Question

(2) Consider the following examples. (a) Let A = n.) for each neN. What is Ain A, for any i,j EN? What is an An? (b) Let Bn = (0,1/n) for each neN. What is Bin B; for any i, j e N? What is aan Bn? (c) What's the deal with these examples? Do they relate to Helly's Theorem? What might be going wrong here?

Answer 1

a) An= [n, col for each nE INI. then if inj, then Ai : [ Co), 4 ) : 1,0) and AoA; - (1.60) then Ain Aj = [1,0) If i<j, Consider In An - An A₂ 0 As.... NEN Suppose An & 8, then I KEIR St KENAN nen = K zo. & K6 nAn. NEN lkst (Lkl - greatest integer function But k & - k for n= [n, co) & n An A contradiction NEN 2 nan = 0 NEN 6 Bn = (0 in), for each ne HN Bi= (0,1; and Bj = (0, ';) if is; then ric /j. - Bin Bj = (0, '14) similarly for iss, BinB = (0, 1)

Suppose that non to NEN let k o such that ke nBn NEN By the archimedian property, we can find new st = h <k, so - kq non 4 (0, 'n) at a contradiction - non is = 0. NEN Hellys theorem - Let x,,ta, ixn be a finite collection of convex seibset of Rd with ned if the intersection of every dtl of these set is non-empty then for Infinite collection, we need to assume Compactness In problem Chl, we have intinite intersection, But the subsets Bn are not compact. so Helly's theorem Can not relate to the example (b) In example cal, the subset An are compact But in Rd every Collection of Cardinality at most del has non-empty intersection this case fails. so. Helly's theorem cannot relate to the example (a)
Arbitrary intersection in both examples is empty sets. The Helly's theorem can't relate.