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 ...
Math 320, Real Analysis I Quiz 2 — Solutions 1. Give an example of ...
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<strong>Math</strong> <strong>320</strong>, <strong>Real</strong> <strong>Analysis</strong> I<br />
<strong>Quiz</strong> 2 <strong>—</strong> <strong>Solutions</strong><br />
<strong>1.</strong> <strong>Give</strong> <strong>an</strong> <strong>example</strong> <strong>of</strong> each <strong>of</strong> the following, or state that such a request is impossible by referring<br />
to the proper theorem(s):<br />
(a) a function f : A → R that is continuous on A but not uniformly continuous on A.<br />
Example: Consider the function f : (0, 1) → R given by f(x) = 1/x. Then f is<br />
continuous on (0, 1), since it is a rational function <strong>an</strong>d its denominator is never equal to<br />
zero on (0, 1). However, f is not uniformly continuous on (0, 1) since, for the sequences<br />
(x n ) = ( 1<br />
n+1 ) <strong>an</strong>d (y n) = ( 1<br />
n+2 ), we have |x n − y n | → 0 while |f(x n ) − f(y n )| = 1 > 0 for<br />
all n ∈ N.<br />
(b) a function f : [a, b] → R that is continuous on [a, b] but not uniformly continuous<br />
on [a, b].<br />
Impossible! By the Uniform Continuity Theorem, since [a, b] is a compact set, <strong>an</strong>y<br />
function f that is continuous on [a, b] must be uniformly continuous on [a, b].<br />
(c) a function f : [a, b] → R that is continuous at c ∈ (a, b) but not differentiable at c.<br />
Example: Let’s use our favorite <strong>example</strong> <strong>of</strong> this, which is f(x) = |x| on [−1, 1], which<br />
is continuous, but not differentiable, at c = 0.<br />
(d) a function f : [a, b] → R that is differentiable at c ∈ (a, b) but not continuous at c.<br />
Impossible! Our “differentiable implies continuous” theorem says there c<strong>an</strong> be no such<br />
<strong>example</strong>. If f is differentiable at c, then f must also be continuous at c.<br />
(e) a function f : [a, b] → R that is differentiable on [a, b] but whose derivative f ′ is<br />
not continuous on [a, b].<br />
Example: Consider f : [−1, 1] → R given by<br />
{<br />
x 2 sin(1/x) if x ≠ 0,<br />
f(x) =<br />
0 if x = 0.<br />
Then f is differentiable on [−1, 1], with<br />
{<br />
f ′ 2x sin(1/x) − cos(1/x) if x ≠ 0,<br />
(x) =<br />
0 if x = 0.<br />
However, f ′ is not continuous at c = 0 since lim<br />
x→0<br />
f ′ (x) does not exist.<br />
2. (a) Define what it me<strong>an</strong>s for a set to be compact.<br />
Definition: I will accept either <strong>of</strong> the following:<br />
• (Analytical Version) A set K ⊆ R is compact if every sequence (x n ) in K has a<br />
subsequence (x nk ) that converges to a point in K.<br />
• (Topological Version) A set K ⊆ R is compact if every open cover G = {G λ : λ ∈ Λ}<br />
<strong>of</strong> K has a finite subcover {G λ1 , G λ2 , . . . , G λn }. That is, if G is <strong>an</strong>y collection <strong>of</strong><br />
open sets such that<br />
K ⊆ ⋃ λ∈Λ<br />
G λ ,<br />
then there is a finite subset <strong>of</strong> G, say {G λ1 , G λ2 , . . . , G λn }, for which it is still the<br />
case that<br />
n⋃<br />
K ⊆ G λj = G λ1 ∪ G λ2 ∪ · · · ∪ G λn .<br />
j=1
<strong>Math</strong> <strong>320</strong>, <strong>Real</strong> <strong>Analysis</strong> I<br />
<strong>Quiz</strong> 2 <strong>—</strong> <strong>Solutions</strong><br />
[Note: While I will not accept the Heine-Borel Theorem as the definition <strong>of</strong> a<br />
compact set, I c<strong>an</strong> state it here. According to the Heine-Borel Theorem, a set K ⊆ R is<br />
compact if <strong>an</strong>d only if it is both closed <strong>an</strong>d bounded.]<br />
(b) State the Preservation <strong>of</strong> Compactness Theorem.<br />
Preservation <strong>of</strong> Compactness Theorem: Suppose f : A → R is a continuous function.<br />
If K ⊆ A is <strong>an</strong>y compact subset <strong>of</strong> A, then f(K) = {f(x) : x ∈ K} is a compact<br />
subset <strong>of</strong> R.<br />
(c) Prove the Extreme Value Theorem, which is a corollary <strong>of</strong> the Preservation <strong>of</strong> Compactness<br />
