Math 317 HW #9 Solutions
1. Exercise 4.3.7. Assume h : R → R is continuous on R and let K = {x : h(x) = 0}. Show that
K is a closed set.
Proof. Notice that, since h is defined on all of R,
K c = {x : h(x) 6= 0};
showing that K c is open is equivalent to showing that K is closed. Of course,
K c = h−1 (R\{0}) = h−1 ((−∞, 0) ∪ (0, +∞)).
Since (−∞, 0) ∪ (0, +∞) is open, problem 5 below implies that K c is also open.
2. Exercise 4.3.9 (Contraction Mapping Theorem). Let f be a function defined on all of R, and
assume there is a constant c such that 0 < c < 1 and
|f (x) − f (y)| ≤ c|x − y|
for all x, y ∈ R.
(a) Show that f is continuous on R.
Proof. Let > 0. Then whenever |x − y| < , we have that
|f (x) − f (y)| ≤ c|x − y| < c < since 0 < c < 1. Therefore, f is uniformly continuous on R, which implies that f is
continuous on R.
(b) Pick some point y1 ∈ R and construct the sequence
(y1 , f (y1 ), f (f (y1 )), . . .).
In general, if yn+1 = f (yn ), show that the resulting sequence (yn ) is a Cauchy sequence.
Hence we may let y = lim yn .
Proof. I claim that, if d = |y1 − f (y1 )|, then |yn − yn+1 | ≤ dcn−1 . The proof is by
induction:
Base Case: Note that |y1 − y2 | = |y1 − f (y1 )| = d = dc0 .
Inductive Step: Suppose |yk − yk+1 | ≤ dck−1 . Then
|yk+1 − yk+2 | = |f (yk ) − f (yk+1 )| ≤ c|yk − yk+1 | ≤ c(dck−1 ) = dck ,
as desired.
Thus, by induction, I can conclude that |yn − yn+1 | ≤ dcn−1 for all n.
Now, a suitable modification of Problem 5 from the Take-Home part of Midterm #1
(with dcn−1 taking the place of (1/2)n ) implies that (yn ) is a Cauchy sequence.
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(c) Prove that y is a fixed point of f (i.e., f (y) = y) and that it is unique in this regard.
Proof. Since (yn ) → y and f is continuous, we know that (f (yn )) → f (y). However, for
each n, f (yn ) = yn+1 , so the sequence (f (yn )) is the same as the sequence (yn+1 ), which
converges to y. Therefore, we can conclude that f (y) = y.
To see uniqueness, suppose that f (z) = z. Then, on the one hand,
|f (y) − f (z)| ≤ c|y − z|.
On the other hand, |f (y) − f (z)| = |y − z| since y and z are fixed points. Putting this
together with the above inequality, we see that
|y − z| = |f (y) − f (z)| ≤ c|y − z|.
Since 0 < c < 1, this can only be true if |y − z| = 0, meaning that z = y. Since the
choice of fixed point z was arbitrary, we see that every fixed point of f is equal to y, so
y is the unique fixed point of f .
(d) Finally, prove that if x is any arbitrary point in R then the sequence (x, f (x), f (f (x)), . . .)
converges to y defined in (b).
Proof. The same proof as in (b) implies that (x, f (x), f (f (x)), . . .) converges to some
number z. Moreover, the same proof as in the first part of (c) implies that f (z) = z.
Therefore, the uniqueness part of (c) implies that z = y, so the sequence (x, f (x), f (f (x)), . . .)
converges to y, as desired.
3. Exercise 4.3.12. Let C be the Cantor set constructed in Section 3.1. Define g : [0, 1] → R by
(
1 if x ∈ C
g(x) =
0 if x ∈
/ C.
(a) Show that g fails to be continuous at any point c ∈ C.
Proof. Clearly, for any c ∈ C, g(c) = 1. On the other hand, the set C is totally
disconnected (cf. Exercise 3.4.9). Therefore, for each n = 1, 2, 3, . . ., there exists xn ∈
/C
such that c − 1/n < xn < c + 1/n. Hence, (xn ) is a sequence entirely contained in the
complement of C such that (xn ) → c. Since g(xn ) = 0 for all n, we see that (g(xn )) does
not converge to g(c), so Theorem 4.3.2(iv) implies that g cannot be continuous at c.
