Help one question about analysis

[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.

 

[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...

 

[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.

 

[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 U_n 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.

 

[SiddharthM]

is still NOT compact is what halls of ivy means.

 

[HallsofIvy]

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

 

[mathwonk]

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