# k-spaces, I

There are many examples in general topology of defining a new topology on a set based on a given topology already defined on the set. For example, given a topology $\tau$ on $X$, one can define a finer topology $\tau_s$ consisting of all sequentially open subsets of $X$ (based on the original topology $\tau$). Sequential spaces are precisely those spaces for which the original topology coincides with $\tau_s$ (see the post Sequential spaces, II). A related concept is the notion of k-spaces. We show that the compactly generated open sets form a finer topology and that k-spaces are precisely those spaces for which the compactly generated topology coincides with the original topology. We also give an external characterization of k-spaces, namely, those spaces that are quotient images of locally compact spaces. All spaces under consideration are Hausdorff.

Let $X$ be a space. We say $A \subset X$ is a compactly generated closed set in $X$ if $A \cap K$ is closed in $K$ for any compact $K \subset X$. We say $B \subset X$ is a compactly generated open set in $X$ if $X-B$ is a compactly generated closed set in $X$. The space $X$ is said to be a k-space if

$A \subset X$ is a compactly generated closed set in $X$ if and only if $A$ is a closed set.

The direction $\Leftarrow$ of the above statement always holds. So a space is a k-space if it satisfies the direction $\Rightarrow$ in the above statement. We also want to mention that in the above definition, “closed” can be replaced by “open”.

Suppose $\tau$ is the topology of the space $X$. Then define $\tau_k$ as the set of all compactly generated open sets in $X$. It can be easily verified that $\tau_k$ is a topology defined on the set $X$ and that $\tau_k$ is a finer topology than the original topology $\tau$, i.e. $\tau \subset \tau_k$. It follows that $X$ is a k-space if and only if $\tau=\tau_k$.

Much of the discussion here mirrors the one in Sequential spaces, II. In a sequential space, the topology coincides with $\tau_s$, the open sets generated by convergent sequences (a particular type of compact sets). In a k-space, the topology conincides with $\tau_k$, the open sets generated by the compact sets. A sequential space is the quotient space of a topological sum of disjoint convergent sequences. Any k-space is the quotient space of a topological sum of disjoint compact sets.

We have the following theorem.

Theorem
For any space $X$, the following conditions are equivalent:

1. $X$ is a k-space.
2. $X$ is a quotient space of a locally compact space.

Proof. $1 \Rightarrow 2$ Let $X$ be a k-space. Let $\mathcal{K}$ be the set of all compact subsets of $X$. Let $Y=\oplus_{K \in \mathcal{K}}K$ be the topological sum of all $K \in \mathcal{K}$ where each $K \in \mathcal{K}$ has the relative topology inherited from the space $X$. Then $Y$ is a locally compact space. There is a natural mapping we can define on $Y$ onto $X$. The space $Y$ is a disjoint union of all compact subsets of $X$. We can map each compact set $K \in \mathcal{K}$ onto the corresponding compact subset $K$ of $X$ by the identity map. Call this mapping $f$ where $f:Y \mapsto X$. We claim that the quotient topology generated by this mapping coincides with the original topology on $X$.

Let $\tau$ be the given topology on $X$ and let $\tau_f$ be the quotient topology generated on the set $X$. Clearly, $\tau \subset \tau_f$. We need to show that $\tau_f \subset \tau$. Let $O \in \tau_f$. Since $X$ is a k-space, if we can show that $O$ is a compactly generated open set, then $O \in \tau$.

Let $K \subset X$ be compact. We need to show that $O \cap K$ is open in $K$. Since $O \in \tau_f$, $f^{-1}(O)$ is open in $Y$. Since $K$ is open in $Y$, $f^{-1}(O) \cap K$ is open in $Y$. It is also the case that $f^{-1}(O) \cap K$ is open in $K$. We can consider $f^{-1}(O) \cap K$ as a subset of $K \subset Y$ and as a subset of $K \subset X$. As a subset of $K \subset X$, we have $f^{-1}(O) \cap K=O \cap K$. Thus $O \cap K$ is open in $K$ and $O \in \tau$.

$2 \Rightarrow 1$ Let $Y$ be locally compact and let $f:Y \mapsto X$ be a quotient map. We show that $X$ is a k-space. To this end, we show that if $A \subset X$ is a compactly generated closed set in $X$, $A$ is closed in $X$. Or equivalently, if $A$ is not closed in $X$, then $A$ is not a compactly generated closed set in $X$. Under the quotient map $f$, $A$ is closed in $X$ if and only if $f^{-1}(A)$ is closed in $Y$.

Suppose $A$ is not closed in $X$. Then $f^{-1}(A)$ is not closed in $Y$. Then there is $y \in \overline{f^{-1}(A)}-f^{-1}(A)$. Let $U \subset Y$ be open in $Y$ such that $y \in U$ and $\overline{U}$ is compact. Then $f(\overline{U})$ is compact. It follows that $A \cap f(\overline{U})$ is not closed in $f(\overline{U})$. Note that $f(y) \in f(\overline{U})$ and $f(y) \notin A \cap f(\overline{U})$. However, $f(y) \in \overline{A \cap f(\overline{U})}$. Thus $A$ is not a compactly generated closed set in $X$. $\blacksquare$

Reference

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

# Sequential spaces, II

Sequential spaces are topological spaces in which convergent sequences are sufficient to define the topology. In any topological space $(X,\tau)$, the convergent sequences can generate a finer topology $\tau_s$. Sequential spaces are precisely those spaces whose topology coincide with $\tau_s$. We also present an external characterization of sequential spaces, namely those spaces that are quotient images of first countable spaces. This post is a continuation of Sequential spaces, I. All spaces under consideration are Hausdorff.

For any space $X$, $A \subset X$ is sequentially closed in $X$ if whenever $\left\{x_n\right\}_{n=1}^{\infty}$ is sequence of points of $A$ and $x_n \mapsto x$, then $x \in A$. In other words, $A$ contains all sequential limits of convergent sequences of points of $A$. A space $X$ is sequential if this condition holds: if $A \subset X$ is sequentially closed in $X$, then $A$ is closed in $X$.

Given a topological space $(X,\tau)$ where $\tau$ is the topology. Consider the following topology:

$\tau_s=\left\{U \subset X: X-U \text{ is sequentially closed in }X\right\}$

It is straightforward to verify that $\tau_s$ is a topology on the set $X$ and that it is finer than the original topology $\tau$, i.e. $\tau \subset \tau_s$. It can also be verified that $X$ is a sequential space if and only if $\tau=\tau_s$. In the previous post Sequential spaces, I, all the sequential examples are quotient images of first countable spaces. It turns out that sequential spaces are precisely those spaces that are quotient images of first countable spaces. We have the following theorem.

