Does there exist in an extension R of Q such that the limit, i.e

Motivation Examples

(a) Since there does not exist a rational number p = m

n ∈ Q, gcd(m, n) = 1, satisfying the equation

p² = 2,

we try to find a sequence {p_k} approximation solutions of rational numbers such that lim

k→∞p²_k = 2.

Naturally, one would ask Q1. Does the limit lim

k→∞p_k, i.e. the solution, exist in Q?

Q2. Does there exist in an extension R of Q such that the limit, i.e. the solution, lim

k→∞p_k∈ R?

(b) To find a root of the equation

p²− 4p − 8 = 0, we consider a sequence {p_k} defined by







p₁ = −2

pk+1 = 1 + 11

p_k− 3 for k = 1, 2, . . . . Note that {p_k} is a sequence of rational numbers, but the limit lim

k→∞p_k, i.e. a root of the equation p²− 4p − 8 = 0, is not a rational number.

(c) To solve the initial value problem (IVP)





 dy

dx = f (x, y) for x ∈ I = (a, b) ⊆ R y(x₀) = y₀ for some x₀ ∈ I we consider a sequence {y_k} defined by

(1) Let y₀(x) = y₀ for every x ∈ I (2) and dyk+1

dx = f (x, y_k) for k = 0, 1, 2, . . . . Naturally, one would ask

Q1. Does the limit lim

k→∞yk exist?

Q2. Suppose the limit y = lim

k→∞yk exist, does it solve the IVP? i.e.

Does lim

k→∞

d

dxy_k+1 = d dx lim

k→∞y_k+1? Does lim

k→∞f (x, y_k) = f (x, lim

k→∞y_k)?

(d) To solve the IVP,

(y⁰ = 2x(1 + y) for x ∈ (−∞, ∞) y(0) = 0

we consider a sequence {y_k} defined by

(1) Let y₀(x) = 0 for every x ∈ (−∞, ∞) (2) and dy_k+1

dx = 2x(1 + y_k) for k = 0, 1, 2, . . . . Then we have

y_k+1(x) = Z x

0

2s(1 + y_k(s)) ds and

y₁(x) = Z x

0

2s ds = x² y₂(x) =

Z x 0

2s(1 + s²) ds = x²+ x⁴ 2

· · · · y_k(x) = x²+ x⁴

2 + x⁶

3! + · · · + x^2k

· · · k!

Since

∞

X

n=1

x²ⁿ

n! converges everywhere in R, it is uniformly convergent on any bounded closed subinterval of R and

y(x) = lim

k→∞y_k(x) =

∞

X

n=1

x²ⁿ

n! = e^x2 − 1 is the solution for the IVP.

Definitions

(a) Suppose S is an ordered set, and E ⊂ S. If E there is a real number β such that x ≤ β for all x in E,

we say that E is bounded from above, and call β an upper bound of E.

(b) Suppose S is an ordered set, E ⊂ S, and E is bounded from above. A number α is called the least upper bound (or supremum) of E, denoted by α = lub E or sup E, if it satisfies the following conditions:

(i) x ≤ α for all x in E, i.e. α is an upper bound of E,

(ii) if γ < α then there exists an x ∈ E such that γ < x, i.e. γ is not an upper bound of E.

(c) An ordered set S is said to have the least upper bound (lub) property if the following is true:

If E is a nonempty, bounded from above subset of S, then sup E exists in S, i.e. sup E ∈ S.

Remarks Suppose S is an ordered set and E is a nonempty subset of S.

(a) If α = sup E exists, then it is unique.

(b) If α = sup E exists, then either α ∈ E or α /∈ E.

(c) If E ⊂ F ⊂ S and F is bounded from above, then so is E and

if both sup E and sup F exist, then sup E ≤ sup F.

(d) Examples (i) Let

S = Q

A = {p ∈ Q | p > 0 and p² < 2} 6= ∅ bounded from above B = {p ∈ Q | p > 0 and p² > 2} 6= ∅ bounded from belowe

Since p² = 2 has no rational root =⇒ sup A /∈ Q and Q does not have the lub property.

