1. Oct 9, 2007

ShengyaoLiang

The set A={1/n : n in N} is not compact.
A) prove this by explicitly finding a family of open sets which covers A but has no finite subfamily whcih also covers A.
B) Find another family of open sets which covers A and does have a finite subfamily which cobers A.

2. Oct 9, 2007

ShengyaoLiang

i have no idea on how it works.....
i am learning by myself right now....so could some one help me out?...thank you very much...

3. Oct 9, 2007

eastside00_99

I assume you are working in the standard topology for R. Then just take the family of open sets defined by

{U_n = (1/n,infinity) | n in N}.

This does not have a finite subcover. To see this notice that U_n+1 contains U_n for all n. Then any finite subcollection of these sets will have a maximal lement U_M where M is in N. Clear the point 1/(M+1) is not in any of the sets in the finite subcollection.

B. Take the collection {(-infinty, infinity)}.

Try to extend this argument for the set {(x,y) | |x| + |y| < 1} and to be able to say any open set in R or R^2 is not compact.

4. Oct 10, 2007

HallsofIvy

Here's another example:
The distance from 1/n to 1/(n+1) is 1/(n(n+1)) (the distance from 1/n to 1/(n-1) is larger). Take as your open sets each Un the neighborhood of 1/n with radius 1/(n(n+1)). Each of those contains exactly one point in {1/n} and so no finite subset will cover all of {1/n}.

For B, just take { (0, 2)}. That is a single set that covers all of {1/n} by itself. Of course, that tells us nothing about the compactness. {1/n} is "still" compact because not every open cover has a finite subcover.

5. Oct 11, 2007

SiddharthM

is still NOT compact is what halls of ivy means.

6. Oct 11, 2007

HallsofIvy

Oops, thanks! My fingers don't always connect to my brain.

7. Oct 11, 2007

mathwonk

here is a clue. if you throw in 0, it becomes compact.