# 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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$\copyright \ \ 2012$

# 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 $Y$.

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

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

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

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

Reference

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

# Thinking about the Space of Irrationals Topologically

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

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

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

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

___________________________________________________________________________

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

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

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

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

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

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

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Remark

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

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

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

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

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

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

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

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

# An observation about sequential spaces

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

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

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

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

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

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

Reference

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

# Sequential spaces, III

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Reference

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