Theorem
For any space $X$, the following conditions are equivalent:

1. $X$ is a sequential space.
2. $X$ is a quotient space of a metric space.
3. $X$ is a quotient space of a first countable space.

Proof. We prove $1 \Rightarrow 2$. The direction of $2 \Rightarrow 3$ is clear. The direction $3 \Rightarrow 1$ following from Theorem 2 in Sequential spaces, I.

$1 \Rightarrow 2$ Let $X$ be a sequential space. Let $\mathcal{K}$ be the set of all compact subsets of $X$ that are of the form:

$K=\left\{x\right\} \cup \left\{x_n:n=1,2,3,\cdots\right\}$

where $x_n \mapsto x$. We call such sets compact sequences. We define a topology on each $K \in \mathcal{K}$. We let each point $x_n$ in the convergent sequence be isolated. The open neighborhood at the limit $x$ is made up of $x$ together with all but finitely many $x_n$. Note that the topology just described for each $K$ is the relative topology inherited from $X$.

Let $Y=\oplus_{K \in \mathcal{K}}K$ be the topological sum of all compact sequences in $\mathcal{K}$. This is simply the disjoint union of all the compact sequences $K$ with each $K$ having the relative topology inherited from $X$. The resulting space $Y$ is first countable as well as a metric space. Since each point in $Y$ belongs to some compact sequence $K \subset X$, we can define $f:Y \mapsto X$ such that $f$ maps each point in $Y$ to the corresponding point in the sequence in $X$. We claim that the quotient topology generated by this map coincides with the original topology on $X$.

Let $\tau$ be the given topology on $X$ and let $\tau_f$ be the quotient topology generated on the set $X$. Clearly, $\tau \subset \tau_f$. We need to show that $\tau_f \subset \tau$. Let $O \in \tau_f$. Since $X$ is sequential, if we can show that for any convergent sequence that converges to a point in $O$, all but finitely many terms of the sequence must be in $O$, then $O \in \tau$.

Let $x_n \in X$ such that $x_n \mapsto x \in O$. Let $K=\left\{x\right\} \cup \left\{x_1,x_2,\cdots\right\}$. We know that $f^{-1}(O)$ is open in $Y$ since $O \in \tau_f$. We know that with $x$ as a point in $K \subset Y$, $x \in f^{-1}(O) \cap K$. Since $f^{-1}(O) \cap K$ is open in $Y$, $f^{-1}(O)$ contains all but finitely many $x_n \in K$. Thus $f(x_n)=x_n \in O$ for all but finitely many $n$. This means $O \in \tau$. We just show that the quotient topology on $X$ coincides with the given topology. $\blacksquare$

Reference

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

# Sequential spaces, I

Any topological space where there is a countable base at every point is said to satisfy the first axiom of countability or to be first countable. In this post we discuss several properties weaker than the first axiom of countability. All spaces under consideration are Hausdorff. Any first countable space $X$ satisfies each of the following conditions:

1. If $A \subset X$ and $x \in \overline{A}$ then there is a sequence $\left\{x_n\right\}$ of points in $A$ such that the sequence converges to $x$.
2. The set $A \subset X$ is closed in $X$ if $A$ is sequentially closed in $X$, which means that: if $\left\{x_n\right\}$ is a sequence of points in $A$ such that $\left\{x_n\right\}$ converges to $x \in X$, then $x \in A$.
3. The set $A \subset X$ is closed in $X$ if this condition holds: if $K \subset X$ is compact, then $A \cap K$ is closed in $K$.

Spaces satisfying:

• condition 1 are called Frechet spaces,
• condition 2 are called sequential spaces,
• condition 3 are called k-spaces.

All three of these conditions hold in first countable spaces. We have the following implications:

First countable $\Rightarrow \$ Frechet $\Rightarrow \$ Sequential $\Rightarrow \$ k-space

This post is an introductory discussion of these notions. Each of the above implications is not reversible (see the section below on examples). After we discuss sequential spaces, we take a look at the behavior of these four classes of spaces in terms of whether the property can be passed onto subspaces (the property being hereditary) and in terms of images under quotient maps. For previous posts in this blog on first countable spaces and quotient spaces, see the links at the end of the post. Excellent texts on general topology are [1] and [2].

For a subsequent discussion on sequential space, see Sequential spaces, II.

The Forward Implications

First countable $\Rightarrow \$ Frechet
Suppose $x \in \overline{A}$ and $A \subset X$. Let $U_1,U_2,\cdots$ be a local base at $x$. Then choose $x_n \in A \cap U_n$ and we have $x_n \mapsto x$.

Frechet $\Rightarrow \$ Sequential
Let $A \subset X$ be sequentially closed in $X$. Suppose $A$ is not closed in $X$. Then there is $x \in \overline{A}$ such that $x \notin A$. By Frechet, there is a sequence $x_n \in A$ such that $x_n \mapsto x$. Since $A$ is sequentially closed, $x \in A$, a contradiction. So any sequentially closed set in a Frechet space must be a closed set.

Sequential $\Rightarrow \$ k-space
Suppose $A \subset X$ is not closed in $X$. Since $X$ is sequential, there is a sequence $x_n \in A$ such that $x_n \mapsto x$ and $x \notin A$. The set $K=\left\{x_n:n=1,2,3,\cdots\right\} \cup \left\{x\right\}$ is a compact set. Note that $A \cap K$ is not closed in $K$. This shows that $X$ is a k-space.

Sequential Spaces

Sequential spaces are ones in which the topology can be completely described by convergent sequences. Let $X$ be a space. Let $A \subset X$. The set $A$ is said to be sequentially closed in $X$ if whenever we have a convergent sequence of points in $A$, the sequential limit must be in $A$. In other words, $A$ contains all the limits of the convergent sequences of points in $A$. For $U \subset X$, $U$ is sequentially open in $X$ if this condition holds: if $\left\{x_n \in X: n=1,2,3,\cdots\right\}$ is a sequence of points converging to some $x \in U$, then $x_n \in U$ for all but finitely many $n$. It can be verified that:

$U$ is sequentially open in $X$ if and only if $X-U$ is sequentially closed in $X$.

Theorem 1
A space $X$ is Frechet if and only if every subspace of $X$ is sequential.

Proof
$\Rightarrow$ Suppose $W \subset X$ is not a sequential space. Then there is $A \subset W$ such that $A$ is sequentially closed in $W$ but $A$ is not closed in $W$. We show that $X$ is not Frechet. There is a point $w \in W$ such that $w$ is a limit point of $A$ (in the subspace $W$) and $w \notin A$. Since $A$ is sequentially closed, no sequence of points in $A$ can converge to $w$ (otherwise $w \in A$).

