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.

  • 1
    $\begingroup$ If you take your answer. Do not forget to accept the answer. This will encourage others to help you. $\endgroup$
    – TXC
    Feb 8, 2013 at 7:52

2 Answers 2


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 ...


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.