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 be any Lindelof, Hausdorff and first countable space. Then has cardinality .

**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 be a topological space. A local base at is a collection of open sets (each containing ) with the property that for each open set containing , there is some with . A space is said to be first countable (or satisfies the first axiom of countability) if for each point , there is a countable local base at each . In the real line, one particular countable local base at is defined by the open intervals of the form where .

**Remark**

By , we mean the first infinite ordinal, which can be viewed as the set of all nonnegative intergers. By we mean the first uncountable ordinal. We use 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 be any first countable space. If and , then .

* Proof*. Let . Since there is a local base at , we can find a sequence of points of such that . The mapping is a one-to-one function from into . Thus we have .

**Proof of Theorem**

This is a proof based on a transfinite induction process. We pick points of in each of the many steps in the induction process. We pick in such a way that we exhaust all points of in the process. In each of the steps we pick many points. Thus .

The results of the induction process: For each , we have , a closed subset of , with the following properties:

for all ,

for all .

For each , we fix a local base at .

To start off, we choose and let . The set is a closed subset of and .

Suppose for , we have chosen sets for each such that is closed subset of and and that for all . Now we indicate how to move to the next step.

Case 1. The ordinal is a limit ordinal.

Let . Note that is the union of countably many sets, each of which has cardinality . Thus . By the lemma, .

Case 2. The ordinal is a successor ordinbal, say, .

We now define using . Consider the following collection:

We consider all countable such that and . If no such exists, then covers and , 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 described in the above paragraph, we choose . Let be the set of all points . Note that . Thus where can be considered as the cardinality of the set of all countable subsets of . Thus .

Let . By the lemma, .

Now that the induction process is done, let . Since this is the union of many sets, each of which has cardinality , . We just need to show that . First we show that is closed in .

Let be a limit point of . We claim that for some . There exists a sequence of points of such that . For each , for some . Choose such that for all . Then . Since is closed, we have .

To show that , suppose we have . Since , for each , for some . The set of all covers . Since is closed (thus Lindelof), we have countably many covering . Let be the set of these countably many . Furthermore, the countably many must belong to some where .

We can trace back to the induction step at . Note that , and (since ). In the induction step at , we have chosen a point . Note that . Thus . Note also that (from the induction step). Thus , a contradiction. Thus .

* Links to some previous posts*:

The Cantor set, I

Perfect sets and Cantor sets, II

Closed uncountable subsets of the real line

*Reference*

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