How I'm conceptualizing compactness. Does this make sense? So from what I know about compactness


*

*a compact set, $S$, is defined as a set for which every open cover of $S$ has a finite subcover

*where an open cover, $O$, is a union of open intervals such that $S$ is a subset of $O$

*and a subcover, $C$, is a subset of the open cover $O$ for which $S$ is still a subset of $C$


but it wasn't really clicking for me why $[0, 1]$ was compact while $(0, 1)$ was not. After thinking for a while, here's how I ended up conceptualizing it using the above definitions:


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*$(0, 1)$ has the open cover $(0, 1)$, which is equivalent to itself, and you cannot possibly find a subcover of $(0, 1)$ that still covers $(0, 1)$

*$[0, 1]$ being a closed interval isn't covered by $(0, 1)$, but is covered by infinitely many open intervals just slightly bigger than $(0, 1)$ e.g. $\{(-1,2),(-\frac12, \frac32), (-\frac14,\frac54),...(0 - \frac1n, 1 + \frac1n)\}$ and all those open intervals have subcovers that still cover $[0,1]$


Does that make sense? Also I assume a single interval rather than a union of intervals is an open cover; is that fine?
 A: You're right that $(0,1)$ is non-compact, but you can't use the open cover $\{ (0,1) \}$ to prove that $(0,1)$ is non-compact. Remember, we need to find an open cover of $(0,1)$ such that no FINITE subset of it is an open cover. But $\{ (0,1) \}$ is already finite! It contains only one open set! So there is a finite subset of $\{ (0,1) \}$ that covers $(0,1)$, namely, $\{ (0,1) \}$ itself!
To show that $(0,1)$ is non-compact, you may like to consider the open cover:
$$ \{ (\tfrac 1 2, 1 ), (\tfrac 1 3, 1), (\tfrac 1 4, 1), (\tfrac 1 5, 1) , \dots \}$$
I hope it's clear that this collection of open sets covers $(0,1)$, but no finite subcollection within this collection covers $(0,1)$.
For $[0,1]$, you gave an example of an open cover. Indeed, your open cover admits a finite refinement, but not for the reason you gave. Your open cover admits a finite refinement because $\{ (-1,2) \}$ is a finite subcollection within your open cover  that covers $[0,1]$, and $\{ (-1,2) \}$ is finite, containing only one open set!
Anyway, you can't prove that $[0,1]$ is compact by exhibiting a single open cover that admits a finite refinement. You need to prove that ALL open covers of $[0,1]$ admit a finite refinement. This is quite tricky to prove, and is (a special case of) the Heine-Borel theorem.
