28 1. The Real Numbers
Solution: Clearly, inf A = −1 and sup A = 1. These are finite, sup A belongs
to A, but inf A does not.
Also, inf B = −∞ and sup B = 5. In this case, the inf is not finite. The sup is
finite but does not belong to B.
Since
n2
n + 1

n
2
, the set C is unbounded, and so sup C = ∞. Also, we have
n + 1
n2
+
n2
=
2n2,
and so
1
2

n2
n + 1
for all n N. Thus, 1/2 is a lower bound for C. It is the greatest lower bound,
since it actually belongs to C, due to the fact that
n2
n + 1
=
1
2
when n = 1. Thus,
inf C = 1/2.
Certainly 0 is a lower bound for the set D. It follows from the Archimedean
property (see Example 1.4.9) that there is no x F with x 0 which is a lower
bound for this set, and so 0 is the greatest lower bound. Thus, inf D = 0. Clearly,
sup D = 1.
If A is a set of numbers and sup A actually belongs to A, then it is called the
maximum of A and is denoted max A. Similarly, if inf A belongs to A, then it is
called the minimum of A and is denoted min A.
The following theorem is really just a restatement of the definition of sup, but
it may give some helpful insight. It says that sup A is the dividing point between
the numbers which are upper bounds for A (if there are any) and the numbers
which are not upper bounds for A. A similar theorem holds for inf. Its formulation
and proof are left to the exercises.
Theorem 1.5.4. Let A be a non-empty subset of R and let x be an element of R.
Then
(a) sup A x if and only if a x for every a A;
(b) x sup A if and only if x a for some a A.
Proof. (a) By definition a x for every a A if and only if x is an upper bound
for A. If x is an upper bound for A, then A is bounded above. This implies its sup
is its least upper bound, which is necessarily less than or equal to x.
Conversely, if sup A x, then sup A is finite and is the least upper bound for
A. Since sup A x, x is also an upper bound for A. Thus, sup A x if and only
if a x for every a A.
(b) If x sup A, then x is not an upper bound for A, which means that x a
for some a A. Conversely, if x a for some a A, then x sup A, since
a sup A. Thus, x sup A if and only if x a for some a A.
Example 1.5.5. If A =
4n 1
6n + 3
: n N , find the set of all upper bounds for A.
Solution: Long division yields
4n 1
6n + 3
=
2
3

1
2n + 1

2
3
.
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