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

Links to some previous posts:
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.
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