Clearly, the point $w$ is also a limit point of $A$ with respect to the toplology of $X$, i.e. $w \in \overline{A}$ with respect to $X$. Since no sequence of points in $A$ can converge to $w$, $X$ is not Frechet.

$\Leftarrow$ Suppose $X$ is not Frechet. Then there is $x \in \overline{A}$ such that $A \subset X$ and no sequence of points in $A$ can converge to $x$. Consider the subspace $Y=A \cup \left\{x\right\}$. The set $Y-\left\{x\right\}$ is sequentially closed in $Y$ but is not closed in $Y$. $\blacksquare$

Theorem 2
Every quotient space of a sequential space is always a sequential space.

Proof. Let $X$ be a sequential space. Let $f:X \mapsto Y$ be a quotient map. We show that $Y$ is a sequential space. Suppose that $B \subset Y$ is sequentially closed. We need to show that $B$ is closed in $Y$. Because $f$ is a quotient map, $B$ is closed in $Y$ if and only if $f^{-1}(B)$ is closed in $X$. So we need to show $f^{-1}(B)$ is closed in $X$.

Suppose $x_n \in f^{-1}(B)$ for each $n=1,2,3,\cdots$ and the sequence $x_n$ converges to $x \in X$. Then $f(x_n) \in B$. Since the map $f$ is continuous, $f(x_n) \mapsto f(x)$. Since $B$ is sequentially closed, $f(x) \in B$. This means $x \in B$. Thus $f^{-1}(B)$ is sequentially closed in $X$. Since $X$ is sequential, $f^{-1}(B)$ is closed in $X$. $\blacksquare$

Corollary 3
Every quotient space of a first countable space is sequential. Every quotient space of a Frechet space is sequential.

Some Examples

Example 1. First countable $\nLeftarrow \$ Frechet
The example is defined in the post An example of a quotient space, I. This is a non-first countable example. There is only one non-isolated point $p$ in the space. It is easy to verify it is a Frechet space.

Example 2. Frechet $\nLeftarrow \$ Sequential
We consider the space $Y$ defined in the post An example of a quotient space, II. Note that the space $Y$ is the quotient image of a first countable space. Thus $Y$ is sequential by Corollary 3. Consider the subspace $Z=\left\{(0,0)\right\} \cup V$. Within $Z$, no sequence of points in $V$ can converge to the point $(0,0)$. However, $(0,0)$ is a limit point of $V$. Thus $Z$ is not sequential. By Theorem 1, $Y$ is not Frechet.

Example 3. Sequential $\nLeftarrow \$ k-space
Any compact space is a k-space. Let $\omega_1$ be the first uncountable ordinal. Then $\omega_1+1=[0,\omega_1]$ with the ordered topology is compact. Note that $[0,\omega_1)$ is sequentially closed but not closed. Thus $[0,\omega_1]$ is not sequential.

Example 4. A space that is not a k-space
This example is also defined in the post An example of a quotient space, II. Consider the subspace $Z=\left\{(0,0)\right\} \cup V$. Every compact subset of $Z$ is finite. So $V \cap K$ is closed in $K$ for every compact $K \subset Z$. But $V$ is not closed.

Which of the four properties discussed here are preserved in subspaces? Or which of them are hereditary? It is fairly straightforward to verify that first countability is hereditary and so is the property of being Frechet. By Theorem 1, for any sequential space that is not Frechet has a subspace that is not sequential. Thus the property of being a sequential space is not hereditary. However, closed subspaces and open subspaces of a sequential space are sequential.

The property of being a k-space is also not hereditary. The space $Y$ defined in An example of a quotient space, II is a sequential space (thus a k-space). Yet the subspace $Z=\left\{(0,0)\right\} \cup V$ is not a k-space.

Continuous image of a first countable space needs not be first countable. The other three properties (Frechet, sequential and k-space) are also not necessarily preserved by continuous mappings. A quick example is to consider any space $X$ that does not have any one of the four properties. Then consider $D=X$ with the discrete topology. Then the indentity map from $D$ onto $X$ is continuous.

Example 1 shows that the property of being first countable is not preserved by quotient map. Example 2 shows that the Frechet property is not preserved by quotient map. Theorem 2 shows that the property of being sequential space is preserved by quotient map. We have the following theorem about k-spaces under quotient map.

k-spaces

The spaces that are k-spaces are called compactly generated spaces. In a k-space, the closed sets and open sets are generated by compact sets. For example, for a k-space $X$, $A \subset X$ is closed in $X$ if and only if $A \cap K$ is closed in $K$ for every compact $K \subset X$. Let’s take another look at sequential spaces. The following definition is equivalent to the definition of sequential space given above:

$A \subset X$ is closed in $X$ if and only if $A \cap K$ is closed in $K$ for every compact $K \subset X$ of the form $\left\{x\right\} \cup \left\{x_1,x_2,x_3,\cdots\right\}$ where the $x_n$ are a convergent sequence and $x$ is the sequential limit.

Thus the sequential spaces are compactly generated by a special type of compact sets, namely the convergent sequences.

Theorem 4
Quotient images of k-spaces are always k-spaces.

Proof. Let $X$ be a k-space. Let $f:X \mapsto Y$ be a quotient map. We wish to show that $Y$ is a k-space. Suppose $B \subset Y$ is not closed in $Y$. Since $f$ is a quotient mapping, $f^{-1}(B)$ is not closed in $X$. Since $X$ is a k-space, there is a compact $K \subset X$ such that $f^{-1}(B) \cap K$ is not closed in $K$. Let $x \in K$ such that $x \in \overline{f^{-1}(B) \cap K}-(f^{-1}(B) \cap K)$. We have just produced a compact set $f(K)$ in $Y$ such that $B \cap f(K)$ is not closed in $f(K)$. Note that $f(x) \in f(K)$ and $f(x)$ is a limit point of $B \cap f(K)$. This implies that if $B \cap C$ is closed in $C$ for every compact $C \subset Y$, then $B$ must be closed in $Y$ (i.e. $Y$ is a k-space). $\blacksquare$

The discussion on sequential space continues with the post Sequential spaces, II.