(ii) Let

S = Q

E = {1

n ∈ Q | n = 1, 2, 3, . . .}

Then

sup E = 1 ∈ E and inf E = 0 /∈ E.

(iii) Let

S = Q

E₁ = {r ∈ Q | r < 0}

E₂ = {r ∈ Q | r ≤ 0}

Then

E1 ⊆ E2, sup E1 = sup E2 = 0 ∈ E2 and 0 /∈ E1.

Definitions

(a) Suppose S is an ordered set, and E ⊂ S. If E there is a real number β such that β ≤ x for all x in E,

we say that E is bounded from below, and call β a lower bound of E.

(b) Suppose S is an ordered set, E ⊂ S, and E is bounded from below. A number α is called the greatest lower bound (or infimum) of E, denoted by α = glb E or inf E, if it satisfies the following conditions:

(i) α ≤ x for all x in E, i.e. α is a lower bound of E,

(ii) if α < γ then there exists an x ∈ E such that x < γ, i.e. γ is not a lower bound of E.

(c) An ordered set S is said to havethe greatest lower bound (glb) property if the following is true:

If E is a nonempty, bounded from below subset of S, then inf E exists in S, i.e. inf E ∈ S.

Theorem 1.11 If S is an ordered set with the lub property, then S is an ordered set with the glb property, i.e. if B is a nonempty, bounded from below subset of S, then inf B exists in S.

Since

(i) B is a nonempty, bounded from below subset of S =⇒ ∃ z ∈ S such that z ≤ x ∀ x ∈ B.

This implies that

L = {y ∈ S | y ≤ x, ∀ x ∈ B} = the set of lower bounds of B is a nonempty, bounded from above subset of S, ans since S has the lub property

α = sup L exists in S.

(ii) Note that

if γ < α = sup L =⇒ γ is not an upper bound of L

=⇒ ∃ y ∈ L s.t. γ < y ≤ x ∀x ∈ B

=⇒ γ < x ∀ x ∈ B

=⇒ γ /∈ B which implies that

if x ∈ B then α ≤ x, i.e. α is a lower bound of B.

Also note that

if α < γ =⇒ y ≤ sup L = α < γ ∀ y ∈ L

=⇒ γ /∈ L

=⇒ γ is not a lower bound of B This proves that α = inf B.

Definition An ordered field F is a field which is also an ordered set such that (i) if x, y, z ∈ F and y < z then x + y < x + z (ii) if x, y ∈ F, x > 0, y > 0 then xy > 0

Construction of R from cuts

(a) Definition A subset α of Q is called a cut if the following conditions hold:

(i) α 6= ∅ and α 6= Q. [=⇒ α is a nonempty, bounded above subset of Q.]

(ii) if p ∈ α, q ∈ Q, then q ∈ α. [ ⇐⇒ if p ∈ α and q /∈ α, then p < q. ⇐⇒ if r /∈ α and r < s then s /∈ α.]

(iii) if p ∈ α, then p < r for some r ∈ α. [=⇒ α has no largest element.]

(b) Notation The roman letters p, q, r . . . will denote rational numbers, and the greek letters α, β, γ, . . . will denote cuts.

(c) Definition Define α < β to mean: α ⊂ β, i.e. α is a proper subset of β.

(d) Definition Deinfe R = (R, <) = {α | α is a cut}.

Since

(i) for any α, β ∈ R, one and only one of the following relations holds:

α < β, α = β, β < α, (ii) if α < β and β < γ, then α < γ,

R is an ordered set.

(e) Proposition The ordered set R has the least-upper-bound property.

(f) Proposition R is an ordered field.

Remarks

(i) Define 0^∗ = {p ∈ Q | p < 0}. It is clear that 0^∗ ∈ R.

(ii) For any α, β ∈ R, define α + β by

α + β = {r + s | r ∈ α and s ∈ β}.

It is clear that the axioms for + (in Definition 1.12) hold in R, with 0^∗ playing the role of 0.

