Math 171: Problem Set 2
Caitlin Stanton
October 7, 2014
Problem 1
We first show that this limit exists. We have that {an,n } is bounded above by sup A, and see that
if n > m, then an,n ≥ am,m , so this sequence is increasing. Thus the limit exists.
Since an,n ≤ sup A for all n, we have that limn→∞ an,n ≤ sup A. Suppose this inequality were strict;
in other words, that limn→∞ an,n < sup A. Then there are some m1 , m2 > 0 such that limn→∞ an,n <
am1 ,m2 ≤ sup A. Let N = max{m1 , m2 }. Then we have that aN,N ≥ am1 ,m2 , and since {an,n } is
increasing it follows that limn→∞ an,n ≥ am1 ,m2 . This is a contradiction, so we must have that
limn→∞ an,n = sup A.
Problem 2
xn
Suppose ∑∞
n=1 1+xn converged.
x
We first claim that the function f ∶ R≥0 → R is increasing, where f (x) ∶= 1+x
. If 0 ≤ x < y, then
x + xy < y + xy, and thus x(y + 1) < y(x + 1). Since x + 1 and y + 1 are each strictly greater than 0,
y
x
we have that x+1
< 1+y
.
We next claim that limn→∞ xn = 0. Suppose for sake of contradiction that this limit did not
equal zero. Then there is some ε > 0 such that for all N > 0, there is some n > N such that xn > ε.
Let {xnj } describe the (necessarily infinite) subsequence of terms for which xn > ε. Then we have:
∞
x nj
xn
≥∑
n=0 1 + xn
j=0 1 + xnj
∞
∑
ε
,
j=0 1 + ε
∞
≥∑
this last inequality occuring because we showed that f was increasing. But this series on the right
does not converge, so we have reached a contradiction. Thus limn→∞ xn = 0, and in particular, this
sequence is bounded. Let N denote a bound.
1
1
> 1+y
. Since N is a bound for {xn }, we have that
If x, y > 0, then x < y if and only if 1+x
1
1
> 1+N for all n. Thus we have:
1+xn
∞
xn
xn
≥∑
1
+
x
1
n
n=0
n=0 + N
∞
∑
≥
∞
1
∑ xn .
1 + N n=1
But this series does not converge, and thus we have proved that ∑∞
n=0
1
xn
1+xn
diverges.
Problem 3
Fix ε > 0. Since the sequence of partial sums sn = ∑nj=1 aj is convergent it is also Cauchy, so choose
N > 0 such that ∣sn − sm ∣ < ε/2 for all n, m > N . Now consider the difference ∣s2n − sn ∣ for n > N .
This value, by selection of N , is less than ε/2. But we also have:
RRR 2n
RRR
n
∣s2n − sn ∣ = RRRRR ∑ aj − ∑ aj RRRRR
RRRj=1
RRR
j=1
2n
= ∑ aj .
j=n+1
There are n terms in this summation and each is ≥ a2n , since this sequence is decreasing. Thus we
have that ε/2 > ∣s2n−1 − sn−1 ∣ ≥ na2n , and therefore 2na2n < ε.
We have shown that limn→∞ 2na2n = 0. We need to show that nothing goes wrong when we
introduce the odd terms.
Fix ε > 0. Then there is some N such that for all n > N , 2na2n < ε/2. We see that:
(2n + 1)a2n+1 ≤ (2n + 1)a2n
ε 2n + 1
)
< ( )(
2
2n
≤ ε.
Therefore we have that limn→∞ nan = 0.
s
s
Suppose that for some 0 ≤ s ≤ 1, the series ∑∞
n=1 1/n converged. We see that {1/n } is decreasing,
s
s
1−s
and thus by what we just proved we have that limn→∞ n/n = 0. But n/n = n
≥ 1 for all n. Thus
s
we have reached a contradiction, and must have had that ∑∞
n=1 1/n diverged.
