# Example of cover (of a set) having finite sub-covers in collection.

1. Dec 30, 2012

### experimentX

I think I am not understanding the concept of compactness. Can anyone give me an example of a cover that contains finite sub-covers? for example:- G = {S1,S2, ... }, Sn={(1/n,2/n): n ≥ 2} is an example of cover of set (0,1) but it is an infinite collection.

Last edited: Dec 30, 2012
2. Dec 30, 2012

### Staff: Mentor

As you have shown, (0,1) is not compact.
It is not interesting to have some finite subcover - just add T=(0,1) to your list, and there is a finite subcover (any finite subset with T). Compactness means that every cover has a finite subcover. For compact sets, you cannot create examples like the one on your post.

3. Dec 31, 2012

### experimentX

Then how could we say that cover of compact sets have finite sub-cover if we can't construct them. Shouldn't there be at least one example? I understand the proof given in my book but I can't visualize it.

4. Dec 31, 2012

### Bacle2

It depends. In R^n , there is a result that compact is equivalent to being closed and bounded. Maybe if you go over this theorem (Heine-Borel) , you can see how being closed and bounded implies that every cover has a finite subcover. In metric spaces there are related results you can use, like that every sequence has a convergent subsequence. So you can try to see the connection between these two by going over the proofs. But I don't know how to do this more generally.

5. Dec 31, 2012

### Fredrik

Staff Emeritus
The set of all open intervals with length 1 is an open cover of the closed interval [0,1]. The set $\left\{\left(\frac{-1}{3},\frac{2}{3}\right),\left(\frac{1}{3},\frac{4}{3}\right)\right\}$ is a finite sub-cover.

6. Jan 1, 2013

### experimentX

but this also covers (0,1), this would also be a finite sub cover of (0,1) making (0,1) compact. What I am asking is an example of sub covers of [0,1] that makes [0,1] compact while it does not make (0,1) compact

7. Jan 1, 2013

### Fredrik

Staff Emeritus
It covers (0,1), but it doesn't make (0,1) compact (or [0,1] for that matter...I mean, [0,1] is compact, but it takes more than this to prove it). A set is compact if every open cover has a finite subcover. I only mentioned one open cover that has an open subcover.

Also, every cover of [0,1] will of course also cover (0,1).

So I don't understand what you want. I'm not even sure I understand what you mean by an "example".

8. Jan 1, 2013

### HallsofIvy

Staff Emeritus
The set X= {0, 1, 1/2, 1/3, ..., 1/n,...}, that is, the set containing the sequence 1/n for n any positive integer and 0, is both closed (because it contains the limit of the sequence) and bounded (it obviously lies in [0, 1]) and so, as a subset of the real numbers, is compact. To prove that directly from the definition of "compact" I would do the following:

Suppose $\{U_n\}$ is a open cover for X. Then 0 is in one of those sets- rename it U0. Since U0 is open, there exist some r such that the interval (-r, r) is a subset of U0. There exis some integer N, such that N> 1/r and so 1/N< r. That means that for all $n\ge N$, 1/n< r and so is contained in U0. We then choose a set from the open cover containing 1/n for every n< N, a finite number of sets, and those, together with U0 form a finite sub-cover.

Notice that is essential that 0 be in this set. The set {1, 1/2, ..., 1/n, ...} itself is bounded but not closed and so not compact. For every integer n, Let Un be the interval from (2n+1)/(2n(n+1)) to (2n-1)/(2n(n-1)). The point of that is that (2n+1)/(2n(n+1)) is exactly half way between 1/n and 1/(n+1) while (2n-1)/(2n(n-1)) is exactly half way between 1/n and 1/(n-1). That is, that interval includes ONLY the point 1/n from this set. Since each set in the cover includes only one point of the set, we have to have all of them to cover the set- and we certainly can't reduce to a finite cover.

9. Jan 1, 2013

### experimentX

would this set be finite sub-cover of [0, 1] Sn={(1/n -ε, 2/n + ε): n ≥ 2} ? Choosing n > N ≥ 1/ε where ε > 0 would make it cover [0, 1]?

10. Jan 1, 2013

### Fredrik

Staff Emeritus
The set $\{(1/n-\varepsilon,2/n+\varepsilon):n\geq 2\}$ has infinitely many members. n is just a dummy variable. The set is the same if you use another dummy variable. So it doesn't really make sense to call the set Sn. Also, (this is a bit nitpicky) the notation should specify that n is an integer.

I would just call the set S. It's an open cover of both [0,1] and (0,1). The subset of S that's defined by the condition n≥N contains all but a finite number of members of S. It's another open cover of both [0,1] and (0,1), and it's not finite.