182 Chapter 15

since the viewing points themselves form such a sequence: each next viewing point must

be strictly lower than the previous one. So in that case, we are done.

Alternatively, consider the case that there are only finitely many viewing points. In this

case, pick any a

n

0

beyond all of them. Since it is not a viewing point, there is some later

term with a

n

0

≤ a

n

1

. And since a

n

1

is not a viewing point, there is a further term a

n

1

≤ a

n

2

,

and so on. In this way, we find a monotone nondecreasing subsequence {a

n

i

}, as desired.

So in any case, there is a monotone subsequence.

It follows that every bounded sequence of real numbers has a convergent subsequence,

since one may first pass to the monotone subsequence and then observe that every bounded

increasing sequence converges to its supremum and that every bounded decreasing se-

quence converges to its infimum.

Can one improve the theorem to show that every sequence has a strictly increasing or

strictly decreasing subsequence? No, of course not, because it might be eventually con-

stant. But in exercise 15.14, the reader will explain how to strengthen the conclusion of the

theorem to strictly monotone subsequences.

15.8 The principle of continuous induction

Earlier I made an analogy between the principle of mathematical induction in number the-

ory and the least-upper-bound principle in real analysis. This analogy is stronger than may

be supposed, in light of the principle of continuous induction (see below), an induction-

like formulation of the least-upper-bound principle, which can be used to derive many

fundamental results in real analysis.

Theorem 140 (Continuous induction principle). If B is a set of nonnegative real numbers

such that

1. 0 ∈ B;

2. whenever x ∈ B, then there is some δ>0 with [x, x + δ) ⊆ B; and

3. whenever x is a nonnegative real number and [0, x) ⊆ B, then x ∈ B,

then B is the set of all nonnegative real numbers.

One can view statement (1) as an instance of (3), if you interpret [0, 0) as the empty set,

which would satisfy the hypothesis of (3) vacuously. This is exactly analogous to the fact

that the principle of strong induction on the natural numbers does not need to be anchored.

Proof. Suppose toward contradiction that B does not contain all nonnegative real numbers.

So there is some nonnegative real number d that is not in B. Let x be the supremum of the

numbers a for which [0, a] ⊆ B. Since 0 ∈ B, we know that 0 ≤ x. Also, [0, x) ⊆ B, and

so x ∈ B. But then, there is some positive δ>0 with [x, x + δ) ⊆ B. But in this case, the

supremum defining x would have been bigger, a contradiction.