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I've trying to prove the next statement: If $$A=\{1/n:n\in\mathbb{N}\}$$ show that $A'=\{0\}$ on $(\mathbb{R},|\cdot|)$ as a metric space. I've already shown that $\{0\}\subset A'$ but I can't prove that if $0

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    You wrote $\{0\}\subset A'$ but what about the title of your post? – 2017-02-27
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    sorry, I have always trouble for the titles, yes... thanks. – 2017-02-27
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    Maybe the title is, Show that $0$ is the only limit point. – 2017-02-27
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    Notation clarification: $(\mathbb{R}, |\cdot|)$ refers to the Euclidean metric? – 2017-02-27
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    yes, that's right. – 2017-02-27

2 Answers 2

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You already showed that $0\in A'$. Suppose $x\in\Bbb R$ such that $x\neq 0$. The case $x<0$ or $x\geq 1$ is easy to handle. Lets take at the case (where you got a problem) where $0x>\frac{1}{n+1}$. So, we take $G=\bigg(\frac{1}{n+1},\frac{1}{n}\bigg)$. In this case, $$G\cap A=\emptyset.$$ Hence, $x\notin A'$.

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Choose the smallest $N$ such that $1/N