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I have seen examples of sets that have these properties, like:

$$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}$$ And it is clear that 0 and all 1/n are limit points. However, how does one show that there are no other limit points?

I am completely stuck here. I have found many examples of sets that have these properties, but always run into trouble showing there are no other limit points. For example, I also tried:

$$A=\{0\}\cup\left\{\frac1n:n\in\Bbb N\right\}\cup\left\{\frac{n}{kn+1}:k,n\in\Bbb N\right\}$$

And showed that 0 and all 1/n are limit points, but I am lacking in how I can show that the $n/(kn+1)$ terms are not. To me, this means showing that there is some deleted ball around each of them which contains no element of A.

Thank you so much for your help!

mb7744
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4 Answers4

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Consider the first set $A$ in your question. Pick a sequence of points of $A$ that converges to some point in $\mathbb R$. Either the sequence has infinite intersection with some interval $[1/n, 1/(n+1))$, or it does not.

If the first option holds, check that the sequence converges to $1/n$, because a subsequence of it does (namely, the subsequence whose values are in the interval).

The other option is that each of these intervals contains only finitely many points of the sequence. Check then that the sequence converges to $0$.

By the way, much more complicated countable compact subsets of $\mathbb R$ are possible. The typical examples are modeled after the countable successor ordinals. See also this answer.

Community
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Andrés E. Caicedo
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  • Thank you very much, but please see my above comment. – mb7744 Oct 21 '13 at 00:39
  • It is easy to adapt the language of sequences to this setting. Anyway, let's do this directly: For the first $A$ in your question, argue as you suggest: For any $p\notin A$, show that there is a neighborhood of $p$ (an interval centered at $p$) that misses $A$ entirely. For $p\in A$ but not $0$ and not $1/n$ for some $n$ show that some neighborhood of $p$ only meets $A$ at $p$. For example, if the point $p$ lies between $1/n$ and $1/(n+1)$, check that it is $1/n +1/m$ for some $m$, and use $m$ to find $\epsilon$ so that the interval of radius $\epsilon$ centered at $p$ is as wanted. – Andrés E. Caicedo Oct 21 '13 at 00:53
  • Thanks, is this what you meant? For $p=1/a + 1/b$, choose c to be the biggest c such that p < 1/c. Then $1/(c+1) < p < 1/c$, then let the radius

    $$r=\frac{1}{2}\text{min}\left { p-\frac{1}{c+1},\frac{1}{c}-p \right }$$ ?

     – mb7744 Oct 21 '13 at 02:05
  • Wait now I am severely confused. How do I show that no other members of A of the form 1/a + 1/b are in that interval? I've pretty much been at this point all day. – mb7744 Oct 21 '13 at 02:10
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I'll tackle your first question. Let $x$ be a limit point of $A$. Then there exist sequences $(n_i)$ and $(m_i)$ with $1/n_i + 1/m_i \to x$ as $i \to \infty$ and $1/n_i + 1/m_i \neq x$ for all $i$. $(n_i)$ has a subsequence along which $n_i \to \infty$ or $n_i$ is constant. In the former case, along a subsequence, $1/n_i \to 0$, so either $m_i \to \infty$ and $x=0$ or $m_i = m \neq 0$ and $x = 1/m$. In the latter case, assume $n_i = n$ for all $i$ so $x = 1/n + \lim_{i\to\infty}1/m_i$. Either $\lim_{i\to\infty}1/m_i = 0$ or $\lim_{i\to\infty}1/m_i = 1/m$ for some $m \in \mathbb{N}^+$. The latter case is impossible, for then $1/n_i + 1/m_i = x$ for some $i$.

I might have been a little sloppy with sequences vs. subsequences and subscripts but I think the argument is correct.

Stefan Smith
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  • Thank you for your answer, I think I understand the concept behind it. However I am in my first analysis course and we have covered compactness but not quite begun limits and sequences.

    We were taught compactness in terms of "every open cover having a finite subcover" and then showed that in the real numbers, compactness is equivalent to closed & bounded.

    What I'm looking for is a way to do this by showing that A contains all of its limit points, going back to the definition of a limit point as a point with a deleted ball/neighbourhood with an intersection with the set. Is this reasonable?

     – mb7744 Oct 21 '13 at 00:37
  • @mb7744 : I removed some comments because MSE was complaining. You don't need to know anything about compactness, period, to do this problem. Your definition of limit point of $A$ using deleted neighborhoods is 100% equivalent to the definition using a sequence of points in $A$ (at least for metric spaces, such as $\mathbb{R}$. I'm not sure about other topological spaces). You should try to convince yourself of this (that is, prove it). – Stefan Smith Oct 21 '13 at 02:56
  • It is equivalent, yeah, but the point is I'm finding it tricky to show all that without the concepts of sequence, subsequence or limit. In your solution there are a number of concepts that would require a lot of work to show equivalence to my current approach. (This is the work that essentially takes place throughout the next chapter, I'm guessing.) Anyway thank you for your help, I think my goals are specific enough that it is just up to me to structure my solution as needs. – mb7744 Oct 21 '13 at 03:25
  • @mb7744 : the concepts of sequence, subsequence, and limit are extremely important and useful for problems such as these. Trying to avoid them in a problem like this seems like an artificial and pointless exercise. – Stefan Smith Oct 21 '13 at 15:28
  • We had not covered those topics in my course, I was not excluding them by choice. Thank you for your help in any case. I removed some comments as well. Cheers – mb7744 Oct 21 '13 at 16:48
  • @mb7744 : I find it bizarre that you are expected to prove anything about the limit points of $A$ without using sequences or subsequences. – Stefan Smith Oct 21 '13 at 17:15
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This doesn't answer your questions about the set $A$, but rather gives you an alternative way to construct a set with the desired properties. For each $m\in\Bbb n,$ let $$C_m=\left\{\frac1m+\frac1{nm(m+1)}:n\in\Bbb N\right\},$$ and let $$C=\{0\}\cup\bigcup_{m\in\Bbb N}C_m$$ You should be able to show that the sets $C_m$ lie in pairwise-disjoint intervals of the form $$\left(\frac1m,\frac2m-\frac1{m+1}\right],$$ and that each $C_m$ has only $\frac1m$ as a limit point. From these facts, you can show that $0$ is the only other limit point of the set $C$.

Cameron Buie
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Take a sequence $(a_n)$ in $A$ that converges. For every $\epsilon$ there are only finitely many elements $a\in A$ with $|a|>\epsilon$, so either the sequence becomes eventually constant, or $a_n \to 0$.

nullUser
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