It seems very intuitive and simple, but how would I go about proving something like this? Thanks.
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$\begingroup$ This is a duplicate of Show uncountable set of real numbers has a point of accumulation, which has a solution. The answers to this question show other ways to prove it, even getting it to be in the uncountable set. This answer proves that you can choose the point so that the set clusters on both sides of it. $\endgroup$ – Brian M. Scott Oct 10 '13 at 3:29
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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
