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I'm struggling a bit to understand some aspects of compactness in infinite sets.

We say that $[0,1]$ is compact by Heine-Borel, but that means that for all open covers of $[0,1], \exists$ a finite subcover of $S$. Does that mean that no infinite collection of open sets between $0$ and $1$ inclusive exists? Furthermore, if that's true does that mean there's a minimum non-empty size/interval for an open set? How is that possible?

What about the case of $$\mathscr{F} = \{ (\frac{1}{(n+1)}, \frac{1}{n})  \mid n \in \mathbb{N} \} \cup \{ (\frac{1}{n} - \frac{1}{n^3 + 1}, \frac{1}{n} + \frac{1}{n^3 + 1})  \mid n \in \mathbb{N} \}$$?

It seems to me that here that the only subcover that would exist for $[0,1]$ would be infinite.

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    $\begingroup$ Your sets do not cover $0$ and $1$. $\endgroup$ – Kavi Rama Murthy Jan 28 at 8:33
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    $\begingroup$ Existence doesn't imply necessity. An infinite collection of open sets can exist, but a compact space can always be covered by a finite subset of that infinite collection. Both things can be true at the same time. $\endgroup$ – Amaan M Jan 28 at 8:35
  • $\begingroup$ @KaviRamaMurthy I think the second part of my union does mean 1 is included given $n=1$ produces the interval $(\frac{1}{2},\frac{3}{2})$, but I see now that I probably wouldn't be able to get the 0 into there, as the final union doesn't work for that purpose. $\endgroup$ – Paul Jan 28 at 8:42
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No, it does not mean that "no infinite collection of open sets between 0 and 1 inclusive exists". It means that for every (also infinite) cover (of open sets) that contains $[0,1]$, we can find a finite number of them (open sets) that contains $[0,1]$.

Important remark: the condition for compactness is that every open cover has a finite subcover, not just that a single given open cover. A proof that [0,1] is compact with respect to the euclidean topology can be found here.

I suggest that you try to understand the main idea of the proof: it contains all the ingredients to understand what you have to do when dealing with compactness.

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  • $\begingroup$ I understand that, but I felt that I had defined with my $\mathscr{F}$ an infinite cover that doesn't contain a finite subcover. $\endgroup$ – Paul Jan 28 at 8:38
  • $\begingroup$ As already commented by Kavi, your set does not contain 0 and 1, and indeed (0,1) is not compact. The fact you want to cover [0,1] takes away your example of cover as one that needs to admit a finite subcover. $\endgroup$ – Son Gohan Jan 28 at 8:38
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    $\begingroup$ I think it does contain 1, but now I see that it cannot cover 0. $\endgroup$ – Paul Jan 28 at 8:43
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To take one of your questions.

Suppose we take as a cover of the closed unit interval the set containing every every open sub-interval.

Note that the endpoints have to be included in a cover so there will be sets of the form $[0, a)$ and $(b, 1]$ included as well as the whole set itself $[0,1]$. These are open in the topology we are considering.

This has a finite sub-cover by Heine Borel.

Note that any such sub-cover can be refined to a cover with smaller intervals. eg Take $(a, b)$ with $a \lt b$ (where it may involve $[0, .$ or $.,1]$ to the two intervals $(a, \frac {2a+3b}5)$ and $(\frac {3a+2b}5, b)$. The cover remains finite, but the length of the largest interval is reduced. We can make the cover as fine as we choose while it remains finite.

The ability to refine covers in this way becomes significant in the theory of measure/integration.

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  • $\begingroup$ This is extremely helpful. I was missing the detail about those $[0,a]$ and $(b,1]$ endpoint sets in my understanding. I also see how there isn't so much a minimum interval size as there is always a specified (and thus finite) granularity to any existing interval, meaning you can build up at whatever granularity enough sets to cover the original bounded interval (e.g. $[0,1]$) $\endgroup$ – Paul Jan 28 at 8:56

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