1. Exercise 1.4.5. Prove that T
∞ n =1
(0 , 1 /n ) = ∅ . Notice that this demonstrates that the intervals in the Nested Interval Property must be closed for the conclusion of the theorem to hold.
Proof.
Since the elements of (0 , 1 /n ) are positive for all n , any potential element of the set
T
∞
(0 , 1 /n ) must be positive. Now, suppose a > 0. Using the Archimedean Property, choose n =1
N ∈
N such that N > 1 /a . Then
1
< a,
N meaning that a / (0 , 1 /N ). Hence, ∈ T
∞ n =1
(0 , 1 /n ). However, since our choice of a > 0 was arbitrary, this implies that no positive real numbers can be in the set T
∞ n =1
(0 , 1 /n ). Since we already saw that only positive reals have a chance of being in this set, we can, therefore,
T
∞ conclude that the set n =1
(0 , 1 /n ) is empty.
2. Complete the proof (part of which was done in class) that
√ b ∈
R for every b ≥ 0.
Proof.
Recall that we defined show that α
2
A := { x ∈
R
: x
2
< b } and then
= b , and in class we already showed that α
2 ≥ b
α := sup A . Our goal is to
. Now, we intend to show that
α
2
> b is impossible.
To that end, suppose the Archimedean Property, we can choose
2 α
N
< α
2 − b .
α
2
> b . This implies α
2
N
− b > 0, so the real number
∈
N such that N >
α
2 α
2 − b
Then we see that α − 1
N
< α and
α
2 α
2 − b
> 0. Using or, equivalently,
α −
1
N
2
= α
2
−
2 α
N
+
1
N 2
> α
2
−
2 α
N
> α
2
− ( α
2
− b ) = b, so α − 1
N is a smaller upper bound for the set be a smaller upper bound for A .
A than α . However, α = sup A , so there cannot
From this contradiction, we conclude that α 2 then, is that α
2
= b , as desired.
cannot be > b . The only remaining possibility,
3. Exercise 1.4.10. Show that the set of all finite subsets of
N is a countable set.
Proof.
Let
F be the set of all finite subsets of
N
. There are a variety of ways of proving that
F is countable (including the inductive proof I suggested in the discussion sections). The simplest that I could come up with is the following. For any finite set S ⊆
N
, write
S = { s
1
, . . . , s n
} .
Then we can sum the elements of S (i.e. we can compute s
1 sum will also be finite.
+ . . .
+ s n
); since S is finite, this
This suggests the following partition of the set of all finite subsets of
N
: for each n ∈
N
, define
A n to be the set of finite subsets of
N whose elements add up to n . Each A n is certainly finite
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(although it’s not necessary, we can derive a crude upper bound for the size of A n as follows: if n = a
1
+ . . .
+ a k
, then each a i
≤ n and there can be no more than n of the a i
’s. Since there are only n + 1 natural numbers ≤ n , this means there are no more than ( n + 1) n ways of summing distinct natural numbers and getting n . Hence there are no more than ( n + 1) n elements of A n
).
Since each A n is finite, we can write it as
A n
= { S n 1
, S n 2
, . . . , S nj n
} .
Therefore, we can list all the finite subsets of
N
:
S
01
, . . . , S
0 j
0
, S
11
, . . . , S
1 j
1
, S
21
, . . . , S
2 j
2
, . . . .
Finally, define f :
N
→
F by setting f ( i ) = the ( i + 1)st element in the above list (I use i + 1 rather than i because we want f (0) to be the first element in the list). By construction this map f is bijective, so we conclude that
F is countable.
4. Exercise 1.4.12. A real number x ∈
R is called algebraic if there exist integers a
0
, a
1
, a
2
, . . . , a n
∈
N
, not all zero, such that a n x n
+ a n − 1 x n − 1
+ . . .
+ a
1 x + a
0
= 0 .
Said another way, a real number is algebraic if it is the root of a polynomial with integer coefficients. Real numbers that are not algebraic are called transcendental numbers. Reread the last paragraph of Section 1.1. The final question posed here is closely related to the question of whether or not transcendental numbers exist.
(a) Show that
√
2,
√
3 2, and
√
3 +
√
2 are algebraic.
Proof.
√
2 is one of the roots (along with −
√
2) of the polynomial x
Likewise,
√
3 2 is the unique real root of the polynomial x
3 − 2.
2 − 2.
Finally, notice that x −
√
2 +
√
3 x −
√
2 −
√
3 x −
√
3 −
√
2 x − −
√
2 −
√
3 = x
4
− 10 x
2
+1 and that
√
2 +
√
3 is clearly a root of this polynomial.
(b) Fix n ∈
N
, and let A n be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree n . Using the fact that every polynomial has a finite number of roots, show that A n is countable.
Proof.
To prove that A n is countable, we will make use of the following lemma:
Lemma 0.1.
Let P n
P n is countable.
be the set of polynomials of degree n with integer coefficients. Then
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Proof of Lemma.
Define the set Q m by
Q m
:= { a n x n
+ . . .
+ a
1 x + a
0
: | a n
| + . . .
+ | a
1
| + | a
0
| = m } .
In other words, Q m is the set of all polynomials of degree absolute value) to m . If a n x n
+ . . .
