TITLE: Subset of Cantor set that isn't compact
QUESTION [5 upvotes]: How to prove that the Cantor set has a subset that is not compact? Actually, I want to prove that every infinite set $X\subset\mathbb{R}^n$ has a subset $Y$ that is not compact. If $X$ isn't bounded, then $X$ has a unbounded subset $Y$ that is not compact. If $X$ is bounded and includes some ball, then $X$ includes a open ball $Y$ that is not compact. But if $X$ is bounded and includes no ball, like the Cantor set, I don't know. I think can be easier start by Cantor set, but I'm not sure. Can you help me?
Thanks.

REPLY [5 votes]: HINT: If no point of $X$ is a limit point of $X$, then you can take $Y=X$. If there is a point $p\in X$ that is a limit point of $X$, show that there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X\setminus\{p\}$ that converges to $p$, and let $Y=\{x_n:n\in\Bbb N\}$.
This works with any infinite $X\subseteq\Bbb R^n$.