Let $D\subset \mathbb{R}^{n}$ be a not countable set. Prove that D has limit points. We want to show that if $D$ is an not countable subset of the euclidean n-dimensional space (whit its usual topology), then $D^{'}\not =\emptyset$.
My best guess is to prove it whit a contradiction.
But the only thing I get is an infinite number of isolated points, which leads me to nothing.
Any help?
 A: Suppose that $A\subseteq\mathbb R^n$ has no limit points in $\mathbb R^n.$ Then (in particular) for each $a\in A,$ there is some real number $r(a)>0$ such that $B_{r(a)}(a)\cap A=\{a\}$ (here $B_r(x)$ denotes an open ball with radius $r$ and center $x$).  Now, for each $a\in A,$ let $q(a)\in B_{r(a)/2}(a)\cap\mathbb Q^n$ and let $\rho(a)$ be a rational number such that $\|a-q(a)\|<\rho(a)<r(a)/2.$ Then $a\in B_{\rho(a)}(q(a))$ and it follows that the set $\{B_{\rho(a)}(q(a)):a\in A\}$ is at most countable and it is such that $\bigcup\limits_{a\in A}B_{\rho(a)}(q(a))\supseteq A.$ Therefore $A$ can be covered by an at most countable collection of open balls, each of which contains exactly one element from $A$ and hence $A$ is at most countable. Taking the contrapositive, if $A$ is uncountable then it has a limit point in $\mathbb R^n.$ Note this shows that if no point of a set $E\subseteq\mathbb R^n$ is a limit point of $E$ then $E$ must be at most countable.
A: Suppose $D$ has no limit points. 
Intersect $D$ with $K_1=[-1,1]^n$. This must be finite. If this was not the case, then by compactness we would have a limit point for $D$.
Now, let $K_i=[-i,i]^n$. For the same reason as before, $D \cap K_i$ is finite for every $i$. But $\cup_{i=1}^{\infty} D \cap K_i=D$. Since it is a enumerable union, $D$ is enumerable.
