Download Completeness of real numbers

Math. 501, Spring 2017
Existence of real numbers, decimal representation of real numbers
Bolzano-Weierstrass Theorem, Cauchy sequences, and Cauchy
Completeness Theorem
Jacek Polewczak
Existence of real numbers
Definition 1. An ordered set is a set S 6= ∅ with relation < satisfying two properties:
(a) If x ∈ S and y ∈ S then one and only one of the statements
x < y,
x = y,
y<x
is true
(b) If x, y, z ∈ S, if x < y and y < z, then x < z.
For example Q, the set of rational numbers, is an ordered set if r < s, for r, s ∈ Q, is defined to mean that s − r
is a positive rational number.
Remark 1. The statement “x < y” may be read as “x is less than y” or “x is smaller than y” or “x precedes y”.
The notation x ≤ y indicates that x < y or x = y, without specifying which of these two is to hold. Equivalently,
x ≤ y is the negation of x > y.
Definition 2. Let E ⊂ S be a subset of an ordered set S. If there exists a β ∈ S such that x ≤ β for every
x ∈ E, we say that E is bounded above, and call β an upper bound of E.
Remark 2. Lower bounds are defined similarly, with ≥ in place of ≤.
Definition 3. Let E ⊂ S be a bounded above subset of an ordered set S. If there exists an α ∈ S satisfying the
following properties:
(i) α is an upper bound of E.
(ii) If γ < α then γ is not an upper bound of E.
Then α is called the least upper bound of E or the supremum of E and we write α = sup E. The least upper
bound is unique.
Remark 3. When E ⊂ R (R denotes here the set of real numbers), then α is the least upper bound of E if and
only if
(i) α is an upper bound of E, and
(ii) given any > 0, there is a number x() ∈ E for which x() > α − .
Remark 4. Q is an ordered set that does not have the least-upper-bound property. Indeed, the set
E = {r ∈ Q : r > 0 and r2 < 2}
√
contains no largest number in Q. Notice that E as a subset of R has the least upper bound equal to 2.
1 + xn
As another (complementary) example, consider set E = {xn }, where x1 = 1 and xn+1 = 2
, for n ≥ 2.
2 + xn
Then, the sequence {xn } is an increasing sequence of rational numbers (i.e., xn ∈ Q with xn < xn+1 , for every
1 + xn
n ≥ 1) such that xn = 2
≤ 2, for n ≥ 1. (Can you check it? ) Next, one shows (Can you do it? ) that
2 + xn
√
lim xn = 2 6∈ Q. Thus, E does not posses the least upper bound in Q.
n→∞
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Definition 4. A field is a set F with two operations, addition and multiplication, which satisfy so called “field
axioms” (A), (M), and (D).
(A) Axioms for addition
(A1) If x ∈ F and y ∈ F , then their sum x + y is in F .
(A2) Addition is commutative: x + y = y + x for all x, y ∈ F .
(A3) Addition is associative: (x+y)+z=x+(y+z) for all x, y, z ∈ F .
(A4) F contains an element 0 such that 0 + x = x for every x ∈ F .
(A5) To every x ∈ F corresponds and element −x ∈ F such that x + (−x) = 0.
(M) Axioms for multiplication
(M1) If x ∈ F and y ∈ F , then their product xy is in F .
(M2) Multiplication is commutative: xy = yx for all x, y ∈ F .
(M3) Multiplication is associative: (xy)+z=x(yz) for all x, y, z ∈ F .
(M4) F contains an element 1 6= 0 such that 1x = x for every x ∈ F .
(M5) To every x ∈ F corresponds and element 1/x ∈ F such that x(1/x) = 1.
(D) The distributive law
x(y + z) = xy + xz
holds for all x, y, z ∈ F .
Definition 5. An ordered field is a field F which is also and ordered set, such that
(i) x + y < x + z if x, y, z ∈ F and y < z,
(ii) xy > 0 if x, y ∈ F , x > 0 and y > 0.
If x > 0, we call x positive; if x < 0, x is negative.