Reference

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

# The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets $A_\alpha$ where $\alpha \in S$. Let $W=\bigcup \limits_{\alpha \in S} A_\alpha$. The product $\prod \limits_{\alpha \in S} A_\alpha$ is the set of all functions $f:S \mapsto W$ such that for each $\alpha \in S$, $f(\alpha) \in A_\alpha$. If the index set $S=\left\{1,2,\cdots,n\right\}$ is finite, the functions $f$ can be regarded as sequences $(f_1,f_2,\cdots,f_n)$ where each $f_i \in A_i$. If the index set $S=\mathbb{N}$, we can think of elements $f$ of the product as the sequence $(f_1,f_2,\cdots)$ where each $f_i \in A_i$. In general we can regard $f \in \prod \limits_{\alpha \in S} A_\alpha$ as functions $f:S \mapsto W$ or as sequences $f=(f_\alpha)_{\alpha \in S}$.

Consider the topological spaces $X_\alpha$ where $\alpha \in S$. Let $X=\prod \limits_{\alpha \in S} X_\alpha$ be the product as defined above. The product space of the spaces $X_\alpha$ is $X$ with the topology defined in the following paragraph.

Let $\tau_\alpha$ be the topology of each space $X_\alpha$, $\alpha \in S$. Consider $Y=\prod \limits_{\alpha \in S} O_\alpha$ where for each $\alpha \in S$, $O_\alpha \in \tau_\alpha$ (i.e. $O_\alpha$ is open in $X_\alpha$) and $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in S$. The set of all such sets $Y$ is a base for a topology on the product $X=\prod \limits_{\alpha \in S} X_\alpha$. This topology is called the product topology of the spaces $X_\alpha$, $\alpha \in S$.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let $\mathcal{B}_\alpha$ be a base for the space $X_\alpha$. Consider $B=\prod \limits_{\alpha \in S} B_\alpha$ such that there is a finite set $F \subset S$ where $B_\alpha \in \mathcal{B}_\alpha$ for each $\alpha \in F$ and $B_\alpha=X_\alpha$ for all $\alpha \in S-F$. The set of all such sets $B$ is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each $\alpha \in S$, consider the map $\pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha$ such that $\pi_\alpha(f)=f_\alpha$ for each $f$ in the product. In words, the function $\pi_\alpha$ maps each point in the product space to its $\alpha^{th}$ coordinate. The mapping $\pi_\alpha$ is called the $\alpha^{th}$ projection map. For each set $U \subset X_\alpha$, $\pi_\alpha^{-1}(U)$ is the following set:

$\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}$

Consider sets of the form $\bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha)$ where $F \subset S$ is finite and $U_\alpha$ is open in $X_\alpha$ for each $\alpha \in F$. The set of all such sets is another equivalent base for the product topology. If we only require that each $U_\alpha \in \mathcal{B}_\alpha$, a predetermined base for the coordinate space $X_\alpha$, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For $i=1,2,3,\cdots$, let $X_i$ be a first countable space. We show that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Let $\mathbb{N}$ be the set of positive integers. For each $n \in \mathbb{N}$, let $[n]=\left\{1,2,\cdots,n\right\}$ and let $\mathbb{N}^{[n]}$ be the set of all functions $t:[n] \mapsto \mathbb{N}$. For each $i$ and each $x \in X_i$, let $\mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\}$ be a countable local base at $x$.

Let $f \in X=\prod \limits_{i=1}^{\infty}X_i$. We wish to define a countable local base at $f$. For each $n \in \mathbb{N}$, define $W_n$ to be:

$W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}$

Let $W$ be the set of all subsets of the product space $X$ of the following form:

$\prod \limits_{j=1}^{\infty}V_j$ where there is some $n \in \mathbb{N}$ such that $\prod \limits_{j=1}^{n}V_j \in W_n$ and for all $j>n$, $V_j=X_j$.

Each $W_n$ is countable and $W$ is essentially the union of all the $W_n$. Thus $W$ is countable. We claim that $W$ is a local base at $f$. Let $O \subset X$ be an open set containing $f$. We can assume that $O=\prod \limits_{i=1}^\infty O_i$ where there is some $n \in \mathbb{N}$ such that for each $i \le n$, $O_i$ is open in $X_i$ and for $i>n$, $O_i=X_i$.

For each $i \le n$, $f(i) \in O_i$. Choose some $B_{f(i)}(i,t(i))$ such that $f(i) \in B_{f(i)}(i,t(i)) \subset O_i$. Let $V=\prod \limits_{i=1}^{\infty} V_i$ such that $\prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i))$ and $V_i=X_i$ for all $i>n$. Then $V \in W$ and $f \in V \subset O$. This completes the proof that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Uncountable Product
Let $S$ be an uncountable index set. For $\alpha \in S$, let $X_\alpha$. We want to avoid the situation that all but countably many $X_\alpha$ are one-point space. So we assume each coordinate space $X_\alpha$ has at least two points, say, $p_\alpha$ and $q_\alpha$ with $p_\alpha \ne q_\alpha$. We show that $X=\prod \limits_{\alpha \in S}X_\alpha$ is not first countable.

Let $f \in \prod \limits_{\alpha \in S}X_\alpha$. Let $U_1,U_2, \cdots$ be open subsets of the product space such that for each $i$, $f \in U_i$. We show that there is some open set $O$ such that $f \in O$ and each $U_i \nsubseteq O$. For each $i$, there is a basic open set $B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i})$ such that $f \in B_i \subset U_i$.

Let $F=F_1 \cup F_2 \cup \cdots$. Since $S$ is uncountable and $F$ is countable, choose $\gamma \in S-F$. Since $X_\gamma$ has at least two points $p_\gamma$ and $q_\gamma$, choose one of them that is different from $f_\gamma$, say, $p_\gamma$. Choose two disjoint open subsets $M_1$ and $M_2$ of $X_\gamma$ such that $f_\gamma \in M_1$ and $p_\gamma \in M_2$ of $X_\gamma$. Let $O=\prod \limits_{\alpha \in S}O_\alpha$ such that $O_\gamma=M_1$ and $O_\alpha=X_\alpha$ for all $\alpha \ne \gamma$. We have $f \in O$. For each $i$, there is $g_i \in B_i \subset U_i$ such that $g_i(\gamma)=p_\gamma$. Thus each $g_i \notin O$. Thus there is no countable local base at $f$. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

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

# An example of a quotient space, II

This post presents another example of a quotient space of a first countable space. The resulting quotient space is not first countable. After we first define the space without referring to the concepts of quotient space, we show that this example is a quotient space. This example will be further discussed when we discuss sequential spaces.

For a previous discussion on quotient space in this blog, see An example of a quotient space, I. For more information on quotient spaces in general, see [2].

