In $\mathbb{R}^n$ we prove that a set is compact (using the definition about open covers) if and only if it's closed and bounded. It is pretty clear that if $\mathcal{O}$ is an open cover of one unbounded set $X$, then $\mathcal{O}$ cannot have a finite subcover, it'll clearly need in general infinitely many sets to cover the set $X$.
Now, if a set isn't closed, I cannot see in which way it fails to be compact. For instance, if $X$ is the closed unit ball centered at the origin, then it is compact. If on the other hand we consider $Y=X\setminus\{0\}$, then it's not compact anymore, because $Y$ isn't closed (the point $0$ is a limit point of $X$ and so, $0 \in \operatorname{Cl}(X)$ and on the same time $0 \notin Y$.
So, what should be the intuition about this? How can we intuitively see that $Y$ isn't compact?
Thanks very much in advance!
Best Answer
Let $p$ be a limit point of the set $Y$ that is not contained in $Y$. Then $(U_n)_{n\in\mathbb{Z}^+}$ with
$$U_n := \{ y \in Y : \lvert y - p\rvert > \frac1n\}$$
forms an open cover of $Y$ which has no finite subcover.