{ c \in I: \forall \ h \in H, \ W_c(h)=f_h(c)=k(h) \ne 0 \} \\&=\{ c \in I: \forall \ h \in H, \ c \in I-O_{h,k(h)} \} \\&=\bigcap_{h \in H} (I-O_{h,k(h)}) \\&=I-\bigcup_{h \in H} O_{h,k(h)}=I-I =\varnothing \end{aligned}

The last step is $\varnothing$ because $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. The fact that $F=\varnothing$ means that $G$ is an open subset of $Y$ containing the point $k$ such that $G$ contains no point of $W$.

Case 2. $k(r) = 0$ for some $r \in I$.
Since $k \notin W$, $k \ne W_x$ for all $x \in I$. In particular, $k \ne W_r$. This means that $k(t) \ne W_r(t)$ for some $t \in I$. Define the open set $G$ as follows:

$G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}$

Clearly $k \in G$. Observe that $W_r \notin G$ since $W_r(t) \ne k(t)$. For each $p \in I-\{ r \}$, $W_p \notin G$ since $W_p(r) \ne 0$. Thus $G$ is an open set containing $k$ such that $G \cap W=\varnothing$.

Both cases show that $W$ is a closed subset of $Y=\omega^I$.

Proofs of Key Steps for Exercise 2.B

Suppose that $\mathbb{P}$, the set of all irrational numbers, is $\sigma$-compact. That is, $\mathbb{P}=A_1 \cup A_2 \cup A_3 \cup \cdots$ where each $A_i$ is a compact space as a subspace of $\mathbb{P}$. Any compact subspace of $\mathbb{P}$ is also a compact subspace of $\mathbb{R}$. As a result, each $A_i$ is a closed subset of $\mathbb{R}$. Furthermore, prove the following:

Each $A_i$ is a nowhere dense subset of $\mathbb{R}$.

Each singleton set $\{ r \}$ where $r$ is any rational number is also a closed and nowhere dense subset of $\mathbb{R}$. This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus $\mathbb{P}$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.C

The Sorgenfrey line $S$ is a Lindelof space whose square $S \times S$ is not normal. This is a famous example of a Lindelof space whose square is not Lindelof (not even normal). For reference, a proof is found here. An alternative proof of the non-normality of $S \times S$ uses the Baire category theorem and is found here.

If the Sorgenfrey line is $\sigma$-compact, then $S \times S$ would be $\sigma$-compact and hence Lindelof. Thus $S$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.D

Suppose that $X$ is Lindelof and that $Y$ is compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\{ x \} \times Y$. Putting it another way, $\{ x \} \times Y \subset \cup \mathcal{U}_x$. By the Tube lemma, for each $x \in X$, there is an open $O_x$ such that $\{ x \} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is Lindelof, there exists a countable set $\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $\{ O_{x_1},O_{x_2},O_{x_3},\cdots \}$ is a cover of $X$. Then $\mathcal{U}_{x_1} \cup \mathcal{U}_{x_2} \cup \mathcal{U}_{x_3} \cup \cdots$ is a countable subcover of $\mathcal{U}$. This completes the proof that $X \times Y$ is Lindelof when $X$ is Lindelof and $Y$ is compact.

To complete the exercise, observe that if $X$ is Lindelof and $Y$ is $\sigma$-compact, then $X \times Y$ is the union of countably many Lindelof subspaces.

Proofs of Key Steps for Exercise 2.E

Proof of Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that \$latex $\overline{O_{n+1}} \subset O_n$ for each $n$. Show that the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Suppose that $\bigcap_{i=1}^\infty O_i=\varnothing$. Choose $x_1 \in O_1$. There must exist some $n_1$ such that $x_1 \notin O_{n_1}$. Choose $x_2 \in O_{n_1}$. There must exist some $n_2>n_1$ such that $x_2 \notin O_{n_2}$. Continue in this manner we can choose inductively an infinite set $A=\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $x_i \ne x_j$ for $i \ne j$. Since $X$ is compact, the infinite set $A$ has a limit point $p$. This means that every open set containing $p$ contains some $x_j$ (in fact for infinitely many $j$). The point $p$ cannot be in the intersection $\bigcap_{i=1}^\infty O_i$. Thus for some $n$, $p \notin O_n$. Thus $p \notin \overline{O_{n+1}}$. We can choose an open set $U$ such that $p \in U$ and $U \cap \overline{O_{n+1}}=\varnothing$. However, $U$ must contain some point $x_j$ where $j>n+1$. This is a contradiction since $O_j \subset \overline{O_{n+1}}$ for all $j>n+1$. Thus Fact E.1 is established.

