1. Jan 28, 2009

### alvielwj

I am studying Heine-Borel theorem.
I finally get the sense of this theorem after i read a lots posts here.
But I found a question posted by Rach3
why for S={1,1/2,1/3……} there is no subcollection of it is a cover of S
but every open cover of T={0,1,1/2,1/3……} has a finite subcover?

2. Jan 28, 2009

### alvielwj

I know that I can find a cover e.g Ia=(1/n,1+1/n),union of Ia from n=1 to inifinte is a cover for S but not finite

3. Jan 28, 2009

### alvielwj

but how to prove that every open cover of T has a finite subcover

4. Jan 28, 2009

### quasar987

There is no universally recognized statement of the Heine-Borel theorem so you'd better tell people what you call the "Heine-Borel" theorem when you speak about it.

I will assume that you take it to mean "A subset of R is compact if and only if it is closed and bounded."

So the answer to your question is that every open cover of T admits a finite subcover because it is compact. And why is it compact? Because it is closed and bounded. On the other hand, the set S above is not compact because it is not closed.

5. Jan 28, 2009

### alvielwj

I need to prove every every open cover of T has a finite subcover.
My version of Heine-Borel theorem is every open cover of a closed interval has a finite subcover.
It is easy to prove if it is a closed interval. I mean like [a,b]
not a interval like T.

6. Jan 28, 2009

### quasar987

Don't forget the word "bounded": your version of the Heine-Borel theorem is that every open cover of a closed and bounded interval has a finite subcover.

It is important not to forget "bounded" because for instance, the whole real line R is a closed interval but clearly there is no finite subcover of the open cover $\{(-n,n)\}_{n\in\mathbb{N}}$.

So, why does your version of the H-B theorem implies that every open cover of T has a finite subcover?

Well, notice that T is a closed subset of the closed and bounded interval [0,1]. Now consider $\{U_i\}_{i\in I}$ an open cover of T. Then add to that cover the open set V:=R\T so that $\{V\}\cup \{U_i\}_{i\in I}$ is an open cover of [0,1]. By the H-B theorem then, there exists a finite subcover of that open cover. Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover), we get a finite subcover of T.

7. Jan 28, 2009

### alvielwj

Thank you.
I think I need some times to understand the last step" Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover)"

8. Jan 28, 2009

### alvielwj

I think you are using the result to prove the question .
you mentioned that t is closed and bounded,then processing to prove in the way T is bounded and closed...
If you using S instead of T , your prove still hold.
"Now consider an open cover of T. Then add to that cover the open set V:=R\T so that is an open cover of [0,1]. By the H-B theorem then, there exists a finite subcover of that open cover. Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover), we get a finite subcover of T.
"

9. Jan 28, 2009

### quasar987

If I understood correctly, the question was "Use the Heine-Borel theorem to prove that every open cover of T has a finite subcover.", and it was not "Show that T is closed and bounded."

However, you're right that in solving the problem, I did use the fact that T closed. But that is easy to show.