# 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 $\lvert X \lvert \le \lvert \mathbb{R} \lvert$.

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 each $x \in X$. 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 $X=X^*$. $\blacksquare$