Proof of (A5) Given α ∈ R, need to show that there exists β ∈ R such that {p ∈ Q | p < 0} = 0^∗ = α + β = {r + s | r ∈ α and s ∈ β}

which implies that

if s ∈ β then r < −s for each r ∈ α =⇒ −s /∈ α.

Also since a cut does have the largest element, we define β = {p ∈ Q | ∃ r > 0 s.t. − p − r /∈ α}.

To show that 0^∗ = α + β, we prove

∀ p ∈ β, ∃ s > 0 s.t. − p − s /∈ α

=⇒ −p /∈ α since − p > −p − s

=⇒ r < −p or r + p < 0 ∀ r ∈ α

=⇒ α + β ⊂ 0^∗.

On the other hand, if v ∈ 0^∗, then v < 0 and v ∈ Q. By setting w = −v

2 =⇒ w ∈ Q, w > 0 and either w ∈ α or w /∈ α So,

∃ n ∈ Z s.t. nw ∈ α and (n + 1)w /∈ α.

α

w nw (n + 1)w

α w

nw (n + 1)w

Put p = −(n + 2)w. Since (n + 1)w = (n + 2)w − w = −p − w ∈ α, we prove that p ∈ β and v = −2w = nw + p ∈ α + β =⇒0^∗ ⊂ α + β.

(iii) For any α ∈ R, define −α to be the cut such that α + (−α) = 0^∗. It follows that α > 0^∗ if and only if −α < 0^∗.

(iv) Let R⁺ = {α ∈ R | α > 0^∗}. If α, β ∈ R⁺, we define αβ by

αβ = {p | p ≤ rs for some choice of r ∈ α, s ∈ β, r > 0, s > 0},

and define 1^∗ = {p ∈ Q | p < 1}. Then the axioms (M ) and (D) of Definition 1.12 hold, with R⁺ in place of F , and with 1^∗ in the role of 1.

(v) On R, define the multiplication by setting

α0^∗ = 0^∗α = 0^∗, and by setting

αβ =







(−α)(−β) if α < 0^∗, β < 0^∗,

−[(−α)β] if α < 0^∗, β > 0^∗,

−[α · (−β)] if α > 0^∗, β < 0^∗, (g) Proposition Q is isomorphic to a subfield of R.

Observe that

(i) for each r ∈ Q, letting r^∗ = {p ∈ Q | p < r}, then r^∗ ∈ R, i.e. r^∗ is a cut.

(ii) if Q^∗ = {r^∗ | r ∈ Q}, then Q^∗ is a subfield of R and Q^∗ is isomorphic to Q.

Theorem 1.20

(a) [Archimedean Property of R] If x, y ∈ R and x > 0, then there is a positive integer n such that

nx > y.

(b) [Q is dense in R] If x, y ∈ R and x < y, then there exists a p = m

n ∈ Q, n ∈ N, such that x < p < y ⇐⇒ x < m

n < y ⇐⇒ nx < m < ny.

Remarks

(a) This implies that the set of real numbers R can be defined as R = {r | ∃ {pⁿ} ⊂ Q s.t. lim_n→∞pn= r}.

(b) For each x ∈ R and x > 0. Let

n₀ = the largest integer s.t. n₀ ≤ x

n₁ = the largest integer s.t. n₁ ≤ 10(x − n₀)

· · · ·

n_k = the largest integer s.t. n_k ≤ 10^k(x − n₀− n₁

10− · · · − n_k−1 10^k−1)

· · · ·

and let

E = {n₀+ n₁

10+ · · · + n_k

10^k | k = 0, 1, 2, . . .}.

Then E ⊂ Q, x = sup E and the decimal expansion of x is n0.n1n2n3· · · .

Conversely, for any infinite decimal n₀.n₁n₂n₃· · · , 0 ≤ n_j ≤ 9 for all j ≥ 1, the set

E = {

k

X

j=0

n_j

10^j | k = 0, 1, 2, . . .} is bounded above and n₀.n₁n₂n₃· · · ∈ R is the decimal expansion of sup E.