Let $\mathbb{N}$ be the set of all positive integers. For each $i \in \mathbb{N}$ and $j \in \mathbb{N}$, let $V_{i,j}=\left\{(\frac{1}{i},\frac{1}{k}):k \ge j\right\}$. Let $V=\bigcup \limits_{i=1}^{\infty} V_{i,1}$. Let $H=\left\{(\frac{1}{i},0):i \in \mathbb{N}\right\}$. For each $i \in \mathbb{N}$, let $V_i=V_{i,1} \cup \left\{(\frac{1}{i},0)\right\}$. Define the space $Y$ as:

$Y=\left\{(0,0)\right\} \cup H \cup V$.

The set $H$ is the horizontal part of the space and the set $V$ is the vertical part of the space. The origin $(0,0)$ is an additional point. In the topology for the space $Y$, each point in $V$ is isolated. Each point $(\frac{1}{i},0) \in H$ has an open neighborhood of the form

$\left\{(\frac{1}{i},0)\right\} \cup V_{i,j}$ for some $j \in \mathbb{N}$

The open neighborhoods at $(0,0)$ are obtained by removing finitely many $V_i$ from $Y$ and by removing finitely many isolated points in the $V_i$ that remain. The open neighborhoods just defined form a base for a topology on the set $Y$, i.e. by taking unions of these open neighborhoods, we obtain all the open sets for this space.

We wish to discuss $Y$ and its subspace $Z=\left\{(0,0)\right\} \cup V$. If we consider $Z$ as a topological space in its own right (open sets in $Z$ are of the form $U \cap Z$ where $U$ is open in $Y$), then in $Z$ there are no infinite compact sets (all compact subsets of $Z$ are finite). In particular, no sequence of points $\left\{x_n \in V:n \in \mathbb{N}\right\}$ can converge to the point $(0,0)$. However, with respect to $Z$, $(0,0)$ is a limit point of $V$. This implies that the space $Y$ is not first countable.

A space $W$ is said to be a Frechet space if $w \in \overline{A}$ where $A \subset W$, then there is some sequence $\left\{w_n \in A:n \in \mathbb{N}\right\}$ that converges to the point $w$. Any first countable space is a Frechet space. If there is a local base $\left\{U_n: n \in \mathbb{N}\right\}$ at $w$ and if $w \in \overline{A}$, then whenever we pick $w_n \in A \cap U_n$, $w_n$ would converge to $w$.

The space $Y$ defined above is not a Frechet space. Note that $(0,0) \in \overline{V}$ and no sequence of points in $V$ can converge to $(0,0)$.

Though the space $Y$ defined above is not a first countable space, $Y$ is a quotient space of a first countable space $X$ (in fact, a subspace of the Euclidean plane). Let $E=H \cup V$ where $H$ and $V$ are defined as in the definition of the space $Y$ above. Let $H_{-1}$ be the set $H_{-1}=\left\{(\frac{1}{i},-1):i \in \mathbb{N}\right\}$. Let $X=E \cup H_{-1} \cup \left\{(0,-1)\right\}$ with the Euclidean topology inherited from the Euclidean plane.

Define a quotient space from $X$ by collapsing each pair of points $\left\{(\frac{1}{i},0),(\frac{1}{i},-1)\right\}$ into one point $\left\{(\frac{1}{i},0)\right\}$. This is done for each $i \in \mathbb{N}$. For convenience, the point $\left\{(0,-1)\right\}$ is shifted to $\left\{(0,0)\right\}$. The resulting quotient space is the same as the space $Y$ defined above.

The quotient space we just describe can also be described by the quotient map $f:X \mapsto Y$:

$f((0,-1))=(0,0)$,

$f((\frac{1}{i},-1))=(\frac{1}{i},0)$ for each $(\frac{1}{i},-1) \in H_{-1}$,

$f((\frac{1}{i},0))=(\frac{1}{i},0)$ for each $(\frac{1}{i},0) \in H$,

$f(x)=x$ for each $x \in V$.

Then the following topology coincides with the topology on $Y$ that we define earlier in the post:

$\tau_f=\left\{U \subset Y: f^{-1}(U) \text{ is open in } X\right\}$

Reference

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

# An example of a quotient space, I

It is well known that in a compact Hausdorff space, if every point in the space is a $G_\delta$ point, then the space is first countable (see The cardinality of compact first countable spaces, II). Outside of compact spaces, this is not true. We present an example of a space where every point is a $G_\delta$ point but has no countable local base at any one point. This is an opportunity to introduce the notion of quotient space. Our example can be viewed as the image of a first countable space under a quotient map. We first define the example and then present a brief discussion of quotient spaces.

Let $\mathbb{N}$ be the set of all positive integers. Let $W=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}$. Let $S_1=\mathbb{N} \times W$ and $S_2=\mathbb{N} \times \left\{0\right\}$. Consider $X=S_1 \cup S_2$ as a subspace of the Eucliean plane. In the Euclidean topology, each point is $S_1$ is isolated and an open neighborhood at each point $(i,0)$ in $S_2$ is of the form:

$\left\{(i,0)\right\} \cup \left\{(i,\frac{1}{k}):k \ge n\right\}$ for some $n \in \mathbb{N}$

We use $X$ to define a space $Y$ by considering $S_2$ as one point (call this point $p$). Thus $Y=S_1 \cup \left\{p\right\}$. Points in $S_1$ remain isolated. An open neighborhood of $p$ is of the form $\left\{p\right\} \cup U^-$ where $U^-=U-S_2$ and $U$ is a Euclidean open subset of $X$ such that $S_2 \subset U$.

To facilitate the discussion, we describe the open sets at the point $p$ in more details. Let $\mathbb{N}^{\mathbb{N}}$ be the set of all functions $f:\mathbb{N} \mapsto \mathbb{N}$. For each $i \in \mathbb{N}$ and $j \in \mathbb{N}$, let $V_{i,j}=\left\{(i,\frac{1}{k}): k \ge j\right\}$. An open neighborhood of $p$ is of the following form:

$B_f=\left\{p\right\} \cup \bigcup \limits_{i=1}^{\infty} V_{i,f(i)}$

To describe in words, each open neighborhood at $p$ is obtained by removing finitely many points in each vertical segment $V_{i,1}$.

The resulting space $Y$ is not first countable. We show that any countable set of open neighborhoods at $p$ cannot be a local base at $p$. To this end, we consider $B_{f_1},B_{f_2},B_{f_3},\cdots$ where each $f_i \in \mathbb{N}^{\mathbb{N}}$. We wish to find an open set $O$ such that $p \in O$ and each $B_{f_i}$ is not a subset of $O$.

For each $i \in \mathbb{N}$, pick some $g(i) \in \mathbb{N}$ such that $g(i)>f_i(i)$. Consider $O=B_g$. Note that for each $i \in \mathbb{N}$, $(i,\frac{1}{f_i(i)}) \in B_{f_i}$ and $(i,\frac{1}{f_i(i)}) \notin O=B_g$. This shows that no countable number of open neighborhoods can form a base at the point $p$.

