Given the set E:= {0, 1,1 3 2 5 ..}. Then the number 1 is an upper bound for the set E. The supremum of E is sup E = 1. Proof: Since the number 1 is greater than all the elements of the set E, it is an upper bound for E. Suppose that 1 is not the supremum of E. Then there is a real number m <1 that is also an upper bound for E. By the Density of the Rationals, there is a fraction such that m < <1 for some natural number n eN for every n+1 natural number n. Notice that the fraction 1e E. This contradicts the assumption that m is an upper bound for E. Therefor m = 1 = sup E. n+1 True O False

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