Perfect sets of a complete metric space

compactnessgeneral-topologyreal-analysisset-theorysolution-verification

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?

Best Answer

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 $\mathrm{int}\,S=S$.

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