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Theorem $\hspace{5 pt}$ Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

Proof $\hspace{5 pt}$ Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows. Let $V_1$ be any neighborhood of $\mathbf{x_1}$. If $V_1$ consists of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| \leq r$. Suppose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $\mathbf{x_n} \notin \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed. Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact.

I've understood things up until this point.

Since $\mathbf{x_n} \notin K_{n+1}$, no point of $P$ lies in $\bigcap_1^\infty K_n$.

1) I understand that $\{\mathbf{x_n}\}$ is a subset of $P = \{\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots\}$. If this subset was equal to $P$ then I can see how no point of $P$ would lie in the intersection. But I don't think this is true (since $P$ would have to be bounded in order for $V_1 \supset P$ so that $\mathbf{x_n}\in V_n \subset V_{n-1} \ldots \subset V_1$).

What's the actual reason why $\mathbf{x_n} \notin K_{n+1}\implies\forall p \in P, p \notin \bigcap_1^\infty K_n$ ?

Update I've been Trying to understand Quang Hoang's answer for 1)

"Every element of $P$ is of the form $x_n$ for some $n$, hence doesn't lie in $K_n$."

But doesn't $\{\mathbf{x_n}\}$ only includes elements within the first neighborhood? (Since each $\mathbf{x_n}\in V_n$ for a particular $V_{n}\subset V_{n-1}$, surely $\{\mathbf{x_n}\}$ can't contain every point in $P$?)

What am I missing?

Since $K_n \subset P$, this implies that $\bigcap_1^\infty K_n$ is empty. But each $K_n$ is nonempty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.

Resolved 2) In order to apply Theorem 2.36, $K_n$ must be compact. How do we know that $K_n$ is compact? EDIT: For 2) I understand the answer given by Quang Hoang, thanks!

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  • $\begingroup$ I had the same exact confusion as you, now this makes so much sense! Now just how to imitate this proof for Exercise 2.30... $\endgroup$
    – MT_
    Feb 20, 2015 at 4:06

2 Answers 2

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  1. Every element of $P$ is of the form $x_n$ for some $n$, hence doesn't lie in $K_n$.
  2. $K_n$ is a closed subset (intersection of two closed sets) of a compact set.
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  • $\begingroup$ 1.But doesn't $\{\mathbf{X_n}\}$ only includes elements within the first neighborhood? 2. Ah I see, thanks! $\endgroup$
    – ignoramus
    Sep 12, 2014 at 6:18
  • $\begingroup$ (Since each $\mathbf{X_n}$ corresponds to a particular $V_{n} \subset V_{n-1}$, surely $\{\mathbf{x_n}\}$ can't contain every point in $P$?) $\endgroup$
    – ignoramus
    Sep 12, 2014 at 6:26
  • $\begingroup$ Note that $P=\{x_1,x_2,\dots\}$ and that for every $n$, $x_n\not\in K_n$. It follows that for every $n$, $x_n\not\in \cap K_j$. Thus $P\cap (\cap K_j)=\varnothing$. $\endgroup$ Sep 12, 2014 at 13:48
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Ok, I've figured out the cause of my confusion:

1) I understand that $\{\mathbf{x_n}\}$ is a subset of $P = \{\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots\}$. If this subset was equal to $P$ then I can see how no point of $P$ would lie in the intersection. But I don't think this is true (since $P$ would have to be bounded in order for $V_1 \supset P$ so that $\mathbf{x_n}\in V_n \subset V_{n-1} \ldots \subset V_1$).

Let $V_n$ be the neighborhood of $\mathbf{y_n}$. Then $\mathbf{x_1} = \mathbf{y_1}$. I realise now that even if $\mathbf{x_n} \neq \mathbf{y_n}$, we can still construct $V_{n+1}$ so that (ii) is satisfied:

Since $\mathbf{x_n}$ is an arbitrary point, and there are infinitely many limit points of $P$ in $V_n$, we can choose $\mathbf{y_{n+1}}\in P, V_n$ such that $\mathbf{y_{n+1}} \neq \mathbf{y_{n}}$ and $\mathbf{y_{n+1}}\neq \mathbf{x_{n}}$.

Now since every point of $P$ is a limit point, we can pick a small enough $r>0$ such that $V_{n+1}$ satisfies (i)-(iii).

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