# Is its possible to modify the theorem like this let $P$ be a nonempty perfect set in $\mathbb{R}^K$ .Then $P$ is countable

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 $$|y−x_1|, the closure $$\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.

If $$P$$ is uncountable you can't enumerate its points as $${\bf x_1, x_2,x_3,...}$$ - the argument above requires that every point in $$P$$ is one of the $${\bf x_n}$$s, and that means $$P$$ has to be countable.