Every non-empty perfect set in $\mathbb{R}^{k}$ is uncountable

Question

In the Principles of Mathematical Analysis, 3rd Ed. by Rudin, he mentions in a Theorem 2.43 on page 41 that every non-empty perfect set $P$ in $\mathbb{R}^{k}$ is uncountable.

In the proof, he assumes contrariwise that the points of $P$ are denumerable and labelled $\overline{x_{1}}, \overline{x_{2}}, \overline{x_{3}}, ...$ . He then constructs a sequence of neighborhoods $\{V_{n}\}$ by induction as follows:

Let $V_{1}$ be the neighborhood of $\overline{x_{1}}$, defined as $V_{1} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| < r\}$, and the closure of $V_{1}$ denoted $\overline{V_{1}}$ defined $\overline{V_{1}} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| \leq r\}$.

Now, suppose $V_{n}$ has been constructed, with $V_{n} \cap P \neq \emptyset$. Then, there exists $V_{n+1}$ such that:

1. $\overline{V_{n+1}} \subset V_{n}$ (the closure of $V_{n+1}$ is a proper subset of $V_{n}$)
2. $\overline{x_{n}} \notin \overline{V_{n+1}}$
3. $\overline{V_{n+1}} \cap P \neq \emptyset$.

The third condition ensures our induction hypothesis holds and allows for the construction to continue.

Now, let $K_{n} = \overline{V_{n}} \cap P$. Since $\overline{V_{n}}$ is closed and bounded, it must be compact as well (proved earlier). Since $\overline{x_{n}} \notin K_{n+1}$ by construction, no point of $P$ is in $\cap_{1}^{\infty}K_{n}$. Since $K_{n} \subset P$, this implies that $\cap_{1}^{\infty}K_{n} = \emptyset$. This fact in turn contradicts a corollary of an earlier theorem that such an intersection must be non-empty.

Where my understanding falls short is in the statement that is in bold. The contradiction relies on it. Now, since since $P$ is perfect, $P$ is closed and every point of $P$ must be a limit point of $P$. Further, since $P$ is non-empty, it must have infinitely many points (or else, it cannot have any limit points). Now, when I take a point $x_{n+1}$ inside the neighborhood $V_{n}$ of $x_{n}$, and construct a neighborhood $V_{n+1}$ in a manner that doesn't include $x_{n}$, $V_{n+1}$ still contains infinitely many points; it just doesn't contain $x_{n}$. So, when I continue this induction, doesn't each subsequent neighborhood $V_{m}$ have points in it? And, when I take an infinite intersection as suggested, it needn't contain the specific labelled points at the centre of the neighborhood; but it still does contain infinitely many points.

Where am I going wrong?

For some visual context, I am visualizing my argument as a set of circles with each successive circle (denoting $V_{n}$) drawn within the previous one but drawn so that it doesn't contain the centre of the previous one.

Answer 1

You are making things too complicated. $\overline {x_n} \notin K_{n+1}=\overline {V_{n+1}} \cap P$ by property 2). If $\overline {x_i}$ is a point of $P$ which belongs to every $K_n$ then $\overline {x_i} \in K_{i+1} \subset \overline {V_{i+1}} $ again contradicting 2).