# Compact sets and Open sets in a metric space

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?

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'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 $n = 1, 2, 3, ...$