Completeness of the Reals

NeighborhoodsFor any real number a, if δ is a positive real number, the δ-neighborhood of a is the set N(a, δ) = {x ∈ℝ: | x - a | < δ } = (a -δ, a +δ).

That is N(a, δ) is the open interval centered at a with length 2δ.

So, N(5, .03) is the open interval (4.97, 5.03).

Open Sets

For a set A ⊆ ℝ, a point x is an interior point of A iff there exists a δ > 0 such that N(a, δ) ⊆ A. The set A is open iff every point of A is an interior point of A.

Example: Any open interval is an open set. Any union of open intervals is an open set.

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Closed SetsThe set A is closed iff its complement is open.

Example: Any closed interval is a closed set. Any intersection of closed intervals is a closed set.

A set of reals need not be either open nor closed.

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CoversLet B be a set. A collection of sets A is a cover of B iff B is contained in the union of all the sets of A .

A subcover of A for B is a subcollection of the sets of A which also cover B.

Example: Let B = (0,1/2). Let A = {A

n} where A

n = [-1/n, 1/n)

A is a cover of B. {A

1, A

2} is a subcover of A for B.

Compact SetsA set A is compact iff every cover of A by open sets has a finite subcover.

Examples: The empty set is compact. Any finite set of points is a compact set.

The set B = {0} ∪ {1/n : n ∈ ℕ} is a compact set.

Heine-Borel TheoremHeine-Borel Theorem: A subset A of ℝ is compact iff A is closed and bounded.

Pf: (⇒) Suppose that A is a compact set in ℝ. The family of open sets (-n,n) for n ∈ ℕ is an open cover of ℝ and hence of A. By compactness, this open cover has a finite subcover. Since these open sets contain one another, the largest one in the finite subcover must contain A. If this is (-N,N) then -N is a lower bound and N is an upper bound for A, so A is bounded. Now consider a point y ∈ Ac. For each point x ∈ A there is a positive distance to y, let δ

x = half that distance = ½|x-y| and

consider the neighborhoods Nx = N(x, δ

x). They form an open

cover of A, and by compactness, this open cover has a finite subcover. From this finite set, choose the x which is closest to y and notice that N(y, δ

x) contains no element of A. So, Ac is open.

Heine-Borel Theorem(⇐) Suppose that A is a closed and bounded set and C is an open cover of A. For each x ∈ ℝ, let A

x = {a ∈ A: a ≤ x}. Also, let D =

{x ∈ ℝ: Ax is contained in a finite subcover of C}.

Since A is bounded, it is bounded from below by, say L. For x < L, A

x = ∅ and so, contained in any single set of C. That is, x ∈ D

and we have (-∞, L) ⊆ D, so D is not empty. We claim that D has no upper bound (and give no proof of this). If this is true then we can pick some x ∈ D with x > sup(A) which exists since ℝ is complete. Then A

x = A and since x ∈ D, A is

contained in a finite subcover of C. Thus, A is compact.

Accumulation PointsAn element x is an accumulation point of the set A iff for all δ > 0, N(x,δ) contains a point of A distinct from x.

Example: Any interior point of a set is an accumulation point of the the set. For the set (3,5], both 3 and 5 are accumulation points. So, an accumulation point need not belong to the set. Any real number is an accumulation point of the set of rationals.

The set of accumulation points of a set A is called the derived set of A and denoted A'.

Example: If A = (3,5] then A' = [3,5]. If A = ℚ then A' = ℝ.

Closed SetsTheorem: A set is closed iff it contains all its accumulation points. (i.e., B is closed iff B' ⊆ B.)

Pf: Suppose that set B is closed and x is an accumulation point of B.BWOC assume that x ∉ B. Then x ∈ Bc which is an open set.Since Bc is open, there is a δ-neighborhood centered at x contained entirely in Bc, and so, contains no point of B. →← Thus, x ∈ B. Now suppose that B' ⊆ B. Again, BWOC assume that Bc is not open. Then, there is a point y ∈ Bc which is not an interior point of Bc. Every δ-neighborhood of y, must contain points of B and so, y is an accumulation point of B. →← Thus, Bc is open and so, B is closed. ÿ

Bolzano-Weierstrass TheoremBolzano-Weierstrass Theorem: Every bounded infinite subset of ℝ has an accumulation point in ℝ.

Pf: Suppose that A is bounded and infinite, but has no accumulation points. Since A' = ∅, A is closed. By the Heine-Borel theorem, A is compact. Since A has no accumulation points, for each x ∈ A there exists a δ

x-neighborhood which contains no element of A other

than x. The set of all these neighborhoods is an infinite open cover of A (with each open set containing only one point of A). This open cover can have no finite subcover, contradicting the compactness of A. Thus, A must have an accumulation point. ÿ

SequencesA sequence is a function whose domain is ℕ. We shall be talking only about real sequences, which are sequences whose codomain is ℝ.

A sequence is bounded iff its range is a bounded subset.

Example: If f is a sequence, then f: ℕ → ℝ and if f(n) = xn we

usually just refer to the sequence {xn}.

Thus, {xn} = {1/n} = {1, 1/2, 1/3, 1/4, ...} is a bounded sequence

with upper bound 1 and lower bound 0.

ConvergenceA sequence {x

n} has a limit L, or {x

n} converges to L, iff for every

real ε > 0, there exists a natural number N such that if n > N, then | x

n - L | < ε. If no such L exists, the sequence diverges.

Examples: The sequence {1/n} converges to 0.

The sequence {n} diverges.

The sequence {3 – 1/n} converges to 3.

Increasing and DecreasingA sequence {x

n} is increasing iff for all n, m ∈ ℕ, x

n ≤ x

m

whenever n < m.