Proof of Fact E.2
Let $X$ be a compact space. Let $U_1,U_2,U_3,\cdots$ be open subsets of $X$ such that each $U_i$ is also a dense subset of $X$. Let $V$ a non-empty open subset of $X$. We wish to show that $V$ contains a point that belongs to each $U_i$. Since $U_1$ is dense in $X$, $O_1=V \cap U_1$ is non-empty. Since $U_2$ is dense in $X$, choose non-empty open $O_2$ such that $\overline{O_2} \subset O_1$ and $O_2 \subset U_2$. Since $U_3$ is dense in $X$, choose non-empty open $O_3$ such that $\overline{O_3} \subset O_2$ and $O_3 \subset U_3$. Continue inductively in this manner and we have a sequence of open sets $O_1,O_2,O_3,\cdots$ just like in Fact E.1. Then the intersection of the open sets $O_n$ is non-empty. Points in the intersection are in $V$ and in all the $U_n$. This completes the proof of Fact E.2.

Proof of Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Show that $Y$ is not a $\sigma$-compact space.

Suppose $Y$ is $\sigma$-compact. Let $Y=\bigcup_{n=1}^\infty B_n$ where each $B_n$ is compact. Each $B_n$ is obviously a closed subset of $X$. We claim that each $B_n$ is a closed nowhere dense subset of $X$. To see this, let $U$ be a non-empty open subset of $X$. Since $X-Y$ is dense in $X$, $U$ contains a point $p$ where $p \notin Y$. Since $p \notin B_n$, there exists a non-empty open $V \subset U$ such that $V \cap B_n=\varnothing$. This shows that each $B_n$ is a nowhere dense subset of $X$.

Since $Y$ is a dense $G_\delta$-subset of $X$, $Y=\bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open and dense subset of $X$. Then each $A_n=X-O_n$ is a closed nowhere dense subset of $X$. This means that $X$ is the union of countably many closed and nowhere dense subsets of $X$. More specifically, we have the following.

(1)………$X= \biggl( \bigcup_{n=1}^\infty A_n \biggr) \cup \biggl( \bigcup_{n=1}^\infty B_n \biggr)$

Statement (1) contradicts the fact that $X$ is a Baire space. Note that all $X-A_n$ and $X-B_n$ are open and dense subsets of $X$. Further note that the intersection of all these countably many open and dense subsets of $X$ is empty according to (1). Thus $Y$ cannot not a $\sigma$-compact space.

Proof of Fact E.4
The space $X=[0,1]^\omega$ is compact since it is a product of compact spaces. To see that $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X$, note that $Y=\bigcap_{n=1}^\infty U_n$ where for each integer $n \ge 1$

(2)………$U_n=(0,1) \times \cdots \times (0,1) \times [0,1] \times [0,1] \times \cdots$

Note that the first $n$ factors of $U_n$ are the open interval $(0,1)$ and the remaining factors are the closed interval $[0,1]$. It is also clear that $X-Y$ is a dense subset of $X$. This completes the proof of Fact E.4.

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Concluding Remarks

Exercise 2.A
The exercise is to show that the product of uncountably many $\sigma$-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that $\mathbb{R}^{c}$ has uncountable extent where $c$ is continuum. Having uncountable extent (i.e. having an uncountable subset that is both closed and discrete) implies the space is not Lindelof. The uncountable extent of the product space $\mathbb{R}^{\omega_1}$ is discussed in this post.

