Kenneth Kuttler

4.10. CAUCHY SEQUENCES AND COMPLETENESS 75

sequence is similar. Alternatively, you could consider the sequence {−an} and apply whatwas just shown to this decreasing sequence.

By Theorem 4.10.2 the following definition of completeness is equivalent to the originaldefinition when both apply. However, note that convergence of Cauchy sequences does notdepend on an order to it applies to much more general situations. Recall from Theorem4.8.14 that C and R are complete. Just apply that theorem to the case where p = 1.

4.10.1 DecimalsYou are all familiar with decimals. In the United States these are written in the form.a1a2a3 · · · where the ai are integers between 0 and 9.5 Thus .23417432 is a number writ-ten as a decimal. You also recall the meaning of such notation in the case of a terminatingdecimal. For example, .234 is defined as 2

10 +3

102 +4

103 . Now what is meant by a nonter-minating decimal?

Definition 4.10.4 Let .a1a2 · · · be a decimal. Define

.a1a2 · · · ≡ limn→∞

∑k=1

10k .

Proposition 4.10.5 The above definition makes sense. Also every number in [0,1] canbe written as such a decimal.

Proof: Note the sequence{

∑nk=1

ak10k

}∞

n=1is an increasing sequence. Therefore, if there

exists an upper bound, it follows from Theorem 4.10.3 that this sequence converges and sothe definition is well defined.

∑k=1

10k ≤n

∑k=1

910k = 9

∑k=1

110k .

Now 910

(∑

nk=1

110k

)= ∑

nk=1

110k − 1

10 ∑nk=1

110k = ∑

nk=1

110k −∑

n+1k=2

110k = 1

10 −1

10n+1 and so

∑nk=1

110k ≤ 10

(110 −

110n+1

)≤ 10

( 110

)= 1

9 . Therefore, since this holds for all n, it followsthe above sequence is bounded above. It follows the limit exists.

Now suppose x ∈ [0,1). Let a110 ≤ x < a1+1

10 where a1 is an integer between 0 and 9.If integers a1, · · · ,an each between 0 and 9 have been obtained such that ∑

nk=1

ak10k ≤ x <

∑n−1k=1

ak10k +

an+110n (∑0

k=1 ≡ 0). Then from the above, 10n(

x−∑nk=1

ak10k

)< 1 and so there

exists an+1 such that

an+1

10≤ 10n

(x−

∑k=1

10k

an+1 +110

which shows that an+110n+1 ≤

(x−∑

nk=1

ak10k

an+1+110n+1 . Therefore,

x = limn→∞

∑k=1

10k

5In France and Russia they use a comma instead of a period. This looks very strange but that is just the waythey do it.

4.10. CAUCHY SEQUENCES AND COMPLETENESS 75sequence is similar. Alternatively, you could consider the sequence {—a,} and apply whatwas just shown to this decreasing sequence.By Theorem 4.10.2 the following definition of completeness is equivalent to the originaldefinition when both apply. However, note that convergence of Cauchy sequences does notdepend on an order to it applies to much more general situations. Recall from Theorem4.8.14 that C and R are complete. Just apply that theorem to the case where p = 1.4.10.1 DecimalsYou are all familiar with decimals. In the United States these are written in the form.a\aa3--- where the a; are integers between 0 and 9.° Thus .23417432 is a number writ-ten as a decimal. You also recall the meaning of! such notation in the case of a terminatingdecimal. For example, .234 is defined as 4 + iw + Now what is meant by a nonter-minating decimal?ioDefinition 4.10.4 Ler .a\a2+++ be a decimal. Definea2 = = lin ip TorProposition 4.10.5 The above definition makes sense. Also every number in {0,1] canbe written as such a decimal.coProof: Note the sequence 4 Yy_ Tg is an increasing sequence. Therefore, if thereexists an upper bound, it follows from Theorem 4.10.3 that this sequence converges and sothe definition is well defined.ndn ak n 9Dio SY ioe 9% 0k9 n 1 _yn 1 lyn 1 n+l _ 1 1Now TO (ra ix) —_ yk=1 Tok To Lk=1 Tok = Ye 1 1d >a 2 rd = 10 ~ jot and soVe 71d < y (4- OF ot) < (+ | ) = 5: Therefore, since this holds for all n, it follows10 = 9 \10the above sequence is bounded above. It follows the limit exists.Now suppose x € [0,1). Let <x< a where a, is an integer between 0 and 9.If integers a),--- ,a, each between 0 and 9 have been obtained such that )7_, Tok <x<ye tor + + att (y°_, =0). Then from the above, 10" (x —Ve 4, < 1 and so thereexists d,+1 such thatan+1 n Ak anit< — _io =” (- py “) < "10which shows that +74, < (« — Year tte) < “u-1** . Therefore,n: ak= im, Tok5In France and Russia they use a comma instead of a period. This looks very strange but that is just the waythey do it.