An uncountable subset of $\mathbb{R}$ contains a convergent sequence.

Question

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!

Answer 1

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.

Answer 2

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)

Answer 3

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.