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:

  • $(0, 1)$ 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?

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    $\begingroup$ A sub-cover can be the original cover itself, therefore the proof of non-compactness of $(0,1)$ is not correct. Also to prove compactness, you need to consider all posible open covers. In your proof of the compactness of $[0,1]$ you only dealt with open covers with a particular property, namely all intervals in the open cover contains $[0,1]$, but there are more open covers. As long as the union of intervals contains $[0,1]$, then these intervals form an open cover of $[0,1]$. $\endgroup$ – Frank Lu May 2 '17 at 19:31
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    $\begingroup$ In fact if you want to prove compactness by definition, in most cases it could be complicated. That's why we characterise compactness via the Heine-Borel theorem, which states that a subset of the Euclidean space is compact if and only if it is closed and bounded. $\endgroup$ – Frank Lu May 2 '17 at 19:33
  • $\begingroup$ Probably of interest: Finding open covers that do not contain finite subcovers. $\endgroup$ – Andrew D. Hwang May 2 '17 at 19:35
  • $\begingroup$ @FrankLu thanks a lot this was very helpful. $\endgroup$ – m0meni May 2 '17 at 19:41
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    $\begingroup$ An open cover of $S$ is not a union of open sets. It is a collection of open sets whose union contains $S.$ $\endgroup$ – zhw. May 2 '17 at 21:22

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.

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    $\begingroup$ Thank you for your answer. I get it now. Regarding the proof for $[0, 1]$ being compact I found this really nice and simple one just now, which doesn't require Heine-Borel math.stackexchange.com/a/189053/173829. $\endgroup$ – m0meni May 2 '17 at 19:41
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    $\begingroup$ Here is (what I believe to be) an even simpler argument: Suppose for contradiction that $\{ U_\alpha \}$ is an open cover of $[0,1]$ such that $[0,1]$ cannot be covered by finitely many $U_\alpha$'s. Let's bisect the interval $[0,1]$ into $[0, \tfrac 1 2]$ and $[\tfrac 1 2, 0 ] $. Then either $[0, \tfrac 1 2]$ cannot be covered by finitely many $U_\alpha$'s or $[0, \tfrac 1 2]$ cannot be covered by finitely many $U_\alpha$'s. Suppose without loss of generality that $[0, \tfrac 1 2]$ cannot be covered by finitely many $U_\alpha$'s... $\endgroup$ – Kenny Wong May 2 '17 at 19:47
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    $\begingroup$ Then we divide $[0, \tfrac 1 2]$ into $[0, \tfrac 1 4]$ and $[\tfrac 1 4, \tfrac 1 2]$ and do the same thing again. If we keep doing this, then eventually, we will get a sequence of closed intervals $I_n$ of length $1 / 2^n$, with $[0,1] = I_0 \supset I_1 \supset I_2 \supset \dots$ such that each $I_n$ cannot be covered by finitely many $U_\alpha$'s. $\endgroup$ – Kenny Wong May 2 '17 at 19:48
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    $\begingroup$ Since the $I_n$'s are CLOSED intervals, the intersection of all $I_n$'s is a single point, which I'll call $x$. Clearly, $x$ is inside some $U_\alpha$ (since the $U_\alpha$'s form an open cover for $x$). But then, since $U_\alpha$ is OPEN, there is some $\delta > 0$ such that $(x - \delta, x + \delta ) \subset U_\alpha$... $\endgroup$ – Kenny Wong May 2 '17 at 19:50
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    $\begingroup$ But then, if we pick an $n$ such that $1 / 2^n < \delta$, we see that $I_n \subset U_\alpha$. And this contradicts that statement that $I_n$ cannot be covered by finitely many $U_\alpha$'s. $\endgroup$ – Kenny Wong May 2 '17 at 19:50

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