What is the corresponding set?

proof:

it's easy to see that $X \subset X'$ . Because $\forall x \in X, \exists r > 0 \Rightarrow (B_r (x) \setminus \{x\}) \cap X \neq \emptyset$ .

in particular for $r = \frac{5}{12}$ we have $(B_r (x) \setminus \{x\}) \cap X \neq \emptyset, \forall x \in X$ .

Beyond that, $\{0\} \subset X'$ because by the Archimedean property of the real numbers, we have, $\forall r > 0, \exists n \in \mathbb{N}^*,$ such that, $B_r (0) \supset B_{\dfrac{1}{2n}}[0] \setminus B_{\dfrac{1}{2n + 1}}(0) \Rightarrow B_r (0) \cap X \neq \emptyset \Rightarrow 0 \in X' \Rightarrow \{0\} \subset X'$ .

With this, $X \cup \{0\} \subset X'$ .

On the other hand, $\forall x \in X', \exists r > 0 \Rightarrow (B_r (x) \setminus \{x\}) \cap X \neq \emptyset \Rightarrow x \in X \cup \{0\} \Rightarrow X' \subset X \cup \{0\}$

$\therefore X' = X \cup \{0\}$

$X$ consists of an infinite sequence of closed concentric rings (annuli) centered at $0$ with outer radius $1/2n$ and inner radius $1/(2n + 1)$ . We are asked to find the closure $X'$ of this set. First, we have $X \subseteq X'$ for any set $X \subseteq \mathbb{R}^2$ . Also, since $X \subset B := B_{1/2}[0]$ (the closed ball of radius $1/2$ centered at $0$ ), we have $X' \subseteq B' = B$ . Therefore we only have to check $B \setminus X$ for potential accumulation points in $X' \setminus X$ .

Let $p \in B \setminus X$ . If $p = 0$ , then $\{(1/2, 0), (1/4, 0), (1/8, 0), / \dots, (1/2n, 0), \dots\}$ is a sequence of points in $X$ whose limit is $0$ , so $0 \in X'$ . If $p \neq 0$ , then there exists $n \in \mathbb{N}$ such that $1/2n < |p| < 1/(2n - 1)$ . Choose $\epsilon > 0$ such that $\epsilon < \min\{|p| - 1/2n, 1/(2n - 1) - |p|\}$ . Then $B_\epsilon(p)$ is an open neighborhood of $p$ that does not intersect $X$ , so $p$ is not an accumulation point of $X$ . Thus the only accumulation point of $X$ outside of $X$ is $0$ , and $X' = X \cup \{0\}$ .