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We define a set $X$ as compact if: "for every open cover of $X$, there exists a finite subcover."

An Open Cover $C$ of $X$ is defined as the union of a collection of Open Sets: $$C = \bigcup_{i \in I} A_i$$ and a finite subcover as the union over a finite subcollection: $$F=\bigcup_{i \in F} F_i$$

If we define the collection of of open sets to contain only one element e.g. the reals $A=\{\Bbb R\}$. Then there exists a finite subcover of $\Bbb R$, since we have a finite subcover consisting of only one element i.e. $\Bbb R$. Thus, $\Bbb R$ is Compact (which is obviously incorrect.)

Take another example: consider any bounded set $E \subset \Bbb R$. Hence one could construct a superset of E given by: $F \supset E $. Let the collection of open sets consist of only element e.g. $A = \{F\}$. Hence there exists a finite subcover of $E$ consisting of only one element i.e. $F$. Hence $E$ is Compact (regardless of whether or not $E$ is closed or open, which is again incorrect.)

I am struggling to understand the concept of a finite subcover. We could always construct the Open Cover to be the union of a collection of a single set, hence a finite subcover will always exist. Is there a constraint on what a finite subcover should be, other than it should be a finite collection of sets? Do the sets in the collection have to be bounded for example, or is there something else? From what I've read, there is no imposed constraint on the finite subcover other than for it to consist of finitely many sets.

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  • $\begingroup$ If $X$ is compact, then there is a finite subcover of any open cover, not just of those examples where it's easy to find a finite subcover $\endgroup$ – J. W. Tanner Jul 24 at 10:41
  • $\begingroup$ But how do we construct such an open cover? There are infinitely many ways to construct an Open Cover? $\endgroup$ – Joeseph123 Jul 24 at 10:42
  • $\begingroup$ Your subcover has to take sets from your cover $\{A_i\colon i\in I\}$, not some new sets. In other words, there exists a finite subset $F$ of $I$ such that $X=\bigcup_{i\in F}A_i$. You are not allowed to change the $A_i$ once they are given to you. $\endgroup$ – user10354138 Jul 24 at 10:42
  • $\begingroup$ There may be infinitely many open covers, but to show it's compact, you must show that any one of them has a finite subcover $\endgroup$ – J. W. Tanner Jul 24 at 10:43
  • $\begingroup$ You have emboldened a lot of words, but you didn't embolden "for every" in the first line. That's a universal quantifier, "for every" means "for every". It doesn't mean, "for some", "for one", "for whichever one I first think of" etc. It does mean "for every". Did I mention that "for every" means "for every"? $\endgroup$ – Angina Seng Jul 24 at 10:57
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It seems to me that you understood perfectly the meaning of “finite subcover”. The problem lies at “for every open cover”. Yes, $\{\Bbb R\}$ is an open cover of $\Bbb R$ which has a finite subcover. However, $\Bbb R$ is not compact since, for instance, the cover$$\bigl\{(n,n+2)\mid n\in\Bbb Z\bigr\}$$is another open cover of $\Bbb R$, and this cover has no finite subcover.

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  • $\begingroup$ I completely agree with you, but as far as I understand, the construction of an Open Cover is arbitrary, so how does one go about proving Compactness? In the lectures I see, the lecturer seems to sort of pull out an Open Cover from a hat, without generalizing over what the collection constituting the the Open Cover could be. The case you mention is easy to see, but let's say generally, how does one prove the "for <b>any</b> Open Cover" part? $\endgroup$ – Joeseph123 Jul 24 at 10:52
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    $\begingroup$ When a space is not compact, there is no general method to prove that it is not compact by constructing an open cover without a finite subcover. That is done case by case. For instance, the set $\{(x,y)\in\Bbb R^2\mid x^2+y^2<1\}$ is not a compact subset of $\Bbb R^2$ because the set$$\left\{\{(x,y)\in\Bbb R^2\mid(x-1)^2+y^2>r\}\,\middle|\,r>0\right\}$$is an open cover without a finite subcover. $\endgroup$ – José Carlos Santos Jul 24 at 11:05
  • $\begingroup$ But you could consider the set $0<r<2$, wouldn't this be a finite subcover? $\endgroup$ – Joeseph123 Jul 24 at 11:13
  • $\begingroup$ No, since there are infinitely many numbers $r$ such that $0<r<2$ and if $r_1$ and $r_2$ are two such numbers, then$$\{(x,y)\in\Bbb R^2\mid(x-1)^2+y^2>r_1\}\ne\{(x,y)\in\Bbb R^2\mid(x-1)^2+y^2>r_2\}.$$ $\endgroup$ – José Carlos Santos Jul 24 at 11:16
  • $\begingroup$ That's not what I meant. There are infinitely many numbers yes, but we could put them in a single set: $\{\{(x,y)∈R^2∣(x−1)^2+y^2>r\}∣∣0<r<2\}$ hence a finite subcover. $\endgroup$ – Joeseph123 Jul 24 at 11:22

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