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Definition 15.0.6 Define [x]> [0] means that there exists δ > 0,δ ∈Q and a subse-quence of x

{xnk

}∞

k=1 with the property that xnk > δ for all k. Since x is a Cauchy sequence,this requires that for all n large enough, xn > δ/2. Thus we could also say [x]> [0] meansxk > η for some positive η whenever k is large enough. It has already been shown that[x] = [0] means limk→∞ xk = 0.

Theorem 15.0.7 The above definition makes R into an ordered field.

Proof: The first two order axioms are obvious. As to the third, if [x] ̸= [0] , it followsthat for all k large enough, xk > δ for some δ > 0 or −xk < δ for all k large enough. Inthe first case, [x] > [0] and in the second case, − [x] > [0] . The only other case is where[x] = [0] . Thus R is an ordered field.

Now that we know this is an ordered field, the usual notions apply to <.

Observation 15.0.8 To say [x]< ε ∈Q is to say that [ε]− [x] = [ε−x]> [0] where ε

is the constant sequence every term equal to ε ∈ Q. Thus ε > [x] says that for all k largeenough, xk < ε .

Lemma 15.0.9 Q is dense in R in the sense that given ε > 0 where ε ∈ Q, and x ∈ R,there is r ∈ Q such that [|x− r|] < ε where r is the constant sequence always equal tor ∈Q.

Proof: Note that |x− r|∈ R from the triangle inequality. It remains to verify Q isdense in R. I need to verify that if x ∈ R, then there is r ∈Q such that for large enough k,r− xk < ε ∈ Q and xk− r < ε . It suffices to get an r such that |xk− r| < ε . Since x ∈ R,|xk− xm|< ε/4 for all k,m≥ n for some n. Let r≡ xn so k≥ n implies |xk− r|< ε/4. Thenε−|xk− r|> 3ε/4 so [|x− r|]< ε .

Definition 15.0.10 Define |[x]| ≡ [|x|] . This makes sense because of the triangleinequality ||xk|− |yk|| ≤ |xk− yk| . Thus |x| ≡ {|xk|}k ∈ R and if [x] = [y] then [|x|] = [|y|] .

Theorem 15.0.11 |·| as just defined satisfies the usual properties of the absolutevalue. Also, if [q]> 0 and if [x] ∈ R then there is r ∈Q such that |[x]− [r]|< [q]. Thus Qis dense in R. Also R is complete.

Proof: If [x] ̸= [0] , then eventually, for large enough k either xk ≥ δ > 0 or xk < −δ

for some rational δ > 0. Therefore, |[x]|= [|x|]> δ .

|[x] [y]| ≡ |[xy]| ≡ [|xy|] and |[x]| |[y]| ≡ [|x|] [|y|]≡ [|xy|]

Thus |[x] [y]|= |[x]| |[y]|. Also [x]2 = |[x]|2 and |[x] [y]| ≥ [x] [y].

|[x]+ [y]|2 = ([x]+ [y])2 = [x]2 +[y]2 +2 [x] [y]≤ |[x]|2 + |[y]|2 +2 |[x]| |[y]|= (|[x]|+ |[y]|)2

and so |[x]+ [y]| ≤ |[x]|+ |[y]| . The usual properties of absolute value are obtained.Now since [q] > 0, we have qk > δ > 0 for some δ ∈ Q whenever k is large enough.

Let n be still larger such that for k,m ≥ n, |qk−qm| < δ/3. Thus k ≥ n implies qk− qn >− δ

3 ,qk > qn− δ

3 > 2δ/3. From Theorem 15.0.7, if [x] ∈ R then there is r ∈ Q such that[|x− r|]< δ/2 < 2δ/3 < [q] . It follows Q is dense in R.