4 1. The Real Line

my readers to have substantial experience with the use and manipulation of

limits.

Throughout this section F will be an ordered field. The reader will miss

nothing if she simply considers the two cases F = R and F = Q. She will,

however, miss something if she fails to check that everything we say applies

to both cases equally.

Definition 1.5. We work in an ordered field F. We say that a sequence a\,

a2, .. . tends to a limit a as n tends to infinity, or more briefly

an — a as n —• oo,

if, given any e 0, we can find an integer no(e) [read 'no depending on e']

such that

\an — a\ e for all n no(e).

The following properties of the limit are probably familiar to the reader.

Lemm a 1.6. We work in an ordered field F.

(i) The limit is unique. That is, if an — a and an — b as n — * oo, then

a — b.

(ii) If an — » a as n — oo and 1 n(l) n(2) n ( 3 ) . . . , then an^ — a

as j — oo.

(Hi) If an = c for all n, then an — » c as n — oo.

(iv) If an — a and bn — + b as n — oo, then an + bn — a + 6.

( ^ If an ^ a and bn ^ b as n — oo, t/ien a

n

6

n

— afr.

( ^ Suppose that an —» a as n — oo. If an ^ 0 for each n and a ^ 0,

then

a~l

— a

- 1

.

fmi,) If an A for each n and an — » a as n — oo, t/ien a A IfbnB

for each n and bn — b as n — oo, then b B.

Proof. I shall give the proofs in detail, but the reader is warned that similar

proofs will be left to her in the remainder of the book.

(i) By definition:-

Given e 0 we can find an n\{e) such that \an — a\ e for all n n\{e).

Given e 0 we can find an 712(e) such that \an — 6| e for all n 712(e).

Suppose, if possible, that a ^ b. Then setting e = \a — 6|/3 we have e 0.

If T V = max(ni(e), 712(e)) then

\a — b\

\ON

— o\ +

\ON

— b\ e + e = 2\b — a|/3 ,

which is impossible. The result follows by reductio ad absurdum.