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Math 554 - Fall 08
Lecture Note Set # 1
Defn. From the introductory lectures, an ordered set is a set S with a relation ‘<’ which satisfies two properties:
- (Trichotomy property) for any two elements a, b ∈ S, exactly one of the following hold
a < b, a = b, or b < a.
- (Transitive property) for any three elements a, b, c ∈ S, if a < b and b < c, then a < c.
In this case the relation ‘<’ is called an order.
Defn. Suppose that S is an ordered set and A ⊆ S. An element β ∈ S is said to be an upper bound for A if a ≤ β, ∀a ∈ A.
An element α is said to be a least upper bound for A if
- α is an upper bound for A
- if β is any upper bound for A, then α ≤ β.
In this case, the supremum of A (=: sup A) is defined as α. The definitions are similar for lower bound, greatest lower bound and inf A, respectively. Note that we have already shown that the least upper bound (for a nonempty set bounded from above) is unique.
Defn. A set S is said to have the least upper bound property if each nonempty subset of S which is bounded from above, has a least upper bound.
Theorem. Suppose the ordered set S has the least upper bound property, then it has the greatest lower bound property (i.e. each nonempty subset of S which is bounded from below has a greatest lower bound).
Proof. Suppose that a nonempty set A has a lower bound, call it ℓ. Define L as the set of all lower bounds of A, then L is nonempty (ℓ ∈ L). Observe that each member of the nonempty set A is an upper bound of L so by the least upper bound property, L has a least upper bound. Call this element α. First observe that α is a lower bound for A. Otherwise, there exists an element b ∈ A with b < α, but each element of A is an upper bound for L, so this element b is an upper bound of L which is smaller than α, the least upper bound of L. This would be a contradiction. Therefore, α ∈ L. Also, if ℓ is any lower bound of A, then ℓ ≤ α since α = sup L. Hence α is the greatest lower bound of A. 2
Defn. The real numbers are defined to be a set IR with two binary operations (+, ·) which satisfy the following properties: Given any a, b, c in IR
- a + (b + c) = (a + b) + c.
- a + b = b + a.
- ∃ 0 ∈ IR ∋ a + 0 = a, ∀a ∈ IR.
- for each a ∈ IR, ∃(−a) ∈ IR, so that a + (−a) = 0.
- a · (b · c) = (a · b) · c.
- a · b = b · a.
- ∃ 1 ∈ IR ∋ 1 6 = 0 and a · 1 = a, ∀a ∈ IR.
- for each a ∈ IR with a 6 = 0, ∃a−^1 ∈ IR, so that a · a−^1 = 1.
- a · (b + c) = (a · b) + (a · c).
(Note: These properties just say that IR is a nontrivial field.)
Moreover, there is a distinguished subset IP (the positive cone) of IR with the following proper- ties: Given any a, b in IP ,
a. a + b ∈ IP , b. a · b ∈ IP , c. For each a in IR, exactly one of the following properties holds: i) a ∈ IP , ii) −a ∈ IP , iii) a = 0.
Finally, IR must satisfy the least upper bound property, that is, each nonempty subset of IR which has an upper bound has a least upper bound. These terms are defined shortly.
Defn. For the real numbers, define a < b as b − a ∈ IP.
Lemma. Using the field properties of IR, the following properties hold and are all assigned home- work problems (see p. 3 of this lecture set):
(1) The additive and multiplicative identities are unique. (2) The additive and multiplicative inverses are unique. (3) If a ∈ IR, then a · 0 = 0.
Theorem. The positive cone IP induces an order on the real numbers, i.e. IR equipped with the relation ‘<’, is an ordered set.
Proof. Assigned homework Problem 2.5. 2
Notation:
‘b − a’ is defined as b + (−a). ‘a ≤ b’ means either a < b or a = b. The fraction ‘ a b ’ means a · b−^1.
Lemma. For each a, b, c ∈ IR,
(i.) (−a) = (−1) · a. (ii.) 0 < 1. (iii.) if 0 < a, then (−a) < 0. (iv.) if a < b and 0 < c, then a · c < b · c. (v.) if a < b, then a + c < b + c. (vi.) if 0 < a, then the multiplicative inverse of a is positive, i.e. 0 < a−^1. (vii.) the product of two negative real numbers is positive, while the product of a negative real number and a positive real number is negative.
