14 1. From the Real Numbers to the Complex Numbers

all three values are the same. By contrast, let xn = (-1)n. Then lim inf(xn) = -1

and lim sup(xn) = 1. Occasionally in the subsequent discussion we can replace

limit by lim sup and things still work.

Next we turn to the concept of continuity, which we also deﬁne in terms of

sequences.

Deﬁnition 3.8. Let f : R → R be a function. Then f is continuous at a if

whenever {xn} is a sequence and limn→∞xn = a, then limn→∞f(xn) = f(a). Also,

f is continuous on a set S if it is continuous at each point of the set. When S is

R or when S is understood from the context to be the domain of f, we usually say

“f is continuous” rather than the longer phrase “f is continuous on S”.

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Exercise 1.19. Prove that the sum and product of continuous functions are

continuous. If c is a constant and f is continuous, prove that cf is continuous.

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Exercise 1.20. Prove that f is continuous at a if and only if the following holds.

For each 0, there is a δ 0 such that |x - a| δ implies |f(x) - f(a)| .

We close this section by showing how the completeness axiom impacts the

existence of square roots. First we recall the standard fact that there is no rational

square root of 2, by giving a somewhat unusual proof. See Exercise 1.22 for a

compelling generalization. These proofs are based on inequalities. For example,

the order axioms yield the following: 0 a b implies 0

a2

ab

b2;

we use

such inequalities without comment below.

Proposition 3.4. There is no rational number whose square is 2.

Proof. Seeking a contradiction, we suppose that there are integers m,n such that

(m

n

)2

= 2. We may assume that m and n are positive. Of all such representations

we may assume that we have chosen the one for which n is the smallest possible

positive integer. The equality

m2

=

2n2

implies the inequality 2n m n. Now

we compute

(10)

m

n

=

m(m - n)

n(m - n)

=

m2

- mn

n(m - n)

=

2n2

- mn

n(m - n)

=

2n - m

m - n

.

Thus

2n-m

m-n

is also a square root of 2. Since 0 m - n n, formula (10) provides

a second way to write the fraction

m;

n

the second way has a positive denominator,

smaller than n. We have therefore contradicted our choice of n. Hence there is no

rational number whose square is 2.

Although there is no rational square root of 2, we certainly believe that a

positive real square root of 2 exists. For example, the length of the diagonal of the

unit square should be

√

2. We next prove, necessarily relying on the completeness

axiom, that each positive real number has a square root.

Theorem 3.1. If t ∈ R and t ≥ 0, then there is an x ∈ R with

x2

= t.

Proof. This proof is somewhat sophisticated and can be omitted on ﬁrst reading.

If t = 0, then t has the square root 0. Hence we may assume that t 0. Let S

denote the set of real numbers x such that x2 t. This set is nonempty, because

0 ∈ S. We claim that M = max(1,t) is an upper bound for S. To check the claim,