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I want to find $\overline{A}=\overline{\{\frac1n, n\in\mathbb{N}\}}$ in $(\mathbb{R},|.|)$

i know that $\{\frac1n, n\in\mathbb{N}\}\subset \overline{A}$

let $x\in \mathbb{R}\setminus{A}$, $x\in \overline{A}\Leftrightarrow \forall \varepsilon>0, ]x-\varepsilon,x+\varepsilon[\cap A\neq \emptyset$

How to find $x$ ?

1 Answers 1

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Try a few cases.

If $x < 0$, then there is plenty of space between $x$ and any element of $A$, so you can find an $\varepsilon > 0$ so that $(x - \varepsilon, x + \varepsilon) \cap A = \varnothing$.

If $x > 0$, then one element $a \in A$ is closer to $x$ than any other (and $a \neq x$). Use that fact to find an $\varepsilon$ so that $(x - \varepsilon, x + \varepsilon) \cap A = \varnothing$.

The only remaining case is $x=0$. But is there an $\varepsilon > 0$ so that $(-\varepsilon, \varepsilon) \cap A = \varnothing$?

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    If $x<0$ how to choose $\varepsilon$ to get that $]x-\varepsilon, x+\varepsilon[\cap A=\emptyset$ ? thank you – 2017-02-14
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    Every thing in $A$ is bigger than zero. So you just need to make sure $x+\varepsilon < 0$. – 2017-02-14
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    If $x>0$ , if $x>1$ hen we choose $\varepsilon – 2017-02-14
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    Sure, that works for $x > 1$ – 2017-02-14
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    and for $\frac1n$\varepsilon $ ? – 2017-02-14
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    Well, $x$ is closer to one of those than it is the other. So use that smaller distance. – 2017-02-14
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    So i can take $\varepsilon< \frac{\frac1n-\frac{1}{n+1}}{2}$ for example, and please how we explain that $\forall \varepsilon>0, ]-\varepsilon,\varepsilon[\cap A\neq \emptyset$ without using the fact that $\frac1n$ converge to 0 in $(\mathbb{R},|.|)$ – 2017-02-14
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    Please, is $\overline{A}$ compact ? – 2017-02-14