- Prove that if 0 < a < b, then 0 <
b
a
Notation: Next we define intervals of real numbers.
(a, b) := {x ∈ IR|a < x < b} is called the open interval with endpoints a, b. [a, b] := {x ∈ IR|a ≤ x ≤ b} is called the closed interval with endpoints a, b. (a, b] := {x ∈ IR|a < x ≤ b} and [a, b) := {x ∈ IR|a ≤ x < b} are called the half open intervals with endpoints a, b. The common length, or measure, of these intervals is defined to be b − a.
Theorem. Suppose that I is an interval with endpoints a, b and a < b, then I contains a rational number.
Proof. Define the length of I by ℓ := b − a. By the previous corollary, there exists no ∈ IN such that 0 < 1 /no < ℓ. Let A := {k| k an integer and k/no < a}. A is nonempty, since the negative integers are not bounded from below. Let ko belong to A. Set B := {k| k an integer and k ≥ ko} ∩ A. Also, A is bounded from above by a · no, which shows that B is in fact a finite set of integers. Let K be the largest member of B and therefore of A, then K + 1 6 ∈ A. Let r := (K + 1)/no, then
a <
K + 1
no
K
no
which shows that the rational r ∈ (a, b) ⊆ I. 2
Corollary. Each interval with nonzero length contains an infinite number of rationals.
Defn. A real number is said to be irrational if it is not rational.
Remark: Each interval with nonzero length contains an uncountably infinite number of irrationals. (Proved later.)
We establish a few other facts about irrational numbers and also prove directly that each interval of positive length contains an infinite number of irrationals.
Lemma. The product of a nonzero rational with an irrational is irrational.
Proof. Suppose that q 1 · α = q 2 , where q 1 , q 2 are rational and α is irrational. Since q 1 6 = 0, then α = q 2 /q 1 and it follows that α is rational. Contradiction. 2
Lemma. If m is an odd integer, then m^2 is odd.
Proof. If m is odd, then there exists an integer k such that m = 2k + 1. In this case m^2 = 2(2k^2 + 2k) + 1. 2
Lemma.
2 is irrational.
Proof. Suppose that
2 = m/n where m, n are integers with n > 0. We may assume that the rational is in lowest terms (i.e. m and n have no common factors). Squaring the equation and multiplying by n^2 , we obtain that m^2 = 2n^2. This shows that m^2 is even. By the lemma m must be even and equivalently that it contains 2 as a factor. This shows 4k^2 = 2n^2 for some integer k.
Consequently, n is even and 2 appears as one of its factors. Contradiction, since m/n was supposed to be in lowest terms. 2
Theorem. Each interval with nonzero length contains an infinite number of irrationals.
Proof. Let a, b be the endpoints of the interval I. Consider the interval (a/
2 , b/
2). It has length (b − a)/
2 > 0, and so contains a nonzero rational number q. It follows that q
2 is between a and b and hence belongs to I. 2
Defn. The absolute value of a real number a is defined by
|a| :=
{ a, if a ≥ 0 −a, if a < 0.
Lemma. The absolute value function has the following properties:
- |a| ≥ 0 for all a ∈ IR.
- ±a ≤ |a|
- | − a| = |a|
- |a · b| = |a| · |b|
- Suppose α ≥ 0, then |a| ≤ α if and only if −α ≤ a ≤ α.
- |a + b| ≤ |a| + |b|
- ||b| − |a|| ≤ |b − a|
Proof. To prove property 1, ‘case it out.’ Either a ≥ 0 or a < 0. In the first case |a| = a ≥ 0. In the second case, |a| = −a > 0, since a < 0. Property 2 follows similarly. Properties 3 and 4 are left for the student. To prove property 5, notice that |a| ≤ α is equivalent to the statement ±a ≤ α. Using the ele- mentary order properties, it is easy to see that this last statement is equivalent to both inequalities: a ≤ α and −α ≤ a. This is equivalent to the statement on the right hand side of property 5. To prove property 6, set α = |a| + |b| and apply property 5 after verifying that ±(a + b) ≤ α. To prove property 7, use the fact that |b| = |(b − a) + a| ≤ |b − a| + |a| and subtract |a| from each side. This shows that |b| − |a| ≤ |b − a|. By symmetry in a and b, one proves that |a| − |b| ≤ |a − b| = |b − a|. 2
