Kenneth Kuttler

356 CHAPTER 15. CONSTRUCTION OF REAL NUMBERS

Theorem 15.0.5 With the two operations defined above,R is a field. The operationsare well defined.

Proof: Why are these operations well defined? Consider multiplication because it isfairly obvious that addition is well defined. If x∼ x′ and y∼ y′, is it true that [x′y′] = [xy]?Is {x′ny′n− xnyn}∞

n=1 ∈ N?∣∣x′ny′n− xnyn∣∣≤ ∣∣x′ny′n− x′nyn

∣∣+ ∣∣x′nyn− xnyn∣∣≤C

(∣∣y′n− yn∣∣+ ∣∣x′n− xn

∣∣)where C is a constant which bounds all terms of all four given Cauchy sequences, the con-stant existing because these are all Cauchy sequences. See Theorem 4.5.2. By assumption,the last expression converges to 0 as n→ ∞ and so {x′ny′n− xnyn}∞

n=1 ∈ N which verifiesthat [x′y′] = [xy] as hoped. The case for addition is similar but easier.

Now it is necessary to verify that the two operations are binary operations on R. Thisis obvious for addition. The question for multiplication reduces to whether xy is a Cauchysequence.

|xnyn− xmym| ≤ |xnyn− xmyn|+ |xmyn− xmym| ≤C (|xn− xm|+ |yn− ym|)

for some constant which is independent of n,m. This follows because x,y Cauchy impliesthat these sequences are both bounded. See Theorem 4.5.2.

Commutative and associative laws for addition and multiplication are all obvious be-cause these hold pointwise for individual terms of the sequence. So is the distributive law.The existence of an additive identity is clear. You just use [0] . Similarly [1] is a multiplica-tive identity. For [x] ̸= [0] , let yn = 1 if xn = 0 and yn = x−1

n if xn ̸= 0. Is y ∈ R? Since[x] ̸= [0] , Lemma 15.0.4 implies that there exists δ > 0 and N such that |xk| > δ for allk ≥ N. Now for m,n > N,

|yn− ym|=∣∣∣∣ 1xn− 1

∣∣∣∣= |xn− xm||xn| |xm|

≤ 1

δ2 |xn− xm|

which shows that {yn}∞

n=1 ∈ R. Then clearly [y] = [x]−1 because [y] [x] = [yx] and yx is aCauchy sequence which equals 1 for all n large enough. Therefore, [xy] = [1] as required,because xy−1 ∈ N. It is obvious that an additive inverse [−x] ≡ − [x] exists for each[x] ∈ R. Thus R is a field as claimed.

It might be of interest to note that with the operations described, R is a commutativering and N is a maximal ideal. Thus from algebra R/N is a field. Showing N is maximal isessentially done above where if [x] ̸= [0] , then the multiplicative inverse exists which gets1 in any ideal containing N making N maximal. You do the same thing with algebraic fieldextensions but the argument is harder there.

Of course there are two other properties which need to be considered. Is R ordered? IsR complete? First recall what it means for R to be ordered. There is a subset of R calledthe positive numbers, such that

The sum of positive numbers is positive.

The product of positive numbers is positive.

[x] is either positive [0] , or − [x] is positive.

356 CHAPTER 15. CONSTRUCTION OF REAL NUMBERSTheorem 15.0.5 With the wo operations defined above, R is a field. The operationsare well defined.Proof: Why are these operations well defined? Consider multiplication because it isfairly obvious that addition is well defined. If x ~ x’ and y ~ y’, is it true that [x’y’] = [xy]?Is {x.y —XnYn},—1 € N?In —XnYn| < nV —x,¥n| + [x0 —XnYn| <C ([y, —Yn| + Ix, —Xn|)where C is a constant which bounds all terms of all four given Cauchy sequences, the con-stant existing because these are all Cauchy sequences. See Theorem 4.5.2. By assumption,the last expression converges to 0 as n — oo and so {x),y/, —xnyn};_, € N which verifiesthat [x’y’] = [xy] as hoped. The case for addition is similar but easier.Now it is necessary to verify that the two operations are binary operations on R. Thisis obvious for addition. The question for multiplication reduces to whether xy is a Cauchysequence.IXnYn —XmYm\| S |XnYn —XmYn| + |XmYn — XmYm| < C (|Xn — Xml + Lyn — Yl)for some constant which is independent of n,m. This follows because x, y Cauchy impliesthat these sequences are both bounded. See Theorem 4.5.2.Commutative and associative laws for addition and multiplication are all obvious be-cause these hold pointwise for individual terms of the sequence. So is the distributive law.The existence of an additive identity is clear. You just use [0]. Similarly [1] is a multiplica-tive identity. For [x] 4 [0], let y, = 1 if x, =0 and y, =x,! if x, 40. Is y € R? Since[x] 4 [0], Lemma 15.0.4 implies that there exists 6 > 0 and N such that |x;| > 6 for allk > N. Now for m,n >N,1 1Xn XmXn —Xm| 1= ——— < —_ —xn | Xml - & bin nllyn — Ym =which shows that {y, };_, € R. Then clearly [y] = [x] | because [y] [x] = [yx] and yx is aCauchy sequence which equals | for all n large enough. Therefore, [xy] = [1] as required,because xy—1 EN. It is obvious that an additive inverse [—x] = — [x] exists for each[x] € R. Thus R isa field as claimed.It might be of interest to note that with the operations described, R is a commutativering and N is a maximal ideal. Thus from algebra R/N is a field. Showing N is maximal isessentially done above where if [x] 4 [0], then the multiplicative inverse exists which gets1 in any ideal containing N making N maximal. You do the same thing with algebraic fieldextensions but the argument is harder there.Of course there are two other properties which need to be considered. Is R ordered? IsR complete? First recall what it means for R to be ordered. There is a subset of R calledthe positive numbers, such thatThe sum of positive numbers is positive.The product of positive numbers is positive.[x] is either positive [0], or — [x] is positive.