Question

Please help verify my proof..

Answer 1

Prove that the intersection of nonempty compact sets K1, K2,...... is nonempty.

Answer 2

where \[K1 \supset K2 \supset K3 \supset .....\]

Answer 3

I was told to work on compact metric space (K1,d|_{K1xK1}) and consider the set Vn:=K1\Kn which is open on K1. Then assume such intersection is empty for contradiction, and use theorem 9 in page 6 of Week 2 notes of http://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/

Answer 4

Sorry my proof using Theorem 9 was wrong. It appears I'm stuck on this one.... But I could prove it without Theorem 9. Theorem 9 complicates the proof for me....

my proof:
suppose intersection of such set is empty. Then we have two cases.
Case 1:there could exists an empty set among each, which contradicts the hypothesis that we're dealing with non-empty compact sets.
Case 2:also there exists set KN such that KN whose elements are not in other sets Kn for all n>=1. This would mean there exists M>N such that KM doesn't contain any element in KN. But by induction and the definition of such sets, KM is a subset of KN whenever N>M. Hence for all M>N, KM will contain all elements from KN. The other case is that there exists M<N such that KM doesn't contain any element in KN. But by induction again, one can show that KM is a subset of KN when M<N. So KM will contain elements in KN for all M<N. But how if KM contains elements in KN for M<N which is not in KM' for M'>N. Since KM' contains all elements from KN for M'>N, and KM contains elements in KN for M<N, then KM' would also contain elements in M for all M<N and M'>N. So such set KN cannot exist.

we're done

Answer 5

I think your proof works. I can't find anything immediately wrong with it. However, as fair warning, this subject material is not my strongest area.

Answer 6

@KingGeorge, thanks!!!