Understanding compactness of metric spaces I have been studying a basic course in metric spaces. I really get confused whenever the notion of compactness arrives. I know that any subset of $\mathbb{R}^n$ is compact iff it is closed and bounded. Is this theorem true for any arbitrary metrics defined on space or is it true only under the metric induced from $\mathbb{R}^n$? I am really sorry if the question is of too low standard as per standards of maths stackexchange. 
 A: The general definition is via open covers, this remains valid in any topological space: A metric space $X$ is compact, if any open cover of $X$ has a finite subcover. In other words, if you take any collection of open sets $\mathcal{U} \subset \mathcal{P}(X)$ s.t. $\cup \mathcal{U} =X$ there is a finite set $F \subset \mathcal{U}$, such that $\cup F = X$.
Let me share some of my personal heuristics on this: Compact sets are "small" in a very specific sense.  A metric space $X$ is compact if and only if it is sequentially compact (in the context of real numbers, this is Bolzano–Weierstrass theorem). This means that if you take any countable set of points $(x_n)$ in a compact metric space $X$, they will inevitably stack around a point in $X$. I like to think that there is no room in the set $X$ to spread an infinite number of points to the set in a way that they would not stack anywhere. In this sense e.g. the unit ball in the sequence space $l^2$ is not small, since you can put an infinite number of points there that are a constant $c > 0$ away from each other.
Hope this helps.
