Perfect sets of a complete metric space

Question

So I am aware something like this question had been asked, so I would appreciate if you could direct me to a duplicate, if there is any.

In particular, I've read the post

Proof that a perfect set is uncountable,

And of course there is a classical proof when the metric space is $\mathbf{R}^n$ endowed with the usual Euclidean metric, the crux of the proof of which is very well surmised here:

Rudin 2.43 Every nonempty perfect set in $\mathbb{R}^k$ is uncountable.

My question is, in seeing Rudin's proof, it seems the only place where he actually used the fact that the metric space is $\mathbf{R}^n$ (and not just any old complete metric space) is when he invoked the Heine-Borel theorem to conclude that the closure of each $V_n$ is compact. But we know in general that in a complete metric space, if $(F_n)$ is a nested sequence of closed sets whose diameters converge to $0$, then the intersection $\bigcap_{n=1}^{\infty}F_n$ is nonempty, and that is all we need! (Of course, Rudin's proof doesn't explicitly mention that the diameter of $V_n$ tends to $0$, nor did he have to, but we can always choose the nested sequence so that the radii decay exponentially.)

Am I missing a critical detail here?

Answer 1

You are right. First observation is that the completeness of the space seems important. It's easy to see that the conclusion is false in the space $\Bbb Q$. Namely the whole space is a perfect subset.

A little enhanced proof can show that this is true in any complete metric space (at the end of the answer I provide the second proof based on Baire theorem). Consider a perfect set $S\subset X$ where $X$ is a complete metric space. Since $S$ is perfect, it's closed so it's itself a complete metric space (as a closed subset of a metric space). Therefore we can proceed without supspace $X$, that is we can assume $S$ is the whole space.

First proof

Observation: any nonempty open subset of $S$ (I recall that we consider the relative topology) is infinite. Especially, all balls have got infinitely many points.

Now observe that for any ball $B(x,r)$ and any point $z\in S$ the set $B(x,r)\setminus\{z\}$ is nonempty and open, so for any $n\in \Bbb N$ there is a ball $B(y,r')$ with $r'\leq 1/n$ such that $B(y,r')\subset \overline{B(y,r')}\subset B(x,r)\setminus\{z\}$. To make the proof easy to proceed let's denote the family of such balls $B(y,r')$ as $\mathcal I_n(B(x,r),z)$.

Assume $S$ is countable: $S=\{x_1,x_2,x_3,\ldots\}$.

• Define $V_1 := B(x_1,1)$.
• Let $V_2$ be any element of $\mathcal I_2(V_1,x_1)$.
• Let $V_3$ be any element of $\mathcal I_3(V_2,x_2)$.
• Let $n\in\Bbb N$ and assume we have defined sets $V_1,V_2,\ldots,V_n$. Let $V_{n+1}$ be any element of $\mathcal I_{n+1}(V_n,x_n)$.

Mathematical induction asserts that the sequence of $V_n$ is well defined for all $n\in \Bbb N$. This sequence satisfies the condition $x_n\notin V_{n+1}$.

Now we define $F_n:=\overline{V_n}$. We see that:

• $F_1\supset F_2\supset F_3\supset\cdots$.
• $\mathrm{diam}{F_{n}}=\mathrm{diam}{V_{n}}\leq 2\cdot \frac 1{n}$, so $\mathrm{diam} F_n\to 0$.
• All the sets $F_n$ are closed and nonempty.

Therefore the set $F:=\bigcap_{n=1}^\infty F_n$ is nonempty, that is $x_m\in F\subset F_{m+1}$ for some $m\in\Bbb N$. On the other hand, from the construction we know that $x_m\notin V_{m+1}\subset F_{m+1}$. A contradiction.

This proof is very similar to the proof of Baire Theorem. This suggests that we can use this theorem to prove our fact. This led me to another proof:

Second proof

Let $S$ be a countable complete space $S=\{x_1,x_2,x_3,\ldots\}$. If $S$ is perfect then all the closed sets $\{x_n\}$ for $n\in\Bbb N$ have got empty interiors. Therefore their sum $S=\bigcup_{n=1}^\infty\{x_n\}$ has also got empty interior, which is impossible, as $\mathrm{int}\,S=S$.

Answer 2

Preliminaries:

$\textbf{Theorem:1}$ If $\{K_{\alpha} :\}$ is a family of non empty compact subsetsets of a metric space $(X, d) $ such that the intersection of any finite subfamily is non empty , then $\bigcap_{\alpha} K_{\alpha}\neq \emptyset$

$\text{Theorem} \space 2.36$ (Principles of mathematical analysis, Rudin)

$\textbf{Corollary:1} $ If $(K_n) $ be any descending sequence of non empty compact subsets in a metric space $(X, d) $ , then $\bigcap_{n} K_n\neq \emptyset$

$\bigstar$ Note 1: The same is not true for any sequence of non empty closed sets in a metric space.