For $\mathbb{R}^{c}$ and $\mathbb{R}^{\omega_1}$, there is another way to show non-Lindelof. For example, both product spaces are not normal. As a result, both product spaces cannot be Lindelof. Note that every regular Lindelof space is normal. Both product spaces contain the product $\omega^{\omega_1}$ as a closed subspace. The non-normality of $\omega^{\omega_1}$ is discussed here.

Exercise 2.B
The hints given above is to show that the set of all irrational numbers, $\mathbb{P}$, is not $\sigma$-compact (as a subspace of the real line). The same argument showing that $\mathbb{P}$ is not $\sigma$-compact can be generalized. Note that the complement of $\mathbb{P}$ is $\mathbb{Q}$, the set of all rational numbers (a countable set). In this case, $\mathbb{Q}$ is a dense subset of the real line and is the union of countably many singleton sets. Each singleton set is a closed and nowhere dense subset of the real line. In general, we can let $B$, the complement of a set $A$, be dense in the real line and be the union of countably many closed nowhere dense subsets of the real line (not necessarily singleton sets). The same argument will show that $A$ cannot be a $\sigma$-compact space. This argument is captured in Fact E.3 in Exercise 2.E. Thus both Exercise 2.B and Exercise 2.E use a Baire category argument.

Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not $\sigma$-compact. In this case, the suggested space is $\mathbb{R}^{\omega}$, the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product space $\mathbb{R}^{\omega}$ is embedded in the compact space $[0,1]^{\omega}$, which is a Baire space. As mentioned earlier, Fact E.3 is essentially the same argument used for Exercise 2.B.

Using the same Baire category argument, it can be shown that $\omega^{\omega}$, the product of countably many copies of the countably infinite discrete space, is not $\sigma$-compact. The space $\omega$ of the non-negative integers, as a subspace of the real line, is certainly $\sigma$-compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not $\sigma$-compact. With the product space $\omega^{\omega}$, there is a connection with Exercise 2.B. The product $\omega^{\omega}$ is homeomorphic to $\mathbb{P}$. The idea of the homeomorphism is discussed here. Thus the non-$\sigma$-compactness of $\omega^{\omega}$ can be achieved by mapping it to the irrationals. Of course, the same Baire category argument runs through both exercises.

Exercise 2.C
Even the non-$\sigma$-compactness of the Sorgenfrey line $S$ can be achieved by a Baire category argument. The non-normality of the Sorgenfrey plane $S \times S$ can be achieved by Jones’ lemma argument or by the fact that $\mathbb{P}$ is not a first category set. Links to both arguments are given in the Proof section above.

See here for another introduction to the Baire category theorem.

The Tube lemma is discussed here.

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Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2019 – Dan Ma

# The Michael Line and the Continuum Hypothesis

There exist a Lindelof space and a separable metric space such that their Cartesian product is not normal (discussed in the post “Bernstein Sets and the Michael Line”). The separable metric space is a Bernstein set, a subspace of the real line that is far from being a complete metric space. However, this example is constructed without using any additional set theory axiom beyond the Zermelo-Fraenkel axioms plus the axiom of choice (abbreviated ZFC). A natural question is whether there exists a Lindelof space and a complete metric space such that their product is not normal. In particular, does there exist a Lindelof space $L$ such that the product of $L$ with the space of all irrational numbers is not normal? As of the writing of this post, it is still unknown that such a Lindelof space can exist in just ZFC alone without applying additional set theory axiom. However, such a Lindelof space can be constructed from various additional axioms (e.g. continuum hypothesis or Martin’s axiom). In this post, we present an example of such construction using the continuum hypothesis (the statement that the cardinality of the real line is the same as the first uncountable cardinal $\aleph_1$).

Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace $L$ of $\mathbb{M}$ such that $L$ contains the set $\mathbb{Q}$ of rational numbers. The same proof that $\mathbb{M} \times \mathbb{P}$ is not normal will show that $L \times \mathbb{P}$ is not normal.

