let $S=\{ \frac{1}{m}+\frac{1}{n}:m,n\in \mathbb{N}\} \cup \{0\}$. then $S$ is compact set in $\mathbb{R}$. True or False.

Question

let $S=\{ \frac{1}{m}+\frac{1}{n}:m,n\in \mathbb{N}\} \cup \{0\}$.

then $S$ is compact set in $\mathbb{R}$. True or False.

since derived set of $S=\{\frac{1}{n}:n\in \mathbb{N}\} \cup\{0\}$ . and this is not a subset of $S$. Hence $S$ is not closed set and not compact set in $\mathbb{R}$

But in my book it is given to compact .Is my argument valid ?

Answer 1

and this is not a subset of $S$.

What makes you think so? Note that every (nonzero) element of the derived set can be written as $\frac{1}{n}=\frac{1}{2n}+\frac{1}{2n}$ and hence it belongs to $S$.

So $S$ actually is closed. And so it is compact because it is bounded.

Answer 2

$S$ is obviously bounded, and it is closed because $\operatorname{acc}(S)=\{1/m\,|\,m\in\mathbb{N}\}\cup\{0\}\subset S$, so it is compact by Bolzano's theorem.