2

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!

Martin Sleziak
  • 53,687
EQJ
  • 4,369

1 Answers1

5

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

mookid
  • 28,236