However, every point in the space $Y$ is a $G_\delta$ point. For the point $p$, we have $\left\{p\right\}=\bigcap \limits_{i=1}^\infty B_{h_i}$ where $h_i \in \mathbb{N}^{\mathbb{N}}$ is a constant function mapping to $i$.

Quotient Spaces
We now present one definition of quotient spaces. Let $(X,\tau)$ be a topological space. Let $q:X \mapsto Y$ be a surjection, i.e. $q(X)=Y$. Consider the following collection of subsets of $Y$:

$\tau_q=\left\{O \subset Y: q^{-1}(O) \text{ is open in }X\right\}$

The set $\tau_q$ is a topology for the set $Y$. With this topology $\tau_q$ on $Y$, the function $q$ is continuous. In fact, $\tau_q$ is the finest (largest or strongest) topology on $Y$ that makes the function $q$ continuous, i.e. if $\tau_1$ is another topology on $Y$ making $q$ a continuous function, then $\tau_1 \subset \tau_q$.

The topology $\tau_q$ is called the quotient topology induced on $Y$ by the mapping $q$. When $Y$ is given such a quotient topology, it is called a quotient space of $X$ (most of the times just called quotient space). The induced map $q$ is called a quotient map. If $Y$ has a quotient topology defined by a quotient map, then $Y$ is said to be the quotient image of $X$. We plan to discuss quotient space and quotient topology in greater details in this blog. For further information, see [2].

In the example discussed in this post, the quotient map is to map each point in $S_1$ to itself and to map each point in $S_2$ to a single point $p$. In essence, we collapse the whole x-axis $S_2$ into one single point. The open sets for the point $p$ are simply the Euclidean open sets containing $S_2$. This example shows that the quotient image of a first countable space needs not be first countable. Though the quotient image of a first countable space may not be first countable, it has the property that sequences suffice to define the topology (these are called sequential spaces, see [1]).

Reference

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

# The cardinality of compact first countable spaces, III

In the real line every uncountable closed set has cardinality continuum, the cardinality of the real line (see Closed uncountable subsets of the real line). It follows that every uncountable compact subset of the real line has cardinality continuum. In this post we show that in any abstract topological space that is compact, Hausdorff and first countable, every uncountable compact space has cardinality continuum. In the previous post The cardinality of compact first countable spaces, I, we show that continuum is an upper bound of the cardinality of any compact, Hausdorff, first countable space. In this post, we show that continuum is a lower bound for any compact, Hausdorff, first countable space that is uncountable. This is accomplished by defining a subset that has the same candinalty as the Cantor set. Finding this “Cantor” subset is facilitated by the following lemmas. All spaces under consideration are Hausdorff.

Definitions
Let $Y$ be a topological space. Let $A \subset Y$ and $p \in Y$. The point $p$ is said to be a complete accumulation point of $A$ if for every open set $O$ containing $p$, $\lvert O \cap A \lvert=\lvert A \lvert$. Since we are only interested in $A$ being uncountable without requiring $\lvert A \lvert$ to be any specific cardinal, we call $p$ a complete accumulation point of $A$ if for every open $O$ containing $p$, $O \cap A$ is uncountable. In particular, the point $p$ a complete accumulation point of the space $Y$ if for every open set $O$ containing $p$, $O \cap Y$ is uncountable.

Lemma 1
Let $Y$ be a Lindelof space such that $Y$ is uncountable. Then $Y$ has a complete accumulation point.

Proof. Suppose $Y$ is Lindelof and suppose that every point of $Y$ is not a complete accumulation point. Then for each $y \in Y$, there is an open set $O_y$ containing $y$ such that $O_y \cap Y$ is countable. Then the set $\mathcal{O}$ of all $O_y$ is an open cover of $Y$. Since $Y$ is Lindelof, $\mathcal{O}$ has a countable subcover. This means that $Y$ is countable. $\blacksquare$

Lemma 2
Let $Y$ be a Lindelof space such that $Y$ is uncountable and first countable. Then $Y$ has at least two complete accumulation points.

Proof. Suppose $Y$ is Lindelof, uncountable and first countable. By Lemma 1, there is $p \in Y$ that is a complete accumulation point of $Y$. Let $\left\{U_n:n=1,2,3,\cdots\right\}$ be a local base at $p$. Observe that there is some $n$ such that $Y-U_n$ is uncountable. Otherwise, the whole space $Y$ would be countable. By Lemma 1, $Y-U_n$ would have a complete accumulation point $q$. $\blacksquare$

Lemma 3
Let $Y$ be a compact space such that $Y$ is uncountable and first countable. Then there are disjoint compact subsets $K_0$ and $K_1$ such that both $K_0$ and $K_1$ are uncountable.

Proof. Suppose $Y$ is a compact space such that $Y$ is uncountable and first countable. By Lemma 2, there are two complete accumulation points $p_0$ and $p_1$ in $Y$. Choose disjoint open sets $U_0$ and $U_1$ containing $p_0$ and $p_1$, respectively, such that $\overline{U_0} \cap \overline{U_1}=\phi$. The desired compact sets are $K_0=\overline{U_0}$ and $K_1=\overline{U_1}$. $\blacksquare$

Theorem 4
Let $X$ be a compact space such that $X$ is uncountable and first countable. Then there is a compact subset $C$ of $X$ such that $\lvert C \lvert=2^\omega$.

Proof. The induction process is similar to the process in constructing the middle third Cantor set in The Cantor set, I. We start with the compact space $X$ and find two disjoint compact sets $B_0$ and $B_1$ that are uncountable using Lemma 3. For $B_0$, we apply Lemma 3 to obtain uncountable disjoint compact subsets $B_{00}$ and $B_{01}$. Likewise obtain uncountable disjoint compact subsets $B_{10}$ and $B_{11}$ of $B_1$.

In each compact set in each subsequent step, we find two disjoint compact subsets that are uncountable. Suppose that at step $n$, we have obtained compact sets $B_g$ for each $n$-length sequence of zeros and ones. Then we apply Lemma 3 to obtain two uncountable disjoint compact sets $B_{g0}$ and $B_{g1}$ of $B_g$.

Let $A_0=X$. For each integer $n>0$, let $A_n$ be the union of all $B_g$ over all $n$-length sequences $g$ of zeros and ones. Let $C^*=\bigcap \limits_{n=0}^{\infty}A_n$. We choose a subset of $C^*$ of cardinality continuum.