The set of rational numbers, Q, is an ordered field.
Theorem 1. There exists an ordered field R which has the least-upper-bound property.
R is unique in the following sense. If R1 and R2 are such fields, there exists an isomorphism of ordered fields
between them, i.e., there exists a bijective mapping φ : R1 → R2 which preserves the structure: for any x, y ∈ R1 ,
we have φ(x + y) = φ(x) + φ(y) and φ(xy) = φ(x)φ(y); for any x, y ∈ R1 with x < y, we have φ(x) < φ(y).
Moreover, R contains Q as a subfield.
Proof. For details of the proof see, for example, Principles of Mathematical Analysis by Walter Rudin, or Mathematical Analysis An Introduction by Andrew Browder. The idea behind the proof is to construct members of
R as certain subsets of Q, called Dedekind cuts. A (Dedekind) cut is, by definition, any set α ⊂ Q with the
following three properties:
(i) α 6= ∅ and α 6= Q,
(ii) If p ∈ α, q ∈ Q, and q < p, then q ∈ α,
(iii) If p ∈ α, then p < s for some s ∈ α (i.e., α has no largest member).
Moreover, every subset α ⊂ Q that satisfies (i)-(iii) has the form {r ∈ Q : r < a} for some a ∈ R; in fact,
a = sup α.
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Corollary 1. Suppose that R is an ordered field with the least-upper-bound property.
(a)
Let ∅ =
6 B ⊂ R and B is bounded below. Let L be the set of all lower bounds of B. Then α = sup L exists
in R, and α = inf B. α is the greatest lower bound of B and B has the greatest-lower-bound property.
(b)
If x, y ∈ R and x > 0, then there exists n ∈ N (N denotes here the set of natural numbers) such that
nx > y.
(c)
If x, y ∈ R and x < y, then there exists a r ∈ Q such that x < r < y.
proof of (a). Since B is bounded below, L 6= ∅. Since L consists of exactly those y ∈ R which satisfy the
inequality y ≤ x for every x ∈ B, we see that every x ∈ B is an upper bound of L. Thus L is bounded above.
By our assumption about R, L has a supremum in R; call it α.
If γ < α then (see Definition 3) γ is not an upper bound of L, hence, γ 6∈ B. It follows that α ≤ x for every
x ∈ B. Thus, α ∈ L.
If α < β then β 6∈ L, since α is an upper bound of L. Thus we have shown that α ∈ L but β 6∈ L if β > α. In
other words, α is lower bound of B and, but β is not if β > α. This means that α = inf B.
proof of (b). Let A = {nx : n ∈ N}. If (b) were false, then y would be an upper bound of A. Then A has a least
upper bound in R, α = sup A. Next, x > 0 implies α − x < α and α − x is not an upper bound of A. Thus,
α − x < mx for some m ∈ N. But then α < (m + 1)x, which is impossible, since α is an upper bound of A.
proof of (c). Combining x < y (i.e., y − x > 0) with (b) yields n ∈ N such that n(y − x) > 1. Applying (b) again
(with x = 1), we obtain m1 , m2 ∈ N such that m1 > nx and m2 > −nx, or equivalently
−m2 < nx < m1 .
Hence there exists an integer m (with −m2 ≤ m ≤ m1 ) such that
m − 1 ≤ nx < m. (Do you know why? )
Combining these inequalities we obtain
nx < m ≤ 1 + nx < ny.
However, n > 0, thus
x<
m
< y.
n
This proves (c) with r = m/n.
Remark 5. Part (a) shows that an ordered field with the least-upper-bound property has also the greatest-lowerbound property. Part (b) is referred to as the Archimedean property of R, while part (c) states that Q is dense
in R: between any two real numbers there is a rational number.
Theorem 2. For every real x > 0 and every integer n > 0 there is one and only one positive real y such that
y n = x.
√
This number y is written as n x or x1/n .
Proof. For a proof see, for example, Principles of Mathematical Analysis by Walter Rudin.
