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Let $E'$ denote the set of the limit points of $E$.

Prove: If $E$ is a subset of $\mathbb{R}^n$, and $E'$ is countable, then $E$ is countable.

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  • $\begingroup$ If you take your answer. Do not forget to accept the answer. This will encourage others to help you. $\endgroup$ – TXC Feb 8 '13 at 7:52
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Any point of $E$ that is not a limit point of $E$ is an isolated point of $E$. If $x$ is an isolated point of $E$, there is a point with rational coordinates that is closer to $x$ than to any other point of $E$. This implies that there are at most countably many isolated points of $E$. The union of two countable sets is countable, so ...

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Suppose that $E$ is uncountable but $E'$ is countable, say $E'=\{x_n\mid n=1,2,\dots\}$. We "thin out" $E$ by recursion: There is some integer $l$ such that $E_0=\{x\in E\mid |x|<l\}$ is uncountable. Suppose $E_n\subseteq E$ has been defined and is uncountable. Let $E_{n+1}=\{y\in E_n\mid |y-x_n|>1/m\}$ where the integer $m=m_n$ is such that $E_{n+1}$ is uncountable. Note that $E\supseteq E_0\supseteq E_1\supseteq\dots$

Now for each $i\in\mathbb N$, pick $y_i\in E_i$. By construction, the sequence of $y_i$ cannot converge to $x_n$ for any $n$, because all $y_i$ for $i> n$ are at distance larger than $1/m_n$ from $x_n$. But the sequence is bounded, since all $y_i$ are in $E_0$, so it has a convergent subsequence, whose limit is not in $E'$, which of course is a contradiction.

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