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Math Help - find all limit points of a set

  1. #1
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    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?

    Thanks in advance
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    Re: find all limit points of a set

    EDIT: forget it
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    Re: find all limit points of a set

    Quote Originally Posted by Nikita2011 View Post
    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 \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 (x-\tfrac12\delta,x+\tfrac12\delta), and deduce that x cannot be a limit point of A.
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    Re: find all limit points of a set

    Quote Originally Posted by Opalg View Post
    Let x be a real number that is not of the form 1/n, and let \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 (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 \frac1{n} and \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 \delta= \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.
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    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 \tfrac1p<x<\tfrac1{p-1} for some natural number p>1.

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

    Thus m\leqslant 2/\delta. The same argument with m and n interchanged shows that 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.
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    Re: find all limit points of a set

    Quote Originally Posted by Opalg View Post
    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 \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.
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    Re: find all limit points of a set

    Quote Originally Posted by Nikita2011 View Post
    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 \tfrac1p<x<\tfrac1{p-1}, except that in this case you would want to take p=1. But then \tfrac1{p-1} = \tfrac10, which isn't allowed. So you just have to make some small adjustments to the argument, replacing the inequalities \tfrac1p<x<\tfrac1{p-1} by the single inequality 1<x.
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