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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?

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  • $\begingroup$ What is your definition of "numerable"?. I guess that you mean that $D$ is neither finite nor countably infinite?, because the fact that $D\neq\varnothing$ follows directly from the definition (because the empty set is finite). $\endgroup$ – CIJ Sep 24 '15 at 1:59
  • $\begingroup$ I would recommend working out the details for the case $n=1$, the real number line, and then transferring the argument to a higher-dimensional setting. $\endgroup$ – hardmath Sep 24 '15 at 1:59
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    $\begingroup$ @C.I.J. Numerable must be countable. Numerable is used in spanish. $\endgroup$ – DonQuixote Sep 24 '15 at 2:08
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    $\begingroup$ See here for the $n=1$ case: math.stackexchange.com/questions/218949/… $\endgroup$ – Bungo Sep 24 '15 at 2:10
  • $\begingroup$ See @user264983 math.stackexchange.com/questions/218949/… $\endgroup$ – DonQuixote Sep 24 '15 at 2:14
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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.

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  • $\begingroup$ Just one question. How can I assure that $D\cap K_{i}$ is compact? I remember a result that establishes: If $A$ is closed and $B$ is compact, then $A\cap B$ is compact. Then, $D$ must be closed, or am I getting something wrong? $\endgroup$ – eNR Sep 24 '15 at 3:14
  • $\begingroup$ You don't. I'm just using the fact that an infinite subset of a compact set must have limit points. $\endgroup$ – Aloizio Macedo Sep 24 '15 at 3:19
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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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Sep 24 '15 at 6:47

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