Some book is using An=(1/n,11/n) as the open subsets. It reaches a conclusion that is not finite.
I am wondering can we just use open subsets like (0,1) itself or (0,0.5)U(0.5,1)to cover (0,1)? what's wrong with this idea?
THanks,
why the open interval (0,1) in R is not compact
Re: why the open interval (0,1) in R is not compact
Back up a second. What's your definition of compactness? Judging from your approach, I assume it's "a space is compact if any open cover has a finite subcover"?
Yes, you can cover (0,1) by itself. Or by (0,0.55) union (0.45,1). Both are finite open covers, but that does not tell you anything about the compactness of (0,1).
What you need to be worried about are infinite open covers. If a space is compact, any and every infinite open cover admits a finite subcover. So let's look at the open cover An=(1/n,11/n). Does it admit a finite subcover? If not, (0,1) is not compact.
Yes, you can cover (0,1) by itself. Or by (0,0.55) union (0.45,1). Both are finite open covers, but that does not tell you anything about the compactness of (0,1).
What you need to be worried about are infinite open covers. If a space is compact, any and every infinite open cover admits a finite subcover. So let's look at the open cover An=(1/n,11/n). Does it admit a finite subcover? If not, (0,1) is not compact.
Re: why the open interval (0,1) in R is not compact
Oh. right. I see. Every open covering has to contain a finite sub collection. This counter example will fail it.owlpride wrote:Back up a second. What's your definition of compactness? Judging from your approach, I assume it's "a space is compact if any open cover has a finite subcover"?
Yes, you can cover (0,1) by itself. Or by (0,0.55) union (0.45,1). Both are finite open covers, but that does not tell you anything about the compactness of (0,1).
What you need to be worried about are infinite open covers. If a space is compact, any and every infinite open cover admits a finite subcover. So let's look at the open cover An=(1/n,11/n). Does it admit a finite subcover? If not, (0,1) is not compact.
Thank you very much!!

 Posts: 1
 Joined: Thu Oct 13, 2011 7:48 pm
Re: why the open interval (0,1) in R is not compact
If you're doing the Math GRE you'll definitely want to take the Heine Borel theorem for granted:
"A subset of a metric space is compact if and only if it is closed and bounded."
Or for Euclidean space:
"A subset of a R^n is compact if and only if it is closed and bounded."
The only if direction essentially can be proven with the same trick you use to provide a counterexample for (0,1).
"A subset of a metric space is compact if and only if it is closed and bounded."
Or for Euclidean space:
"A subset of a R^n is compact if and only if it is closed and bounded."
The only if direction essentially can be proven with the same trick you use to provide a counterexample for (0,1).

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: why the open interval (0,1) in R is not compact
Your HeineBorel statement for metric spaces is not correct. You need completeness and total boundedness