Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $M\setminus L$ ?

share|improve this question
    
In general limit point of $M$ doesn't had to belong to $M$. In fact $0$ is a limit point too. –  userNaN 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

1 Answer 1

up vote 4 down vote accepted

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 $M_n$.

share|improve this answer
    
great, thanks for proof of generalized problem –  Qbik May 13 '12 at 13:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.