Theorem.<br />
Extreme Value Theorem: If f : [a, b] → R, then there are points x ∗ <strong>an</strong>d x ∗ in [a, b]<br />
such that<br />
f(x ∗ ) ≤ f(x) ≤ f(x ∗ )<br />
for all x ∈ [a, b].<br />
Pro<strong>of</strong>. Since the set [a, b] is both closed <strong>an</strong>d bounded, by the Heine-Borel Theorem we<br />
may conclude that it is compact. Thus, by the Preservation <strong>of</strong> Compactness Theorem,<br />
we may say that f([a, b]) is also a compact set. Therefore, f([a, b]) is bounded, so it is<br />
bounded above <strong>an</strong>d below. Moreover, f([a, b]) is not empty, since f(a) is in it. Thus,<br />
both sup f([a, b]) <strong>an</strong>d inf f([a, b]) exist.<br />
Yet, f([a, b]) is compact also implies it is closed, so both sup f([a, b]) <strong>an</strong>d inf f([a, b]),<br />
which are limit points <strong>of</strong> the set f([a, b]) as we have shown in our homework, belong<br />
to f([a, b]). Therefore, by definition <strong>of</strong> the set f([a, b]) = {f(x) : x ∈ [a, b]}, since<br />
sup f([a, b]) ∈ f([a, b]), there exists <strong>an</strong> element x ∗ ∈ [a, b] such that sup f([a, b]) = f(x ∗ ).<br />
Hence f(x ∗ ) is <strong>an</strong> upper bound for the set f([a, b]). Similarly, inf f([a, b]) = f(x ∗ ) for<br />
some x ∗ ∈ [a, b], so f(x ∗ ) is a lower bound for the set [a, b]. Thus, there exists elements<br />
x ∗ , x ∗ ∈ [a, b] such that<br />
f(x ∗ ) ≤ f(x) ≤ f(x ∗ )<br />
for all x ∈ [a, b], since f(x ∗ ) is a lower bound <strong>an</strong>d f(x ∗ ) is <strong>an</strong> upper bound for f([a, b]),<br />
which is what we w<strong>an</strong>ted to show.<br />
3. (a) Define what it me<strong>an</strong>s for a set to be connected.<br />
Definition: A set E is connected if it is not disconnected.<br />
A set E is disconnected if there are nonempty sets A <strong>an</strong>d B such that E = A ∪ B <strong>an</strong>d<br />
A ∩ B = A ∩ B = ∅. (In this case, we say the sets A <strong>an</strong>d B are separated.)<br />
(b) <strong>Give</strong> <strong>an</strong> <strong>example</strong> <strong>of</strong> each <strong>of</strong> the following, or state that such a request is impossible by<br />
referring to the proper theorem(s):<br />
i. a continuous function f : [a, b] → R such that f([a, b]) is not <strong>an</strong> interval.<br />
Impossible! Since the domain is <strong>an</strong> interval, it is connected. As f is continuous,<br />
f([a, b]) must be connected by the Preservation <strong>of</strong> Connectedness Theorem.<br />
Finally, as f([a, b]) is a connected set in R, it is <strong>an</strong> interval by Theorem 3.4.7.<br />
ii. a continuous function f : [a, b] → R such that f([a, b]) ≠ [f(a), f(b)].<br />
Example: Consider the function f : [−1, 2] → R given by f(x) = x 2 . Then<br />
f(−1) = 1 <strong>an</strong>d f(2) = 4, but<br />
f([−1, 2]) = {f(x) : x ∈ [−1, 2]} = {x 2 : −1 ≤ x ≤ 2} = [0, 4]
<strong>Math</strong> <strong>320</strong>, <strong>Real</strong> <strong>Analysis</strong> I<br />
<strong>Quiz</strong> 2 <strong>—</strong> <strong>Solutions</strong><br />
is not equal to [f(−1), f(2)] = [1, 4].<br />
iii. a continuous function f : [a, b] → R that does not have the intermediate value<br />
property.<br />
Impossible! By the Intermediate Value Theorem, every continuous function<br />
f : [a, b] → R has the intermediate value property.<br />
iv. a function f : [a, b] → R with the intermediate value property that is not continuous.<br />
Example: Consider the function f : [−π, π] → R given by<br />
{<br />
sin(1/x) if x ≠ 0,<br />
f(x) =<br />
0 if x = 0.<br />
Then f has the intermediate value property, but f is not continuous at c = 0, <strong>an</strong>d<br />
hence is not continuous.<br />
v. a differentiable function f : [a, b] → R such that f ′ does not have the intermediate<br />
value property on [a, b].<br />
Impossible! By Darboux’s Theorem, if f : [a, b] → R is differentiable <strong>an</strong>d α is<br />
<strong>an</strong>y number betweeen f ′ (a) <strong>an</strong>d f ′ (b), then there is a number c ∈ (a, b) such that<br />
f ′ (c) = α. Hence, f ′ has the intermediate value property.