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Problem:find the number of limit points {$ \frac{1}{m} +\frac{1}{n}:m,n \in \Bbb N$}

Solution:$$ \lim_{(m,n)\to\infty}\frac{1}{m} +\frac{1}{n}$$

$=0$

So there is one limit point

Am I doing right ?

Answer is infinitely many limit points

Community
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rst
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    We're missing the limit points $$\frac{1}{m}+\lim_{n \rightarrow \infty} \frac{1}{n}=\frac{1}{m}$$ for all $m \in \mathbb{N}$. It's not apparent to me if there are others missing. – Rebecca J. Stones Sep 18 '13 at 12:01
  • Actually the hard part is where you have to prove that there are no other limit points except ${ 1/n: n \in\mathbb{N} } \cup {0}$. – user66733 Sep 18 '13 at 12:30
  • @some, not all that hard, as you know at least one summand must approach zero. – Gerry Myerson Sep 18 '13 at 12:32
  • Answer is infinitely many limit points – rst Sep 18 '13 at 12:32
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    @GerryMyerson: I don't know what approach you have in mind, but I solved this question as a problem in my Analysis I exam by showing that if $0<x<1$ is a real number not of the form $1/n$ we can find a neighborhood that doesn't contain any points other than itself and the proof wasn't really trivial. Now would you please tell me what you have in mind? – user66733 Sep 18 '13 at 12:48
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    Pick some large number $N$. There are only finitely many numbers $(1/m)+(1/n)$ where both $m$ and $n$ are less than $N$, so eventually one of the two summands must be less than $1/N$. But $N$ was arbitrary, so the smaller summand must approach zero. That just leaves the larger summand which, if it's approaching a limit, must be approaching zero, or eventually constant. – Gerry Myerson Sep 18 '13 at 13:00
  • A COMPLETE PROOF IS HERE: http://math.stackexchange.com/questions/930646/to-find-limit-points-of-the-set-frac1n-frac1m-n-m-1-2-3/1217297#1217297 – Gregory Grant Apr 02 '15 at 14:00

1 Answers1

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the limit points are only $\{\frac{1}{n}: n \in \mathbb N\} \cup 0$

as $1+\frac{1}{n} \rightarrow 1$

similarly $\frac{1}{m}+\frac{1}{n}\rightarrow \frac{1}{m}$ where $m$ is fixed and $n$ varies.

jim
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