Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.
What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$?
Best Answer
You are right that $G$ is already a finite subcover of itself, so that isn't particularly useful. Perhaps a more interesting characterization of compactness is that every infinite cover contains a finite subcover. (Do you see why they are equivalent?)
Your second example cover, $$\left\{\left(\frac1n,1\right):n\geq 1\right\},$$ is an example of an infinite cover from which no finite subcover can be extracted. This shows that $(0,1)$ is not compact.