Math 320, Real Analysis I Quiz 1 — Solutions 1. (a) Define what it ...

Math 320, Real Analysis I
Quiz 1 — Solutions
1. (a) Define what it means for a set to be countable.
Definition: We say that a set A is countable if N ∼ A, i.e., if there is a function
f : N → A that is one-to-one and onto.
(b) Give an example of each of the following:
i. a set that is countable.
Example: N, Z, Q are all examples of countable sets.
algebraic numbers is countable.
Additionally, the set of
ii. a set that is uncountable.
Example: R and every interval (a, b) = {x ∈ R : a < x < b} are examples of uncountable
sets. Moreover, the set of irrational numbers and the set of transcendental
numbers are uncountable.
iii. a set that is neither countable nor uncountable.
Example: Any finite set is neither countable nor uncountable, so {1}, {π, e, 22/7},
{1, 2, 3, . . . , 50} are all examples of sets that are neither countable nor uncountable.
2. (a) Define what it means to say that M is the supremum of a set S of real numbers.
Definition: We say that a real number M is the supremum of S if it satisfies the
following conditions:
i. M is an upper bound for the set S;
ii. if b is any upper bound for the set S, then M ≤ b.
(b) State the Axiom of Completeness.
Axiom of Completeness: Every nonempty subset A ⊂ R that is bounded above has
a supremum in R.
(c) Name three theorems that are equivalent to the Axiom of Completeness. [You don’t
have to prove that they are equivalent right now. You already did so in the homework.]
• Nested Interval Property
• Monotone Convergence Theorem
• Bolzano-Weierstrass Theorem
• Cauchy Criterion for Sequences
3. (a) Define what it means for a sequence to converge.
Definition: We say that a sequence (a n ) converges to a if, for every positive real number
ε > 0, there is a natural number N ∈ N such that
|a n − a| < ε
whenever n ≥ N. In this case, we call a the limit of the sequence and write a = lim a n .
(b) Define what it means for a sequence to be Cauchy.
Definition: We say that a sequence (a n ) is Cauchy if, for every positive real number
ε > 0, there is a natural number N ∈ N such that whenever m, n ≥ N it is true that
|a m − a n | < ε.

Math 320, Real Analysis I
Quiz 1 — Solutions
(c) Prove that every convergent sequence is a Cauchy sequence.
Proof. Let (a n ) be a convergent sequence with limit a = lim a n . Let ε > 0 be arbitrary.
Then ε/2 > 0 as well, so there is a natural number N ∈ N such that
|a n − a| < ε 2
whenever n ≥ N. Now suppose m, n ∈ N with m, n ≥ N and consider
|a m − a n | = |a m − a + a − a n | = |(a m − a) + (a − a n )| ≤ |a m − a| + |a − a n | < ε 2 + ε 2 = ε
by the Triangle Inequality. Therefore, |a m −a n | < ε whenever m, n ≥ N, so the sequence
(a n ) is Cauchy.
4. (a) Define what it means for an infinite series to converge.
Definition: We say that the infinite series ∑ b n converges to B if its sequence of partial
sums (s m ), whose m-th term is given by s m = b 1 + b 2 + b 3 + · · · + b m for all m ∈ N,
converges to B.
(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 convergent infinite series that is commutative.
Example: By the Rearrangement Theorem, every series that is absolutely convergent
is commutative. So the series
∞∑ 1
n 2 = 1 + 1 4 + 1 9 + 1 16 + · · · ,
n=1
which converges since it is a p-Series with p = 2 > 1 and is absolutely convergent
since ∑ ∣ ∣ ∣∣
1 ∣∣∣
n 2 = ∑ 1 again which converges, is an example of a convergent infinite
n2 series that is commutative.
ii. a convergent infinite series whose sequence of partial sums is unbounded.
Impossible! If ∑ b n is a convergent infinite series, then its sequence of partial sums,
(s m ), must converge by the definition of convergence of an infinite series. Yet, if (s m )
converges, then (s m ) is bounded, for every convergent sequence is bounded. Thus
it is impossible to find an example of a convergent infinite series whose sequence of
partial sums is unbounded.
iii. a convergent infinite series that is not absolutely convergent.
Example: The Alternating Harmonic Series,
∞∑ (−1) n+1
= 1 − 1 n 2 + 1 3 − 1 4 + · · · ,
n=1
is a convergent infinite series by the Alternating Series Test, but it is not absolutely
convergent since the associated series of absolute values, ∑ ∞
n=1 ∣ (−1)n+1
n
∣ =
∑ ∞
n=1 1 n = 1 + 1 2 + 1 3 + 1 4
+ · · · , is the Harmonic Series, which diverges.
iv. an absolutely convergent infinite series that is not convergent.
Impossible! By the Absolute Convergence Test, every absolutely convergent
series is convergent.