See the following posts for a basic discussion of the Michael line:

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Luzin Sets

The Lindelof space $X$ we want to find is a subset of the real line that is called a Luzin set. Before defining Luzin sets, recall some definitions. Let $Y$ be a space. Let $A \subset Y$. The set $A$ is said to be nowhere dense in $Y$ if for every non-empty open subset $U$ of $Y$, there is a non-empty open subset $V$ of $Y$ such that $V \subset U$ and $V$ misses $A$ (equivalently, the closure of $A$ has no interior). The set $A$ is of first category in $Y$ if it is the union of countably many nowhere dense sets.

To define Luzin sets, we focus on the Euclidean space $\mathbb{R}$. Let $A \subset \mathbb{R}$. The set $A$ is said to be a Luzin set if for every set $W \subset \mathbb{R}$ that is of first category in the real line, $A \cap W$ is at most countable. The Russian mathematician Luzin in 1914 constructed such an uncountable Luzin set using continuum hypothesis (CH). A good reference for Luzin sets is [4]. We have the following theorem.

Theorem 1
Assume CH. There exists an uncountable Luzin set.

Proof of Theorem 1
There are continuum many closed nowhere dense subsets of the real line. Since we assume the continuum hypothesis, we can enumerate these sets in a sequence of length $\omega_1$. Let $\left\{F_\alpha: \alpha < \omega_1 \right\}$ be the set of all closed nowhere dense sets in the real line. Choose a real number $x_0 \notin F_0$ to start. For each $\alpha$ with $0 < \alpha <\omega_1$, choose a real number $x_\alpha$ not in the following set:

$\left\{x_\beta: \beta<\alpha \right\} \cup \bigcup \limits_{\beta<\alpha} F_\beta$

The above set is a countable union of closed nowhere dense sets of the real line. As a complete metric space, the real line cannot be of first category. In fact, according to the Baire category theorem, the complement of a set of first category (such as the one described above) is dense in the real line. So such an $x_\alpha$ can always be selected at each $\alpha<\omega_1$. Then $X=\left\{x_\alpha: \alpha<\omega_1 \right\}$ is a Luzin set. $\blacksquare$

Now that we have a way of constructing an uncountable Luzin sets, the following observations provide some useful facts for our problem at hand.

Nowhere dense sets and sets of first category are "thin" sets. Any "thin" set can intersect with a Luzin set with only countably many points. Thus any "co-thin" set contains all but countably many points of a Luzin set. For example, let $A$ be an uncountable Luzin set. Then if $F$ is a closed nowhere dense set in the real line, then $\mathbb{R}-F$ contains all but countably many points of $A$. Furthermore, if $F_1,F_2,F_3,\cdots,$ are closed nowhere dense subsets of the real line, then $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ contains all but countably many points of the Luzin set $A$.

Note that the set $\mathbb{R}-F$ in the preceding paragraph is a dense open set. Thus the complement of a closed nowhere dense set is a dense open set. Note that the set $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ in the preceding paragraph is a dense $G_\delta$-set. Thus the complement of the union of countably many closed nowhere dense sets is a dense $G_\delta$-set. Thus the observation in the preceding paragraph gives the following proposition:

Proposition 2
Given an uncountable Luzin set $A$ and given a dense $G_\delta$ subset $H$ of the real line, $H$ contains all but countably many points of $A$.

In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.

Proposition 3
Let $A$ be an uncountable Luzin set. Let $Y \subset \mathbb{R}$ be uncountable and dense in the real line such that $A \cap Y$ is uncountable. Given a dense $G_\delta$ subset $H$ of $Y$, $H$ contains all but countably many points of $A \cap Y$.

