Download MIDTERM EXAM II Problem 1. Suppose that A is a subset of the

MIDTERM EXAM II
Problem 1. Suppose that A is a subset of the closed interval [0, 1]. Prove
that if A has only one limit point, then A is countable.
Proof. Let a be the limit point of A. Set
An =
½
1
x ∈ A : |x − a| ≥
n
¾
for each n ∈ N.
As An is a subset of A, a limit point of An must be a if it exists. But every
point of An has the distance to a at least 1/n. Thus a can not be a limit
point of An . The Bolzano - Weierstrass Theorem yields that the set An must
be finite otherwise An has to have a limit point. Now we have
A⊂
Ã
∞
[
An
n=1
!
∪ {a}.
Therefore, we conclude that A is contained in a countable union of finite
sets. Hence A is countable.
♥
1
2
MIDTERM EXAM II
Problem 2. Let {an } be the sequence defined by
an+1
a2n + 2
=
3
and
a1 = a ∈ R.
Determine the set of real numbers a ∈ R such that the sequence {an } converges and find the limit.
Answer. The first remark to make is that the sign of a is irrelevant, i.e.,
changing signe of a does not affect on the sequence after a2 . Namely, the
change from a to −a gives the same a2 and hence produces the same sequence
{an } for n ≥ 2. So we assume that a ≥ 0. Now let f be the function defined
by
x2 + 2
f(x) =
, x ∈ R.
(1)
3
Then we have
an+1 = f (an ) and
a1 = a.
(2)
In order to determine whether the sequence {an } is increasing or decreasing,
we compare the function f with the identity function x ∈ R 7→ x ∈ R:
f (x) − x =
x2 + 2
x2 − 3x + 2
(x − 1)(x − 2)
−x=
=
≤0
3
3
3
exactly when 1 ≤ x ≤ 2. Also if 1 ≤ x ≤ 2, then
x2 + 2
x2 − 1
(x − 1)(x + 1)
−1=
=
≥ 0;
3
3
3
x2 + 2
x2 − 4
(x − 2)(x + 2)
f(x) − 2 =
−2=
=
≤ 0,
3
3
3
f(x) − 1 =
i.e., 1 ≤ f (x) ≤ 2 if 1 ≤ x ≤ 2. Therefore if 1 ≤ a ≤ 2, then the sequence
{an } is non-increasing because an ≥ an+1 = f (an ) ∈ [1, 2] if an ∈ [1, 2] and
therefore the mathematical induction gives that if a1 = a ∈ [1, 2], then the
rest of the sequence stays in the interval [1, 2] and non-increasing. Hence if
1 ≤ a ≤ 2, then the sequence {an } converges.
Since f is a continuous function, if {an } converges to α, then we have
f(α) = lim f (an ) = lim an+1 = α.
n→∞
n→∞
Therefore, f (α) = α, i.e.,
α=
α2 + 2
.
3
MIDTERM EXAM II
3
Hence the limit α must be a solution of the equation:
α2 − 3α + 2 = 0,
hence α = 1 or 2.
(3)
So the candidates for the limit of {an } are limited to only two numbers 1
and 2.
If a = 2, then f (a) = f (2) = 2. Therefore we have
2 = a1 = a2 = a3 = · · · = an = · · · ,
so that
lim an = 2 if
n→∞
a = 2.
If 1 ≤ a < 2, then the sequence {an } is decreasing. Therefore the limit α is
less than or equal to a1 = a, i.e., the limit α cannot be 2. Thus α = 1 in this
case.
If 0 ≤ x < 1, then f (x) > x and 0 ≤ f (x) ≤ 1 as seen above. Hence if
0 ≤ a < 1, then 0 ≤ a2 < 1. If 0 ≤ an < 1 then 0 ≤ an < f(an ) = an+1 < 1.
The mathematical induction yields that the sequence {an } is increasing and
satisfies 0 ≤ an < 1 for all n. Thus the sequence {an } converges and the
limit α must be 1 in this case.
If x > 2, then f(x) > x as seen above and f(x) > 2. Hence the sequence
{an } is an increasing sequence by the mathematical induction. Since the
sequence {an } cannot converge to any number larger than 2, the sequence
{an } diverges to infinity.
Thus we summerize the consequence as follows:


 1 if |a| < 2;
lim an =
2 if |a| = 2;
n→∞


+∞ if |a| > 2.
♥
4
MIDTERM EXAM II
Problem 3. Let f be the function on the unit interval [0, 1] defined by the
following:
f(x) =
½
1
n
0
if x is rational and x =
if x is irrational.
m
n
in the reduced fraction;
a) Prove that f is discontinuous at every rational number in the unit
interval [0, 1].
b) Prove that f is continous at every irrational number in the unit interval [0, 1].
Prove. a) Fix a rational number x0 = p/q ∈ [0, 1] in the reduced form.
Choose ε = 1/(2q) > 0. Then no matter how small δ > 0 is we can find an
irrational number x ∈ (x0 − δ, x0 + δ) ∩ [0, 1] and therefore f (x) = 0 and
|f (x) − f (x0 )| = f(x0 ) =
1
1
>
= ε.
q
2q
Hence for this ε there is no δ > 0 such that |x − x0 | < δ and x ∈ [0, 1]
guarantee the inequality |f(x) − f (x0 )| < ε.
b) Fix an irrational number x0 ∈ [0, 1]. Take an arbitrary ε > 0. Choose
n ∈ N so large that 1/n < ε. Consider the set
An =
½
p
: p = 0, 1, · · · , q, q = 1, 2, · · · , n
q
¾
of all rational numbers p/q with q ≤ n appear in the set An . This set An is
a finite set and does not contain x0 as x0 is irrational. Hence
δ = min{|a − x0 | : a ∈ An } > 0
If x = p/q is a rational number in [0, 1] such that |x − x0 | < δ, then x is not
a member of An . Hence q > n, so that 1/q < 1/n < ε. Thus we have
¯ µ ¶
¯
¯
¯
¯f p − f(x0 )¯ = 1 < ε
¯
¯ q
q
p
for every
q
and consequently
|f(x) − f (x0 )| < ε
¯
¯
¯p
¯
¯
such that ¯ − x0 ¯¯ < δ
q
whenever |x − x0 | < δ.
Therefore the function f is continuous at irrational numbers in [0, 1]. The
graph of the function f looks like the following figure:
MIDTERM EXAM II
5
♥