# Prove or disprove: $x \in \overline{S} \iff B_{\delta}(x) \cap S \neq \emptyset$ for any $\delta > 0$

Let $$S\neq \emptyset \subseteq \mathbb{R}^n$$, prove or disprove: $$x \in \overline{S} \iff B_{\delta}(x) \cap S \neq \emptyset$$ for any $$\delta > 0$$

Here $$\overline{S}$$ denotes the closure of set $$S$$, $$\overline{S} = S \cup S_{L}$$ . $$S_{L}$$ denotes the set of all limit/accumulation points of $$S$$.

My attempt : $$(\Rightarrow)$$ Let $$x \in \overline{S}$$. Which means $$x \in S$$ or $$x \in S_{L}$$, which further implies that $$\exists \, B_{\delta}(x)$$ such that $$B_{\delta}(x) \cap S \neq \emptyset$$ for any $$\delta > 0$$. $$(\Leftarrow )$$ Now let $$B_{\delta}(x) \cap S \neq \emptyset$$ for any $$\delta > 0$$. This implies $$x \in S$$ or $$x \in S_{L} \implies x \in \overline{S}$$.

I am not sure why $$B_{\delta}(x) \cap S \neq \emptyset$$ implies that $$x$$ should be in $$S$$ or $$\overline{S}$$. Is this proof correct ? Are there any other ways to solve this ?

If $$B(x,\delta)\cap S\neq\emptyset$$ for all $$\delta>0$$, you can find $$(x_n)$$ such that $$|x-x_n|\leqslant\frac{1}{n+1}$$ and $$x_n\in S$$ for all $$n\in\mathbb{N}$$. I have been taught that the definition of $$\overline{S}$$ is $$\overline{S}=\{x\in\mathbb{R}^n\ |\ \forall\varepsilon >0,\,B(x,\varepsilon)\cap S\neq\emptyset \}$$, you definition seems kinda strange to me.
• $x \in \mathbb{R}^n$ is a limit point of $S$ if for every $\delta > 0$, $B_{\delta}(x) \cap S$ should have infinitely many points of $S$ – Sabhrant Sep 10 '19 at 18:16
• Or in other words, $(B_{\delta}(x)-\{x\}) \cap S \neq \emptyset$ for any $\delta >0$ – Sabhrant Sep 10 '19 at 18:17