# Thread:

1. ## what is a finite subcover?

I'm trying to write up a proof and getting really stuck on this idea of "finite subcover."

The problem is to prove that given any closed and bounded interval [a,b] in R, any cover of [a,b] with open intervals contains a finite subcover.

I saw a proof by contradiction sketched out that went something like this:
Assume that there exists a cover of [a,b] that does not contain a finite subcover.

Divide the interval [a,b] in half. We know that one of those subintervals contains a cover that does not have a finite subcover. Divide in half again.... repeat process until we have a series of nested intervals all with the property that they are assumed to have an open cover that does not have a finite subcover.

The nested interval principle tells us that that the union of infinitly many nested subintervals will contain a single point. And here there is a contradiction that rests on an understanding of finite subcover. And that is where I'm stuck.

Can anyone please explain what a finite subcover is and why there is a contradition here? Is it that any open cover of a single point must have a finite subcover? Is that all it is?

2. A finite subcover is an open cover that is formed out of a finite number of sets. An open cover of E is a collection of open sets whose union contains E.
In fact, if E contain a finite number of element, then for any open cover of E, you can find a finite subcover. For example if E = {1,2,3}, then given any open cover (can be infinite of course), choose three sets from it in such a way that the first set contains 1, the second contains 2 and the third contain 3. You are sure that these three exist since we started from an open cover covering all E. Hence, the collection of the three sets found constitute a finite subcover. In you example, yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).

Hope this helps

3. Originally Posted by Beckeroo
I'm trying to write up a proof and getting really stuck on this idea of "finite subcover."
The problem is to prove that given any closed and bounded interval [a,b] in R, any cover of [a,b] with open intervals contains a finite subcover.
Suppose that $[a,b]$ is covered by a collection of open intervals.
Let $P=\{x\in (a,b]: [a,x]\text{ is covered by a finite subcollection.}\}$
1) Show that $P$ is not empty. Hint $a+\delta\in P$.
2) Let $q=\sup(P)$ then show that $q=b$. Hint what if $q<b?$

4. Originally Posted by mohammadfawaz
yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).

Hope this helps
I thought this must be the case but wasn't sure I was understanding correctly. Thanks!

5. Originally Posted by mohammadfawaz
A finite subcover is an open cover that is formed out of a finite number of sets.
Be careful here! Not just a "an open cover that is formed out of a finite number of sets". If that were true, since every set can be covered by a single open set, the entire space, every set would be compact! A finite open subcover is an open cover, consisting of a finite number of sets from a given open cover.

An open cover of E is a collection of open sets whose union contains E.
In fact, if E contain a finite number of element, then for any open cover of E, you can find a finite subcover. For example if E = {1,2,3}, then given any open cover (can be infinite of course), choose three sets from it in such a way that the first set contains 1, the second contains 2 and the third contain 3. You are sure that these three exist since we started from an open cover covering all E. Hence, the collection of the three sets found constitute a finite subcover. In you example, yes, any open cover of a single point must have a finite subcover since the given set contain a finite number of points (in fact only one point).

Hope this helps

6. Thank you HallsofIvy. I really appreciate your clarification.
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