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I was wondering whether it was true that an uncountable subset of $\mathbb{R}$ contains a convergent sequence. I was thinking about a proof by contradiction but did not manage to complete it.

I tried: Let A be a subset of $\mathbb{R}$ with no limit points. Then each point of A has a neighborhood containing finitely many points of A. (Then what???)

I feel like showing that A must be at most countable should be easy, but I can't do it.

Hints would be more apprieciated than solutions :)

Cheers!

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If $U$ denotes a uncountable subset of $\mathbb R$ then $U_n:=U\cap[-n,n]$ will be uncountable for some positive integer $n$.

This because $U=\bigcup_{n=1}^{\infty}U_n$ so that countability of each $U_n$ would imply countability of $U$.

This $U_n$ is subset of the set $[-n,n]$ wich is compact, so that every sequence in that set has a convergent subsequence.

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  • $\begingroup$ I think I get it. Thank you! $\endgroup$ – I want to make games Sep 28 '15 at 17:04
  • $\begingroup$ Actually I made a small mistake wich is repaired now. I said that $U_n$ would be compact but that does not have to be the case. However, it is enough that $U_n$ is subset of the compact set $[-n,n]$. A sequence in $U_n$ is also a sequence in $[-n,n]$ hence will have a convergent subsequence. $\endgroup$ – drhab Sep 29 '15 at 7:38
  • $\begingroup$ Wait, how can a subset of a compact set not be compact? I thought that, at least in $\mathbb{R}$, compactness was equivalent to being a subset of a closed interval. $\endgroup$ – I want to make games Sep 29 '15 at 17:40
  • $\begingroup$ In $\mathbb R$ compactness is equivalent to being bounded (i.e. being a subset of an interval $(a,b)$ where $-\infty<a<b<\infty$) and being closed. E.g. the open interval $(a,b)$ is not compact. It has an open cover in the sets $(a+\frac1{n},b-\frac1{n})$ wich has no finite subcover. $\endgroup$ – drhab Sep 29 '15 at 18:00
  • $\begingroup$ Oh, ok. I understand. Thanks for the clarification. $\endgroup$ – I want to make games Sep 29 '15 at 18:14
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Your basic idea will work. Use the fact that $\Bbb R$ has a countable base $\mathscr{B}$, for instance, the family of open intervals with rational endpoints. If $A$ has only isolated points, then each $x\in A$ has a nbhd $B_x\in\mathscr{B}$ such that $B_x\cap A$ is finite (in fact such that $B_x\cap A=\{x\}$). Now usse the countability of $\mathscr{B}$.

This question has three answers to your question.

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  • $\begingroup$ Thank you! I tried to look for a similar question, but I did not find that one. $\endgroup$ – I want to make games Sep 28 '15 at 17:12
  • $\begingroup$ @Iwanttomakegames: You’re welcome! $\endgroup$ – Brian M. Scott Sep 28 '15 at 17:17
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You can split the real line into subintervals of length $1/2^n$ for increasing $n$ and show that there is a nested sequence of intervals of decreasing length which are all containing uncountably many points, which is basically all you need. (Note that the number of intervals is countable, which is key)

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  • $\begingroup$ Hmm, I get the reasoning in general, but I've never really dealt with countability etc. Could you explain why this reasoning fails for a countable-infinite subset A? $\endgroup$ – I want to make games Sep 28 '15 at 16:57

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