# Compact Space: “every open covering” Vs “an open covering”

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?

• 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. – J. W. Tanner Feb 25 at 15:46

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 $$[a, b]$$".

• Ok. The proof is for "any" open cover, right? Also, are open covers defined using "converging sequences" ? Do sequences have any significance here? – Satheesh Paul Feb 25 at 16:14
• @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. – Mariah Feb 25 at 17:11
• Thank you for the answer and the pointer ma'am, I'll continue with Munkres. – Satheesh Paul Feb 25 at 17:41
• @Mariah only in metric spaces, not in all spaces though. – Henno Brandsma Feb 25 at 18:38
• @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. – Henno Brandsma Feb 25 at 21:53

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.