The usual proof that the rationals have measure zero in the reals has always bugged me. If you surround every point in a dense subset of [0,1] with a finite interval, however small, the intervals ought to overlap like crazy and cover the unit interval many times over, no? And if, instead of surrounding the nth rational (for some ordering) by epsilon/2^n, you surround every rational n/d by epsilon/d^2, the overcovering ought to be even worse, no? No. Admittedly, pictures can deceive, but check out the epsilon=1 case on page 11 of http://www.tweedledum.com/rwg/rectarith12.pdf . Clearly every n/d can be surrounded (by how many intervals?), and the the area sum is finite. --rwg