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

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