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

Question

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

Answer 1

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.

Answer 2

Yep the set is indeed not open.

As for compactness, notice that your set is a closed subset of the compact set $[0,2]$.

Answer 3

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.