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Well it comes down to word-play again. I'm confused to the core of my bones as to why the following isn't equivalent to saying that a space is compact

Every open cover is finite.

A compact set must have every open cover in which there is a finite subcover.

Well, the statement tells me that every single possible open cover is finite. So...why doesn't this qualify?

Subcover needn't be a proper subset/cover so if I have every open cover being finite...the open cover itself can act as a subcover(which is of course finite) so voila, we're done..no?

Elaborate explanation much appreciated on this

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  • $\begingroup$ You are quite right that, if every open cover is finite, then every oven cover has a finite subcover. You haven't explained why you think that, if every open cover has a finite subcover, then every open cover is finite. $\endgroup$ – bof Apr 22 '16 at 22:12
  • $\begingroup$ Well, isn't it because I can let the open cover itself be a subcover of itself? It needn't be a proper subcover like I said so I must be allowed to do this. That's why every open cover has a finite subcover in this case. $\endgroup$ – John Trail Apr 22 '16 at 22:17
  • $\begingroup$ When you say two things are equivalent, it means that "A if and only if B". In this case, you're saying a set is compact if and only if every open cover is finite. The example below presents the case of a finite subcover, but with an infinite cover. You can't just deny the existence of infinite covers because you'd rather deal with the finite covers. They're still there, so a compact set necessarily doesn't have only finite subcovers. $\endgroup$ – Sean I Apr 22 '16 at 22:17
  • $\begingroup$ As for the example...indeed, it presents an infinite open cover. But, the statement says if a particular space has every possible open cover be finite (therefore, we need not worry nor talk about finding a infinite open cover) it is compact. So we know that we have finite open covers for every possible open covers. None infinite. Considering the case of an infinite open cover's existence already violates the statement's finite open cover bit so I don't see how this would answer my confusion... $\endgroup$ – John Trail Apr 22 '16 at 22:21
  • $\begingroup$ Then I guess you['d say there is no difference between the statements "every integer greater than one has a prime divisor" and "every integer greater than one is a prime number", right? Because, an integer is a divisor of itself, so the prime divisor could be the number itself? $\endgroup$ – bof Apr 22 '16 at 22:48
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A compact space is one in which every open cover has a finite subcover. Not every subcover has to be finite. So, for example, the cover $C = \{(-\varepsilon,\varepsilon)\,|\, 0 < \varepsilon\}$ is an open cover of $[-1,1]$, and has as a finite subcover the singleton $\{(-2,2)\}$. The cover is still infinite, however.

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  • $\begingroup$ I still don't get it; a subcover, so if I can find at least one finite subcover of each open cover, it's compact yes? The statement says every open cover is finite. Thus, if I let open cover=open subcover for each, I've found one finite subcover for each case. Therefore, it must be compact. I think in your example the open cover is infinite but the statement in question assumes that every open cover IS FINITE so it doesn't really relate to my question? $\endgroup$ – John Trail Apr 22 '16 at 22:15
  • $\begingroup$ The point of compactness is that every open cover, whether infinite or not, has a finite subcover. That means that for every open cover you can throw away all but finitely many sets and still have a cover. So for example, $[0,1]$ is compact, and one open cover of this is the collection I mentioned. Since $[0,1]$ is compact, there must be a finite subcover of this; I have mentioned one possible finite subcover. $\endgroup$ – Sir Jective Apr 22 '16 at 22:23
  • $\begingroup$ I'm sweating right now, I still don't understand(sorry)... I mean, aren't we considering a space which every open cover is finite? I think the statement's saying "if the space you're looking at happens to have every open cover to be finite, it's compact". In your words, in a sense, I need not throw away anything from each open cover. So shouldn't it be compact? I guess I'm missing a point you're trying to make but I just can't see the relation to my question..... $\endgroup$ – John Trail Apr 22 '16 at 22:28
  • $\begingroup$ No, compactness does not mean that every open cover is finite. The condition of compactness is this: "Every open cover has a finite subcover". Nowhere does compactness say every open cover is finite. Compactness does not say that. $\endgroup$ – Sir Jective Apr 22 '16 at 22:31
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    $\begingroup$ @JohnTrail So instead of saying "you're quite right that $A\implies B$ but you haven't explained why $B\implies A$ I should have said "you're quite right that $A\implies B$ but you're wrong about $B\implies A$"? Got it. I'll try and remember to be more direct next time. $\endgroup$ – bof Apr 22 '16 at 23:01

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