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It seems very intuitive and simple, but how would I go about proving something like this? Thanks.

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Yes, this is true, and it follows from Weirstrass' Thm : every bounded , infinite subset has a limit point. And this generalizes to $\mathbb R^n$

More formally, cover $\mathbb R^n$ by balls $B(0,n) ; n=1,2,3,... $ , i.e., for all

$n $ in $\mathbb N$ Now, by a straight-forward cardinality argument (countable union of countable is countable) , there will be at least one ball $B(0, n_k)$containing infinitely-many points. Then apply Weirstrass' theorem.

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  • $\begingroup$ In fact, an uncountable set $S$ of real numbers has a so-lled condensation point, meaning a point whose every neighborhood contains uncountably many elements of $S$. In fact, all but countably many of the points in $S$ are condensation points of $S$. (In the last formulation, we don't have to assume that $S$ is uncountable.) $\endgroup$ – bof Oct 10 '13 at 3:40

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