Problem 4
Suppose first that {an } is unbounded. Suppose further that is has no upper bound. We construct
a subsequence inductively as follows. Let n1 = 1. Suppose we have chosen n1 < n2 < . . . < nj such
∞
that an1 ≤ . . . ≤ anj . Then since {an }∞
n=1 is unbounded, so is the sequence {an }n=nj +1 . Therefore we
can choose some nj+1 ≥ nj + 1 such that anj+1 > aj . By construction, we have that {anj }∞
j=1 is an
increasing subsequence of {an }.
If {an } is unbounded below, then we have that {−an } is unbounded above, and thus there is
∞
some increasing subsequence {−anj }∞
j=1 . Therefore {anj }j=1 would be a decreasing subsequence of
our original sequence.
Now let that {an } be bounded. Suppose that {an } contains no increasing subsequence (otherwise,
we are done). Define:
bn = sup aj .
j≥n
Since {an } is bounded, bn is well-defined.
We claim that bn = aj for some j ≥ n. Suppose for sake of contradiction that this weren’t true.
Then we construct an increasing subsequence of {an } as follows. Let nj ≥ n be the smallest index
for which ∣anj − bn ∣ < 1/2j . Then we have that n1 ≤ n2 ≤ . . ., and that this sequence of integers is
not bounded, since otherwise we would have the final nj appearing satisfying ∣anj − bn ∣ < 1/2m for
all m ≥ j, and thus anj = bn , which is a contradiction.
We now claim that the sequence {anj }∞
j=1 is increasing. Suppose anj > anj+1 . By construction,
we know that ∣anj+1 − bn ∣ < 1/2j+1 . But this is less than 1/2j , and thus nj+1 is a strictly smaller
2
index such that ∣anj+1 − bn ∣ < 1/2j , which contradicts our choice of nj . Therefore this sequence is
increasing.
We have a small problem to fix before we can complete the proof of this claim. We may have
nj = nj+1 for some j, in which case we cannot simply call {anj } a subsequence of {an }. We remedy
this by just removing redundant indices, and realizing that we already showed that the indices
appearing were not bounded, so we do in fact obtain a subsequence. Thus we have contstructed
an increasing subsequence of {an }, which is a contradiction. So we must have had that for each n,
bn = am for some m.
Now we inductively construct a decreasing sequence as follows. Let n1 be the smallest index
such that b1 = an1 . For the inductive defintion, we let nj+1 ≥ nj + 1 be the smallest index such
that bnj +1 = anj+1 . Then we have that n1 < n2 < . . ., and since the bn are decreasing, we have that
{anj }∞
j=1 is also decreasing.
Problem 5
Let {an } be a bounded sequence, and assume that every monotone subsequence has limit L. Suppose
for sake of contradiction that limn→∞ an ≠ L. Therefore there exists some ε > 0 such that for all
N > 0, there is some n > N such that ∣an − L∣ ≥ ε.
We define the subsequence {anj } to contain all the indices such that ∣anj − L∣ ≥ ε. By our
selection of ε, the set of such indices is unbounded, so this really is a subsequence of {an }. By the
previous problem, there exists a monotone subsequence of this subsequence, so assume that {anj }
is monotone (by potentially passing to a subsequence). Then since this sequence is bounded, it has
a limit. By assumption it must have limit L, and thus there is some N > 0 such that for all nj > N ,
∣anj − L∣ < ε. But this contradicts our choice of nj , and so we must have had that limn→∞ an = L.
Problem 6
Let {xn } be a bounded sequence. Let A = lim supn→∞ {xn } as defined in the text, in other words:
A = lub{a ∶ a is a limit of some subsequence of {xn }}.
Let B = lim supn→∞ xn as defined in the Preliminary Notes, in other words:
B = lim bn
n→∞
where bn = supj≥n {xj }.
