Find the closure of $A=\left\{\frac1n+\frac1m : m,n\in\mathbb{N}\right\}$. 
Find the closure of  $A=\left\{\displaystyle\frac{1}{n}+\displaystyle\frac{1}{m} : m,n\in\mathbb{N}\right\}$.

I think that $\bar{A}=\{0\}\cup\left\{\displaystyle\frac{1}{n}:n\in\mathbb{N}\right\} \cup A $. But I don't know how to prove $\bar A\subseteq\{0\} \cup\left\{\displaystyle\frac{1}{n}:n\in\mathbb{N}\right\} \cup A$
Thanks a lot!

Consider a sequence $x_p = \frac 1{n_p}+\frac 1{m_p}$, $p\ge 0$ such as $x_p\to x$.


*

*if either $n_p$ or $m_p$ is bounded: $x\in \{\frac 1n:  n\in\Bbb N\}\cup A$;

*otherwise, $x_p\to 0$.


hence $\bar A\subseteq\{0\} \cup\{\frac{1}{n}:n\in\mathbb{N}\} \cup A$
