Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Exam #1
Solutions to the sample
This is a take-home examination. The exam includes 8 questions. The total mark is 100
points. Please show all the work, not only the answers.
1. [12 points] Fix any d ∈ N and let k · k denote the usual Euclidean norm in Rd . That is,
v
u d
uX
x2i for a vector
x = (x1 , . . . , xd ) ∈ Rd .
kxk = t
i=1
Suppose that u, v ∈ Rd and ku − vk > 1. Find all x ∈ Rd such that
kx − uk2 = kx − vk2 + 1.
Solution: Assume that
kx − uk2 = kx − vk2 + 1.
(1)
Write
x = v − t(v − u) + z⊥ ,
where
(v − u) · z⊥ = 0
and t ∈ R is a suitable constant. Notice that
x = u + (1 − t)(v − u) + z⊥
Thus, plugging the above expressions for x into (1) and using the Pythagoras theorem, we
obtain
(1 − t)2 kv − uk2 + kz⊥ k2 = t2 kv − uk2 + kz⊥ k2 + 1.
It follows that z⊥ can be any vector orthogonal to (v − u) and
(1 − 2t)kv − uk2 = 1,
1
which yields
t=
ku − vk2 − 1
.
2ku − vk2
In other words, x satisfies (1) if and only if it belongs to a hyperplane crossing through the
point
ku − vk2 − 1
v−u
v+u
+
v− v−u
=
2
2ku − vk
2
2ku − vk2
and having the vector v − u as its normal.
2. [12 points] Consider the usual Euclidean spaces R2 equipped with the polar
p coordinates
(r, θ). Thus, as usual, if x = (x1 , x2 ) in the cartesian coordinates, then r = x21 + x22 and
tan θ = x2 /x1 , θ ∈ [0, 2π) unless x1 = 0. If x1 = 0 then θ is either π/2 or 0 or 3π/2, depending
on the sign of x2 .
Call two non-zero vectors x = (rx , θx ) and y = (ry , θy ) (written in the polar coordinates)
equivalent and write x ∼ y if θx = θy .
(a) Show that ∼ is an equivalence relation in R2 \{0}.
(b) Let E denote the set of equivalence classes induced by ∼ in R2 \{0}. Verify that the
elements of E are infinite arrays (open half-lines) starting at zero (the origin itself is
not included).
(c) We can identify the elements in E (arrays) with the corresponding values of θ. Thus,
we identify θ ∈ [0, 2π) with the array
x = (x1 , x2 ) ∈ R2 : x1 = t cos θ and x2 = t sin θ for some t ∈ (0, ∞) .
Correspondingly, we introduce a distance in [0, 2π) by setting
p
d(θ, ϕ) = (cos θ − cos ϕ)2 + (sin θ − sin ϕ)2 .
Give a geometric interpretation of d(θ, ϕ) in R2 .
(d) Show that the distance d(·, ·) satisfies the triangle inequality, that is for any real numbers θ, ϕ, ψ ∈ [0, 2π),
d(θ, ϕ) ≤ d(θ, ψ) + d(ψ, ϕ).
When the equality holds?
Solution:
(a) Follows immediately from the definition of the equivalence relation (reflexivity, symmetry and transitivity of the binary relation ∼ are readily verified).
2
(b) Indeed, by the definition equivalence classes are subsets of R2 \{0} with a common
value of θ.
(c) d(θ, ϕ) = kxθ − xϕ k, where k · k is the usual Euclidean distance in R2 and xθ is defined
as the unique unit vector in R2 which belongs to the equivalence class θ.
(d) The inequality follows immediately from the identity d(θ, ϕ) = kxθ − xϕ k and the
triangle inequality for the Euclidean norm k · k. The equality never holds true for
distinct numbers θ, ϕ, ψ because xθ , xϕ , xψ are distinct R2 -vectors on the unit circle
and hence cannot lie on the same line.
3. [12 points]
(a) Let E be the set of all real numbers x ∈ [0, 1] whose decimal expansion contains only
the digits 3, 4, and 5. Is E dense in [0, 1]? Recall that a set is called dense in [0, 1] if
its intersection with any (open) subinterval of [0, 1] is non-empty.
(b) Let N denote the set of all positive integers, i.e. N = {1, 2, 3, . . .}. Let A and B be two
(not necessarily bounded) subsets of N, and define
C = {z ∈ N : there exist x ∈ A and y ∈ B such that z ≤ x + y}.
Prove that sup C = sup A + sup B.
Solution:
(a) By definition, for any x ∈ E we have
0.33333 . . . ≤ x ≤ 0.55555 . . . .
Thus, for instance, E ∩ (0, 0.3) = ∅ and hence E is not dense in [0, 1]. In fact, E is a
so-called Cantor-type set and is actually dense nowhere (in no interval). We will briefly
study such sets later in the course.
