Compact or open $\{0\}\cup\{\frac1n + \frac1m | m,n \in N\})$ in R? 
Compact or open  $\{0\}\cup\{\frac1n + \frac1m | m,n \in N\})$ in R ?

The question is straight forward
There exists no interval about $2\in S$ that has only elements of S. Not open
What about compact?
EDIT: Sorry my reasoning for not compact was incorrect
 A: Hint: Let
$$S=\{ 0\} \cup \{ \frac{1}{n} \ : \ n\in \mathbb{N}\}$$
and
$$ T = \{0\} \cup \{ \frac{1}{m} + \frac{1}{n} \ : \ n, m \in \mathbb{N}\}.$$
Furthermore, let
$$f: \mathbb{R} \times \mathbb{R}\rightarrow \mathbb{R}, \ (x, y) \mapsto x+y $$
Then we can proceed the following way:
1.) Show that $S$ is compact.
2.) Conclude that $S\times S$ is compact.
3.) Show that $f$ is continuous.
4.) Show that $f(S\times S)=T$.
5.) Deduce from 2.), 3.), and 4.) that $T$ is compact.
A: Yep the set is indeed not open.
As for compactness, notice that your set is a closed subset of the compact set $[0,2]$.
A: Let $\mathcal U$ be an open cover of your set $A$. Then some  $U_0\in\mathcal U$ contains $0$ and also contains an open ball of radius $r_0>0$ around $0$.  For the finitely many $n$ with $n\le\frac 1{r_0}$, there exists  $U_n\in\mathcal U$ with $\frac 1n\in U_n$ (because $\frac1n=\frac1{2n}+\frac1{2n}\in A$). Again, $U_n$ contains an open ball of radius $r_n$ around $\frac 1n$. Let $r=\min\{\frac12r_0,r_1,r_2,\ldots\}$ (only finitely many!).
For each of the finitely many $\frac 1n+\frac1m$ with both $n\le \frac1r$ and $m\le \frac 1r$, pick $U_{n,m}\in\mathcal U$ with $\frac1n+\frac1m\in U_{n,m}$
Consider an arbitrary point $a\in A$. 


*

*If $a=0$, then $a\in U_0$. 

*If $a=\frac1n+\frac1m$ with  $\max\{n,m\}\le \frac1r$, we have $a\in U_{n,m}$

*If $a=\frac1n+\frac1m$ with $n\le \frac1r< m$, we have $a\in B_r(\frac 1n)\subseteq U_n$; similar for $m\le \frac 1r<n$.

*If $a=\frac1n+\frac1m$ with $\min\{n,m\}>\frac1r$, we have $a<2r\le r_0$ and so $a\in U_0$.
Thus we have found a finite subcover of $\mathcal U$.
We conclude that $A$ is compact.
