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?
Best Answer
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.