Prove that the union of finitely many compact subsets is bounded.?

Question

Actually the original problem is "Prove that the union of finitely many compact subsets is compact." I have shown that the union is closed. What is left is to show that it is bounded so that I can conclude that the union is indeed compact.

Answer 1

There's a more general proof that's good in every topological space (from your question, I see you are in euclidean spaces, where Heine Borel theorem holds).

By definition, a set A of a topological space is compact if every open cover of A contains a finite subcover. An open cover of A is any collection of open sets whose union contains A.

Suppose A_1, ...A_n are compact and put A = Union (i=1,n) A_i. If G is an open cover of A, then G is an open cover of each A_i and, by compacteness, each A_i is covered by a finite subcollection G_i of G. This implies G_1 U G_2....U G_n is a finite subcollection of G that covers A. And since this conclusion holds for every open cover of A, it follows A is compact, as desired.

Answer 2

Which definition of 'compact' do you use? The typical one is 'every open cover has a finite subcover'. In this case, if you have a cover of the union of those compact sets, then you have a cover of each one, individually. So each one has a finite subcover. Take all of the sets from each finite subcover and you'll get a finite subcover for the union.

As for your question: Take a bound for each of the finite sets. Then the maximum of this collection of bounds will be a bound for the union. This is finite since there are only finitely many sets.

Answer 3

Well, since a compact set can be covered with a finite number of open sets, obviously the amount of open sets covering the finite union of compact sets must also be finite (finíte times finite = finite). But that exactly means that the union is compact. QED.

Answer 4

enable's call this subset A. %. a think approximately A, say, x. because of the fact A is open, there is an open era containing x totally in A. enable a = inf{y in R: (y,x) is in A} and b = sup{z in R: (x,z) is in A} the two those contraptions are non-empty and bounded, so the sup and inf exist. for sure (a,b) is likewise in A. doing this for each think approximately A yields a (in all probability limitless) number of open periods whose union is A. we could desire to in user-friendly terms tutor that for any 2 factors x1,x2 in A, the two: one million) x1 and x2 lie interior the comparable era, or 2) x1 and x2 lie in distinctive periods. we could assume that x1 is in (a1,b1) and x2 is in (a2,b2). assume x2 is in (a1,b1). then because of the fact b1 is the least larger sure of all z such that (x2,z) is in A, b1 ? b2. further, a2 ? a1. consequently x2 is in (a2,b2), and because b2 is the least larger sure of all z such that (x1,z) is in A, b2 ? b1. further, a1 ? a2, and so a1 = a2, and b1 = b2. on the different hand assume x2 isn't in (a1,b1). i declare that (a1,b1) ? (a2,b2) is then empty. assume no longer, and enable w be an ingredient of (a1,b1) ? (a2,b2) = (c,d). with the aid of the comparable argument as earlier a1 = c, and b1 = d, and likewise a2 = c, and b2 = d. for this reason a1 = a2 and b1 = b2. yet then x2 is in (a1,b1), a contradiction. for this reason (a1,b1) and (a2,b2) could desire to be disjoint.