Proof of Proposition 3
We want to show that $Y-H$ can only contain countably many points of $A$. Let $H=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is open and dense in $Y$. Then for each $i$, let $U_i$ be open in the real line such that $U_i \cap Y=O_i$. Each $U_i$ is open and dense in the real line. Thus $H^*=\bigcap \limits_{i=1}^\infty U_i$ contains all but countably many points of the Luzin set $A$. Note the following set inclusion:

$H=\bigcap \limits_{i=1}^\infty U_i \cap Y=\bigcap \limits_{i=1}^\infty O_i \subset \bigcap \limits_{i=1}^\infty U_i=H^*$

Suppose that $Y-H$ contains uncountably many points of $A$. Then these points, except for countably many points, must belong to $H^*=\bigcap \limits_{i=1}^\infty U_i$. The above set inclusion shows that these points must belong to $H$ too, a contradiction. Thus $Y-H$ can only contain countably many points of $A$, equivalently the $G_\delta$-set $H$ contains all but countably many points of $A \cap Y$. $\blacksquare$

The following proposition follows from Proposition 3 and is a useful fact that will help us see that the product of an uncountable Luzin set and $\mathbb{P}$ is not normal.

Proposition 4
Let $Y$ be an uncountable Luzin set such that $\mathbb{Q} \subset Y$. Then $Y-\mathbb{Q}$ cannot be an $F_\sigma$-set in the Euclidean space $Y$, equivalently $\mathbb{Q}$ cannot be a $G_\delta$-set in the space $Y$.

Proof of Proposition 4
By Proposition 3, any dense $G_\delta$-subset of $Y$ must be co-countable. $\blacksquare$

The following proposition is another useful observation about Luzin sets. Let $A \subset \mathbb{R}$. Let $D \subset \mathbb{R}$ be a countable dense subset of the real line. The set $A$ is said to be concentrated about $D$ if for every open subset $O$ of the real line such that $D \subset O$, $O$ contains all but countably many points of $A$. The following proposition can be readily checked based on the definition of Luzin sets.

Proposition 5
For any $A \subset \mathbb{R}$, $A$ is a Luzin set if and only if $A$ is concentrated about every countable dense subset of the real line.

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Lindelof Subspace of The Michael Line

Let $A$ be an uncountable Luzin set. We can assume that $A$ is dense in the real line. If not, just add a countble subset of $\mathbb{P}$ that is dense in the real line. Let $L=A \cup \mathbb{Q}$. It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus $L$ is also a Luzin set. Now consider $L$ as a subspace of the Michael line $\mathbb{M}$. Then points of $L-\mathbb{Q}$ are discrete and points in $\mathbb{Q}$ have Euclidean open neighborhoods. By Proposition 5, the set $L$ is concentrated about every countable dense subset of the real line. In particular, it is concentrated about $\mathbb{Q}$. Thus as a subspace of the Michael line, $L$ is a Lindelof space, since every open set containing $\mathbb{Q}$ contains all but countably many points of $L$.

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The Non-Normal Product $L \times \mathbb{P}$

We highlight the following two facts about the Luzin set $L=A \cup \mathbb{Q}$ as discussed in the preceding section.

• $L-\mathbb{Q}$ is not an $F_\sigma$-set in $L$ (as Euclidean space).
• $A=L-\mathbb{Q}$ is dense in the real line.

The first bullet point follows from Proposition 4. The second bullet point is clear since we assume the Luzin set $A$ we start with is dense. Recall that when thinking of $L$ as a subspace of the Michael line, $L-\mathbb{Q}$ are isolated and $\mathbb{Q}$ retains the usual real line open sets. Because of the above two bullet points, $L \times \mathbb{P}$ is not normal. The proof that $L \times \mathbb{P}$ is not normal is the corollary of the proof that $\mathbb{M} \times \mathbb{P}$ is not normal. Note that in the proof for showing $\mathbb{M} \times \mathbb{P}$ is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an $F_\sigma$-set and are dense in the real line (found in “Michael Line Basics”).

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Michael Space

The example $L \times \mathbb{P}$ that we construct here was hinted in footnote 4 in [6]. In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in [5]). That construction should produce a similar set as the Luzin sets since the approach in [5] is a mirror image of the Luzin set construction. The approach in the Luzin set construction in Theorem 1 is to pick points not in the union of countably many closed nowhere dense sets, while the approach in [5] was to pick points in dense $G_\delta$-sets in a transfinite induction process.