We first show that A ≥ B. Define an increasing sequence of positive integers n1 < n2 < . . .
inductively as follows. Let n1 = 1. Now for the inductive step, suppose we have already chosen
n1 < . . . < nj . We choose nj+1 to be some integer ≥ nj + 1 such that ∣bnj +1 − xnj+1 ∣ < 1/2j+1 . We can
see that {xnj }∞
j=1 converges to B as follows. Choose any ε > 0. Then there exists some N > 0 such
that ∣bn − B∣ < ε/3 for all n > N . Since {bn } is convergent then it is also Cauchy, so assume that
we chose N large enough so that ∣bn − bm ∣ < ε/3 for all n, m > N . We also choose M > 0 such that
1/2M < ε/3. Then for all j > max{N, M } we have:
∣B − xnj ∣ ≤ ∣B − bj ∣ + ∣bj − bnj−1 +1 ∣ + ∣bnj−1 +1 − xnj ∣
< ε,
(using the fact that nj ≥ j for all j). Thus limj→∞ xnj = B, we have shown that A ≥ B.
Now we show that A ≤ B. Let {xnj }∞
j=1 be a convergent subsequence of {xn }. We know that
xnj ≤ bnj for all j, and thus limj→∞ xnj ≤ limj→∞ bnj = B. Thus we have that A ≤ B.
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Problem 7
Define a f ∶ R → R as:
⎧
⎪
⎪0 if x rational
f (x) = ⎨
⎪
1 if x irrational.
⎪
⎩
Fix a ∈ R and suppose that limx→a f (x) = L for some L ∈ R. Since the rationals are dense in R,
there exists a strictly increasing sequence of rationals {an } converging to a. Fix ε > 0. Then there
exists some δ > 0 such that for all x ∈ R with 0 < ∣b − a∣ < δ, we have that ∣L − x∣ < ε.
Since {an } converges to a, there is some M > 0 such that for all m > M , ∣am − a∣ < δ. We also
notice that since {an } is strictly increasing, we never have that ∣am − a∣ = 0. Thus we have that for
all m > M , ∣L − f (am )∣ < ε. But by selection of {an } we have that f (an ) = 0, and therefore we have
shown that L = 0.
An almost identical argument (by instead choosing a strictly increasing sequence of irrational
numbers converging to a) shows that L = 1. This is a contradiction, so limx→a f (x) cannot exist.
Problem 8
Define the function f on [0, 1] as:
⎧
⎪
⎪x
f (x) = ⎨
⎪
0
⎪
⎩
if x rational
if x irrational.
First we prove that f is continuous at 0. Fix some ε > 0, and suppose we have some y ∈ [0, 1]
with 0 < ∣y∣ < ε. If y is rational, then we have that ∣f (y)∣ = ∣y∣ < ε. If y is irrational, then we have
∣f (y)∣ = 0 < ε. Thus we have shown that limx→0 f (x) = 0.
Now choose some non-zero a ∈ [0, 1]. Clearly a is an accumulation point of [0, 1], so assume for
sake of contradiction that limx→a f (x) = f (a). Let ε = a/2. Then there exists some δ > 0 such that
for any x ∈ [0, 1] with 0 < ∣x − a∣ < δ, we have that ∣f (x) − f (a)∣ < a/2.
If a is rational, choose some irrational number x0 ∈ [0, 1] with 0 < ∣x0 − a∣ < δ. Therefore we
must have that ∣f (x0 ) − f (a)∣ < a/2. But in this case we have that f (a) = a and f (x0 ) = 0, and thus
∣f (x0 ) − f (a)∣ = a > a/2, which is a contradiction.
We can make a very similar argument to obtain a contradiction when a is irrational. Choose
some rational x1 ∈ [0, 1] with 0 < ∣x0 − a∣ < min{δ, a/2}. Then we have that ∣f (x1 ) − f (a)∣ < a/2. But
the lefthand side is just equal to ∣f (x1 )∣ = ∣x1 ∣ > a/2, so we have reached a contradiction.
Therefore we have shown that f is only continuous at 0.
Problem 9
Let d be a metric on a set M Fix x, y, z ∈ M . Assume first that d(x, z) ≥ d(y, z). We know by the
definition of a metric space that d(x, z) ≤ d(x, y) + d(y, z), and thus d(x, z) − d(y, z) ≤ d(x, y). Since
the lefthand side is non-negative, we have shown that ∣d(x, z) − d(y, z)∣ ≤ d(x, y).