(b) Show that g is continuous at every point c ∈
/ C.
Proof. Let c ∈ C c ∩ [0, 1]. Let > 0. Since C is a closed set, we know that C c is open.
Therefore,
C c ∩ [0, 1] = C c ∩ (0, 1)
is open, since C c and (0, 1) are both open. Hence, there exists δ > 0 such that Vδ (c) ⊆
C c ∩ [0, 1]. Therefore, if |x − c| < δ, we know that x ∈ Vδ (c) ⊆ C c ∩ [0, 1], meaning that
|g(x) − g(c)| = |0 − 0| = 0 < .
Since the choice of was arbitrary, we conclude that g is continuous at c. In turn, since
the choice of c was arbitrary, we see that g is continuous on all of C c ∩ [0, 1].
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4. Exercise 4.4.9. A function f : A → R is called Lipschitz if there exists a bound M > 0 such
that
f (x) − f (y) ≤M
x−y
for all x, y ∈ A. Geometrically speaking, a function f is Lipschitz if there is a uniform bound
on the magnitude of the slopes of lines drawn through any two points on the graph of f .
(a) Show that if f : A → R is Lipschitz, then it is uniformly continuous on A.
Proof. Suppose f is uniformly continuous with Lipschitz constant M . Let > 0. Then
whenever |x − y| < /M , we have that
|f (x) − f (y)| ≤ M |x − y| < M
= ,
M
where the first inequality is equivalent to the Lipschitz condition.
Since the choice of was arbitrary, we conclude that f is uniformly continuous.
(b) Is the converse statement true? Are all uniformly continuous functions necessarily Lipschitz?
Answer. No, not all uniformly continuous functions are Lipschitz. The basic example
of a uniformly continuous function which is not Lipschitz is a continuous function on
a compact domain which has a vertical tangent line somewhere. For example, consider
√
the function f (x) = 3 x on [−1, 1]. We proved in class that f is continuous on all of R,
so it is certainly continuous on [−1, 1]; since [−1, 1] is compact, Theorem 4.4.8 implies
that f is uniformly continuous on [−1, 1]. However, for xn = −1/n3 and yn = 1/n3 we
have that
f (xn ) − f (yn ) −1/n − 1/n −2/n 2
=
=
xn − yn
−1/n3 − 1/n3 −2/n3 = n .
Clearly, there is no M such that n2 ≤ M for all n, so the function f cannot be Lipschitz.
5. Exercise 4.4.11 (Topological Characterization of Continuity). Let g be defined on all of R. If
A is a subset of R, define the set g −1 (A) by
g −1 (A) = {x ∈ R : g(x) ∈ A}.
Show that g is continuous if and only if g −1 (O) is open whenever O ⊆ R is an open set.
Proof. (⇒) Suppose g is a continuous function. Let O ⊆ R be an open set and let x ∈ g −1 (O),
which means that g(x) ∈ O. Now, since O is open, there exists > 0 such that V (g(x)) ⊆ O.
Moreover, since g is continuous, there exists δ > 0 such that y ∈ Vδ (x) implies that
g(y) ∈ V (g(x)) ⊆ O.
This implies that whenever y ∈ Vδ (x) we have that g(y) ∈ O, meaning that y ∈ g −1 (O).
Since the choice of y ∈ Vδ (x) was arbitrary, we see that Vδ (x) ⊆ g −1 (O). Since the choice of
x was arbitrary, we see that every element of g −1 (O) has some δ-neighborhood contained in
g −1 (O), so we conclude that g −1 (O) is open.
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(⇐) On the other hand, suppose g −1 (O) is open whenever O ⊆ R is open. Pick c ∈ R and let
> 0. Then V (g(c)) is open, so, by hypothesis, g −1 (V (g(c))) is open. Since c ∈ g −1 (V (g(c))),
opennness implies that there exists δ > 0 such that Vδ (c) ⊆ g −1 (V (g(c))). In other words,
if y ∈ Vδ (c), then g(y) ∈ V (g(c)). Since the choice of > 0 was arbitrary, we see that g is
continuous at c. Since the choice of c was arbitrary, we conclude that g is continuous wherever
it is defined, namely on all of R.
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