1.5. Sup and Inf 29
Thus, 2/3 is an upper bound for A. If x 2/3, then = 2/3 x is positive, and
the Archimedean property implies we can choose n large enough that
1
2n + 1
1
n
.
Then
x
2
3

1
2n + 1
=
4n 1
6n + 3
for such an n, which means that x is not an upper bound for A.
We conclude that 2/3 is the least upper bound for A that is, sup A = 2/3.
By the previous theorem, the set of all upper bounds for A is the interval [2/3, ∞).
Example 1.5.6. If A =
n2
n + 1
: n N , find sup A and the set of all upper
bounds for A.
Solution: Long division yields
n2
n + 1
= n 1 +
1
n + 1
n 1.
Then the Archimedean property implies that there are no upper bounds for A,
since, for every x R, there is an n N for which n 1 is larger than x. Thus, the
set of upper bounds for A is the empty set and sup A = ∞.
Properties of Sup and Inf. The next theorem uses the following notation con-
cerning subsets A and B of R:
−A = {−a : a A};
A + B = {a + b : a A, b B};
A B = {a b : a A, b B}.
Theorem 1.5.7. Let A and B be non-empty subsets of R. Then
(a) inf A sup A;
(b) sup(−A) = inf A and inf(−A) = sup A;
(c) sup(A + B) = sup A + sup B and inf(A + B) = inf A + inf B;
(d) sup(A B) = sup A inf B;
(e) if A B, then sup A sup B and inf B inf A.
Proof. We will prove (a), (b), and (c) and leave (d) and (e) to the exercises.
(a) If A is non-empty, then there is an element a A. Since inf A is a lower
bound and sup A an upper bound for A, we have inf A a sup A.
(b) A number x is a lower bound for the set A (x a for all a A) if and
only if −x is an upper bound for the set −A (−a −x for all a A). Thus, if
L is the set of all lower bounds for A, then −L is the set of all upper bounds for
−A. Furthermore, the largest member of L and the smallest member of −L are
negatives of each other. That is, inf A = sup(−A). This is the first equality in
(b). If we apply this result with −A replacing A, we have inf(−A) = sup A. If
we multiply this by −1, we get the second equality in (b).
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