A Michael space is a Lindelof space whose product with $\mathbb{P}$ is not normal. The example shown here shows that under CH, there exists a Michael space. However, the question of whether there exists a Michael space in ZFC is still unsolved. This is called the Michael problem. A recent mention of this unsolved problem is [3] (page 160). A Michael space can also be constructed using Martin’s axiom (see [1]).

A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is $\mathbb{P}$ a productively Lindelof space? As we see here, under CH the answer is no. Another way of looking at the Michael problem: is it possible to show that $\mathbb{P}$ is not productively Lindelof in ZFC alone?

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Reference

1. Alster, K., The product of a Lindelof space with the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., 110 (1990) 543-547.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
5. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
6. Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
7. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# 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. 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}. 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}. 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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$\copyright \ \ 2012$

# A Characterization of Baire Spaces

We present a useful characterization of Baire spaces. A Baire space is a topological space $X$ in which the conclusion of Baire category theorem holds, that is, for each countable family $\left\{U_1,U_2,U_3,\cdots \right\}$ of open and dense subsets of $X$, the intersection $\bigcap \limits_{n=1}^\infty U_n$ is dense in $X$. This definition is equivalent to the statement that every non-empty open subset of $X$ is of second category in $X$. An elementary discussion of Baire spaces is found in this blog post. Baire spaces can also be characterized in terms of the Banach-Mazur game (see Theorem 1 in this post). We add one more characterization in terms of point-finite open cover and locally finite open family (from [2] and [3]). We prove the following theorem.

A collection $\mathcal{S}$ of subsets of a space $X$ is said to be point-finite if every point in the space $X$ belongs to at most finitely many members of $\mathcal{S}$. A collection $\mathcal{S}$ of subsets of $X$ is said to be locally finite at the point $x \in X$ if there is an open set $V \subset X$ such that $x \in V$ and $V$ meets at most finitely many members of $\mathcal{S}$. The collection $\mathcal{S}$ is said to be locally finite in the space $X$ if it is locally finite at every $x \in X$. For any terms and concepts not explicitly defined here, refer to [1] (Engelking) or [4]) (Willard).

Theorem
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a Baire space.
2. For any point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.
3. For any countable point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.

Proof
$1 \Rightarrow 2$
Let $\mathcal{U}$ be a point-finite open cover of $X$. Let $O$ be a non-empty open subset of $X$. We wish to show that $O \cap D \ne \varnothing$ where $D$ is the set defined in condition 2. For each $n$, define

$\displaystyle . \ \ \ \ \ F_n=\left\{x \in O: x \text{ belongs to exactly n members of } \mathcal{U} \right\}$.

Note that $O=\bigcup \limits_{n=1}^\infty F_n$. Since $X$ is a Baire space, $O$ must be of second category in $X$. None of the sets $F_n$ can be a nowhere dense set. Thus for some $n$, $F_n$ has non-empty interior. Choose some non-empty open set $W$ such that $W \subset F_n$.

Pick $y \in W$. Since $y \in F_n$, let $U_1,U_2,\cdots,U_n$ be the $n$ members of $\mathcal{U}$ that contain $y$. Let $V=W \cap \bigcap \limits_{j=1}^n U_n$. Note that $V \subset W \subset F_n \subset O$. Observe that $V$ is a non-empty open set that meets exactly $n$ members of $\mathcal{U}$. Therefore $\mathcal{U}$ is locally finite at points of $V$, leading to the conclusion that $V \subset D$ and $O \cap D \ne \varnothing$.

The direction $2 \Rightarrow 3$ is immediate.

$3 \Rightarrow 1$
Suppose condition 3 holds. We claim that $X$ is a Baire space. Suppose not. Let $U$ be a non-empty open subset of $X$ such that $U=\bigcup \limits_{n=1}^\infty K_n$ where each $K_n$ is nowhere dense in $X$. Let $\mathcal{U}$ be defined as the following:

$\displaystyle . \ \ \ \ \ \mathcal{U}=\left\{X \right\} \cup \left\{U_n: n=1,2,3,\cdots\right\}$,

where $U_n=U - (\overline{K_1} \cup \cdots \cup \overline{K_n})$. Clearly, $\mathcal{U}$ is a point-finite open cover of $X$. By condition 3, $D$ is dense in $X$ ($D$ is defined in condition 3). In particular, $U \cap D \ne \varnothing$. Choose $y \in U \cap D$. Since $\mathcal{U}$ is locally finite at $y$, we can choose some open set $V \subset U$ such that $y \in V$ and such that $V$ meets only finitely many $U_j$, say only up to $U_1,\cdots, U_m$ (so $V \cap U_j = \varnothing$ for all $j > m$).

