Chapter 5: Cauchy Sequences and Completeness

Question 1

DEF 5.1: CAUCHY SEQUENCE

Answer 1

We say that a sequence x_1, x_2, . . . in a metric space
X is a Cauchy sequence if for all ε > 0, there exists N such that
d(xₘ, xₙ) < ε whenever m, n > N.

Question 2

Lemma 5.2: CAUCHY and convergence?

Answer 2

In a metric space (X, d), every convergent sequence

is Cauchy.

Question 3

Do cauchy sequences always converge?

Answer 3

Cauchy sequences do not always converge:
e.g. in Q, sequence 3/1, 31/10, 314/100,…. where the nth term is π to n−1 decimal places,
is cauchy since its a convergent sequence in R but limit π isn’t in Q. So sequence isn’t convergent in the metric space Q despite being a Cauchy sequence.

Question 4

EXAMPLE: metric space and sequence properties

Let (X, d) be a metric space, let (xn) be a sequence
in X, and suppose that
d(x_(n+1), x_(n+2)) ≤ 0.5 d(x_n, x_(n+1)) for all n > 1.

Answer 4

This says that the distance between consecutive terms
x_m, x_{m+1} halves every time m increases. Then (x_n) is cauchy. Show this:

we have d(xn, xn−1) ≤(1/2^{n−2})d(x_2, x_1), so that if n > m we have
d(x_m, x_n) ≤ d(xm, xm+1) + · · · + d(xn−1, xn)
≤ d(x1, x2) ×[1/(2^{m−1})+ · · · +1/(2^{n−2})
≤ d(x1, x2) × [1/(2m−1) + 1/2m+ · · · ] = d(x_2, x_1)/(2^{m−2})

where we used the formula for geometric series on the last line. This last quantity converges to 0 as m → ∞. So for ε > 0, take
N to be sufficiently large so that d(x2,x1)/(2^{m−2} < ε for all m > N. Then d(xn, xm) < ε whenever n, m > N. This proves that (xn) is Cauchy.

Question 5

Prop 5.10

Let X be a complete metric space, and let A be
a closed subset of X. Then A is complete.

Proof*****

Answer 5

Proof

Let a1, a2, . . . be a Cauchy sequence in A. Then it is also
a Cauchy sequence in X, and therefore converges to some x ∈ X. Since A is a closed subset of X, x ∈ A, so A is complete

Question 6

Prop 5.11

Let A be a complete subset of a metric space X.
Then A is closed.

Proof *****

Answer 6

Proof
Let (xn) be a sequence in A that converges to x ∈ X. We
need to show that x ∈ A. By Lemma 5.2, (xn) is a Cauchy sequence and so, since A is complete, it converges to some a ∈ A. Since limits
are unique, a = x and x ∈ A as required.

Question 7

Theorem 5.14

C[a, b], d∞

Answer 7

(C[a, b], d∞) is complete.

Question 8

Lemma 5.2: CAUCHY and convergence?

Proof****

Answer 8

In a metric space (X, d), every convergent sequence
is Cauchy.

Proof. Suppose that the sequence (x_n) → x. Let ε > 0. Then there exists N such that d(x_n, x) less than ε/2 for all n > N. For n, m > N, we then have
d(x_n, x_m) ≤ d(x_n, x) + d(x,x_m) < ε/2 + ε/2 = ε,
so that (x_n) is Cauchy.

Question 9

Example 5.4. Consider the space C[0, 1] of continuous functions on the closed interval [0, 1], with the metric d1. Define the sequence (f_n) by

f_n(x)=
{0 if x ≤ 0.5
{n(x-0.5) if 0.5 ≤ x≤ 0.5 +1/n
{1 if 1/2 +1/n ≤ x.

cauchy?

Answer 9

d_1(fn, fm) = (1/2)(1/m−1/n)
less than 1/2m for m less than n.

(work this out for practice)

Then given ε bigger than 0, take N to be the smallest natural number exceeding 1/2ε. If m,n bigger than N then assume n bigger than m st

d_1(f_n,f_m) is less than 1/2m is less than 1/2N is less than ε
and therefore (f_n) is cauchy.

Question 10

Definition 5.5. Completeness

Answer 10

The metric space X is complete if every Cauchy
sequence (xn) in X converges to a limit x ∈ X. More generally, a subset A ⊆ X is complete if every Cauchy sequence in A converges
to a limit in A.

Question 11

example 5.6: is Q complete

Answer 11

Q is not complete. In Q, the sequence 1/1, 14/10 ,141/100 , . . .,
where the nth term is √2 to n − 1 decimal places, is Cauchy (since it is a convergent sequence in R) but its limit is √2, which is not
in Q. So the sequence is not convergent in the metric space Q, even though it is a Cauchy sequence.

Question 12

THM 5.7

cauchy sequence and limit of subsequence

Answer 12

Let (xn) be a Cauchy sequence in a metric space(X, d) and let a ∈ X. If (xn) has a subsequence (x_nk) converging to a, then (x_n) converges to a

Question 13

Bolzano-Weierstrass Theorem, remark

Answer 13

Bolzano-Weierstrass Theorem, MAS221). Let (xn)
be a sequence of real numbers which is bounded, i.e. there are some
α, β such that α 6 xn 6 β for all n. Then the sequence (xn) has a
convergent subsequence.

Question 14

Theorem 5.9. is R complete

Answer 14

R is complete.

Question 15

Proposition 5.10.

subsets of complete metric spaces

Answer 15

Let X be a complete metric space, and let A be

a closed subset of X. Then A is complete.

Question 16

Proposition 5.11.

complete subsets of metric spaces

Answer 16

Let A be a complete subset of a metric space X.

Then A is closed.

Question 17

Theorem 5.12

euclidean metric is it complete?

Answer 17

. For k ≥ 1, R^k, with the Euclidean metric d2, is

complete.

Question 18

important property

of the Riemann integral:

Answer 18

important property
of the Riemann integral: if g is a non-negative continuous function on [a, b] and ∫ _[a,b] g(x)dx = 0, then g(x) = 0 for all a ≤ x ≤ b.

Question 19

Example 5.13. C[a, b], with the metric d1, is not complete.

Answer 19

We’ll demonstrate this in the case of C[0, 1]. we had cauchy

f_n(x)=
{0 if x ≤ 0.5
{n(x-0.5) if 0.5 ≤ x≤ 0.5 +1/n
{1 if 1/2 +1/n ≤ x.

Suppose for a contradiction that this Cauchy sequence does converge,
i.e. that fn → f for some continuous function f. Then
d1(fn, f) → 0 as n → ∞. But for all n ∈ N

d1(fn, f) = ∫ _[0,1] |f_n(t)-f(t)|.dt ≥
∫ [0,1/2] |f_n(t)-f(t)|.dt = ∫[0,1/2] |f(t)|.dt ≥0

so we must have ∫_[0,1/2]|f(t)|.dt =0.

Since f is continuous, this
implies that f(t) = 0 for 0 ≤ t ≤1/2

The same argument shows that
d1(fn, f) ≥ ∫ _[0.5+ 1/n,1] |f_n(t)-f(t)|.dt tends to ∫ _[0.5,1] |f(t)-1|.dt ≥ 0

and since f is continuous, then f(t) = 1 for all 1/2≤ t ≤ 1. No such
f can exist, since we have f(1/2) = 0 and f(1/2) = 1. Thus C[0, 1] is
not complete with respect to the metric d_1.

Question 20

Theorem 5.14.

is C[a,b] on d_infinity complete

Answer 20

(C[a, b], d∞) is complete.

not d_1 remember