Math 235: Assignment 1 Solutions
1.1
: For
n
∈
N
not zero, let
A
n
= [

n,n
] (The closed interval in
R
containing all
real numbers
x
satisfying

n
≤
x
≤
n
). It is easy to see that we have the chain of
inclusion
A
1
⊂
A
2
⊂
A
3
⊂
...
, and therefore
N
[
n
=1
A
n
=
A
N
= [

N,N
]
and
N
\
n
=1
A
n
=
A
1
= [

1
,
1]
.
1.2
: Notice that the consecutive closed intervals, [
n,n
+ 1] and [
n
+ 1
,n
+ 2] do
in fact overlap (This is because we have taken the
closed
interval. What if we had
taken the
open
intervals instead?) We conclude that the
ﬁnite
union is
N
[
n
=1
[
n,n
+ 1] = [1
,N
+ 1]
.
Since any real number
x
≥
1 will eventually lie inside the interval [1
,N
+ 1] for
some suﬃciently large
N
∈
N
, we conclude
∞
[
n
=1
[
n,n
+ 1] = [1
,
∞
)
.
(Note we have the open bracket ”)” on the right side of the above interval. Why?).
Similarly, the consecutive open intervals (
n,n
+ 2) and (
n
+ 1
,n
+ 3) also overlap,
so
N
[
n
=1
(
n,n
+ 2) = (1
,N
+ 2)
∴
∞
[
n
=1
(
n,n
+ 2) = (1
,
∞
)
.
(
∴
means ”therefore”)
1.3
: It is helpful to say in words what each
A
n
is.
A
1
is the set of all natural
numbers,
N
.
A
2
is the set of all
squares
of natural numbers.
A
3
is all
cubes
, etc. Of
course, a square of a natural number is in particular a natural number, so
A
2
⊆
A
1
.
Likewise for cubes, 4
th
powers, etc. We conclude
A
1
⊇
A
n
for all
n
∈
N
,
∴
∞
[
n
=1
A
n
=
A
1
=
N
.
Again it is helpful to ask in words, what does
T
∞
n
=1
A
n
mean? Since each
A
n
is the subset of natural numbers consisting of
n
th
powers, their intersection must
be all natural numbers
x
such that
x
is a square, a cube, a 4
th
power, a 5
th
, etc.
There are only 2 such numbers:
1