Completeness Axiom: Definitions, Examples, and Applications, Lecture notes of Calculus

Download Completeness Axiom: Definitions, Examples, and Applications and more Lecture notes Calculus in PDF only on Docsity! Completeness Axiom Definitions. Let S be a nonempty subset of R. An upper bound for S is any real number b such that if x ∈ S then x ≤ b. A least upper bound (lub) for S is any real number c such that c is an upper bound for S and if b is another upper bound of S then c < b. If S contains its least upper bound c then we say c is the maximum of S. Examples. Let S = (−1, 1). Then 17 is an upper bound of S and 1 is the least upper bound of S. In the case S does not have a maximum. Let T = (0, 1)∪{2}. Then 2 is the least upper bound and the maximum of T . The terms lower bound, greatest lower bound (glb) and min- imum are defined similarly. Examples. Let A = (0, 3) ∪ N. Then A does not have an upper bound. Any negative number is a lower bound and 0 is the greatest lower bound. The set A does not have a minimum. Let B = { n+1 n |n ∈ N } . Then 2 is the maximum and 1 is the greatest lower bound. The set B does not have a minimum. The Completeness Axiom for R. If S is a nonempty subset of R that is bounded from above then S has a least upper bound. In MATH 352 the Completeness Axiom is assumed to be true for R. In MATH 452 we prove the Completeness Axiom is true for R. The Completeness Axiom is false for Q. Let A = {r ∈ Q | r2 < 2}. Then A is not empty since 1 ∈ A. You can show that 3 is an upper bound for A. But A does not have a least upper bound in Q. If it did, say r∗ is the lub, then it can be shown that (r∗) 2 = 2, but we know there is no such rational number. If we regard A as a subset of R then is does have a lub that is called √ 2. In MATH 452 we prove that ( √ 2)2 = 2. In MATH 352 we assume this. Corollary. Using the Completeness Axiom it is easy to prove that if S is a nonempty subset of R that is bounded from below then S has a greatest lower bound. See textbook for proof. Definitions. Let S ⊂ R. Then the supremum and infimum of S are defined as follows. supS =  +∞ if S 6= ∅ and is not bounded above, lub S if S 6= ∅ and is bounded above, −∞ if S = ∅. inf S =  −∞ if S 6= ∅ and is not bounded below, glb S if S 6= ∅ and is bounded below, +∞ if S = ∅. Theorem. The Archimedean Property. Let a and b be positive real numbers. Then ∃ n ∈ N such that na > b. The proof uses the Completeness Axiom and is harder than you would think! Proof. Suppose not. Then ∃ a > 0 and b > 0 such that ∀ n ∈ N na ≤ b. Let S = {na |n ∈ N}. Since b is an upper bound for S, S must have a lub. Call it s0. Since 0 < a we have s0 < s0 + a and hence s0 − a < s0. ∃ n0 ∈ N such that s0− a < noa because s0− a is less than the least upper bound of S. Hence, s0 < (n0 + 1)a. But this means (n0 + 1)a /∈ S. Hence, our supposition was foolish! The Archimedean Property has been vindicated!!