Find a set of limit point of $A=\{ \frac{1}{m}+\frac{1}{n} : m,n \in \Bbb{N} \}$ in $\Bbb{R}$ by using the following way.

general-topology

Find the set of limit points of $A=\{ \frac{1}{m}+\frac{1}{n} : m,n \in \Bbb{N} \}$ in $\Bbb{R}$.

I want to solve this question by the following way, but I need help.

We know that $0< \frac{1}{m}+\frac{1}{n} \leq 2$, for all $m,n \in \Bbb{N}$.

If $x<0$, then $x\notin A’,$ because $(x-1,0)$ is a neighborhood of $x$ and $(x-1,0) \cap A=\emptyset$
If $x>2$, we have $(2,x+1)\cap A=\emptyset$, so $x\notin A’$.
If $x=2$, $(\frac{3}{2},3) \cap (A\setminus \{2\})=\emptyset $, then $x\notin A’$.
If $x=\frac{1}{n}$, where $n\in \Bbb{N} \setminus \{1\}$, let $\epsilon =\min \{x,1-x\}>0$, then we have $x+\frac{\epsilon}{2}\in (x-\epsilon,x+\epsilon)\cap (A\setminus \{x\}) \neq \emptyset$, so $x\in A’$.

What about other cases:

If $x=0$
If $0<x<1$ and $x\neq \frac{1}{n}$
If $x=1$
If $1<x<2$

Thanks.

Solution 1:

Let $L$ denote the set of limit points of $A$ and $B = \{\frac{1}{k} \mid k \in \mathbb{N} \}$. For $x \in \mathbb {R}$ and $r > 0$ let $U_r(x) = (x-r,x+r)$. Finally, let $a_{m,n} = \frac{1}{m} + \frac{1}{n} \in A$.

We shall show that $L = \{ 0 \} \cup B$. This will be done in five steps.

1) $B \subset L$.

For any $k \in \mathbb{N}$ and $r > 0$ the open interval $U_r(\frac{1}{k})$ contains infinitely many elements $a_{k,n}$, hence $\frac{1}{k} \in L$. You have already proved this for $k > 1$.

2) $0 \in L$.

For any $r > 0$ the open interval $U_r(0)$ contains infinitely many $a_{m,n}$; simply take $m, n > \frac{2}{r}$.

3) $(-\infty, 0) \cap L = \emptyset$.

You have proved this.

4) $(1,\infty) \cap L = \emptyset$.

Let $x > 1$ and $r = \frac{x-1}{2} > 0$. Then $x = 1 + 2r$ and $U_r(x) = (1+r, 1+3r)$. There are at most finitely many $a_{m,n} > 1 +r$. In fact, if both $m,n > 1$, then $a_{m,n} \le 1$. Hence we may assume w.l.o.g. that $m = 1$. But then $a_{1,n} \le 1 + r$ for $n \ge \frac{1}{r}$. This shows that $x \notin L$.

5) $(\frac{1}{k+1},\frac{1}{k}) \cap L = \emptyset$ for all $k \in \mathbb{N}$.

Let $x \in (\frac{1}{k+1},\frac{1}{k})$ and $r > 0$ such $\frac{1}{k+1} < x - r < x + r < \frac{1}{k}$. What are the requirements for $a_{m,n} \in U_r(x)$? We get $a_{m,n} \ge \frac{1}{k}$ if $m \le k$ or $n \le k$. Thus we must have $m,n > k$. Moreover, if both $m, n \ge 2(k+1)$, then $a_{m,n} \le \frac{1}{k+1}$. Hence we may assume w.l.o.g. that $m < 2(k+1)$. Thus $k+1 \le m \le 2k+1$. But for each of these $m$ there are at most finitely many $n$ such that $a_{m,n} > x - r$. In fact, for $n \ge \frac{1}{x - r - \frac{1}{k+1}}$ we get $a_{m,n} \le a_{k+1,n} \le x - r$. This shows $x \notin L$.