# Thread: find all limit points of a set

1. ## find all limit points of a set

Hi all,
the problem is to find all the limit points of the set A = { 1/n + 1/m : n,m are positive integers }. I already know that 0 and {1/n : n is a positive integer} are limit points of A. but Are there any other limit points or not? if yes, what are they? If no, how can I show that there is no other limit point of A?

2. ## Re: find all limit points of a set

EDIT: forget it

3. ## Re: find all limit points of a set

Originally Posted by Nikita2011
Hi all,
the problem is to find all the limit points of the set A = { 1/n + 1/m : n,m are positive integers }. I already know that 0 and {1/n : n is a positive integer} are limit points of A. but Are there any other limit points or not? if yes, what are they? If no, how can I show that there is no other limit point of A?
Let x be a real number that is not of the form 1/n, and let $\displaystyle \delta$ be the distance from x to the nearest point of the form 1/n. Show that only finitely many points of A can lie inside the interval $\displaystyle (x-\tfrac12\delta,x+\tfrac12\delta)$, and deduce that x cannot be a limit point of A.

4. ## Re: find all limit points of a set

Originally Posted by Opalg
Let x be a real number that is not of the form 1/n, and let $\displaystyle \delta$ be the distance from x to the nearest point of the form 1/n. Show that only finitely many points of A can lie inside the interval $\displaystyle (x-\tfrac12\delta,x+\tfrac12\delta)$, and deduce that x cannot be a limit point of A.
Thank you for responding. Yes, of course I know I want that. it suffices to show that there exists one neighborhood that contains only a finite number of points of A because It's easy to show that if a neighborhood contains a finitely many points of A then by choosing the minimum distance between the center and those points I can find a neighborhood that contains no points of A. I've been trying to do that since 5 days ago and can't believe that It hasn't been solved yet. I tried your suggestion but still I'm not 100% sure if I have it correctly.

first of all I've shown that there is no point of A between $\displaystyle \frac1{n}$ and $\displaystyle \frac1{n+1}$. my argument is that if there is a point of A between these two fractions then since any point of A is of the form 1/n+1/m we must have 1/n<= 1/n+1/m <= 1/(n+1) or 1/(n+1)<= 1/n+1/m <= 1/n. both cases lead to an absurd statement, therefore, contradiction.(the former case is obviously wrong because 1/n>1/n+1 and the later case forces that m must be negative which is a contradiction because m is a natural number) that means no point of A can lie between 1/n and 1/(n+1). now if I choose $\displaystyle \delta$=$\displaystyle \frac12(\frac1n-\frac1{n+1})$ and take a neighborhood centered at any point that is not of the form 1/n the neighborhood will be somewhere between 1/n and 1/(n+1) for some n and it can't contain any points of A because if it did, that would mean that there exists a point in A in the interval (1/(n+1),1/n) which violates what we've obtained earlier.
Is this a valid argument? Now what I can't understand is that why 0 is a limit point of A! I already have shown that there exists a non-constant sequence in A that converges to 0 and therefore 0 must be a limit point but by the argument above 0 can't be a limit point because it is not of the form 1/n.

5. ## Re: find all limit points of a set

We want to know whether x can be a limit point of the set A. In my previous comment, I ignored the case x=0 because you said that you had already shown that 0 is a limit point. We can also ignore the possibilities x<0 and x>1, since it is fairly obvious that these cannot be limit points of A. Any other point x that is not of the form 1/n must satisfy $\displaystyle \tfrac1p<x<\tfrac1{p-1}$ for some natural number p>1.

Let $\displaystyle \delta = \min\{\tfrac1{p-1}-x,x-\tfrac1p\}$, so that $\displaystyle \delta$ is the distance from x to the nearest point of the form 1/n. Suppose that $\displaystyle \tfrac1n+\tfrac1m$ lies in the interval $\displaystyle J = (x-\tfrac12\delta,x+\tfrac12\delta)$. If n<p then $\displaystyle 1/n\geqslant 1/(p-1)$, which is already greater than anything in J. So $\displaystyle \tfrac1n+\tfrac1m\notin J$. Therefore $\displaystyle n\geqslant p$, and so $\displaystyle \tfrac1n\leqslant\tfrac1p\leqslant x-\delta.$ If $\displaystyle m>2/\delta$ then $\displaystyle 1/m<\delta/2$, and therefore $\displaystyle \tfrac1n+\tfrac1m < (x-\delta)+\tfrac12\delta = x-\tfrac12\delta$. Hence $\displaystyle \tfrac1n+\tfrac1m$ is too small to belong to J.

Thus $\displaystyle m\leqslant 2/\delta.$ The same argument with m and n interchanged shows that $\displaystyle n\leqslant 2/\delta.$ So both m and n have to lie within a finite range, showing that only finitely many points of A lie in J.

6. ## Re: find all limit points of a set

Originally Posted by Opalg
We want to know whether x can be a limit point of the set A. In my previous comment, I ignored the case x=0 because you said that you had already shown that 0 is a limit point. We can also ignore the possibilities x<0 and x>1, since it is fairly obvious that these cannot be limit points of A. Any other point x that is not of the form 1/n must satisfy $\displaystyle \tfrac1p<x<\tfrac1{p-1}$ for some natural number p>1.
I don't understand why It's fairly obvious that x>1 and x<0 can't be limit points of A. I have no experience in Analysis and I'm self-studying Rudin, so I'm sorry if I'm asking naive questions. I understand rest of the argument. Thank you.

7. ## Re: find all limit points of a set

Originally Posted by Nikita2011
I don't understand why It's fairly obvious that x>1 and x<0 can't be limit points of A. I have no experience in Analysis and I'm self-studying Rudin, so I'm sorry if I'm asking naive questions. I understand rest of the argument. Thank you.
The result is obvious for x<0 because every point in A is positive, and so A can't have a negative limit point.

The result for x>1 is a bit more tiresome. The proof is essentially the same as in the case where $\displaystyle \tfrac1p<x<\tfrac1{p-1}$, except that in this case you would want to take p=1. But then $\displaystyle \tfrac1{p-1} = \tfrac10$, which isn't allowed. So you just have to make some small adjustments to the argument, replacing the inequalities $\displaystyle \tfrac1p<x<\tfrac1{p-1}$ by the single inequality $\displaystyle 1<x.$

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