# Find all limit points of $M=\left \{ \frac{1}{n}+\frac{1}{m}+\frac{1}{k} : m,n,k \in \mathbb{N} \right \}$ in space $(M,\rho_{e})$

Find all limit points of $M=\left \{ \frac{1}{n}+\frac{1}{m}+\frac{1}{k} : m,n,k \in \mathbb{N} \right \}$ in $(M,\rho_{e})$. I've founded, by building proper sequences, that points in the set$L=\left \{ \frac{1}{a} : a \in \mathbb{N} \right \}\cup \left \{ \frac{1}{a}+\frac{1}{b} : a,b \in \mathbb{N} \right \}$ are limit points, but what about other points from the set $M\setminus L$ ?

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In general limit point of $M$ doesn't had to belong to $M$. In fact $0$ is a limit point too. – Norbert May 9 '12 at 10:29
of course, but I'm interested only in limit points from the set M – Qbik May 9 '12 at 10:31

Your list is complete. And if you don't restrict your limit points to those belonging to $M$, you can make it complete just by adding $0$ to the list. This is the special case $n=3$ of the following:

For integer $n \ge 0$, let $M_n$ denote the set of real numbers expressible as the sum of exactly $n$ reciprocals of positive integers:

$$M_n = \left\{\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}:a_i \in \mathbb N\right\}$$

(with $M_0 = \{0\}$). Then the set of limit points of $M_n$ is equal to $$L_n = \bigcup_{0 \le r < n}M_r$$

(with $L_0 = \emptyset$).

Proof: If $x \in L_n$, then clearly it is a limit point of $M_n$, because $x \in M_r$ for some $r < n$, so $x$ is the limit as $t \to \infty$ of

$$\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_r} + \frac{1}{t} + \ldots + \frac{1}{t} (n-r \text{ times})$$ Now we use induction to show the converse, namely that if $x \notin L_n$, then $x$ is not a limit point of $M_n$. This is true for $L_0$, which has no limit points at all. So suppose it is true for $r < n$, i.e. for $0 \le r < n$, the set of limit points of $M_r$ is equal to $L_r$.

Now take $x \notin L_n$. We want to show that it can't be a limit point of $M_n$.

For $r < n$, $x \notin L_r$, so by the inductive hypothesis, $x$ is not a limit point of $M_r$. Also, $x \notin M_r$ by the definition of $L_r$. So there exists $\epsilon_r > 0$ such that the interval $[x-\epsilon_r,x+\epsilon_r]$ has no point in common with $M_r$. Putting $\epsilon = \min_{r<n}\epsilon_r$, we get $\epsilon>0$ such that for all $r < n$, the interval $[x-\epsilon,x+\epsilon]$ has no point in common with $M_r$. In words: any sum of up to $n-1$ reciprocals is at least $\epsilon$ distant from $x$.

Thus any sum of $n$ reciprocals

$$\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}$$

at least one of which is less than $\epsilon/2$, must be at least $\epsilon/2$ distant from $x$. So the only such numbers in the interval $[x-\epsilon/2,x+\epsilon/2]$ must have $1/a_i \ge \epsilon/2$ for all $i$, and so $a_i \le 2/\epsilon$ for all $i$. Hence there are only a finite number of such sums in this interval, and $x$ is not a limit point of $M_n$.

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great, thanks for proof of generalized problem – Qbik May 13 '12 at 13:00
It's a little late, so it is a bit redundant, but $M_r$ for $r\neq 0$ form an increasing sequence (because $1/a=1/(2a)+1/(2a)$, for instance), so actually $L_n=\{0\}\cup M_{n-1}$. – tomasz Feb 28 '15 at 5:47
Great proof, it could maybe read easier but overall hard to complain. This is the only complete and rigorous proof I can find anywhere online. In particular nobody wrote down the $n=2$ case specifically. – Gregory Grant Apr 2 '15 at 13:39