Question on Compactness Let the metric space be the real numbers with the usual distance formula. Let $E$ be an open interval from $1/8$ to $2$. Then $E$ would be compact if every open cover of $E$ has a finite cover. I know $E$ is not compact because it is not closed, but I'm not sure how to use the cover definition of compactness to prove it. The sets $(1/n, 1+1/n)$ where $n \in \{1, \ldots, 9\}$ is a finite open cover for $E$, but I am not sure how to find a cover where this does not hold.

How about taking the open set $(1/8, 3/2)$ along with the sets $(1, 1 + \frac{n}{n+1})$ for $n \geq 1$?