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Decimal representation of real numbers
Let x > 0 be real and let n0 be the largest integer such that n0 ≤ x. We denote it by [x] and call it the integer
part of x and x − [x] the fractional part of x.
Observe that existence of such n0 is guaranteed by the Archimedean property of R (see part (b) of Corollary 1).
We will represent x−[x] in the form a1 a2 a3 . . ., where each ai is a digit, i.e., ai is from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
The process of selecting ai is done inductively as follows.
Divide the interval [0, 1), which contains x − [x], into ten open equal parts
1 2
2 3
9
1
,
,
,
,
, ...,
,1
0,
10
10 10
10 10
10
and take as a1 the digit (numerator of left endpoint) corresponding to the subinterval that contains x − [x].
Next, divide [a1 /10, (a1 + 1)/10) into ten equal parts
1
a1
1 a1
2
a1
9 a1 + 1
a1 a1
,
+
,
+
,
+
, ...,
+
,
10 10 102
10 102 10 102
10 102 10
and choose as a2 the digit (multiplier of 1/102 for left endpoint) corresponding to the subinterval that contains
x − [x].
Continue, to get a1 a2 a3 . . . an hence the decimal of x:
x = n0 .a1 a2 a3 . . . .
If x happens to be an endpoint of one of the subintervals, the digits for x are all 0 from some stage on, and
conversely. In this case x is just the finite sum
x = n0 +
a1
a2
ak
+ 2 + ··· + k
10 10
10
and the decimal can be viewed as an abbreviation for this sum. If x is not an endpoint, the decimal has infinitely
many nonzero digits and a resulting sum requires a further consideration.
To this end, let E be the set
{n0 +
a1
a2
ak
+ 2 + · · · + k : ak ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, k = 0.1, 2, 3, . . . .}
10 10
10
Since set E is bounded above by n0 + 1 (Do you know why? ), sup E exists (see Theorem 1) and
x = sup E = n0 .a1 a2 a3 . . .
Proposition 1. Distinct numbers have distinct decimal representations.
Proof. Suppose x < y. Since the set {10k : k = 1, 0, 1, 2, . . . } is unbounded, the Archimedean property implies
that there is k such that
1
0<
< 10k .
y−x
Hence, clearly, the choices of ak for x and y are different.
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Remark 6. (a) The endpoints of the intervals of subdivision are assigned decimals that we call terminating, i.e.,
they conclude with an infinite repetition of the digit 0. For example,
1
125
=
= 0.125000 . . .
8
1000
If we settled on left-open, right-closed intervals for the subdivisions, the only differences in the decimal representations would be at the endpoints, where a decimal is of the form xxx999 . . . would result, such as
1
= 0.124999 . . .
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Another alternative is to consider closed subintervals and accept an ambiguity in the decimal representation of
some numbers; each endpoint would belong to two subinetervals at some stage, and two choices ak 000 . . . and
(ak − 1)999 . . . would result, amounting to
0.125000 . . . = 0.124999 . . . .
(b) Instead of dividing intervals in ten equal parts and assigning digits from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9},
one can make division into b equal parts (2 ≤ b ∈ N) and use the digits 0, 1, . . . b − 1. The result is the base b
expansion of numbers.
We observe that assuming the closed interval method in the subdivision process, the successive digits designate
a succession of closed intervals In (in which the numbers is to be sought) with the following property:
I1 ⊃ I2 ⊃ I3 . . .
and length In → 0
Such a sequence of closed intervals In is called a nest.
Note that for any nest of closed intervals in an ordered field R with Archimedean property, the intersection
∞
T
In
n=1
is either empty or a single point. The property of R that guarantees the existence of a number in R for every
base b expansion is
∞
T
Nested Intervals Property: For every nest of closed intervals in R the intersection
In is nonempty.
n=1
Bolzano-Weierstrass Theorem
Definition 6. For a given a set S 6= ∅ (the set S does NOT have to be a subset of real numbers), a sequence
is a function whose domain is a set of the form {n ∈ Z : n ≥ m} and its range is S. m is often 1 or 0. It is
customary to denote a sequence by a letter such as x and to denote its value at n as xn rather than x(n).
