MATH 271
Homework 1
Solutions
1. Let X be a set, B ⊆ X, I an index set and Ai ⊆ X for all i ∈ I. Prove that
"
#
\
\
B∪
Ai =
(B ∪ Ai )
i∈I
"
Proof : Pick x ∈ B ∪
i∈I
#
\
Ai . Then x ∈ B or x ∈
i∈I
i ∈ I and so x ∈
\
\
Ai . Suppose first that x ∈ B. Then x ∈ B ∪ Ai for all
i∈I
(B ∪ Ai ). Suppose next that x ∈
i∈I
i ∈ I and so x ∈
\
\
Ai . Then x ∈ Ai for all i ∈ I. Hence x ∈ B ∪ Ai for all
i∈I
(B ∪ Ai ).
i∈I
"
Pick x ∈
#
\
\
i∈I
i∈I
(B ∪ Ai ). Then x ∈ B ∪ Ai for all i ∈ I. If x ∈ B, then x ∈ B ∪
"
#
\
\
Then x ∈ Ai for all i ∈ I. Hence x ∈
Ai and so x ∈ B ∪
Ai .
i∈I
Ai . So assume that x ∈
/ B.
2
i∈I
2. Let X be a set, I an index set and Ai ⊆ X for all i ∈ I. Prove that
\
^ [f
Ai =
Ai
i∈I
i∈I
Proof : Let x ∈ X. Then we have
x∈
\
^
Ai
⇐⇒ x ∈
/
i∈I
\
Ai
i∈I
!
⇐⇒ ¬ x ∈
\
A
i∈I
⇐⇒ ¬ (∀i ∈ I : x ∈ Ai )
⇐⇒ ∃i ∈ I : x ∈
/ Ai
fi
⇐⇒ ∃i ∈ I : x ∈ A
[
fi
⇐⇒ x ∈
A
i∈I
\
^ [f
Hence
Ai =
Ai .
i∈I
2
i∈I
−1 [B].
e = f^
3. Let f : X → Y and B ⊆ Y . Prove that f −1 [B]
1
Proof : Let x ∈ X. Then we have
e
x ∈ f −1 [B]
e
⇐⇒ f (x) ∈ B
⇐⇒ f (x) ∈
/B
⇐⇒ ¬(f (x) ∈ B)
⇐⇒ ¬(x ∈ f −1 [B])
⇐⇒ x ∈
/ f −1 [B]
−1 [B]
⇐⇒ x ∈ f^
2
−1 [B].
e = f^
Hence f −1 [B]
4. We define the symmetric difference of two sets Ω and Σ as
Ω∆Σ = (Ω \ Σ) ∪ (Σ \ Ω)
Let A, B, C ⊆ X. Prove that
(A∆B) ∩ C = (A ∩ C)∆(B ∩ C)
e Hence we get
Proof : Recall that Ω \ Σ = Ω ∩ Σ.
(A∆B) ∩ C
= [(A \ B) ∪ (B \ A)] ∩ C
e ∪ (B ∩ A)]
e ∩C
= [(A ∩ B)
e
e ∩ C]
= [(A ∩ B) ∩ C] ∪ [(B ∩ A)
e
e
= (A ∩ B ∩ C) ∪ (B ∩ A ∩ C)
definition of ∆
e
Ω\Σ=Ω∩Σ
Distributive Law
Notice that
(A ∩ C) \ (B ∩ C)
^
= (A ∩ C) ∩ B
∩C
e
e
= (A ∩ C) ∩ (B ∪ C)
e
e
= [(A ∩ C) ∩ B] ∪ [(A ∩ C) ∩ C]
e
= A∩C ∩B
e
Ω\Σ=Ω∩Σ
De Morgan’s Law
Distributive Law
e=∅
C ∩C
Similarly, we get that
e
(B ∩ C) \ (A ∩ C) = B ∩ C ∩ A
Hence it follows from the definition of ∆ that
e ∪ (B ∩ C ∩ A)
e = (A∆B) ∩ C 2
(A ∩ C)∆(B ∩ C) = [(A ∩ C) \ (B ∩ C)] ∪ [(B ∩ C) \ (A ∩ C)] = (A ∩ C ∩ B)
5. (a) Let X be a set and f : X → P(X) a function. Prove that f is not onto.