On the other hand, all sets $K_j$ are nowhere dense. So we can choose some open set $V_0 \subset V$ such that $V_0$ misses the nowhere dense set $\overline{K_1} \cup \cdots \cup \overline{K_m} \cup \overline{K_{m+1}}$. In particular, this means that $V_0 \cap U_{m+1} \ne \varnothing$, contradicting that $V \cap U_j = \varnothing$ for all $j > m$. So $X$ must be a Baire space if condition 3 holds. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Fletcher, P., Lindgren, W. F., A note on spaces of second category, Arch, Math., 24, 186-187, 1973.
3. McCoy, R. A., A Baire space extension, Proc. Amer. Math. Soc., 33, 199-202, 1972.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Baire Category Theorem and the Finite Intersection Property

A Baire space is a topological space in which the intersection of any countable family of open and dense sets is dense (equivalently every non-empty open subset is of second category). One version of the Baire category theorem states that every complete metric space is a Baire space. Another common version states that every compact Hausdorff space is a Baire space. Another version states that every locally compact Hausdorff space is a Baire space. The commonality among these versions is the finite intersection property (whenever a collection of a certain type of sets satisfies the property that any finite subcollection has non-empty intersection, the whole collection has non-empty intersection). For each of these classes of spaces, in addition to countably compact spaces and pseudocompact spaces, Baire category theorem is derived from having one specific form of the finite intersection property. In this post, we explore this relationship.

In each of the following theorem pairs, the B Theorem follows from the A theorem. The A theorem is a form of the finite intersection property and the B theorem is a version of Baire category theorem.

Another interesting observation is that the finite intersection properties discussed here can give a stronger property than being a Baire space. This stronger property is defined by the Banach-Mazur game.

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Theorem 1A
Let $(X, \rho)$ be a metric space. Then the following conditions are equivalent.

1. $(X, \rho)$ is a complete metric space.
2. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ such that the diameters of the sets $C_n$ converge to zero, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 1B
Every complete metric space is a Baire space.

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Theorem 2A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a compact space.
2. For every family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.

Theorem 2B

• Every compact Hausdorff space is a Baire space.
• Every locally compact Hausdorff space is a Baire space.

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Theorem 3A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a countably compact space.
2. For every countable family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.
3. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 3B
Every countably compact Hausdorff space is a Baire space.

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Theorem 4A
Let $X$ be a regular space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$, then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$.
3. If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Theorem 4B
Every regular pseudocompact space is a Baire space.

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Remark
Theorem 1A (the Cantor Theorem) can be found in Engelking (page 269 in [1]). Theorem 2A and Theorem 3A can also be found in Engelking (they are also proved in this post). Theorem 4B is also found in Engelking (Theorem 3.10.23 in page 207 of [1]) and is proved this post.

We would like to explicitly point out that between Thoerem 1A and Theorem 2A, none of the two theorems implies the other. For example, even though both complete metric spaces and compact Hausdorff spaces are Baire spaces, complete metric spaces are not necessarily compact and there are compact spaces that are not even metrizable. However, the finite intersection property of Theorem 2A implies that of Theorem 3A, which in turn implies the finite intersection property of Theorem 4A.
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Baire Category Theorem

The proofs of all four B theorems are amazingly similar. It is a matter of exploiting the fact that whenever a decreasing sequence of open sets satisfying the condition that each closure is a subset of the previous open set (and satisfying some other condition), the sequence of open sets has non-empty intersection. For example, for complete metric space, make sure that the closures of the open sets have diameters going to zero. For any reader who is new to this material, it will be very instructive to walk through the arguments of these Baire category theorems. The proof of Theorem 1A can be found this post. We prove Theorem 4B.

