# Compact metrizable scattered spaces

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$, i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$. A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space $X$, the following conditions are equivalent:

• The function space $C_p(X)$ is a Frechet-Urysohn space.
• The function space $C_p(X)$ is a k space.
• $X$ is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$, let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$. Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$, a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$. Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$, a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$. Let $p \in X$. We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$, all but countably many points of $A$ are limit points of $A$.

We now discuss the main result.

Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$. Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$.

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$. One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$) decrease to zero, then the sets $H_n$ collapse to one point.

The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$. By assumption, both points are limit points of the space $X$. Choose open sets $U_0,U_1 \subset X$ such that

• $p_0 \in U_0$,
• $p_1 \in U_1$,
• $K_0=\overline{U_0}$ and $K_1=\overline{U_1}$,
• $K_0 \cap K_1 = \varnothing$,
• the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.

Note that each of these open sets contains infinitely many points of $X$. Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the $n$th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$. Now we continue the picking for the $(n+1)$st stage. For each $\sigma$, an $n$-length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that

• $p_{\sigma 0} \in U_{\sigma 0}$,
• $p_{\sigma 1} \in U_{\sigma 1}$,
• $K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$,
• $K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$,
• the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$.

For each positive integer $m$, let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$-length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$. Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that

• $C$ is uncountable (in fact of cardinality continuum),
• every point of $C$ is a limit point of $C$.

To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$. Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$, let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$. For each $\tau \in \left\{0,1 \right\}^N$, let $f(\tau)$ be the unique point in the following intersection:

$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$. At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$. Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$. Then the point $x$ must belong to a nested sequence of sets $K$‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$, which has cardinality continuum.

To see the second point, pick $x \in C$. Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$. Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$. Note that $x \in U_{\tau \upharpoonright n}$ for each $n$. Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$. Thus, the point $x$ is a limit point of $C$. This completes the proof. $\blacksquare$

Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.

Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$. By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.

$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$. After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$. Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$. This means that $X$ is not scattered. $\blacksquare$

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2015 \text{ by Dan Ma}$

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