Proving Compactness of Sets Using Open Covers

[Frillth]

Homework Statement

Suppose X ⊂ R^n is a compact set, and U_1, U_2, U3, ... ⊂ R^n are open sets whose union contains X. Prove that for some n ∈ N (the natural numbers) we have X ⊂ U_1 ∪ ... ∪ U_n.

Homework Equations

A set is called compact if it is both closed and bounded.

The Attempt at a Solution

This problem seems trivial to me. If, as stated in the problem, U_1, U_2, U3, ... ⊂ R^n are open sets whose union contains X, does that mean that for some n we have X ⊂ U_1 ∪ ... ∪ U_n? I don't understand how there is anything to prove here. Any help would be appreciated.

 

[d_leet]

You have an infinite union of open sets which cover some compact set X, you wish to show that finitely many of them suffice to cover X.

 

[Frillth]

Ah, ok. That makes a lot more sense. Would this proof suffice?

We will prove by contradiction. Assume that no finite number of the U's will contain all of X. Let x_k ∈ X be one such element that is cannot be contained in a finite number of the U's. But if x_k ∈ X, then it must be contained in at least one of U_1, U_2, U_3 ... So assume that x ∈ U_k. But if we adjoin U_k to our list of U's that earlier did not contain x_k, then it will still be finite and will now contain x_k. We can repeat this procedure for all x_k to show that the list of U's necessary to contain x_k must be finite.

Upon further review, I don't think that this can be right. If we have an infinite number of x_k's, then our set of U's would also be infinite...

 

[Frillth]

This makes intuitive sense to me, but I'm not sure if it would work as a proof:

Since each U is open, there is an open ball around each point in U. But an open ball must have some finite volume, and since the set X is compact, it can be contained by a ball around the origin, and it must thus have finite volume also. Since X has finite volume and each U has finite volume, there must be a finite number of U's that can completely contain X.

Can I use the term "Volume" to describe the space contained by a ball in R^n? Also, how can I make this intuitive proof rigorous?

 

[mutton]

each U has finite volume

You don't know this. The cylinder \{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < 1 \} is an example of an open set that doesn't have finite volume. There are just infinitely many of those "finite-volume" open balls.

You'll need to use the fact that X is closed (and bounded), because the condition given in the problem is equivalent to compactness in R^n.

 

[HallsofIvy]

By the way, not only is this not a "trivial" problem, it is not a problem in set theory: set theory does not include the concepts of "open", "closed", or "compact" sets. This is topology.