Proving Compactness of Sets Using Open Covers

In summary, the problem asks for a proof that for some n there is a compact set X that is contained within a finite number of open sets U_1, U_2, U_3, ...
  • #1
Frillth
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0

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.
 
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  • #2
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.
 
  • #3
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...
 
  • #4
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?
 
  • #5
Frillth said:
each U has finite volume

You don't know this. The cylinder [tex]\{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < 1 \}[/tex] 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.
 
  • #6
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.
 

1. What is a trivial set?

A trivial set is a set that contains no elements or is empty. It is denoted by the symbol ∅ or { }.

2. How is a trivial set different from other sets?

A trivial set is different from other sets because it does not contain any elements, whereas other sets have at least one element.

3. What is the cardinality of a trivial set?

The cardinality of a trivial set is 0, since it contains no elements.

4. Why is a trivial set important in set theory?

A trivial set is important in set theory because it serves as the basis for defining other sets and operations on sets. It also helps to distinguish between empty sets and non-empty sets.

5. Can a set be both trivial and non-trivial at the same time?

No, a set cannot be both trivial and non-trivial at the same time. A set is either empty (trivial) or contains at least one element (non-trivial).

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