# Understanding compactness

I am trying to grasp the concept of compactness in measure theory.

In many examples they use [0,1] as an example of a compact space and (0,1) as a non-compact space. Arugments for the latter is that it does not have a finite subcover.

However, I can just choose the interval (-1,2) which covers (0,1) and is finite. This is wrong I know but can someone explain why? Do the covers need to be stated as a sequence or can I just choose the arbitrarily?

• Any cover needs to have a finite subcover. See if you can find one that doesn‘t. – Lukas Kofler Aug 13 at 14:30
• Also it's an open cover. $\{[1/(n+1),1/n]\}_{n=1}^\infty \cup \{\{0\}\}$ covers $[0,1]$ but has no finite subcover. – eyeballfrog Aug 13 at 14:34
• $[-1,2]$ is not open first of all. Second, that’s just one specific cover. As already mentioned more is needed: one must be able to extract a finite open subcover out of any open cover collection. – Nap D. Lover Aug 13 at 14:45
• Thanks for the comments, I can now see that I missed the "any" part – Dennis Aug 14 at 8:18

The covers need not be stated as a sequence (they could, in fact, be uncountable). However, in $$\mathbb R$$, the union of any collection of open sets is equal to the union of a countable subset of that collection. So it's sufficient to use sequences of open sets for $$\mathbb R$$.
It's also important to have an intuition for why the cover must be open. Nonempty open sets necessarily have nonempty interior. In $$\mathbb R$$, this means they have nonzero extent. Additionally, to cover any set with nonempty interior, the open sets inevitably must have a nonempty overlap, which is itself an open set and has nonzero extent. To handle the endpoints of a closed set, the open sets also overhang the edges. This is what forces the existence of a finite subcover for closed, bounded sets in $$\mathbb R$$--the overlaps cut off any sequence of open intervals that approaches a point but never reaches it.