# 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