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In my efforts to improve my mathematical rigour I am trying to understand each precise step in the proofs I attempt as well as look at the solution for. In this instance I am trying to understand the process in establishing whether a point is a limit point in $\mathbb{R}$. As the title says I'm trying to understand how finiteness is concluded in the following proof: Find the limit points of the set $\{ \frac{1}{n} +\frac{1}{m} \mid n , m = 1,2,3,\dots \}$

In the first and second solution both authors arrive at the conclusion: $$0<\frac{1}2(x-\frac{\epsilon}2)<\frac1m$$ and $$0<\frac{\epsilon}2<\frac1n$$

How do these establish that there are only a finite number of elements in the respective sets? What am I missing in understanding finiteness?

And to follow this to make sure my reasoning is correct, once we establish that there is only a finite number of elements, then by definition we can find a $\delta$ - neighbourhood around any of the finite points such that it contains no other points of $S$ which then implies the point $x$ is not a limit point.

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D.C. the III
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1 Answers1

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Both of the answers to your questions identify the set $A=\{0\}\cup\{\frac{1}{n} : n \in \mathbb{N}\}$ as the complete list of limit points of your set. Checking that these are limit points is easy, but it remains to show there are no others.

The explanations choose a point $x$ not in $A$, build a special interval around it, and then conclude the inequalities you mention for any point in this special interval which has the form $\frac{1}{n}+\frac{1}{m}$. The inequalities bound $m$ and $n$ above, which means there is a strictly finite number of points of the form $\frac{1}{n}+\frac{1}{m}$ in this special interval, and therefore no point in it can be a limit point; it it were, there would be an infinite sequence of terms of that form converging to it, which is not possible.

I hope that helps!

R_B
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  • This does give a bit more clarity yes thank you. But just because a set is bounded it doesn't mean it is finite.....or in this instance that is the case?...... About your first statement: showing that ${0 }$ is a limit point, how do we show it? I understand perfectly what it means visually but I've been thinking about it all day and the only idea that comes to my mind is to do something using contradiction, but I still get stuck... – D.C. the III Mar 03 '19 at 23:55
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    Using the definition of limit point: "$x$ is a limit point of a set $S$ if and only if there exists an infinite sequence of points $x_k\in S$ such that $x_k$ converges to $x$". We can construct one such sequence by letting the first term of the sequence be $x_1=\frac{1}{m}+\frac{1}{n}$ for any choice of positive integers $m$ and $n$ . The next terms are constructed by alternatively increasing the values of $m$ and $n$ by $1$: $x_2=\frac{1}{m+1}+\frac{1}{n}$, $x_3=\frac{1}{m+1}+\frac{1}{n+1}$, $x_4=\frac{1}{m+2}+\frac{1}{n+1}$... All the terms are in $S$, but the sequence converges to $0$. – R_B Mar 04 '19 at 00:07
  • So I'm going through Baby Rudin and up to this point we haven't covered sequences. So in regards to your characterization I fully agree that would be a way to show that 0 is a limit point. Is there a way to show 0 is a limit point without the sequential characterization? i.e by definition that: "$x$ is a limit point of the set $E$ if for every neighbourhood of $x$ there exists a point $q \in E$ such that $x \neq q$" ? I could see how it gets complicated really quickly.... – D.C. the III Mar 04 '19 at 00:26
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    Let's pick any neighbourhood of zero, and show that there is a point of your set $E$ in it. A neighbourhood is an open set containing zero, so there exists (by definition) an open ball of radius $r$ which contains zero and is contained in the neighbourhood. Now find $n$ large enough such that $\frac{1}{n}+\frac{1}{n}<r$; this point is in $E$, it is not $0$, and it is also in the ball of radius $r$, therefore in the neighbourhood too. – R_B Mar 04 '19 at 00:33
  • Awesome!!!! Thank you for all your help.....this makes clear sense to me and since it is for any $n\ in \mathbb{N}$ this shows me what is needed to generalize for these existence proofs. – D.C. the III Mar 04 '19 at 00:39