Math 320, Real Analysis I Quiz 2 — Solutions 1. Give an example of ...

Math 320, Real Analysis I
Quiz 2 — Solutions
1. Give an example of each of the following, or state that such a request is impossible by referring
to the proper theorem(s):
(a) a function f : A → R that is continuous on A but not uniformly continuous on A.
Example: Consider the function f : (0, 1) → R given by f(x) = 1/x. Then f is
continuous on (0, 1), since it is a rational function and its denominator is never equal to
zero on (0, 1). However, f is not uniformly continuous on (0, 1) since, for the sequences
(x n ) = ( 1
n+1 ) and (y n) = ( 1
n+2 ), we have |x n − y n | → 0 while |f(x n ) − f(y n )| = 1 > 0 for
all n ∈ N.
(b) a function f : [a, b] → R that is continuous on [a, b] but not uniformly continuous
on [a, b].
Impossible! By the Uniform Continuity Theorem, since [a, b] is a compact set, any
function f that is continuous on [a, b] must be uniformly continuous on [a, b].
(c) a function f : [a, b] → R that is continuous at c ∈ (a, b) but not differentiable at c.
Example: Let’s use our favorite example of this, which is f(x) = |x| on [−1, 1], which
is continuous, but not differentiable, at c = 0.
(d) a function f : [a, b] → R that is differentiable at c ∈ (a, b) but not continuous at c.
Impossible! Our “differentiable implies continuous” theorem says there can be no such
example. If f is differentiable at c, then f must also be continuous at c.
(e) a function f : [a, b] → R that is differentiable on [a, b] but whose derivative f ′ is
not continuous on [a, b].
Example: Consider f : [−1, 1] → R given by
{
x 2 sin(1/x) if x ≠ 0,
f(x) =
0 if x = 0.
Then f is differentiable on [−1, 1], with
{
f ′ 2x sin(1/x) − cos(1/x) if x ≠ 0,
(x) =
0 if x = 0.
However, f ′ is not continuous at c = 0 since lim
x→0
f ′ (x) does not exist.
2. (a) Define what it means for a set to be compact.
Definition: I will accept either of the following:
• (Analytical Version) A set K ⊆ R is compact if every sequence (x n ) in K has a
subsequence (x nk ) that converges to a point in K.
• (Topological Version) A set K ⊆ R is compact if every open cover G = {G λ : λ ∈ Λ}
of K has a finite subcover {G λ1 , G λ2 , . . . , G λn }. That is, if G is any collection of
open sets such that
K ⊆ ⋃ λ∈Λ
G λ ,
then there is a finite subset of G, say {G λ1 , G λ2 , . . . , G λn }, for which it is still the
case that
n⋃
K ⊆ G λj = G λ1 ∪ G λ2 ∪ · · · ∪ G λn .
j=1

Math 320, Real Analysis I
Quiz 2 — Solutions
[Note: While I will not accept the Heine-Borel Theorem as the definition of a
compact set, I can state it here. According to the Heine-Borel Theorem, a set K ⊆ R is
compact if and only if it is both closed and bounded.]
(b) State the Preservation of Compactness Theorem.
Preservation of Compactness Theorem: Suppose f : A → R is a continuous function.
If K ⊆ A is any compact subset of A, then f(K) = {f(x) : x ∈ K} is a compact
subset of R.
(c) Prove the Extreme Value Theorem, which is a corollary of the Preservation of Compactness
Theorem.
Extreme Value Theorem: If f : [a, b] → R, then there are points x ∗ and x ∗ in [a, b]
such that
f(x ∗ ) ≤ f(x) ≤ f(x ∗ )
for all x ∈ [a, b].
Proof. Since the set [a, b] is both closed and bounded, by the Heine-Borel Theorem we
may conclude that it is compact. Thus, by the Preservation of Compactness Theorem,
we may say that f([a, b]) is also a compact set. Therefore, f([a, b]) is bounded, so it is
bounded above and below. Moreover, f([a, b]) is not empty, since f(a) is in it. Thus,
both sup f([a, b]) and inf f([a, b]) exist.
Yet, f([a, b]) is compact also implies it is closed, so both sup f([a, b]) and inf f([a, b]),
which are limit points of the set f([a, b]) as we have shown in our homework, belong
to f([a, b]). Therefore, by definition of the set f([a, b]) = {f(x) : x ∈ [a, b]}, since
sup f([a, b]) ∈ f([a, b]), there exists an element x ∗ ∈ [a, b] such that sup f([a, b]) = f(x ∗ ).
Hence f(x ∗ ) is an upper bound for the set f([a, b]). Similarly, inf f([a, b]) = f(x ∗ ) for
some x ∗ ∈ [a, b], so f(x ∗ ) is a lower bound for the set [a, b]. Thus, there exists elements
x ∗ , x ∗ ∈ [a, b] such that
f(x ∗ ) ≤ f(x) ≤ f(x ∗ )
for all x ∈ [a, b], since f(x ∗ ) is a lower bound and f(x ∗ ) is an upper bound for f([a, b]),
which is what we wanted to show.
3. (a) Define what it means for a set to be connected.
Definition: A set E is connected if it is not disconnected.
A set E is disconnected if there are nonempty sets A and B such that E = A ∪ B and
A ∩ B = A ∩ B = ∅. (In this case, we say the sets A and B are separated.)
(b) Give an example of each of the following, or state that such a request is impossible by
referring to the proper theorem(s):
i. a continuous function f : [a, b] → R such that f([a, b]) is not an interval.
Impossible! Since the domain is an interval, it is connected. As f is continuous,
f([a, b]) must be connected by the Preservation of Connectedness Theorem.
Finally, as f([a, b]) is a connected set in R, it is an interval by Theorem 3.4.7.
ii. a continuous function f : [a, b] → R such that f([a, b]) ≠ [f(a), f(b)].
Example: Consider the function f : [−1, 2] → R given by f(x) = x 2 . Then
f(−1) = 1 and f(2) = 4, but
f([−1, 2]) = {f(x) : x ∈ [−1, 2]} = {x 2 : −1 ≤ x ≤ 2} = [0, 4]

Math 320, Real Analysis I
Quiz 2 — Solutions
is not equal to [f(−1), f(2)] = [1, 4].
iii. a continuous function f : [a, b] → R that does not have the intermediate value
property.
Impossible! By the Intermediate Value Theorem, every continuous function
f : [a, b] → R has the intermediate value property.
iv. a function f : [a, b] → R with the intermediate value property that is not continuous.
Example: Consider the function f : [−π, π] → R given by
{
sin(1/x) if x ≠ 0,
f(x) =
0 if x = 0.
Then f has the intermediate value property, but f is not continuous at c = 0, and
hence is not continuous.
v. a differentiable function f : [a, b] → R such that f ′ does not have the intermediate
value property on [a, b].
Impossible! By Darboux’s Theorem, if f : [a, b] → R is differentiable and α is
any number betweeen f ′ (a) and f ′ (b), then there is a number c ∈ (a, b) such that
f ′ (c) = α. Hence, f ′ has the intermediate value property.