Limit point of set $\{\sqrt{m}-\sqrt{n}:m,n\in \mathbb N\} $ How can I calculate the limit points of set $\{\sqrt{m}-\sqrt{n}\mid m,n\in \mathbb N\} $? 
 A: Intuition says that every real number is a limit point. So given a real number $a$, we want to show that there are integers $m$ and $n$ such that $\sqrt{m}-\sqrt{n}$ is close to $a$. Without loss of generality we may assume that $a\ge 0$.
Given $\epsilon \gt 0$, we want to produce $m$ and $n$ such $|(\sqrt{m}-\sqrt{n})-a|\lt \epsilon$.
One idea is to note that $\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$. So there is an integer $d=d(\epsilon)$ such that $0\lt \sqrt{d+1}-\sqrt{d} \lt \epsilon$.
Now consider the numbers $k(\sqrt{d+1}-\sqrt{d})=\sqrt{k^2d+k^2}-\sqrt{k^2d}$, as $k$ ranges over the positive integers.  For every $a\ge 0$, there is a positive integer $k$ such that $k(\sqrt{d+1}-\sqrt{d})$ is at distance less than $\epsilon$ from $a$. 
A: Let $a$ be a real number. To show that $a$ is a limit point, it's enough to show that for any $N>0$, some $\sqrt{m} - \sqrt{n}$ is within $1\over N$ units of $a$.


*

*Choose $m=M^2$ large enough so that the consecutive differences in
the sequence
$M=\sqrt{m},\sqrt{m+1},\sqrt{m+2},\dots,\sqrt{m+2M+1}=M+1$ are all
less than $1\over N$. This will be the case if the derivative of $\sqrt{x}$ at $x=m$ is less than ${1\over N}$, or when $M>{\lceil{N\over2}\rceil}$.

*Let $n=(M-\lfloor{a}\rfloor)^2$. Then $\sqrt{m} - \sqrt{n}=\lfloor{a}\rfloor$ and $\sqrt{m+2M+1} - \sqrt{n}=\lfloor{a}\rfloor+1$

*Then $\sqrt{m+i} - \sqrt{n}$ must be within $1\over N$ of $a$ for some $i$.

A: The answer is $\mathbb{R}$, as we can see here, for $x\in (0,\infty)$ and $\epsilon >0$, there are $n_0 , N \in \mathbb{N}$ such that $\sqrt{n_0 +1}-\sqrt{n_0} <1/N<\epsilon /2$. Now we can divide $(0,\infty)$ to pieces of length $1/N$, so there is $k\in \mathbb{N}$ such that $k(\sqrt{n_0 +1}-\sqrt{n_0})\in N_{\epsilon} (x)$.
The proof for $(-\infty , 0)$
is the same.
