Conjecture: Given an open set O that for each n >=1 can contain disjoint open squares of sides 1, 1/2, 1/3,...,1/n , then O can contain disjoint squares of all sides 1/n, n >= 1 simultaneously.

If O is bounded, then the closure of O can contain disjoint squares of all sides 1/n, n >= 1. Proof: For all n, form the infinite sequence of points (p(n,1),p(n,2), p(n,3),...), where for k <= n, p(n,k) is the center of the square of side 1/k in the nth partial packing. (It won't matter how we define p(n,k) for k>n.) By compactness, we can find a subsequence of partial packings such that all coordinates converge. This gives the desired packing of all the squares in the closure of O. Jim Propp