Definition 7. An infinite sequence of real numbers {xn }, n ≥ 1 converges to x as n → ∞ if and only if for each
> 0 there is N ∈ N such that |xn − x| < for all n > N .
We will use the notation
lim xn = x or xn → x as n → ∞.
n→∞
Definition 8. Given as sequence of real numbers {xn }, consider a sequence {nk } of positive integers such that
n1 < n2 < n3 < · · · . The sequence {xnk } is called a subsequence of {xn }. If {xnk } converges as k → ∞, its limit
is called a subsequential limit of {xn }.
Remark 7. Note that {xn } converges to x if and only if every subsequence of {xn } converges to x. (Prove it ! )
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Definition 9. A sequence of real numbers {xn }, n ≥ 1 is increasing if xn ≤ xn+1 for all n ∈ N and strictly
increasing if xn < xn+1 for all n ∈ N.
A sequence of real numbers {xn }, n ≥ 1 is decreasing if xn ≥ xn+1 for all n ∈ N and strictly decreasing if
xn > xn+1 for all n ∈ N.
All four kinds of these special sequences are called monotone.
Lemma 1. Every infinite sequence of real numbers has a monotone subsequence.
Proof. It is enough to consider sequences that have no increasing subsequence and show that they each have
necessarily a decreasing subsequence.
Suppose {xn } has no increasing subsequence. We argue that there is a largest value among {xn : n ∈ N}, in
other words, max xn exists.
If not, then x1 is not the largest, so there is a first index n1 with x1 < xn1 . But xn1 is not the largest, so there
is a first index n2 giving xn1 < xn2 and n2 > n1 . Continuing, we find an increasing subsequence contrary to
assumption.
Now, choose as xm1 the first occurrence of the largest value. Consider {xn : n > m1 }, which has no increasing
subsequence, and take as xm2 the first occurrence of its largest value. We have xm1 ≥ xm2 . Next, we continue by
induction to obtain a decreasing subsequence.
Definition 10. A sequence {xn } of real numbers is bounded if there exists 0 < M < ∞ such that |xn | ≤ M for
all n ∈ N.
Remark 8. Every convergent sequence of real numbers is bounded. (Can you prove it? )
Theorem 3. A bounded and monotone infinite sequence of real numbers converges.
Proof. It is enough to consider a bounded increasing sequence. The case of a bounded from below and decreasing
sequence follows a very similar line of arguments as below, if one applies part (a) of Corollary 1.
Let {xn } be a bounded increasing sequence. If E = {xn : n ∈ N}, then Theorem 1 implies that x = sup E exists
and x ∈ R. We will show that lim xn = x. Let > 0. Since x − is not an upper bound of E (see (ii) of
n→∞
Definition 3), there exists N ∈ N such that xN > x − . Since {xn } is increasing, we have xN ≤ xn for all n > N .
This together with xn ≤ x for all n ∈ N yields
x − ≤ xn ≤ x < x + ,
The above inequality implies that |xn − x| < for all n > N . Equivalently, lim xn = x.
n→∞
Theorem 4 (Bolzano-Weierstrass Theorem). Any bounded infinite sequence {xn } of real numbers contains a
convergent subsequence.
Proof. Lemma 1 implies that {xn } contains a bounded and monotone subsequence, while Theorem 3 shows that
such a subsequence is convergent.
Cauchy sequences
Definition 11. A sequence of real numbers {xn }, n ≥ 1 is called a Cauchy sequence if and only if for each > 0
there is N ∈ N such that |xn − xm | < for all n > N and all m > N .
Lemma 2. Every convergent sequence is a Cauchy sequence.
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Proof. Suppose lim xn = x and let > 0. Then from the definition of convergence (see Definition 7) there is
n→∞
N ∈ N such that
1
|xn − x| < for all n > N .
2
Let xm be any element of {xn } with m > N . We have
1
1
|xn − xm | = |xn − x + x − xm | ≤ |xn − x| + |xm − x| < + = .
