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Let be $A$ subset of a metric space $(X,d)$

Definiton. Point $x\in X$ is adherent point (it can also have any other definition but sorry and forgive me if I wrong) of set $A$ if $$T(x,r)\cap A\neq \phi, $$ for all r>0.

Set of all adherent points of the set A is called slosure and is denoted by $\overline A.$

Please if you can help me to find the closure of $\mathbb{Z}$ and $\mathbb{Q}$ in $\mathbb{R}$.

Previously, thank you for your solution

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  • $\begingroup$ What you’ve defined is normally called the closure of the set $A$. $\endgroup$ – Brian M. Scott Nov 6 '13 at 13:05
  • $\begingroup$ ok sir thanks, but if you can please help me $\endgroup$ – Madrit Zhaku Nov 6 '13 at 13:06
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HINTS:

  • If $x\in\Bbb R\setminus\Bbb Z$, then there is a unique integer $n$ such that $n<x<n+1$; can you find an $r>0$ such that $T(x,r)\cap\Bbb Z=\varnothing$?

  • For any $x\in\Bbb R$ you know that $T(x,r)=(x-r,x+r)$. Does that open interval contain a rational number?

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  • $\begingroup$ sir if you have time and opportunity was jutem tell me a complete and detailed because I do not understand metric spaces very well, thank you very much $\endgroup$ – Madrit Zhaku Nov 6 '13 at 13:10
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Alternatively you can think that $\Bbb Z $ is closed in $\Bbb R$(more easily think that $\Bbb R-Z$ is open) and thus $\overline Z=Z$.

Also for $\Bbb Q$ you have the density of rationals (also irrationals) in $\Bbb R$ due to the fact that if $a,b\in \Bbb R$ then there is a $q\in \Bbb Q:a<q<b$. So $\Bbb Q$ is dense in $\Bbb R$ and thus $\overline {\Bbb Q}=\Bbb R$

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