(b) Prove that P(N) is uncountable.
Proof : (a) Suppose that f is onto. Put S = {x ∈ X | x ∈
/ f (x)}. Pick a ∈ X with f (a) = S. If a ∈ S, then by
definition of S, a ∈
/ f (a) = S, a contradiction. So a ∈
/ S = f (a). Again by definition of S, we get that a ∈ S, a
contradiction.
Hence f is not onto.
2
(b) Suppose that P(N) is countable. Then there exists an onto map θ : N → P(N) by Lemma 1.2, a contradiction
to (a).
2
Hence P(N) is uncountable.
6. Let S be a countable set. Prove that the set of all finite sequences with elements from S, is countable.
Proof : For k ∈ N, we get that
Sk = S × S × · · · × S
|
{z
}
k times
is the set of all finite sequences with k elements from S. Since S is countable, we have seen in class that S k is
countable (Proposition 1.5).
So the set of all finite sequences with elements from S is
+∞
[
Sk
k=1
which is countable since the countable union of countable sets is countable (Corollary 1.7).
2
7. Prove that the set S = {[a, b] : a, b ∈ Q and a < b} is countable (so S is the set of all closed intervals with
rational numbers as endpoints).
Proof : Note that the map
θ : S → Q × Q : [a, b] → (a, b)
is one-to-one. Since Q is countable, we get that Q × Q is infinite and countable by Proposition 1.5. So S is
countable by Lemma 1.3.
2
8. Let E be the set of all infinite sequences with elements from {0, 1}.
(a) Let f : N → E be a function. For all k ≥ 1, we put f (k) = {akn }n≥1 . Show that {1 − ann }n≥1 ∈
/ f (N).
(b) Show that E is not countable.
Proof : (a) Note that for all k ∈ N, we have that the k-th term of f (k) is akk while the k-term of {1 − ann }n≥1
is 1 − akk . So {1 − ann }n≥1 6= f (k) for all k ∈ N. Hence {1 − ann }n≥1 ∈
/ f (N).
(b) Suppose that E is countable. Then there exists an onto map f : N → E by Lemma 1.2, a contradiction to
(a). Hence E is not countable.
2
9. Let P be a subset of the field Q such that
B1 ∀x, y ∈ P : x + y ∈ P
B2 ∀x, y ∈ P : x · y ∈ P
B3 ∀x ∈ P : −x ∈
/P
B4 ∀x ∈ Q : x = 0 or x ∈ P or − x ∈ P
3
Prove that P =
nm
n
o
: m, n ∈ N .
/ P . Then −1 ∈ P by B4. Hence −(−1) = 1 = (−1) · (−1) ∈ P by B2, a contradiction
Proof : Suppose that 1 ∈
to B3. So 1 ∈ P .
Let n ∈ N. Since
n = 1 + ··· + 1
| {z }
n times
it follows from B1 that n ∈ P . If
to B3. Hence
1
1
1
∈
/ P then − ∈ P by B4 and so −1 = n · −
∈ P by B2, a contradiction
n
n
n
1
∈ P.
n
Let m, n ∈ N. Then m ∈ P and
m
1
m
1
∈ P . So
= m · ∈ P by B2. It follows from B3 that − ∈
/ P.
n
n
n
n
Since −0 = 0, it follows from B3 that 0 ∈
/ P . Since
n m
o
nm
o
Q = − : m, n ∈ N ∪ {0} ∪
: m, n ∈ N
n
n
nm
o
we get that P =
: m, n ∈ N .
n
4
2