Let $(M,d)$ be a metric space. A set $A \subset M$ is said to be compact if every open cover of $A$ has a finite subcover.
Why do we use this definition, rather than the other "definition" which holds in $\mathbb{R}^n$, that is, a set is compact if it is closed and bounded? It is a more intuitive definition, and it is hard for me to think of compact sets being separate from "merely" closed and bounded sets (probably because I can only imagine Euclidian spaces).
Is it simply because being closed and bounded (together) is not a topological property?
Best Answer
That definition is the more general definition - it holds in $\mathbb{R}^n$ and in things very different from $\mathbb{R}^n$. (In fact, the more concrete definition isn't even appropriate to arbitrary metric spaces!)
Specifically:
It makes sense in arbitrary topological spaces, even ones which are non-metrizable (that is, which cannot be thought of as coming from a metric). For example, we can say with confidence that the cofinite topology on an infinite set is compact ... even though such a topological space is never metrizable.
Within the context of arbitrary metric spaces, "closed and bounded" doesn't behave the way it should: consider a discrete metric space where every point is at distance $1$ from every other point. Every set in such a space is closed and bounded, but we don't have any of the phenomena associated to compactness in $\mathbb{R}^n$ which we actually want (e.g. we can have an infinite sequence with no convergent subsequence). The open cover definition, by contrast, gets things right (e.g. a subset of a discrete metric space is compact iff it is finite).