(b) For any x ∈ A and y ∈ B we have x + y ≤ sup A + sup B. Thus
sup C ≤ sup A + sup B.
(2)
On the other hand, if both A and B are bounded from above then sup A = max A ∈ A,
sup B = max B ∈ B, and
sup A + sup B = max A + max B ∈ C.
Taking in account (2), this implies sup C = sup A + sup B. Furthermore, if one of the
sets A or B is unbounded from above, then C is also unbounded and hence trivially
sup C = sup A + sup B = +∞.
3
√
4. [14 points] Fix any real number α > 0. Choose x1 > α and define recursively
1
α
xn+1 =
xn +
,
n ∈ N.
2
xn
√
(a) Prove that xn is a monotone decreasing sequence and that lim→∞ xn = α.
√
(b) Put εn = xn − α and show that
ε2n
ε2
< √n .
2xn
2 α
√
n
Conclude that εn+1 < β(ε1 /β)2 , where β = 2 α, and thus the above recursion provides a fast algorithm for computing square roots.
εn+1 =
Solution:
√
(a) First, observe using the inequality x + y ≥ 2 xy that
r
√
1
α
α
= α.
xn+1 =
xn +
≥ xn ·
2
xn
xn
√
√
Furthermore, the equality holds only if xn = α. Next, if xn > α, then
α
α
1
xn +
− xn =
− xn ≤ 0.
xn+1 − xn =
2
xn
xn
Thus xn is a monotone decreasing sequence. It is bounded from below by
hence converges. Let L = limn→∞ xn . Then
α
1
L+
,
L=
2
L
√
and hence L = α.
√
α and
(b) We have
xn+1 −
Thus, since xn >
√
α,
√
√ 2
xn − α
1
α √
α=
xn +
− α=
.
2
xn
2xn
ε2n
ε2n
√
εn+1 =
<
.
2xn
2 α
√
Iterating, and using the notation β = 2 α, we obtain
n
εn+1
ε4n−1
ε8n−1
ε16
ε2n
1 ε4n−1
1 ε8n−2
1 ε16
ε21
n−2
n−1
<
<
=
<
=
<
=
<
·
·
·
<
,
n
β
β β2
β3
β3 β4
β7
β7 β8
β 15
β 22 −1
as required. Of course, · · · in the above inequality really means that the induction is
used to iterate the inequality.
4
5. [12 points] Fix any d ∈ N and let k · k denote the usual Euclidean norm in Rd . That is,
v
u d
uX
kxk = t
x2i for a vector
x = (x1 , . . . , xd ) ∈ Rd .
i=1
A sequence (xn )n∈N of points in Rd is called Cauchy if for any ε > 0 there exists Nε ∈ N
such that
m, n ∈ N and m, n > Nε
⇒
kxn − xm k < ε.
We say that a sequence (xn )n∈N in Rd converges to x ∈ Rd and write limn→∞ xn = x if for
any ε > 0 there exists Nε ∈ N such that
n ∈ N and m, n > Nε
⇒
kxn − xk < ε.
Prove that a sequence in Rn converges if and only if it is a Cauchy sequence.
Solution: Recall that limn→∞ xn = x if and only if every coordinate of xn converges to
the corresponding coordinate
of x. Since the coordinates are elements of R, it follows that
xn = xn (1), . . . , xn (d) converges if and only if the sequence of coordinates xn (k) is Cauchy
for any k = 1, . . . , d. It remains to show that xn is Cauchy in Rd if and only if the sequences
xn (k), k = 1, . . . , d, are Cauchy. This follows immediately from the following inequality and
the definition of the Cauchy sequence in Rd :
|xn (k) − xm (k)| ≤ kxn − xm k ≤
d
X
|xn (i) − xm (i)|,
m, n ∈ N, k = 1, . . . , d.
i=1
6. [14 points]
(a) Let f (x) be a real-valued function such that
|f (x) − f (y)| <
5
· |x − y|
6
for all x, y ∈ R. Prove that the sequence defined by xn+1 = f (xn ) converges for any
x0 ∈ R.
Hint: Using the above inequality, prove by induction that
5 n
|xn+1 − xn | <
· |x1 − x0 |,
n ∈ N.
6
Then use the triangle inequality to evaluate |xn+m − xn | for arbitrary n, m ∈ N. Conclude that (xn )n∈N is a Cauchy sequence in R.
(b) Let (xn )n∈N be a sequence of reals and a ∈ R be a number. Show that xn converges to
a if and only if any it subsequence does.
5
Solution:
(a) Applying the basic inequality
|f (x) − f (y)| <
5
· |x − y|
6
to x = xn and y = xn+1 yields
|xn+2 − xn+1 | = |f (xn+1 ) − f (xn )| <
5
· |xn+1 − xn |.
