Then a 2-a+ 1 = 0 has no real number solution, so there are at least two elements;
Because1(1-a) ∈ a, then1[1-(1-a))] = [(a-1)]
If (a- 1)/a=a, we also get a 2-a+ 1 = 0, which has no real number solution and at least three elements;
Because [(a- 1)/a]∈A, then1[1-(a-1)/a] = a ∈ a is not a new element;
Therefore, if set A is not an empty set, it has at least three elements.