For each $f \in 2^\omega$, let $f_n$ be the $n$-length sequence of zeros and ones that agrees with the first $n$ coordinates of $f$. Then, for each $f \in 2^\omega$, choose a point $x_f$ in $\bigcap \limits_{n=1}^{\infty}B_{f_n}$. Let $C=\left\{x_f: f \in 2^\omega \right\}$. Note that $x_f \ne x_g$ for $f,g \in 2^\omega$ with $f \ne g$. Thus $C$ is a subset of the compact space $X$ such that $\lvert C \lvert=2^\omega$.

The previous posts on compact first countable spaces
The cardinality of compact first countable spaces, I
The cardinality of compact first countable spaces, II

# The cardinality of compact first countable spaces, II

It is a well known fact in topology that in any compact Hausdorff space, if $x \in X$ is a $G_\delta$ point (it is the only point in the intersection of countably many open sets), then there is a countable local base at $x$. See 3.1.F in [1] and 16.A.4 in [2]. Thus for a compact Hausdorff space, if every point is a $G_\delta$ point, the space is first countable. In the previous post The cardinality of compact first countable spaces, I, we show that for every compact, Hausdorff space $X$ that is first countable, $\lvert X \lvert \le 2^\omega$. Combining these two results, we have the following theorem:

Theorem
If $X$ is a compact, Hausdorff and satisfies the condition that every point of $X$ is a $G_\delta$ point, then $\lvert X \lvert \le 2^\omega$, i.e. the cardinality of $X$ is no greater than continuum (the cardinality of the real line).

In this post, we prove that in a compact Hausdorff space, there is a countable local base at every point that is a $G_\delta$.

Definitions
Let $X$ be a topological space. The space $X$ is said to be a Hausdorff space if for $x,y \in X$ with $x \ne y$, there exist disjoint open sets $U \subset X$ and $V \subset X$ such that $x \in U$ and $y \in V$. Hausdorff spaces are also called T2 spaces. The space $X$ is said to be a regular space if it is T2 and for each $x \in X$ and for each closed subset $C \subset X$ with $x \notin C$, there exist disjoint open sets $U \subset X$ and $V \subset X$ such that $x \in U$ and $C \subset V$. Regular spaces are also called T3 spaces.

A local base at $x \in X$ is a collection $\mathcal{B}$ of open sets containing the point $x$ such that if $x \in U \subset X$ where $U$ is open, then $x \in B \subset U$ for some $B \in \mathcal{B}$. The space $X$ satisfies the first axiom of countability if there is a countable local base at each point in the space. The space $X$ is said to be first countable if it satisfies the first axiom of countability.

If the space $X$ is regular and has a countable local base at $x \in X$, then we can have a countable local base $\left\{U_n:n=1,2,3,\cdots\right\}$ at $x \in X$ that satisfies the following:

$\overline{U_{n+1}} \subset U_n$ for each $n=1,2,3,\cdots$

Let $x \in X$. The point $x$ is said to be a $G_\delta$ point in $X$ if $\left\{x\right\}=\bigcap \limits_{n=1}^{\infty}U_n$ where each $U_n$ is open in $X$. Clearly, if there is a countable local base at $x \in X$, then the point $x$ is a $G_\delta$ point. The converse is not true.

The space $X$ is said to be a compact space if every open cover of $X$ has a finite subcover. The space $X$ is said to be countably compact if every countable open cover of $X$ has a finite subcover. Equivalently, $X$ is countably compact if and only if every infinite $A \subset X$ has a limit point in $X$.

Theorem 1
Suppose $X$ is a regular countably compact space in which every point is a $G_\delta$ point. Then $X$ is first countable.

Proof. Let $x \in X$. We can find countably many open sets $U_n$ whose intersection is $\left\{x\right\}$. Since the space is regular, the open sets $U_n$ can be made to satisfy the condition that $\overline{U_{n+1}} \subset U_n$.

We claim that for each open set $U$ containing $x$, $U_n \subset U$ for some $n$. Suppose not. Then there is an open set $U$ with $x \in U$ such that each $U_n$ is not a subset of $U$. For each $n$, choose $x_n \in U_n$ such that $x_n \notin U$. The set $A=\left\{x_n:n=1,2,3,\cdots\right\}$ must be infinite. Otherwise, there is some $j$ such that $x_j=x_k$ for all $k>j$. Then $x_j=x$ since $x_j$ belongs to each $U_n$ (a contradiction).

Since $A$ is infinite, $A$ has a limit point $p$. Note that for each $j$ the point $p$ is a limit point of

$A_j=\left\{x_n:n=j,j+1,j+2,\cdots\right\} \subset U_j$.

Thus $p \in \overline{U_j}$ for each $j$. This mean $p=x$ and the open set $U$ would contain infinitely many $x_n$. But $x_n$ is chosen to be not in $U$, a contradiction. It follows that the open sets $U_n$ that make the point $x$ a $G_\delta$ point must be a local base. $\blacksquare$

Corollary 2
Let $X$ be a compact T2 space such that every point is a $G_\delta$ point. Then $X$ is first countable.

Proof. This follows from theorem 1. Note that any compact T2 space is normal and is thus regular. $\blacksquare$

Reference

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

# The cardinality of compact first countable spaces, I

In the real line, all Cantor sets, all perfect set and all uncountable closed sets have cardinality continuum, the cardinality of the real line (see The Cantor set, I and other links listed below). It follows that every uncountable compact subset of the real line has cardinality continuum. Being subsets of a space of cardinality continuum, the upper bound of the cardinality of compact subsets of the real line is obviously continuum. It turns out that even for an abstract compact topological space, this upper bound of continuum in cardinality applies provided that the compact space is Hausdorff and first countable (both properties hold in the real line). In this post, we prove the following theorem:

Theorem
Let $X$ be any Lindelof, Hausdorff and first countable space. Then $X$ has cardinality at most continuum, i.e. $\lvert X \lvert \le \lvert \mathbb{R} \lvert$.

Remark
This theorem was proved by Arhangelskii in 1969, which answered a half century old question of Alexandroff and Urysohn: is there a compact, first countable space with cardinality greater than the continuum? The proof presented here is due to Roman Pol (see the historical notes in section 3.1 in [1]).

Definitions
Let $X$ be a topological space. A local base at $x \in X$ is a collection $\mathcal{B}$ of open sets (each containing $x$) with the property that for each open set $O$ containing $x$, there is some $B \in \mathcal{B}$ with $x \in B \subset O$. A space $X$ is said to be first countable (or satisfies the first axiom of countability) if for each point $x \in X$, there is a countable local base at $x \in X$. In the real line, one particular countable local base at $x \in \mathbb{R}$ is defined by the open intervals of the form $(x-\frac{1}{n},x+\frac{1}{n})$ where $n=1,2,3,\cdots$.

