1. Suppose that f is uniformly continuous on a set S
R and is a Cauchy
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.
For this > 0 there exists
such that for n ,m > N,
. By (1) it
fol lows that for n , m > N,
2. Let M > 0 and let f : D → R, D
R, satisfy the condition
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
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
whenever . Let r be a rational number with
f(r) = g(r). Moreover,
a contradiction. Hence f(z) = g(z) for all z ∈ [a,
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
above by x* . Since , it follows by the same
is monotonically increasing. Hence by the
(b) Find .
Proof. Let . Then
is the limit point as
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 < p2 < 3. Prove that
(i) E is closed and bounded.
(ii) E is not compact.
(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
either converges in I, and hence or it converges to either
are not in our space. Hence E is closed.
(b) Consider the open cover where
p2 < 3 }. This cover has not finite subcover. Hence it is not compact. Another solution :
Take the sequence . Then
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
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
n → ∞ or → 0 as n → ∞. Hence p(x) = 0.
(b) p : S → R. For
. Thus .
. Given ε > 0, let
= ε. If
8. Suppose that f is continuous on an open interval I containing
that f' is defined
on I except possibly at , and suppose that .
Prove that .
Proof. if the limits exists. Since f is
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
Proof. Let . Since g is
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
that converges to a with.
Since f is continuous, as
n → ∞. But this is not possible as for all n.
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