1. Suppose that f is uniformly continuous on a set S
R and is a Cauchy
sequence in

S. Prove that
is a Cauchy sequence. (f is not as sumed to be continuous outside

S, so you cannot use Theorem 3.2, p. 60).

Proof. Let ε > 0. Since f is uniformly continuous on S, there exists
> 0 such that

lx - yl < implies l f(x) - f(y) l < ε
for all x, y ∈ S.
(1)

For this > 0 there exists
such that for n ,m > N,
. By (1) it

fol lows that for n , m > N,
.
Hence
is Cauchy.

2. Let M > 0 and let f : D → R, D
R, satisfy the condition
for

all x, y ∈ D. Show that f is uniformly continuous.

Proof. Let ε > 0. Put . Then if
,

Hence f is uniformly continuous.

3. Suppose that for some constant M with 0 < M < 1,
,

n = 1, 2, 3,.... Prove that the sequence fang is Cauchy.

Proof. We will first show that

This is true for n = 1 as assumed. Suppose it is true for n = k. Then

Hence (2) is true for all n.

For a fixed N and n = N + r, we have

Given ε > 0, let . Then
. Take logarithms .

4. Suppose that f and g are continuous functions on the
closed interal [a, b] such that f(r) =

g(r) for every rational number r ∈ [a, b]. Prove that f(x) = g(x) for all
x ∈ [a, b].

Proof. Suppose that f(z) ≠ g(z) for some irrational number z in [a, b]. Let

α. For , there exists
> 0 such that
and

whenever . Let r be a rational number with
.
Then

f(r) = g(r). Moreover,

a contradiction. Hence f(z) = g(z) for all z ∈ [a,
b].

5. Let .

(a) Show that is
bounded and monotone.

Proof. Let. Then
for x > -1. Hence f is

increasing. Consider the point . Then f(x*) =
x* . For = 1 < x* ,

. And by induction, we have
. Hence is
bounded

above by x* . Since , it follows by the same
reasoning that

is monotonically increasing. Hence by the
Bolzano-Weierstran Theorem,

must converge.

(b) Find .

Proof. Let . Then
is the limit point as
is discarded

since it is negative .

6. Let S be the space of all rational numbers, with d(p, q) = l p - q l, and E
is the set of all

rational numbers p such that 2 < p^{2} < 3. Prove that

(i) E is closed and bounded.

(ii) E is not compact.

Proof.

(a) E is clearly bounded by
and . Let x ∈ E', then
there is a sequence

in E with → 0 as n → ∞. Now
is either in
or , say
. Hence
it

either converges in I, and hence or it converges to either
or
which

are not in our space. Hence E is closed.

(b) Consider the open cover where

p^{2} < 3 }. This cover has not finite subcover. Hence it is not compact.

Another solution :

Take the sequence . Then
has no
convergent subsequences.

7. Let E be a n onempty subset of a metric space (S, d). Define the distance from
x ∈ S to

the set E by .

(a) Prove that if and only if .

(b) Prove that : S → R is uniformly continuous on S.

Proof. (a) Let p(x) = 0. Then
. Hence there is a sequence
in E

with → 0 as n → ∞. Thus x ∈
. For the converse, let x ∈
.
If x ∈ E,

then as d(x, x) = 0, p(x) = 0. If x ∈ E\E, then there exists
as

n → ∞ or → 0 as n → ∞. Hence p(x) = 0.

(b) p : S → R. For

But or
. Similarly

. Thus .
Hence

. Given ε > 0, let
= ε. If
, then

8. Suppose that f is continuous on an open interval I containing
, suppose
that f' is defined

on I except possibly at , and suppose that .
Prove that .

Proof. if the limits exists. Since f is
continuous,

. Now

9. Let f and g be continuous functions on [a, b], g is
positive and monotonically decreasing

and g'(x) exists on [a, b]. Prove that there exists a point
such that

Proof. Let . Since g is
positive, either

or

In either case, is
between h(a) = 0 and h(b). By the In termediate Value

Theorem, there exists between a and b such that

10. Suppose that f is continuous at x = a such that lf(a)l
< 1. Prove that there exists an

open interval , such that for all x ∈
I, l f(x) l ≤ M < 1, for some

fixed constant M.

Proof. Let L = l f(a) l < 1. If such an M does not exist for any
, there exists a
sequence

that converges to a with.
Since f is continuous, as

n → ∞. But this is not possible as for all n.