2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ What are the accumulation points of $V_1$? $\endgroup$ 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
    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. $\endgroup$ 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
    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
    Jun 29, 2017 at 9:37

1 Answer 1

2
$\begingroup$

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 $V$.

$\endgroup$

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.