Now suppose that d(x, z) < d(y, z). By exchanging the roles of x and y and applying our previous
result, we have that ∣d(y, z) − d(x, z)∣ ≤ d(y, x). The lefthand side equals ∣d(x, z) − d(y, z)∣ and the
righthand side equals d(x, y) (by symmetry of d), and so we are done.
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Problem 10
a)
Suppose {an } ∈ `1 . Then we have that ∑∞
n=1 ∣an ∣ converges to some L. In other words the partial
sums of the series, sn = ∑nj=1 ∣aj ∣, converge to L, and thus this sequence is also Cauchy. Fix some
ε > 0. Then there is some N > 0 such that for all n, m > N , we have that ∣sn − sm ∣ < ε. In particular,
we have that for all n > N + 1, ∣sn − sn−1 ∣ < ε. But ∣sn − sn−1 ∣ = ∣an ∣, and thus we have shown that
{an } ∈ c0 .
Now for the second containment. Suppose that {an } ∈ c0 . By Theorem 13.2 in the textbook,
since {an } is convergent it must also be bounded. Thus we have that {an } ∈ `∞ .
b)
We start by showing that this first containment is proper. Consider the sequence {1/n}. This
sequence converges to 0 and thus it is in c0 , but since ∑∞
/ `1 .
n=1 1/n diverges, we have that {1/n} ∈
To show that the second containment is proper, consider the sequence an = 1 for all n. This
sequence is bounded by 1, but clearly does not converge to 0.
c)
First we make sure that d′ maps to [0, ∞). Let {an }, {bn } ∈ `∞ with bounds A, B > 0 respectively.
Then we have that for any n > 0, ∣an − bn ∣ ≤ ∣an ∣ + ∣bn ∣ ≤ A + B, and thus lub{∣an − bn ∣ ∶ n ∈ P} ≤
A + B. Furthermore every element of this set is non-negative, and thus d′ ({an }, {bn }) ≥ 0 for all
{an }, {bn } ∈ `∞ .
We verify the three conditions for d′ to describe a metric on `∞ . For what follows, let {an }, {bn }, {cn } ∈
∞
` .
i)
Suppose we have that d′ ({an }, {bn }) = 0. This means that 0 ≥ ∣an − bn ∣ for each n, and thus
{an } = {bn }.
Conversely, we see that d′ ({an }, {an }) = lub{∣an − an ∣ ∶ n ∈ P} = 0.
ii)
d′ ({an }, {bn }) = lub{∣an − bn ∣ ∶ n ∈ P}
= lub{∣bn − an ∣ ∶ n ∈ P}
= d′ ({bn }, {an })
iii)
d′ ({an }, {bn }) = lub{∣an − cn ∣ ∶ n ∈ P}
≤ lub{∣an − bn ∣ + ∣bn − cn ∣ ∶ n ∈ P}
≤ lub{∣an − bn ∣ + ∣bm − cm ∣ ∶ n, m ∈ P}
≤ lub{∣an − bn ∣ ∶ n ∈ P} + lub{∣bn − cn ∣ ∶ n ∈ P}
≤ d′ ({an }, {bn }) + d′ ({bn }, {cn }).
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d)
For {an }, {bn } ∈ `0 ⊂ `∞ , define d({an }, {bn }) = ∑∞
n=1 ∣an − bn ∣. Suppose we had that d({an }, {bn }) <
d′ ({an }, {bn }). Then there would be some M > 0 such that ∣aM − bM ∣ > d({an }, {bn }). Since
the partial sums sn = ∑nj=1 ∣aj − bj ∣ are increasing and converge to d({an }, {bn }), we have that
∣aM − bM ∣ > sn for every n. In particular this holds for n = M , in which case we have:
M
∣aM − bM ∣ > ∑ ∣aj − bj ∣,
j=1
−1
′
and thus ∑M
j=1 ∣aj −bj ∣ < 0. But this is impossible, so we must have that d({an }, {bn }) ≥ d ({an }, {bn }).
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