Compact sets and Open sets in a metric space

Question

I have from reading up on things understood that open sets in a metric space is not compact. Though I have no clue why. I would like to know why is it they are not compact? I know that a compact set must have a finite subcover for the set but why does it not occure for open sets when it does for closed?

Answer 1

An open set can be compact, it just also has to be closed. For example, $[0,1]$ is open in $[0,1]$ and is compact under the induced topology.

It can be proven that in any Hausdorff space (and therefore in any metric space), a set can only be compact if it is closed (whether or not it is additionally open).

Answer 2

It's not true for general metric spaces (for example, a space with finitely many points). For $\mathbb{R}^n$, however, it's true. Think about the interval $(-1,1)$. It can be covered by open sets of the form $(-1+\frac{1}{n}, 1-\frac{1}{n})$ for $n = 1, 2, 3, ...$

This cover has no finite subcover. You can do the same thing for open balls in higher dimensions.