6
Iterating the inequality
|xn+2 − xn+1 | <
5
· |xn+1 − xn |,
6
we obtain
|xn+2 − xn+1 | <
5 n+1
6
· |x1 − x0 |.
Therefore, for any n, k ∈ N we have
|xn+k − xn | ≤
n+k−1
X
|xj+1 − xj | <
n+k−1
X j=n
= 6·
j=n
5 n
6
∞
X 5 j
5 j
· |x1 − x0 | ≤
· |x1 − x0 |
6
6
j=n
· |x1 − x0 |.
Thus xn is a Cauchy sequence, and hence converges.
Remark. In general, functions (for instance, linear functions f (x) = cx) that for
some constant c > 0 satisfy the condition
|f (x) − f (y)| ≤ c|x − y|,
∀ x, y,
(3)
are called Lipschitz-continuous. Lipschitz-continuous functions are called contractions
if c ∈ (0, 1). The above given argument can be applied verbatim to any contraction
f : R → R to show that the sequence defined recursively by xn+1 = f (xn ) is a Cauchy
sequence and hence converges for any initial state x0 ∈ R.
(b) Let (nk )k∈N be an arbitrary sequence of integers such that nk < nk+1 for all k ∈ N.
Define yk = xnk , k ∈ N. Thus (yk )k∈N is a subsequence of (xn )n∈N . If a = limn→∞ xn
exists, then for any ε > 0 one can find Nε ∈ N such that |a − xn | < ε when n > Nε .
Let
Kε = min{k ∈ N : nk > Nε }.
Then, nk > Nε when k > Kε , and hence
|a − xnk | = |a − yk | < ε
6
for k > Kε .
It follows that a = limk→∞ yk . Thus if (xn ) converges to a, so does any subsequence of
(xn ). Crudely speaking, if the values of xn for large values of n are close to a then the
same must be true for any subsequence of xn .
On the other hand, if any subsequence of (xn ) converges to a ∈ R, then clearly
a = limn→∞ xn because xn is its own subsequence.
7. [12 points]
(a) Let (sn )n∈N be a sequence of reals and s ∈ R be a number. Prove that if
sn − s
lim
=0
n→∞ sn + s
then limn→∞ sn = s.
(b) Introduce an order in the usual Euclidean space R2 as follows. If x = (x1 , x2 ) and
y = (y1 , y2 ) ∈ R2 we say that x < y if either x1 < y1 or x1 = y1 and x2 < y2 . Such
an order is often called lexicographic. Show that R2 ordered by the lexicographic order
doesn’t possess the least upper bound property, namely there is a bounded from above
set A ⊂ R2 for which sup A cannot be defined as an element of R2 .
Solution:
(a) Let tn =
sn −s
.
sn +s
Then
tn =
sn − s
2s
=1−
,
sn + s
sn + s
and hence
sn =
s(1 + tn )
2s
−s=
.
1 − tn
1 − tn
Thus
s(1 + tn )
s(1 + 0)
=
= s.
n→∞ 1 − tn
1−0
lim sn = lim
n→∞
(b) Consider, for instance, the half-plane
A = x = (x1 , x2 ) ∈ R2 : x1 ≤ 1
Alternatively, if you prefer a set bounded in the usual sense, consider, for instance, the
square
A = x = (x1 , x2 ) ∈ R2 : |x1 | ≤ 1, |x2 | ≤ 1
In either case A is bounded from above by any element in the half-plane
B = x = (x1 , x2 ) ∈ R2 : x1 > 1 .
Clearly, B doesn’t have the least element (is unbounded from below). But sup A, if
exists, should provide a lower bound for the set of all possible upper bounds for A. In
particular, sup A would bound B from below. Thus sup A doesn’t exist.
7
8. [12 points]
(a) Use induction to show that 62n−1 + 1 divides 7 for any n ∈ N.
(b) Show that there is a sequence of reals (xn )n∈N such that for any x ∈ R there exists a
subsequence of xn converging to x. Note, you are not required to specify the values of
xn , just show that such a sequence exists.
Solution:
(a) Let Sn = 62n−1 + 1, n ∈ N. Then S1 = 7 and
Sn+1 = 36(Sn − 1) + 1 = 35(Sn − 1) + Sn = 7 · 5 · (Sn − 1) + Sn ,
and the claim follows by induction.
(b) Since the set Q of rational numbers is countable, there is a bijection f : N → Q
mapping natural numbers into Q. Put xn = f (n). Since the set of rational numbers is
dense in R (in particular, Q ∩ Ik 6= ∅ for any interval Ik = (x − 1/k, x + 1/k)) we can
find a sequence (qk )k∈N of elements of Q converging to any given x ∈ R. The sequence
(qk )k∈N is a desired subsequence of (xn )n∈N converging to x. Formally, the subsequence
is xnk with nk := f −1 (qk ).
8