Question

Prove that if a nonempty set is bounded above, then it has exact upper bound.

Answer 1

In a numeric, non-empty, finite set the elements can be arranged in order. If this order is represented by:

{e₀, e₁, e₂, ...,e_n} where e_i≥e_i-1 for i>0.

e_n is the upper bound, U, and e_n≥e_n-1, e_n-1≤U.

In an infinite series the infinite set consisting of individual terms (a_i) of the series, such that all a_i>a_i-1 for 1<i≤n, then a_n>a_n-1 as n→∞, the series converges to an upper limit U. No element of the series exceeds U. (Many series converge to a limit, but the terms alternate from above to below that limit. These series would not then be bounded above because of the bidirectional approach to the limit. In this question, because the series or set is bounded above, no individual element in the set can exceed the upper bound.)