Theorem 1.21 For every real x > 0 and every integer n > 0, there is one and only one real y such that

yⁿ= x, where y is called the n^th root of x and is written √ⁿ

x or x^1/n. Outline of the proof

Let

E = {t ∈ R | t > 0 and tⁿ < x}.

Show that

y = sup E exists, yⁿ= x and y is unique.

Observe that 1 + x /∈ E =⇒ E is bounded above.

0 x 1 + x

Suppose that yⁿ 6= x, find 0 < h < 1, 0 < k < 1 such that

0 yⁿ (y + h)ⁿ x

0 x (y − k)ⁿ yⁿ

Construction of R from Cauchy sequences

(a) Denition The (archimedean) absolute value of Q is the function Q → Q+∪ {0}

|x| =

(x if x ≥ 0

−x if x < 0.

(b) Denition A Cauchy sequence(of rational numbers) is a sequence {x_n} such that for every

 ∈ Q+ there exists N_ ∈ N such that |xm− x_n| <  for all m, n ≥ N_. (i) Lemma Every Cauchy sequence {x_n} is bounded.

(ii) Lemma If {x_n} and {y_n} are Cauchy sequences, so are {x_n+ y_n} and {x_ny_n}.

(c) Definition A Cauchy sequence {x_n} is equivalent to zero if lim

n→∞|x_n| = 0. Two Cauchy seuqences {x_n} and {y_n} are equivalent if their difference {x_n− y_n} is equivalent to zero.

We can add/multiply equivalence classes of Cauchy sequences by adding/multiplying rep- resentatives (because {x_n} ∼ 0 implies {x_n} + {y_n} ∼ {y_n} and {x_n}{y_n} ∼ 0).

Notation The equivalence class of {x_n} is denoted by [{x_n}].

(i) Lemma Every nonzero equivalence class of Cauchy sequences has a multiplicative in- verse.

(ii) Corollary The set of equivalence classes of Cauchy sequences forms a field.

(d) Definition Let R = (R, | |) = {[{xn}] | {x_n} is a Cauchy sequence in Q}, the field of equivalence classes of Cauchy sequences. We embed Q in R via the map

x 7→ [x] = [{x, x, x, . . .}] = [{x_n | x_n= x ∀ n = 1, 2, . . .}], which is injective because lim

n→∞|x_n− y_n| = |x − y| 6= 0 unless x = y (here we use |x| = 0 ⇐⇒

x = 0). We extend the absolute value of Q to R via

|[{xn}]| := [{|xn|}] and in particular |[x]| := [{x, x, x, . . .}] = |x| ∀ x ∈ Q.

We then have R+ = {x 6= 0 ∈ R | |x| = x} containing Q+ and R = R+∪ {0} ∪ R+. Thus R is an ordered field whose order extends that of Q.

(i) Lemma A Cauchy sequence {x_n} of rational numbers converges to [{x_n}].

(ii) Corollary Every Cauchy sequence of real numbers converges to a real number. Equiv- alently, R is complete.

(e) Theorem (R, <) has the lub property iff every Cauchy sequence {[xn]} ⊂ (R, | |) converges to a point in R.

Proof.

=⇒ Given  > 0, since {[x_n]} = {[{x^k_n | k ∈ N}]} is a Cauchy squence in R, there exists K ∈ N such that

if n, n⁰ ≥ K then |[x_n] − [x_n⁰]| < 

2 =⇒ ∃ L ∈ N such that if k ≥ L then |x^kn− x^k_n0| <  2. For each n ∈ N, since {x^kn| k ∈ N} is a (rational) Cauchy sequence, there exists lnsuch that

if k, k⁰ ≥ l_n then |x^k_n− x^k_n⁰| < 1 n.

Without loss of generality, we can assume that the sequence (l_n) is strictly increasing, as each l_n can be chosen arbitrarily large. Now construct a new sequence in Q by

{b_n} := {x^l_nⁿ}.

Then b := [{b_n}] is a real number and the limit of {[x_n]}.