2
2
Thus {xn } is a Cauchy sequence.
Lemma 3. Every Cauchy sequence is bounded.
Proof. From the definition of Cauchy sequence (see Definition 11) with = 1, there is N0 ∈ N such that
|xn − xm | < 1 for all n > N0 and m > N0 .
In particular, |xn − xN0 +1 | < 1 if n > N0 . Also,
|xn | = |xn − xN0 +1 + xN0 +1 | ≤ |xn − xN0 +1 | + |xN0 +1 | < 1 + |xN0 +1 |,
for all n > N0 . Now, if
M = sup{|x1 |, |x2 |, . . . , |xN0 +1 , 1 + |xN0 +1 |},
then |xn | ≤ M for all n ∈ N.
Theorem 5 (Cauchy criterion for convergence). A necessary and sufficient condition for convergence of a sequence of real numbers {xn } is that it be a Cauchy sequence.
Proof. Lemma 2 shows that a convergent sequence is a Cauchy sequence.
Now, assume that {xn } is a Cauchy sequence. Lemma 3 implies that {xn } is a bounded sequence, while Theorem
4 yields that there is a subsequence {xnk } of {xn } which converges to the limit x. We shall show that the sequence
{xn } itself converges to x. Let > 0. Since {xn } is Cauchy sequence there is N1 ∈ N such that
1
|xn − xm | < for all n > N1 and m > N1 .
2
Next, since lim xnk = x, there is N2 ∈ N such that
k→∞
1
|xnk − x| < for all k > N2 .
2
Now, fix k ≥ N2 such that nk ≥ N1 . (Observe that such a choice is always possible since n1 < n2 < n3 < · · · .)
Then for all n ≥ N1 we have
1
1
|xn − x| = |xn − xnk + xnk | ≤ |xn − xnk | + |xnk − x| < + = .
2
2
Thus lim xn = x.
n→∞
Theorem 5 is very useful because it is often easier to show that a sequence is Cauchy than to show that it
converges. The following example shows that we can prove that a sequence is Cauchy even when we have no idea
what its limit is.
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Example 1. Any sequence of real numbers satisfying
|xn − xn+1 | ≤
1
,
2n
n ∈ N,
is convergent.
Proof. For m > n we have
|xn − xm | = |xn − xn+1 + xn+1 − xn+2 + xn+2 + · · · + xm−1 − xm |
≤ |xn − xn+1 | + |xn+1 − xn+2 | + · · · + |xm−1 − xm |
1
1
≤ n · · · + m−1
2
2
m−n
1
1 X 1
1
= n−1
= n−1 1 − m−n
k
2
2
2
2
k=1
≤
1
2n−1
.
For a given > 0, choose N ∈ N such that n ≥ N implies 1/2n−1 < . Thus, we have proved that {xn } is Cauchy.
By Theorem 5, therefore, it converges to some real number.
Remark 9. A sequence of real numbers satisfying lim (xn+1 −xn ) = 0 is not necessarily Cauchy. Indeed, xn = log n
n→∞
has the property
n+1
xn+1 − xn = log(n + 1) − log n = log
→ log 1 = 0
n
as n → ∞. However, the sequence {log n} does not converge, in fact, it diverges to ∞ as n → ∞.
Cauchy Completeness Theorem
One proves the following result:
Theorem 6 (Cauchy Completeness Theorem). Assume that R is an ordered field with the Archimedean property
(see part (b) of Corollary 1) and consider the following statements:
(a)
R has the least-upper-bound property,
(b)
R has the Nested Intervals Property,
(c)
Every Cauchy sequence in R converges.
Then (a) ⇐⇒ (b) ⇐⇒ (c).
Theorem 6 provides equivalent characterizations of any ordered field with the Archimedean property, while
Theorem 1 exhibits existence of a unique (up to isomorphism) such field, which we call the real numbers and
denote by R. Furthermore, the rational numbers Q ⊂ R are dense in R (see part (c) of Corollary 1) and R \ Q
are called the irrational real numbers.
Can you prove the Cauchy Completeness Theorem ?
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