compactness of closed interval

Question

Using the topological definition of compact, namely that every open cover admits a finite open subcover, I was hoping someone can provide me with the finite sub cover of $\{(\frac{1}{n},1-\frac{1}{n}):n\in \mathbb{N}\}$ of the interval $[0,1]$. (This $\textit{is}$ an open cover of that interval, right?). To me, it seems as though as soon as we choose a particular N to stop at, we wouldn't have covered the whole interval- yet I know the compactness of this interval is a fundamental fact of topology.

Answer 1

You're right that there's no finite subcover. But that's not a problem, because it's not a cover at all! The union of all the sets $(1/n,1-1/n)$ is only $(0,1)$, not $[0,1]$. The points $0$ and $1$ are not in any of your open sets.