Remark
By $\omega$, we mean the first infinite ordinal, which can be viewed as the set of all nonnegative intergers. By $\omega_1$ we mean the first uncountable ordinal. We use $2^\omega$ to denote continuum, the cardinality of the real line. For the basic set theory and notations that we use, see A note on basic set theory.

Lemma
Let $X$ be any first countable space. If $A \subset X$ and $\lvert A \lvert \le 2^\omega$, then $\lvert \overline{A} \lvert \le 2^\omega$.

Proof. Let $x \in \overline{A}$. Since there is a local base at $x \in X$, we can find a sequence $\left\{x_n: n=0,1,2,\cdots\right\}$ of points of $A$ such that $x_n \rightarrow x$. The mapping $x \rightarrow \left\{x_n\right\}$ is a one-to-one function from $\overline{A}$ into $A^\omega$. Thus we have $\lvert \overline{A} \lvert \le \lvert A^\omega \lvert \le \vert (2^\omega)^\omega \lvert = 2^\omega$. $\blacksquare$

Proof of Theorem
This is a proof based on a transfinite induction process. We pick points of $X$ in each of the $\omega_1$ many steps in the induction process. We pick in such a way that we exhaust all points of $X$ in the process. In each of the steps we pick $\le 2^\omega$ many points. Thus $\lvert X \lvert \le 2^\omega$.

The results of the induction process: For each $\alpha < \omega_1$, we have $X_\alpha$, a closed subset of $X$, with the following properties:

$\displaystyle X_\delta \subset X_\gamma$ for all $\delta<\gamma<\omega_1$,
$\displaystyle \lvert X_\alpha \lvert \le 2^\omega$ for all $\alpha <\omega_1$.

For each $x \in X$, we fix a local base $\left\{B(x,n):n=0,1,2,\cdots\right\}$ at $x \in X$.

To start off, we choose $x_0 \in X$ and let $X_0=\left\{x_0\right\}$. The set $X_0$ is a closed subset of $X$ and $\lvert X_0 \lvert \le 2^\omega$.

Suppose for $\alpha < \omega_1$, we have chosen sets $X_{\delta} \subset X$ for each $\delta < \alpha$ such that $\displaystyle X_\delta$ is closed subset of $X$ and $\displaystyle \lvert X_\delta \lvert \le 2^\omega$ and that $X_\delta \subset X_\gamma$ for all $\delta<\gamma<\omega_1$. Now we indicate how to move to the next step.

Case 1. The ordinal $\alpha$ is a limit ordinal.
Let $X_\alpha=\overline{\bigcup \limits_{\delta<\alpha}X_\delta}$. Note that $A=\bigcup \limits_{\delta<\alpha}X_\delta$ is the union of countably many sets, each of which has cardinality $\le 2^\omega$. Thus $\lvert A \lvert \le 2^\omega$ . By the lemma, $\lvert X_\alpha \lvert \le 2^\omega$.

Case 2. The ordinal $\alpha$ is a successor ordinbal, say, $\alpha=\beta+1$.
We now define $X_\alpha$ using $X_\beta$. Consider the following collection:

$\mathcal{B}_\alpha=\left\{B(x,n): x \in X_\beta \text{ and } n=0,1,2,\cdots\right\}$

We consider all countable $g \subset \mathcal{B}_\alpha$ such that $X_\beta \subset \cup g$ and $X-\cup g \ne \phi$. If no such $g$ exists, then $\mathcal{B}_\alpha$ covers $X$ and $X=X_\beta$, in which case the induction process stops and we are done. So we assume that the induction can continue at each step.

For each countable $g \subset \mathcal{B}_\alpha$ described in the above paragraph, we choose $x_g \in X-\cup g$. Let $X_\alpha^*$ be the set of all points $x_g$. Note that $\lvert \mathcal{B}_\alpha \lvert \le 2^\omega$. Thus $\lvert \mathcal{B}_\alpha^{\ \omega} \lvert \le 2^\omega$ where $\lvert \mathcal{B}_\alpha^{\ \omega} \lvert$ can be considered as the cardinality of the set of all countable subsets of $\mathcal{B}_\alpha$. Thus $\lvert X_\alpha^* \lvert=\lvert \mathcal{B}_\alpha^{\ \omega} \lvert \le 2^\omega$.

Let $X_\alpha=\overline{X_\alpha^* \cup X_\beta}$. By the lemma, $\lvert X_\alpha \lvert \le 2^\omega$.

Now that the induction process is done, let $X^*=\bigcup \limits_{\alpha<\omega_1} X_\alpha$. Since this is the union of $\le 2^\omega$ many sets, each of which has cardinality $\le 2^\omega$, $\lvert X^* \lvert \le 2^\omega$. We just need to show that $X^*=X$. First we show that $X^*$ is closed in $X$.

Let $p \in X$ be a limit point of $X^*$. We claim that $p \in X_\alpha$ for some $\alpha<\omega_1$. There exists a sequence $p_n$ of points of $X^*$ such that $p_n \rightarrow p$. For each $n$, $p_n \in X_{\beta(n)}$ for some $\beta(n)<\omega_1$. Choose $\alpha<\omega_1$ such that $\beta(n)<\alpha$ for all $n$. Then $p_n \in X_\alpha$. Since $X_\alpha$ is closed, we have $p \in X_\alpha \subset X^*$.

To show that $X^*=X$, suppose we have $q \in X-X^*$. Since $q \notin X^*$, for each $y \in X^*$, $q \notin B(y,n(y))$ for some $n(y)$. The set of all $B(y,n(y))$ covers $X^*$. Since $X^*$ is closed (thus Lindelof), we have countably many $B(y,n(y))$ covering $X^*$. Let $g$ be the set of these countably many $B(y,n(y))$. Furthermore, the countably many $y \in X^*$ must belong to some $X_\beta$ where $\beta<\omega_1$.

We can trace back to the induction step at $\alpha=\beta+1$. Note that $g \subset \mathcal{B}_\alpha$, $X_\beta \subset \cup g$ and $X-\cup g \ne \phi$ (since $q \in X-\cup g$). In the induction step at $\alpha$, we have chosen a point $x_g \in X-\cup g$. Note that $x_g \notin \cup g$. Thus $x_g \notin X^*$. Note also that $x_g \in X_\alpha$ (from the induction step). Thus $x_g \in X^*$, a contradiction. Thus $X=X^*$. $\blacksquare$

The Cantor set, I
Perfect sets and Cantor sets, II
Closed uncountable subsets of the real line

Reference

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

# A note on basic set theory

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

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

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

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

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

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

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

Reference

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