Serge Ballif
MATH 501 Homework 1
September 26, 2007
In all problems below, measurable means Lebesgue measurable;
m
is the Lebesgue mea
sure and
m
*
is the Lebesgue outer measure.
1. The Cantor Set.
Denote by
C
0
= [0
,
1]
. Let
C
1
be the set obtained by removing the middle
third open interval of
C
0
, that is,
C
1
= [0
,
1
/
3]
∪
[2
/
3
,
1]
. Next, let
C
2
be the set obtained by
removing the middle third open interval from each interval of
C
1
. Continuing this way, we obtain
a decreasing sequence of closed sets
C
0
⊃
C
1
⊃
C
2
⊃
. . .
. The Cantor set
C
is the intersection of
all
C
k
’s,
C
=
∩
∞
k
=0
C
k
. Show:
(a)
C
is totally disconnected, that is, if
x, y
∈ C
and
x < y
, then there is
z
∈
(
x, y
)
such that
z /
∈ C
.
Claim 1.
Every element
x
∈ C
can be written in the form
x
=
∑
∞
i
=1
c
i
3
i
, where
c
i
= 0 or
c
i
= 2 for every whole number
i
.
Proof.
We first note that every number
x
∈
[0
,
1] can be written in the form
x
=
∑
∞
i
=1
c
i
3
i
where
c
i
∈ {
0
,
1
,
2
}
. We adopt the convention, that we express
x
in the form listed in claim 1, if it is possible to do so. This policy removes the
ambiguity of nonuniqueness that arises from the case where
c
i
= 1 for all
i
≥
n
.
The set
C
k
consists precisely of those numbers
x
in for which
c
i
∈ {
0
,
2
}
for each
i
≤
k
(since
c
k
= 1 places
x
in the open middle third of its interval). Therefore,
every
x
of the prescribed form is in
C
n
for all
n
and hence, in
C
. Furthermore, if
x
is not of the form in the claim, then
c
n
= 1 for some
n >
0, and thus,
x
is not
in the set
C
n
or in
C
.
Now if
x, y
∈ C
and
x < y
, then
x
=
∑
∞
i
=1
c
i
3
i
and
y
=
∑
∞
i
=1
d
i
3
i
for
c
i
, d
i
∈ {
0
,
2
}
.
Since
x < y
, then we know that there is a smallest index
j
such that
c
j
= 0 and
d
j
= 2. Define
z
to be the number
z
=
∑
∞
i
6
=
j
c
i
3
i
+
1
3
j
. By claim 1,
z /
∈ C
. Then we
have
x < z < y
such that
z /
∈ C
. Therefore,
C
is totally disconnected.
(b)
C
is perfect, that is, it is closed and without isolated points.
Each set
C
n
is closed as they are defined to be complements of open sets. There
fore,
C
is closed as an intersection of closed sets. To see that
C
has no isolated
points, consider an arbitrary element
x
=
∑
∞
i
=1
c
i
3
i
of
C
. Let
>
0 and consider
the open ball
B
(
x,
). There exists some number
n
such that
1
3
n

1
<
. Therefore,
for any choice of numbers
d
i
∈ {
0
,
1
,
2
}
, we have
∞
X
i
=
n
d
i
3
i
≤
∞
X
i
=
n
2
3
i
=
2
3
n
∞
X
i
=0
1
3
i
=
2
3
n
1
1

1
/
3
=
1
3
n

1
<
.