# Real Analysis: Take Home Final Key

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 assumed 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
follows 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 < p2 < 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
p2 < 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 nonempty 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 Intermediate 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.

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