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 be a topological space. Let and . The point is said to be a complete accumulation point of if for every open set containing , . Since we are only interested in being uncountable without requiring to be any specific cardinal, we call a complete accumulation point of if for every open containing , is uncountable. In particular, the point a complete accumulation point of the space if for every open set containing , is uncountable.

**Lemma 1**

Let be a Lindelof space such that is uncountable. Then has a complete accumulation point.

* Proof*. Suppose is Lindelof and suppose that every point of is not a complete accumulation point. Then for each , there is an open set containing such that is countable. Then the set of all is an open cover of . Since is Lindelof, has a countable subcover. This means that is countable.

**Lemma 2**

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

* Proof*. Suppose is Lindelof, uncountable and first countable. By Lemma 1, there is that is a complete accumulation point of . Let be a local base at . Observe that there is some such that is uncountable. Otherwise, the whole space would be countable. By Lemma 1, would have a complete accumulation point .

**Lemma 3**

Let be a compact space such that is uncountable and first countable. Then there are disjoint compact subsets and such that both and are uncountable.

* Proof*. Suppose is a compact space such that is uncountable and first countable. By Lemma 2, there are two complete accumulation points and in . Choose disjoint open sets and containing and , respectively, such that . The desired compact sets are and .

**Theorem 4**

Let be a compact space such that is uncountable and first countable. Then there is a compact subset of such that .

* 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 and find two disjoint compact sets and that are uncountable using Lemma 3. For , we apply Lemma 3 to obtain uncountable disjoint compact subsets and . Likewise obtain uncountable disjoint compact subsets and of .

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

Let . For each integer , let be the union of all over all -length sequences of zeros and ones. Let . We choose a subset of of cardinality continuum.

For each , let be the -length sequence of zeros and ones that agrees with the first coordinates of . Then, for each , choose a point in . Let . Note that for with . Thus is a subset of the compact space such that .

**The previous posts on compact first countable spaces**

The cardinality of compact first countable spaces, I

The cardinality of compact first countable spaces, II