# Proof: in $\mathbb{R}$, $((0,1),|\cdot|)$ is not compact.

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 $\neq \emptyset$?

• It would better to write $G$ as $G = \{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$ as it is a collection of two open subsets, and not an open subset which is the union of two other open subsets. The set $G$ has cardinality two, so yes, it is a finite cover, and every subcover of it will also be finite. Dec 6 '12 at 16:34
• Corrected. Can we conclude that every open cover we intially take is necessarily a infinite union of open sets? Dec 6 '12 at 16:39
• @pdots1 I was wrong in what I was saying. The proof isn't valid because that open cover contains the trivial subcover (itself) as a finite subcover. Your other example is good. Dec 6 '12 at 16:42

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.
You counter example (the open cover $\cup_{n \in \mathbb N} (1/n,1)$) actually works. It has no finite subcover. Therefore, $(0,1)$ is not compact. Every cover is not necessarily infinite. Again, your $G$ is the counter example (it is a finite cover.)
Note: you can take $G$ as a finite subcover of $G$. So, $G$ does not show that $(0,1)$ is not compact.