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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?

m0meni
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    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 $[0,1]$. – Frank Lu May 02 '17 at 19:31
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    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. – Frank Lu May 02 '17 at 19:33
  • Probably of interest: Finding open covers that do not contain finite subcovers. – Andrew D. Hwang May 02 '17 at 19:35
  • @FrankLu thanks a lot this was very helpful. – m0meni May 02 '17 at 19:41
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    An open cover of $S$ is not a union of open sets. It is a collection of open sets whose union contains $S.$ – zhw. May 02 '17 at 21:22

1 Answers1

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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.

Kenny Wong
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    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 https://math.stackexchange.com/a/189053/173829. – m0meni May 02 '17 at 19:41
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    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... – Kenny Wong May 02 '17 at 19:47
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    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. – Kenny Wong May 02 '17 at 19:48
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    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 $(x - \delta, x + \delta ) \subset U_\alpha$... – Kenny Wong May 02 '17 at 19:50
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    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. – Kenny Wong May 02 '17 at 19:50
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    Note that Heine Borel is not something out of the blue. It is equivalent to completeness of real number system. In fact completeness of real numbers can be expressed in many ways and Heine Borel and nested interval principle are two such ways. What you did was to prove Heine Borel using nested interval principle. The reverse can also be done. – Paramanand Singh May 03 '17 at 06:01