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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.