# Finding the closure of $\mathbb{Z}$ and $\mathbb{Q}$ in $\mathbb{R}$

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

• What you’ve defined is normally called the closure of the set $A$. – Brian M. Scott Nov 6 '13 at 13:05

• 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?
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 $\overline {\Bbb Q}=\Bbb R$