I am reading a general topology book written by Osamu Takenouchi.
Theorem 3:
$\overline{A}$ is the minimum closed set which contains $A$.
I found his proof of the above theorem was not correct. But if I can prove the following proposition below, I can correct his error.
Please prove the following proposition.
Let $A \subset \mathbb{R}^n$.
Let $a$ be a limit point of $A$. ( $a$ is a point of $\mathbb{R}^n$ such that $(B(a ; \epsilon) - \{a\}) \cap A \neq \emptyset$ for any positive real number $\epsilon$.)
Let $\epsilon$ be a positive real number.
Let $B(a ; \epsilon)$ be an open ball in $\mathbb{R}^n$.
Prove that $B(a ; \epsilon)$ contains a point $b$ which is not equal to $a$ and is not an isolated point of $A$.
Your Proposition is not correct:
Consider $A=\{1/n|n\in \mathbb{N}\}\subset \Bbb{R}$. Then $0$ is the only limit point of $A$ but no $B(0,\epsilon)$ contains such $b$ as you claimed.