Recall that $X$ is a Baire space if $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable family of open and dense sets in $X$, $\bigcap \limits_{i=1}^\infty U_i$ is dense in $X$, or equivalently every non-empty open subset of $X$ is of second category in $X$. For more background about the concepts of Baire space and category (see [1] or this post).

Proof of Theorem 4B
Let $X$ be a regular pseudocompact space. Let $\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable family of open and dense sets in $X$. Let $O$ be a non-empty open subset of $X$. We show that $O$ has to contain points of $\bigcap \limits_{n=1}^\infty U_n$. We let $O_1=O \cap U_1$. We find open $O_2$ such that $O_2 \subset U_2$ and $\overline{O_2} \subset O_1$ (using regularity). Continue this inductive process, we have for each $n$, an open $O_n$ such that $O_n \subset U_n$ and $\overline{O_n} \subset O_{n-1}$. Then we have a decreasing sequence of open sets $O_n$ as in condition 2 of Theorem 4A. Then we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$. Since $\overline{O_{n+1}} \subset O_n$ for each $n$, we also have $\bigcap \limits_{n=1}^\infty O_n \ne \varnothing$. It is clear that $\bigcap \limits_{n=1}^\infty O_n \subset \bigcap \limits_{n=1}^\infty U_n$. $\blacksquare$

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Banach-Mazur Game

In proving the above versions of Baire category theorem, we can exploit the appropriate version of the finite intersection property – the situation that any nested decreasing sequence of open sets (under some specified conditions) has non-empty intersection. In fact, the finite intersection property offers more than just Baire category theorem; it can endow the space in question a type of completeness property stronger than Baire space. This completeness property is defined using 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. For more detailed discussion of the game, see this post.

One interesting point that we like to make about the finite intersection property ranging from Theorem 1A to Theorem 4A is that the player $\alpha$ can always win the Banach-Mazur game as long as he/she plays the game according to each specific version of the finite intersection. For example, playing the game in a complete metric space, player $\alpha$ always wins as long as he/she makes the diameters of the closures of the open sets going to zero. In a regular pseudocompact space, player $\alpha$ can always win by making the closure of each of his/her open sets a subset of the previous move of other player.

A topological space in which the player $\alpha$ has a winning strategy is said to be a weakly $\alpha$-favorable space. Thus complete metric spaces, compact Hausdorff spaces, locally compact Hausdorff spaces, countably compact Hausdorff spaces, regular pseudocompact spaces are all weakly $\alpha$-favorable.

There is characterization of Baire spaces in terms of the Banach-Mazur game. A space $X$ is a Baire space if and only if the player $\beta$ has no winning strategy in the Banach-Mazur game played on the space $X$ (see theorem 1 in this post). If the player $\alpha$ can always win, then player $\beta$ can never win. In terms of game terminology, if player $\alpha$ has a winning strategy, then the other player (player $\beta$) has no winning strategy. Thus a space is weakly $\alpha$-favorable implies that it is a Baire space. But the implication is not reversible (see example in this post).

So all the spaces discussed from Theorem 1A to Theorem 4A are all weakly $\alpha$-favorable, a property stronger than Baire spaces. These observations are summarized in the following theorems.

Theorem 1C
Every complete metric space is a weakly $\alpha$-favorable space.

Theorem 2C

• Every compact Hausdorff space is a weakly $\alpha$-favorable space.
• Every locally compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 3C
Every countably compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 4C
Every regular pseudocompact space is a weakly $\alpha$-favorable space.

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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.

# 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.

Let $B=\left\{ x_\alpha: \alpha. 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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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.

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 $Y$.

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

# Michael Line x Irrationals Is Not Normal

The original post was about a proof that the product of the Michael line and the space of irrational numbers is not normal. It was written back in Oct 2009 and is now replaced by a newer post (see the following link).

“Michael Line Basics”