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I am looking at this problem and I decided to attack it by proving the contrapositive.

If $E \subset \mathbb{R}$ has finitely limit points, then $E$ is countable.

My proof: Let $S=\{\text{limit points of E} \}$, then for any $s\in E, \ \exists \delta>0$ such that $B_{\delta}(s)$ contains a point of $E$. Take $\delta_1 >0, s_1 \in S$, then $B_{\delta_1}(s_1)$ contains a point of $E$ say $x_1$. We then take $\delta_2 >0, s_2, B_{\delta_2}(s_2)-{x_1}$ to contain $x_2$. Inductively we have $\{x_n\} \in (B_{\delta_{n}}(s_n)-\{x_1,...,x_n\})$. This gives us our bijective map: $\phi: \mathbb{N} \to E \Rightarrow E$ is countable.

Question: I am unsure if what I am proving is infact the contrapositive. If this argument can be done directly I would like to know, if not and I am not doing the contrapositive, can someone quantify it for me? Am i going in the right direction or do I need to try something else.

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  • The negative of having a limit point is not to have infinitely many. It is to have not any limit point. – mfl Jul 07 '14 at 00:33
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    Here's a much simpler argument: There must be an $n$ such that $A := E \cap [-n,n]$ is infinite. $A$ is an infinite subset of the compact set $[-n,n]$ so it has a limit point in $[-n,n]$. Now argue that such a limit point must also be a limit point of $E$. – kahen Jul 07 '14 at 00:34
  • @mfl: Does proof still hold? – Mr.Fry Jul 07 '14 at 00:34
  • @kahen: Thank you for a simpler, but I would just like to know if the way I am going is remotely correct. I will figure out more clever way on my own once I succeed with one way. – Mr.Fry Jul 07 '14 at 00:35
  • No, you are focusing only on a limit point. This can't give you information about the whole set. – mfl Jul 07 '14 at 00:36
  • $(0,1) \subset \mathbb{R}, {0,1}$ are it's limit points. So then $(0,1)$ is countable, but how? Would we show this by expressing the interval as a union of countable subsets of $\mathbb{R}$. – Mr.Fry Jul 07 '14 at 00:41
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    No, $[0,1]$ is the set of limit points of $(0,1)$. – Ian Jul 07 '14 at 00:42
  • No, the set of limit points of $(1/n,1)$ is $[1/n,1]$. Your result is false if your definition of "limit point" is such that members of the set are not limit points. – Ian Jul 07 '14 at 00:46
  • Sorry, I wasn't thinking, but $\cup_{n=1}^{\infty} (\frac{1}{n},1)$ is countable. But how do I show that? – Mr.Fry Jul 07 '14 at 00:46
  • No, it is not countable, it is $(0,1)$ which is not countable. – Ian Jul 07 '14 at 00:47
  • I'll just think about this some more. – Mr.Fry Jul 07 '14 at 00:50
  • @kahen: I don't see why it should also be a limit point of A, could you help me with it? – luki luk Feb 24 '23 at 13:21

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I think this is the proof by contrapositive that you were getting at.

Suppose that $E$ has no limit points at all. Pick an arbitrary point $x\in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$: $$B_{\delta}(x)\cap E=\{x\}$$ Call this "point $1$". For the next point, take the closest element to $x$ and on its left; that is, choose the point $$\max \Big[E\cap (-\infty,x)\Big]$$ if it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$.

Call this "point $2$". Now take the first point to the right of $x$ for "point $3$". Take the first point to the left of point $2$ for "point $4$". And so on, ad infinitum.

This gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $a<x$ which is never included in the list (picking $a$ on the negative side of $x$ is arbitrary, and the same argument would work for the second case). Then the element closest and to the right of $a$ in $E$ (which exists, by the no-limit-points argument at the beginning) is also not in the list; if it was, $a$ would have been in one of the next two spots. And same with that point (call it $a_1$); there is a closest $a_2>a_1\in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\{a_i\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone Convergence Theorem this sequence has a limit. But that means the sequence $\{a_i\}\subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements.

  • D***, I wanted to try it out myself. But I will attempt the problem again and then give your solution a look. Thanks though. – Mr.Fry Jul 07 '14 at 02:30
  • @Rod No problem... feel free not to read it if you want to try yourself. Even if you can't do it, I'm sure there is a way to make my solution much cleaner and quicker - it's a little awkward as it is, but I think it's correct... cheers! –  Jul 07 '14 at 02:32
  • I will come back to check it off. – Mr.Fry Jul 07 '14 at 02:33
  • The original question should have been, “Prove every uncountable subset of R has at least one limit point.” i.e. “$A \subset R \implies A\ $ has at least one limit point”. It is clear that he negation of “At least one limit point” is “no limit points”. – Adam Rubinson Jan 16 '20 at 13:23