I understand the Rudin theorem $2.43$ .
Suppose if we put uncountable instead of countable
i,e suppose $P$ is uncountable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$
Then this will also contradicts $\bigcap_1^\infty K_n$ is nonempty and it will implies $P$ is countable
My confusion : Is its possible to modify the theorem like this let $P$ be a nonempty perfect set in $\mathbb{R}^K$ .Then $P$ is countable
Here is an outline of Rudin's proof(Theorem $2.43$):
Theorem 2.43 Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Proof 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 $y\in \mathbb{R}^k$ such that $|y−x_1|<r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y\in \mathbb{R}^k$ such that $|y−x_1|≤r$.
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) $x_n\notin \overline{V_{n+1}}$, (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. Since $\mathbf{x_{n}}\notin K_{n+1}$, no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_{n}\subset P$, this implies that $\cap_1^\infty K_n$ 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.
