Questions about Theorem 2.43 in Baby Rudin

Question

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.

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 $V$.