What is the closure of $A$ in $\mathbb{R}$ If the set $A$ has been as follows, find the closure of $A$ in $\mathbb{R}$.
$$A=\left\lbrace \frac{m+n}{2m+n+1}: m,n\in\mathbb{N}\right\rbrace$$

The key is to figure out what fractions can be written in the form $\frac{m+n}{2m+n+1}$ for positive integers $m$ and $n$. Let $a=m+n$ and $b=2m+n+1$, where $m,n\in\Bbb Z^+$. Then $b-a=m+1$, so 
$$n-1=(m+n)-(m+1)=a-(b-a)=2a-b\;.$$
Thus, $2a-b\ge 0$, i.e., $b\le 2a$, and we have $a+2\le b\le 2a$.
Conversely, suppose that $a,b\in\Bbb Z^+$ are such that $2\le a<a+2\le b\le 2a$. Let $n=2a-b+1\ge 1$ and $$m=a-n=a-(2a-b+1)=b-a-1\ge 1\;;$$
then $m,n\in\Bbb Z^+$, $a=m+n$, and $b=2m+n+1$. This shows that
$$A=\left\{\frac{a}b:a,b\in\Bbb Z^+\text{ and }2\le a<a+2\le b\le 2a\right\}\;.$$


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*What is $\inf A$?  

*What is $\sup A$?  

*Show that $A=[\inf A,\sup A)\cap\Bbb Q$. Once you have that, getting $\operatorname{cl}A$ should be immediate.

