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Excerpt from Topology by Munkres

(1)Definition: A collection $\mathbf A$ of subsets of a space X is said to cover X, or to be a covering of X, if the union of the elements of $\mathbf A$ is equal to X. It is called an open covering of X if its elements are open subsets of X. (2)Definition: A space X is said to be compact if every open covering A of X contains a finite subcollection that also covers X.

But later while discussing compact subspaces of real line, he says the below.

Theorem 27.1. Let X be a simply ordered set having the least upper bound property. In the order topology, each closed interval in X is compact.

Proof Step I: Given a < b, let A be a covering of [a, b] by sets open in [a, b] in the subspace topology (which is the same as the order topology). We wish to prove the existence of a finite subcollection of A covering [a, b].

Question: I find the statements contradictory. At first, it is said that "every open covering contains a finite subcollection". But later, to prove compactness of [a,b], we are looking for just one (at least one) finite subcollection. Why does the author say in the beginning "every open cover should have a finite subcover"? Is this related to Cauchy sequences in the set?

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    $\begingroup$ Compact means every open cover has a finite subcover. To show something is compact, we take any open cover and show that it has a finite subcover. $\endgroup$ – J. W. Tanner Feb 25 at 15:46
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He starts the proof of theorem 27.1 by taking an arbitrary open cover of the interval $[a,b]$. To show that the interval is compact it suffices, by definition, to find a finite subcover. Note that $A$ denotes the original collection of open sets covering the interval in the proof. Thus finding a finite subcover corresponds to "existence of a finite subcollection of $A$ covering $[a, b]$".

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  • $\begingroup$ Ok. The proof is for "any" open cover, right? Also, are open covers defined using "converging sequences" ? Do sequences have any significance here? $\endgroup$ – Satheesh Paul Feb 25 at 16:14
  • $\begingroup$ @SatheeshPaul yes the proof is for any open cover. Regarding sequences, keep reading Munkres and you will learn when exactly the condition that any sequence has a convergent subsequence implies the space is compact. $\endgroup$ – Mariah Feb 25 at 17:11
  • $\begingroup$ Thank you for the answer and the pointer ma'am, I'll continue with Munkres. $\endgroup$ – Satheesh Paul Feb 25 at 17:41
  • $\begingroup$ @Mariah only in metric spaces, not in all spaces though. $\endgroup$ – Henno Brandsma Feb 25 at 18:38
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    $\begingroup$ @SatheeshPaul no, open covers are not defined using sequences. In some classes of spaces (including the metric ones) there is an equivalent reformulation of compactness using sequences instead of covers. $\endgroup$ – Henno Brandsma Feb 25 at 21:53
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It's an artifact of the way we prove statements of the form $\forall x: \exists y: \phi(x,y)$: we take some arbitrary element $x$ (temporarily fixing it, as it were) and reason on it to "construct" or prove the existence of , some $y$ such that $\phi(x,y)$ holds, without using anything "specific" about $x$, just the properties it "has to" have (like being an open cover, in this case: so all members are open and its union is the space, say). It looks like we're working on some "specific" cover, but we're working on a "generic" one, really.

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