Download 110.405 Analysis HW 2 Solutions Section 2.1.3 Problem 1: We can

110.405 Analysis
HW 2
Solutions
Section 2.1.3
Problem 1: We can show this by establishing an injection from the power set of the natural
numbers into the set of Cauchy sequences converging to a given number:
Let x be a real number, we fix a Cauchy sequence {xn } converging to x. Now given a set of
natural numbers N , define {yn } as follows:
xn
if n 6∈ N
yn =
otherwise
xn + n1
If N1 and N2 two different sets then there is n ∈ N1 and n 6∈ N2 Hence, in the first case yn = xn
while in the second case yn = xn + n1 and we see that two different sets of natural numbers map to
different sequences, so the inclusion is injective.
Now, we need to show that for each set N we obtain a Cauchy sequence: Note that |xn −yn | ≤ n1 for
1
1
when j > M 0 , and |xj −xk | < 3m
each n. Now, given m there exists M1 , M2 such that |xj −yj | < 3m
for j, k > M2 . Let M be max{M1 , M2 }. Then, for any j.k > M , we have the following:
|yj − yk | ≤ |yj − xj | + |xj − xk | + |xk − yk |
1
1
1
<
+
+
3m 3m 3m
1
<
m
Since we can find such an M for any value of m we see that {yn } is Cauchy.
Problem 4:
If the shuffled sequence is Cauchy, then the individual sequences are equivalent Cauchy sequences:
For the shuffled sequence {zn }, for any m, take an M such that, in the shuffled sequence, |zj −zk | <
1
.
m By the properties of the shuffling, we know that any term xk appears later in the shuffled sequence
than in the individual sequences. For all j, k > M , the following are true:
|xk − yk | = |z2k−1 − z2k |
1
<
m
|xj − xk | = |z2j−1 − z2k−1 |
1
<
m
|yj − yk | = |z2j − z2k |
1
<
m
Therefore, the individual sequences are equivalent Cauchy sequences.
If the individual sequences are equivalent Cauchy sequences, then the shuffled sequence is Cauchy:
For any m, take M1 , M2 ,, and M3 so that the following holds:
1
2
1
2m
1
|xj − xk | <
2m
1
|yj − yk | <
2m
when j and k are greater than M1 , M2 , and M3 respectively. Let M be equal to max{M1 , M2 , M3 }.
Then, we see that for N > 2M , the following are true for all j, k > N :
When j and k are both even:
|xj − yj | <
|zj − zk | = y j − y k 2
2
1
<
2m
When j and k are both odd:
|zj − zk | = x j+1 − x k+1 2
2
1
<
2m
When j is even and k is odd (The case for odd j and even k can be handled in the same way)
|zj − zk | = y j − x k+1 2
2
≤ y j − x j + x j − x k+1 2
2
2
1
<
m
These are true since j and k + 1 are greater than N , so
Therefore, this sequence is Cauchy.
Problem 8:
Yes, take {xn } and {yn } such that xn =
satisfy the desired conditions.
1
n
2
j
2
and
k+1
2
are both greater than M .
and yn = − n1 It is easy to see that these sequences
Section 2.2.4
Problem 2:
We can establish this by constructing an injection f from the real numbers into a power set of
the rationals (since the rationals are countable, we know that the cardinality of the power set of
the rationals is the same as that of the power set of the integers as well as the power set of natural
numbers), then constructing an injection g from a power set of the natural numbers into the real
numbers.
For the first direction, define
f (x) = {r ∈ Q : |r| < |x|}
We can show that this injective by considering two distinct real numbers, x1 and x2 . Without loss
of generality, assume 0 < x1 < x2 . Then, there exists a rational number y so that x1 < y < x2 .
Therefore, y is contained in f (x2 ) but not f (x1 ). Therefore, x1 and x2 map to different sets, so the
map is injective.
For the second direction, we map a set S of natural numbers into the real numbers as follows:
3
g(S) =
X 1
4k
k∈S
(or, more rigorously, into the (increasing) Cauchy sequence of partial sums) Then, we can show
that this is injective: If two power sets S1 and S2 are different, let k be the lowest element at which
they differ (i.e., if S1 denotes the natural numbers, and S2 is the set {1, 2, 3, 5, 6, 7, . . .}, k would be
5. Without loss of generality, assume S1 doesn’t contain k and S2 does. Then, take the number


X
1
+ 2 .

n
4
4k+1
n∈S1 ,n<k
We see that this is greater than g(S1 ), less than g(S2 ), and its difference from g(S2 ) is bounded
2
. Therefore, g(S1 ) 6= g(S2 ), so g is injective.
below by 4k+1
Therefore, the two sets have the same cardinality, so, in particular, the set of real numbers is
uncountable.
Problem 6:
√
For a real number x the construction of x was √
given in class. This was a Cauchy sequence of
rational numbers rn converging to the real number x. Take pqnn = rn .