6 1. The Real Line

Thus setting 723(e) = max(ni(|a|/2),

ni(a2e/2))

we see from (4) that

I 1 1 |

6

\an a\

for all n 713(e). The result follows.

(vii) The proof of the first sentence in the statement is rather similar to

that of (i). By definition, f holds. Suppose, if possible, that a A, that is,

a — A 0. Setting N = n\(a — A) we have

&N = (&N — a) + a a — \ajsj — a\ a — (a — A) = A,

contradicting our hypothesis. The result follows by reductio ad absurdum.

To prove the second sentence in the statement we can either give a similar

argument, or set an — —bn, a = —b and A = —B and use the first sentence.

[Your attention is drawn to part (ii) of Exercise 1.8.] •

Exercise 1.7. Prove that the first few terms of a sequence do not affect

convergence. Formally, show that if there exists an N such that an = bn for

n N, then an — a as n — 00 implies bn — a as n — 00.

Exercise 1.8. In this exercise we work within Q. (The reason for this will

appear in Section 1.5, which deals with the axiom of Archimedes.)

(i) Observe that i / e G Q and e 0; then e = m/N for some strictly pos-

itive integers m and N. Use this fact to show, directly from Definition 1.5,

that (if we work in Q) 1/n — * 0 as n — 00.

(ii) Show, by means of an example, that, if an — a and an b for all

n, it does not follow that a b. (In other words, taking limits may destroy

strict inequality.)

Does it follow that a b? Give reasons.

Exercise 1.9. A more natural way of proving Lemma 1.6 (i) is to split the

argument in two.

(i) Show that if \a — b\ e for all e 0, then a = b.

(ii) Show that if an — • a and an — b as n — oo; then \a — b\ e for all

e 0 .

(Hi) Deduce Lemma 1.6 (i).

(iv) Give a similar 'split proof' for Lemma 1.6 (vii).

Exercise 1.10. Here is another way of proving Lemma 1.6 (v). I do not

claim that it is any simpler, but it introduces a useful idea.

(i) Show from first principles that, if an — a, then can — ca.

(ii) Show from first principles that, if an — • a as n — 00, then a^ —

a2.

(Hi) Use the relation xy = ((x +

y)2

— (x —

y)2)/4

together with (ii), (i)

and Lemma 1.6 (iv) to prove Lemma 1.6 (v).