Download hw4

T.A. Tai Melcher
Office AP&M 6402E
1 1
Problem 2.55:
For
each
natural
number
n,
let
A
=
− n , n and
n
1 1
Bn = − n , n . Find ∩n∈N An and ∩n∈N Bn .
\
\ 1 1
= {0}
An =
− ,
n
n
n∈N
n∈N
Math 109
UCSD Fall 2003
and
Homework 4
\ 1 1
Bn =
= {0}.
− ,
n n
n∈N
n∈N
\
Problem 2.63: Give an example of a collection of sets indexed by a
set consisting of 5 members.
There are infinite such examples, but one collection that would work
would be
A = {(n − 1, n + 1) : n ∈ N, n ≤ 5},
or rather A = {An : n ∈ Λ}, where
An = (n − 1, n + 1) and Λ = {n ∈ N : n ≤ 5}.
Problem 2.64: Give an example of a collection of sets indexed by the
set of prime numbers.
Again, an infinite number of solutions exist, but one example would be
A = {{p − 5, p, p + 2} : p ∈ N, p prime},
or, if you like, A = {Ap : p ∈ Λ}, where
Ap = {p − 5, p, p + 2} and Λ = {p ∈ N : p prime}.
Problem 2.65: Prove Theorem 2.9(c): Let {Aα : α ∈ Λ} be an indexed
family of sets. Then
\
0 [
{Aα : α ∈ Λ} = {A0α : α ∈ Λ}.
Proof. To show the equality of two sets, one must show that each is
a subset of the other. Here, I will show
\
0 [
{Aα : α ∈ Λ} ⊂ {A0α : α ∈ Λ},
and leave the other containment argument to you – it is basically the
same.
So let x ∈ (∩{Aα : α ∈ Λ})0 . By definition of the complement then,
x∈
/ ∩{Aα : α ∈ Λ}. Thus, there exists some α0 ∈ Λ such that x ∈
/ Aα 0 .
0
0
Thus, x ∈ (Aα0 ) , and so x ∈ ∪{Aα : α ∈ Λ} as desired.
1
2
Problem 2.70: Let {Aα : α ∈ Λ} be an indexed family of sets. Prove
that
[
[
\
(Aα − Aβ ) ⊂
Aα −
Aα .
α∈Λ
α,β∈Λ,α6=β
α∈Λ
Proof. Let x ∈ ∪α,β∈Λ,α6=β (Aα −Aβ ). Then there exist some α0 , β 0 ∈ Λ
such that α0 6= β 0 and x ∈ Aα0 − Aβ 0 . By definition of the difference of
sets, x ∈ Aα0 and x ∈
/ Aβ 0 . x ∈ Aα0 implies that x ∈ ∪α∈Λ Aα . x ∈
/ Aβ 0
implies that x ∈
/ ∩α∈Λ Aα . So
[
\
x∈
Aα −
Aα .
α∈Λ
α∈Λ
Since x was an arbitrary element of ∪α,β∈Λ,α6=β (Aα − Aβ ), this implies
that
[
[
\
(Aα − Aβ ) ⊂
Aα −
Aα .
α,β∈Λ,α6=β
α∈Λ
α∈Λ