Theorem 2.43 $\hspace{5 pt}$ 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 $x_1$.

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 $|\mathbf{y} - \mathbf{x_1}| \leq r$. etc...

Here is the problem. I know every neighborhood is an open set from Theorem 2.19, but why the closure of $V_1$ is the set of all y such that $|\mathbf{y} - \mathbf{x_1}| \leq r$? Can someone prove it? Thanks in advance.

  • $\begingroup$ What are the accumulation points of $V_1$? $\endgroup$ Commented Jun 29, 2017 at 8:19
  • $\begingroup$ Sorry, I don't know what is the accumulation point, I think there is no definition in baby rudin or I've not read there.@FrancescoPolizzi $\endgroup$
    – Johnny Ji
    Commented Jun 29, 2017 at 8:21
  • $\begingroup$ You should study Rudin's book more carefully. Accumulation points are considered in Definition 2.18, and in the same place you can also find the definition of closed set, namely a set containing all its accumulation points. $\endgroup$ Commented Jun 29, 2017 at 8:27
  • $\begingroup$ I think you mean limit points, right? A closed set contains all its limit points. $\endgroup$
    – Johnny Ji
    Commented Jun 29, 2017 at 9:24
  • 1
    $\begingroup$ I got it. The limit points of $V_1$ are $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| = r$ and the points such that $|\mathbf{y} - \mathbf{x_1}| < r$. $\endgroup$
    – Johnny Ji
    Commented Jun 29, 2017 at 9:37

1 Answer 1


Let $C$ be the set $\{y~:~\lvert y-x\rvert\leq r\}$. Every point in $C$ is a limit point of $V$, so $C\subset\bar{V}$. On the other hand, any point $z$ in $R^{k}-C$ has a neighborhood disjoint from $C$, so $z$ is not a limit point of $V$ (nor a member of $V$). But $\bar{V}=V\cup V'$, where $V'$ is the set of limit points of $V$.


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