+ a
1 x + a
0 n is an element of whose coefficients sum (in
Q m
, then each coefficient a i must satisfy | a i
| ≤ m . In other words, it must be the case that
− m ≤ a i
≤ m.
Hence, there are only 2 m + 1 possibilities for what a i could be ( m positive possibilities, m negative possibilities, plus 0). In turn, since there are only n + 1 coefficients, this implies that the set Q m contains at most ( n + 1)
2 m +1 elements. In particular, this means that Q m is finite for every m ∈
N
.
Since each Q m is finite, we can write each Q m as the set
Q m
= { q m, 1
( x ) , q m, 2
( x ) , . . . , q m,j m
( x ) } where each q m,i
( x ) is a polynomial of degree n .
Now, starting with Q
1
, line up the Q m
’s back-to-back:
{ q
1 , 1
( x ) , . . . , q
1 ,j
1
( x ) } , { q
2 , 1
( x ) , . . . , q
2 ,j
2
( x ) } , { q
3 , 1
( x ) , . . . , q
3 ,j
3
( x ) } , . . .
Thus, we can list the elements of P n
: q
1 , 1
( x ) , . . . , q
1 ,j
1
( x ) , q
2 , 1
( x ) , . . . , q
2 ,j
2
( x ) , q
3 , 1
( x ) , . . . , q
3 ,j
3
( x ) , . . . .
This implicitly defines a map f :
N
→ P n
, where f ( k ) is just the ( k + 1)st element in the above listing (again, we use k + 1 instead of k so that f (0) is the first term in the list). Since each polynomial of degree n appears in exactly one of the Q m
’s, this map is certainly both injective and surjective. Hence, f :
N
→ P n is bijective, so P n is countable, as desired.
With the lemma in hand, the rest of the proof follows along similar lines. Since P n is countable, we can list all polynomials of degree n as
P n
= { p
1
( x ) , p
2
( x ) , . . .
} .
Now, for each i , p i
( x ) has countably many roots; call them r i, 1
, . . . , r i,` i
. Hence, we can list all possible roots of polynomials of degree n : r
1 , 1
, . . . , r
1 ,`
1
, r
2 , 1
, . . . , r
2 ,`
2
, . . . .
Again, this implicitly defines a map from
N to A n by sending k to the ( k + 1)st term in the above list.
The slight problem is that this map may not be bijective, since numbers can be roots of more than one polynomial. However, note that the above map is certainly surjective, so there are no more elements of A n than of
N
. On the other hand, the set infinite (for example, it contains all the numbers 1 , n 2 ,
√ n
A n is definitely
3 , . . .
, which are roots of the polynomials x n − 1, x n − 2, x n − 3 , . . .
). An infinite set which is no bigger than
N is necessarily countable, so we conclude that A n is countable.
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(c) Now, argue that the set of all algebraic numbers is countable. What may we conclude about the set of transcendental numbers?
Proof.
Let
A be the set of all algebraic numbers. Then
A
=
∞
[
A n
.
n =1
We showed in (b) that each A n is countable. Therefore, by Theorem 1.4.13(ii), the above union of countable sets is countable, meaning that
A is countable. Hence, there are countably many algebraic numbers. Since we know there are uncountably many real numbers, this implies that there are uncountably many transcendental numbers.
5. Exercise 1.5.4. Let S be the set consisting of all sequences of 0’s and 1’s. Observe that S is not a particular sequence, but rather a large set whose elements are sequences; namely
S = { ( a
1
, a
2
, a
3
, . . .
) : a n
= 0 or 1 } .
As an example, the sequence (1 , 0 , 1 , 0 , 1 , 0 , 1 , 0 , . . .
) is an element of S , as is the sequence
(1 , 1 , 1 , 1 , 1 , 1 , . . .
).
Give a rigorous argument showing that S is uncountable.
Proof.
By contradiction.
S is clearly infinite, so we want to rule out the possibility that S is countable. To that end, suppose that S is countable, meaning that there exists a bijective map f :
N
→ S . For notational ease, I will assume, for the sake of this problem, that
N
= { 1 , 2 , 3 , . . .
} (as usual, we can accommodate 0 with an index shift). Let us list the elements f ( n ) for n ∈
N
: f (1) = ( a
11
, a
12
, a
13
, a
14
, a
15
, a
16
, . . .
) f (2) = ( a
21
, a
22
, a
23
, a
24
, a
25
, a
26
, . . .
) f (3) = ( a
31
, a
32
, a
33
, a
34
, a
35
, a
36
, . . .
) f (4) = ( a
41
, a
42
, a
43
, a
44
, a
45
, a
46
, . . .
)
.
..
where each a ij is either 0 or 1. Now, define a new sequence
( b
1
, b
2
, b
3
, b
3
, b
5
, b
6
, . . .
) where b i
:=
(
0 if a ii
= 1
1 if a ii
= 0
.
Note that ( b n
) ∈ S . For each i ∈
N
, b i
= a ii
, so the i th term of the sequence ( b n
) is different from the i th term of the sequence f ( i ). Therefore,
( b n
) = f ( i ) for any i ∈
N
.
This contradicts the assumption that f :
N
→ S is bijective (and, in particular, surjective), so we see that no such f can exist, and so the set S must be uncountable.
4