Ex: Consider $(\Bbb{Q}, d_{\text{sub}}) $ as a metric subspace of $(\Bbb{R}, d_{\text{eucliden}}) $ .Let $\Bbb{Q}=(r_n)_{n\in\Bbb{N}}$ be the enumeration and $C_n=\Bbb{Q}\setminus \{r_n\}$.Then

1. $\emptyset \neq C_n$ closed set.

2. $C_{n+1}\subset C_n$

But $\bigcap_{n} C_n=\emptyset$

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$\bigstar $ Note 2 : $(X, d) $ be a metric space and $A\subset X$ . Let $x\in X$ is a limit point of $A$ and $x\in U$ be a open set. Is it always possible to find an open set $V$ such that $x\in \overline{V}\subset U$ where $\overline{V}$ is compact?

Hint: Consider the metric space $(\Bbb{Q}, d_{\text{sub}}) $ .

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$\textbf{Definition :}$ A topological space $(X, \tau) $ is called a Locally compact Haussdoff space if $\forall x\in X$ and forall open sets $U_x$ containing $x$ there exists an open set $V_x$ such that

1. $x\in V\subset \overline{V}\subset U$

2. Topological closure $\overline{V}$ is closed.

Example 1 : $\Bbb{R^k}$ is locally compact Hausdorff space.For any point $x\in \Bbb{R^k}$ and a open set $U$ containing $x$ , there is a basic open set i.e an open ball $B(x, r) $ such that $x\in \overline{B(x, \frac{r}{2})}\subset B(x, r) \subset U$

Now by Heine-Borel theorem, $B(x, \frac{r}{2}) $ is compact.

Example 2 : Let $(X,d)$ be a metric space where $|X|\ge \aleph_{0}$ i.e $X$ is an infinite set and $d$ is the discrete metric. Then $(X, d) $ is locally compact Hausdorff space as for any $x\in X$ and a open set $U$ containing $x$ , $x\in \{x\}\subset U$ where $\{x\}$ is compact. But $(X, d) $ doesn't have Heine-Borel property as $X$ is closed , bounded but not compact.

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$\textbf{Theorem:}$ Let $\emptyset \neq P$ perfect subset of $\Bbb{R}^k$.

Claim : $P$ is uncountable.

Proof:(Rudin's way) $P\neq \emptyset$ implies $\exists x\in P$ .Since $\Bbb{R^k} $ is Hausdorff space , any open set $U$ containing $x$ contains an infinite subset of $P$. $($ Infact this result is also true for any $T_1$-space $)$ . Hence $P$ is an infinite set.

To show $P$ is not countably infinite . Suppose $P$ is countably infinite and $P=\{x_1,x_2,\ldots\}$.

Let $V_0$ is an open set containing $x_1$. Then by Local compactness of $\Bbb{R}^k$ , $\exists V_1\subset V_0$ open set such that $x_1\in \overline{V_1}\subset V_0$ and since $x_1$ is a limit point of $P$ , $V_1\setminus \{x_1\} \cap P\neq \emptyset$ and $\overline{V_1}$ is compact.

Let $x_2\in V_1\setminus \{x_1\} \cap P$ Again by Local compactness, we can find an open set $V_2\subset V_1$ such that $x_2\in \overline{V_2}\subset V_1$ and $\overline{V_2}$ is compact. Again $V_2\setminus \{x_2\} \cap P\neq \emptyset$

Continuing in that way, we produce a sequence of open sets $(V_n) $ with the properties :

1. $\overline{V_{n+1}}\subset {V_n}$

2. $x_n\notin V_{n+1}$

3. $V_{n+1}\cap P\neq \emptyset$

Put $K_n=\overline{V_n}\cap P$

As $2$ implies $x_n \notin{V_{n+1}}$ implies $\bigcap_{n}K_n=\emptyset$.

But $(K_n) $ is a descending sequence of non empty compact sets and $\bigcap_{n}K_n=\emptyset$ directly contradict $\textbf{Corollary:1}$.

Note : A compact subset of a metric space (infact in any Hausdorff space) is closed.

We can use completeness ,by choosing $V_n$ as the basic open sets i.e open balls in $\Bbb{R^n}$ , $V_n=B(x_n, \frac{r}{2^n})$ and then the Cantor intersection theorem for complete metric space.

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The theorem is still true in every complete metric space as a closed subset of a complete metric space is complete and a complete metric space with no isolated point is uncountable. (an application of Baire Category theorem) .

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$\textbf{Theorem}$ A normed linear space is finite dimensional iff closed unit ball is compact iff closed bounded sets are compact i.e Heine-Borel property is true.

Choose an infinite dimensional Banach space, then it doesn't have the Heine-Borel property but every non empty perfect set is uncountable.