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Consider the set S = {1/n : n ∈ N}. minimum? Does this set have a Definition Given a subset S ⊂ R, and element x ∈ S is said to be the minimum of S if for all y ∈ S, x ≤ y. For a contradiction, assume that 1/N ∈ S is the minimum. Then choose M > N (we need to know 1 > 0 for this). Then 1/M ∈ S, but 1/M < 1/N (we need to prove this). The set S is an example of a set that does not have a minimum. It wants to have 0 as a “minimum”, but 0 is not in the set. We are going to define a concept of “greatest lower bound” that captures this number 0 in this case. 1 Consider the set {1/n : n ∈ N}. Lower bounds: 0, −2, any negative number. Upper bounds: 1, 1729 Maximum: 1 (because it is an element of the set and it is an upper bound) Minimum: there isn’t one Least upper bound: 1 Greatest lower bound: 0 How would you prove any of these statements? 2 Same questions for N. Lower bound: 1 Upper bound: there isn’t one. But how do we prove this? It is tempting to do something like this: Suppose x is an upper bound. Then x ≥ n for all n ∈ N. So take the integer part of x and add 1. The integer part is the greatest integer less than or equal to x. But there is a problem with this, because we have assumed x ≥ n for all n ∈ N. In fact, one cannot prove that the natural numbers are unbounded with the axioms we have so far. 3 Definition The least upper bound of a set S ⊂ R is a number x which satisfies two conditions 1. First, x is a upper bound for the set. That is, for all y ∈ S, we have x ≥ y. 2. If x0 is any other upper bound, then x ≤ x0. Completeness axiom If S ⊂ R which is nonempty and has an upper bound, then S has a least upper bound. 4 Proof that N is unbounded Now suppose that N is bounded above. Then N has a least upper bound, m. Assume that m is not a natural number. Then one can find a smaller upper bound for N by subtracting some small number from m (this needs some work to prove). Therefore m is a natural number. But then m + 1 is also a natural number. Contradiction. So N is unbounded. There is a name for what we have proved: Archimedean axiom Given any real number x, there exists a natural number n such that n > x. 5