36 Chapter 1 • The Real Number System
If A has an upper bound we say that A is bounded above; if A has a
lower bound we say that A is bounded below. If A is bounded above and
below, we say that A is bounded.
Theorem 1.6.2 (a) A set can have more than one upper bound and more than
one lower bound.
(b) A set cannot have more than one maximum nor more than one mini-
mum element.
(c) Every nonempty finite^9 set has both a maximum element and a mini-
mum element.
Proof. (a) Exercise 3.
(b) Part 1: Suppose u = max A and v = max A. Then
(1) u EA, and Vx EA, x:::; u;
(2) v E A, and Vx E A, x :::; v.
In (1) we may take x = v, and so v:::; u. In (2) we may take x = u, and so
u:::; v. Therefore, u = v.
Part 2: Suppose u = minA and v = minA. Show u = v (Exercise 4).
( c) Suppose S is a nonempty finite subset of F. That is , 3 n E N 3
S= {x1,X2,···Xn}·
Part 1: To prove that S has a maximum element, we shall use mathematical
induction on n.
(1) Any set with only one element, S = {xi}, has a maximum element;
X1 = maxS.
(2) Suppose that any set with k elements has a maximum element. Let S
be a set with k + 1 elements, say
S = {x1,X2, · · · Xk+i}·
Then the set T = {x 1 , x 2 , · · · , xk} has a maximum element; say t = maxT.
By the Law of Trichotomy, we may let
u = max{t, Xk+d·
Then u E S, and Vx E S, x :::; u. Therefore, u = max S. Thus, any set with
k + 1 elements has a maximum element.
Therefore, by mathematical induction, any finite set has a maximum ele-
ment.
Part 2: Prove that S has a minimum element (Exercise 5). Therefore, any
finite set has a minimum element. •An infinite set, even when bounded, may or may not have a maximum
or a minimum element. For example, an "open interval" (a, b) has neither a
minimum element nor a maximum element. (Explain.) Yet although a and b- For a rigorous definition of "finite" set, see Definition 2.8.2.