A decreasing sequence requires xn ≥ x

m whenever n < m.

A monotone sequence is one that is either increasing or decreasing.

Examples: The sequence {1/n} is a decreasing sequence. The sequence {n} is an increasing sequence. The sequence {3 + ( – 1)n} is not monotone.

Bounded Monotone Sequence Theorem

Bounded Monotone Sequence Theorem: Every bounded monotone sequence has a limit L in ℝ.

Pf: Let {xn} be a bounded increasing sequence (the proof in the

case that the sequence is decreasing is analogous). If the range of {x

n} is finite, then let L = largest value of x

n. Since

this is in the range, exists N so that L = xN. Because the sequence

is increasing, xn = L for all n > N. Thus, L is the limit of {x

n}.

Now suppose that the range of {xn} is infinite. By the Bolzano-

Weierstrass theorem, this range has an accumulation point, say L. We wish to show that x

n ≤ L for all n ∈ ℕ. BWOC, assume that

there is an N so that xN > L. Since {x

n} is increasing, x

n > L for

Bounded Monotone Sequence ThmPf:(cont.)all n ≥ N. Since {x

n: n < N} is a finite set, L is an accumulation

point of {xn:n ≥ N} (any open set containing L contains an infinite

number of xn, so removing a finite number of elements of the

sequence will still leave an infinite number in the open set). But, let δ = |x

N – L|, and consider the δ-neighborhood N(L,δ). No element

of {xn:n ≥ N} is in this neighborhood. →← So, x

n ≤ L for all n ∈

ℕ. Let ε > 0. Since L is an accumulation point, ∃ M so that x

M ∈

N(L,ε). Thus, L – ε < xM

, and so, for n > M L – ε < x

M ≤ x

n ≤ L < L + ε.

Therefore, for n > M, |xn – L| < ε. Thus L is the limit of {x

n}. ÿ

Completeness againWe require two lemmas that we present without proofs:

Lemma 1: If {xn} and {y

n} are two sequences such that lim y

n = s

and lim (xn – y

n) = 0, then lim x

n = s.

Lemma 2: If {xn} with lim x

n = s and t a real number with t < s,

then ∃ N ∈ ℕ such that xn > t for all n ≥ N.

The main result that we wish to prove is:

Theorem: If an ordered field has the property that every bounded monotone sequence converges to an element of the field, then the field is complete.

Bounded Monotone ⇒ CompletePf: Let A be a non-empty subset of an ordered field F that is bounded above by b ∈ F. To prove that F is complete, we must show that sup(A) exists and is in F. Since A is non-empty we may choose a in A. If a is an upper bound for A then a = sup(A) and we are done. Assume then that a is not an upper bound. We are going to construct two sequences, x

n and y

n, where the x

n's are not upper

bounds of A and the yn's are upper bounds.

If (a+b)/2 is an upper bound of A then let x1 = a and y

1 = (a+b)/2

and if it is not an upper bound then let x1 = (a+b)/2 and y

1 = b. Note

that in either case, y1 – x

1 = (b-a)/2.

If (x1 + y

1)/2 is an upper bound of A then let x

2 = x

1 and

Bounded Monotone ⇒ Complete y

2 = (x

1 + y

1)/2, and if it is not an upper bound let x

2 = (x

1 + y

1)/2

and y2 = y

1. In either case, y

2 – x

2 = (b-a)/4, x

2 ≥ x

1, and y

2 ≤ y

1.

We continue in this fashion and form two sequences, the increasing sequence {x

n} all of whose elements are not upper

bounds of A and a decreasing sequence {yn} all of whose elements

are upper bounds of A. Furthermore, yn – x

n = (b-a)/2n.

Now the sequence {yn} is bounded (above by b and below by a),

so by hypothesis, this sequence has a limit, say s in F. Furthermore, since lim y

n – x

n = lim (b-a)/2n = 0, by lemma 1 we

have lim xn = s.

We now show that s is an upper bound of A. Suppose that there is a z in A with z > s. Since y

n converges to s, there is an N such

Bounded Monotone ⇒ Complete

that z > yN. But this contradicts the fact that y

N is an upper bound.

Now, we show that s is the least upper bound of A. Suppose that t < s, then by lemma 2, there exists an N such that x

n > t for all n ≥ N.

In particular, xN > t, and since x

N is not an upper bound, t can not be

an upper bound. Therefore, s is the least upper bound of A, i.e., s = sup(A). ÿ

Completeness Equivalences

Completeness of ℝ

Heine-BorelBounded Monotone

Bolzano-Weierstrass

The RealsThere are several ways to finish the construction of the reals from the rationals. We mention two of these.

Dedekind Cuts: A Dedekind cut is a partition of the rationals into two sets A and B such that for every a ∈ A and b ∈ B we have a < b. Examples: A = {x ∈ ℚ: x < ¾} B = {x ∈ ℚ: x ≥ ¾}

A = {x ∈ ℚ: x ≤ 0 or x2 < 2} B = {x ∈ ℚ: x > 0 and x2 > 2}

The set of all Dedekind cuts, after defining the appropriate addition, multiplication and order is a complete ordered field.

The RealsCauchy sequences: A sequence x

n is a Cauchy sequence if for every positive rational

number ε, there exists a natural number N so that for all n, m > N, |x

n – x

m| < ε.

Two Cauchy sequences of rational numbers, xn and y

n are

equivalent if for every positive rational number ε, there exists a natural number N so that for all n > N, |x

n – y

n| < ε.

This is an equivalence relation, and the equivalence classes with appropriate definitions of addition, multiplication and order is a complete ordered field.