1. Apr 12, 2012

### cragar

1. The problem statement, all variables and given/known data
If we have the open interval (0,1) my book says that there is no finite sub cover.
2. Relevant equations
$A \subseteq \mathbb{R}$ An open cover for A is a possibly infinite collection of open sets whose union contains the set.
3. The attempt at a solution
By why cant my subcover be $(0,1/2) \bigcup (1/4,1)$
those are 2 open sets whose union is (0,1) and I have 2 sets.
Unless when they say finite sub cover they are talking about elements in the set.
I am probably missing something in the definition. Any help would be much appreciated.

2. Apr 12, 2012

### clamtrox

Perhaps your book is saying that there exists a cover with no finite subcovers? That statement would be true, and it would follow that the interval (0,1) is not compact.

3. Apr 12, 2012

### HallsofIvy

Staff Emeritus
No, it doesn't.

It says that there exist open covers which contain no finite sub cover.

Even simpler is the single open set (0, 1) itself.

However, consider the infinite collection of open sets {1/n, 1} where n can be any integer larger than 1. If x is in (0, 1) it is positive and so 1/x is a positive number. There exist an integer N> 1/x which gives x> 1/N. x is contained in (1/N, 1) for that N so this is an open cover of (0, 1). But there is no finite number of those sets which contain all of (0, 1).

Last edited: Apr 12, 2012
4. Apr 12, 2012

### cragar

What do we mean by finite sub cover. DO our sets have to be indexed by the natural numbers?
Actually is it just saying that there can be an open cover that has no finite sub cover.
Not necessarily that all open covers are infinite.

Last edited: Apr 12, 2012
5. Apr 12, 2012

### Office_Shredder

Staff Emeritus
Your open sets can be indexed by whatever collection of indices you choose - it is common to have a collection of open sets indexed by the points in your topological space when using compactness

6. Apr 12, 2012

### HallsofIvy

Staff Emeritus
First, do you understand what "cover" means? An collection of sets "covers" set A if every member of A is in one of those sets. (Imagine set A as part of a drawing on the floor. You drop pieces of paper on top of it. A is "covered" if every point in A is under at least one of those papers.) An open cover for set A is, of course, a collection of open sets that "covers" A.

In fact, strictly speaking they don't have to be indexed at all. Of course, if there is a finite subcover, you could index that subcover with a finite set of natural numbers.

Yes, that is what both clamtrox and I said.

7. Apr 12, 2012

### cragar

ok thanks for all of your posts, it makes more sense now.

8. Apr 12, 2012

### SteveL27

Of course there's a finite subcover. {(0,1)} is one such. {0,3/4), (1/4, 1)} is another.

What your book probably says is that there exists an open cover that contains no finite subcover. One such would be (0, n/(n+1)) for n = 1, 2, 3, ... Therefore (0,1) is not compact.

But of course it does have a finite subcover, lots of them in fact.

But it's NOT true that every open cover of (0,1) contains a finite subcover. So it's not compact.