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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 ?

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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.

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What definition are you given for limit points? The definition I'm most familiar with is precisely that every open ball around the limit point should intersect S nontrivially. Which is precisely your assumption. But if you have a different definition of limit point, there might be more to do.

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  • $\begingroup$ $ 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 $S$ $\endgroup$ – Sabhrant Sep 10 '19 at 18:16
  • $\begingroup$ Or in other words, $(B_{\delta}(x)-\{x\}) \cap S \neq \emptyset $ for any $ \delta >0 $ $\endgroup$ – Sabhrant Sep 10 '19 at 18:17

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