Choose a natural number K ≥ max{K, L,2

}. Then for all k > k⁰ ≥ K,

|b_k− b_k⁰| = |x^l_k^k− x^l_k^k00|

≤ |x^l_k^k− x^l_k^k0| + |x^l_k^k0 − x^l_k^k00|

<  2+ 1

k⁰ since l_k > l_k⁰ ≥ k⁰ ≥ K

<  2+ 1

k since k > k⁰

≤  2+ 1

K since k ≥ K

<  2+ 

2 since K ≥ 2



This shows that {b_n} is a Cauchy sequence.

Fix n ∈ N, since

∀ k ≥ l_n=⇒ |x^k_n− b_n| = |x^k_n− x^l_nⁿ| < 1 n, we obtain

|[x_n] − b| = [{|x^k_n− x^l_nⁿ|}] < 1 n.

Letting n → ∞, we obtain {[x_n] − b} is a zero sequence in R. Hence

n→∞lim[{x_n}] = b.

⇐= Let S be a nonempty, bounded from above subset of R. Choose a sufficiently large M > 0 such that

M is an upper bound of S and [−M, M ] ∩ S 6= ∅.

Let I₁ = [−M, M ]. Divide I₁ into 2 equal length subintervals and let I₂ be the closed subinterval such that

the right endpoint of I₂ is an upper bound of S and I₂∩ S 6= ∅.

Continuing this process, we obtain a sequence of closed intervals {I_n} such that the right endpoint of I_n is an upper bound of S, I_n∩ S 6= ∅ ∀ n ≥ 1.

and

I₁ ⊃ I₂ ⊃ · · · ⊃ I_n⊃ · · · ; lim

n→∞|I_n| = lim

n→∞

M 2ⁿ⁻² = 0, For each n ∈ N, since In∩ S 6= ∅, let x_n be a point in I_n∩ S.

Since lim

n→∞|I_n| = lim

n→∞

M

2ⁿ⁻² = 0, {x_n} is a Cauchy sequence and hence it converges to a point, say x, in R.

To show that x = lub S, we need to show that

(i) x ≥ s for all s in S, i.e. x is an upper bound of S.

(ii) if u is another upper bound of S then u ≥ x.

Proof of (i): Suppose that x is not an upper bound of S, i.e. suppose that there exists s ∈ S such that s > x. Let  = s − x. Since lim

n→∞x_n = x, x_n ∈ I_n and |I_n| = M

2ⁿ⁻², there exists K ∈ N such that

if n ≥ K then |I_n| = M 2ⁿ⁻² < 

2 and |x_n− x| <  2. This implies that







x_n− 

2 ≤ y ≤ x_n+ 

2 for all y ∈ I_n∩ S ⇐⇒ I_n ⊂ (x_n− 

2, x_n+  2), x − 

2 < x_n< x + 

2 ⇐⇒ x_n∈ (x − 

2, x + 

2) for all n ≥ K.

Combining these, we get

x −  < y < x +  = s for all y ∈ I_n∩ S ⇐⇒ I_n ⊂ (x − , x + ) = (x − , s)

x_n+ ^₂ x_n

x_n− ^₂

x x + ^₂

x − ^₂ x +  = s

x −  y

which implies that the right endpoint of I_n is smaller than s ∈ S. This contradicts to the definition of In. Hence, x is an upper bound of S.

Proof of (ii): Let u be another upper bound of S. Suppose that u < x, then  = x − u 2 > 0.

Since lim

n→∞x_n= x, there exists K ∈ N such that if n ≥ K then |xn− x| < .

Thus 





 = x − u

2 =⇒u = x − 2,

x −  < x_n< x +  ⇐⇒ x_n∈ (x − , x + ) for all n ≥ K.

This implies that

u = x − 2 <x −  <x_n< x +  for all x_n∈ S ∩ I_n, n ≥ K=⇒ u < x_n∈ S

x +  x

x_n

xn− ₂^ xn+ ^₂ x − 

u = x − 2

which contradicts to the assumption that u being an upper bound of S.

Hence u ≥ x.