0
$\begingroup$

Let $S=\{(x,y) \in \mathbb{R}^2: 0<x^2+y^2 \leq 1\}$ be a subset of the metric space $M = \mathbb{R}^2$, prove that $\lim(S) = \{(0,0)\} \cup S$

$\lim(S)$ is the set of limit points of $S$. By definition, a point $p \in M$ is a limit of $S$ if there exists a sequence $(p_n)$ in $S$ that converges to it.

My thought on this question is that I can first prove $\{(0,0)\} \cup S \subseteq \lim(S)$, but I don't know how to show the other way $\lim(S) \subseteq \{(0,0)\} \cup S$.

Can anyone help? Thanks!

$\endgroup$
4
  • $\begingroup$ Are you allowed to use the fact that $\{(0,0)\}\cup S=\{(x,y)\in\mathbb{R}^2:0\leq x^2+y^2\leq1\}$ is a closed ball and therefore closed? That would help. $\endgroup$ Commented Sep 17, 2018 at 20:34
  • $\begingroup$ Assume the inclusion $\{(0,0)\}\cup S\subset\lim(S)$ to be strict. Can you deduce a contradiction? $\endgroup$
    – Uskebasi
    Commented Sep 17, 2018 at 20:34
  • 1
    $\begingroup$ What does $\lim(S)$ mean ? The closure of $S$ ? $\endgroup$
    – idm
    Commented Sep 17, 2018 at 20:35
  • 1
    $\begingroup$ @idm I think the set of accumulation points of $S$. $\endgroup$
    – Uskebasi
    Commented Sep 17, 2018 at 20:36

1 Answer 1

1
$\begingroup$

Usually the definition of limit point is

$p$ is a limit point of $S$ if there exists a sequence $(p_n)$ in $S\setminus\{p\}$ that converges to $p$.

Consider the sequence of the points $p_n=(1/n,0)$ (for $n>0$). This converges to $(0,0)$.

If $p\in S$, consider $$ p_n=\left(1-\frac{1}{n+1}\right)p $$

For the converse inclusion: $\lim(S)$ is closed (prove it) and contained in the closure of $S$ (prove it). What's the closure of $S$?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .