it's a contradction because it's bounded above, i.e there is an upper bound for A, A will converge to some number eventually, however, we don't need to show A has a least upper bound here, we just need to show that there is an element . in A such that . for all S in A.
but why is this has to be in Z though?
That you understand is definitely arguable (and may indeed be argued by those already in the discussion).
Now, it appears that you are introducing a new question.
Consider the intervals [0, 1) and [0, 1] as subsets of ℝ
Both are bounded above. Both have the same least upper-bound (supremum).
One of these intervals, however, does not have a largest element. Indeed - to prove this, you could take and arbitary element and find one that is larger, and still in the interval.