Ex3-3

**Chapter 3 Question 3**

Question: Construct a two-dimensional Cantor set in the unit square

{
(
x
,
y
)
:
0
≤
x
,
y
≤
1
}

{\displaystyle \{(x,y):0\leq x,y\leq 1\}}

as follows. Subdivide the square into nine equal parts and keep only the four closed corner squares, removing the remaining region (which forms a cross). Then repeat this process n times for a suitably scaled version for the remaining squares, ad infinitum. Show that the resulting set is perfect, has plane measure zero, and equals

C
×
C

{\displaystyle C\times C}

.

Proof: Let

C

1

=
[
0
,
1

/

3
]
∪
[
2

/

3
,
1
]

{\displaystyle C\_{1}=[0,1/3]\cup [2/3,1]}

 as in the textbook for constructing Cantor set, let

D

1

{\displaystyle D\_{1}}

be the union of four disjoint closed squares as described in the question, then we find that

(
x
,
y
)
∈

D

1

{\displaystyle (x,y)\in D\_{1}}

if and only if

x
∈
[
0
,
1

/

3
]
∪
[
2

/

3
,
1
]

{\displaystyle x\in [0,1/3]\cup [2/3,1]}

 and

y
∈
[
0
,
1

/

3
]
∪
[
2

/

3
,
1
]

{\displaystyle y\in [0,1/3]\cup [2/3,1]}

, which means precisely

D

1

=

C

1

×

C

1

{\displaystyle D\_{1}=C\_{1}\times C\_{1}}

.

Now let

D

2

{\displaystyle D\_{2}}

 be the result of subdividing each of the four squares, so it is a union of

16

{\displaystyle 16}

disjoint squares. We choose

C

2

=
[
0
,
1

/

9
]
∪
[
2

/

9
,
1

/

3
]
∪
[
2

/

3
,
7

/

9
]
∪
[
8

/

9
,
1
]

{\displaystyle C\_{2}=[0,1/9]\cup [2/9,1/3]\cup [2/3,7/9]\cup [8/9,1]}

, as the same in the textbook, then by looking at the graph visually we may conclude

D

2

=

C

2

×

C

2

{\displaystyle D\_{2}=C\_{2}\times C\_{2}}

. We claim that

D

n

=

C

n

×

C

n

{\displaystyle D\_{n}=C\_{n}\times C\_{n}}

 for all

n

{\displaystyle n}

. For

# n
1

{\displaystyle n=1}

 the assertion is already true. Assuming

D

n
−
1

=

C

n
−
1

×

C

n
−
1

{\displaystyle D\_{n-1}=C\_{n-1}\times C\_{n-1}}

 for some

n

{\displaystyle n}

. From Exercise 2(a), we know both of

x
,
y

{\displaystyle x,y}

 satisfying

(
x
,
y
)
∈

D

n
−
1

{\displaystyle (x,y)\in D\_{n-1}}

 have some triadic expansion

a

1

…

a

n
−
1

a

n

…

{\displaystyle 0.a\_{1}\dots a\_{n-1}a\_{n}\dots }

 such that

a

i

{\displaystyle a\_{i}}

 only takes value from

0
,
2

{\displaystyle 0,2}

 for

1
≤
i
≤
n
−
1

{\displaystyle 1\leq i\leq n-1}

. The process of cancelling crosses from

D

n
−
1

{\displaystyle D\_{n-1}}

 to form

D

n

{\displaystyle D\_{n}}

 means exactly we can further choose

a

n

{\displaystyle a\_{n}}

 to be

0
,
2

{\displaystyle 0,2}

 only, hence

D

n

⊂

C

n

×

C

n

{\displaystyle D\_{n}\subset C\_{n}\times C\_{n}}

. On the other hand if

(
x
,
y
)
∈

C

n

×

C

n

{\displaystyle (x,y)\in C\_{n}\times C\_{n}}

, then

x
,
y

{\displaystyle x,y}

, up to their

n

{\displaystyle n}

th triadic expansions' integer can be expressed using only

0
,
2

{\displaystyle 0,2}

, hence

(
x
,
y
)

{\displaystyle (x,y)}

 is not only contained in some square in

D

n
−
1

{\displaystyle D\_{n-1}}

, but it also positioned at one of the four smaller corner squares, hence

(
x
,
y
)
∈

D

n

{\displaystyle (x,y)\in D\_{n}}

. This proves

C

n

×

C

n

⊂

D

n

{\displaystyle C\_{n}\times C\_{n}\subset D\_{n}}

.

We define the 2-dimensional Cantor set as

# n
1

∞

D

n

{\displaystyle D=\bigcap \_{n=1}^{\infty }D\_{n}}

, and we claim that

# D
C
×
C

{\displaystyle D=C\times C}

. For each

n

{\displaystyle n}

we have

C

n

⊂
C

{\displaystyle C\_{n}\subset C}

, so

D

n

=

C

n

×

C

n

⊂
C
×
C

{\displaystyle D\_{n}=C\_{n}\times C\_{n}\subset C\times C}

. Thus

D
⊂
C
×
C

{\displaystyle D\subset C\times C}

. On the other hand, let

(
x
,
y
)
∈
C
×
C

{\displaystyle (x,y)\in C\times C}

, then

x
,
y

{\displaystyle x,y}

 are both contained in every

C

n

{\displaystyle C\_{n}}

, thus

(
x
,
y
)
∈

D

n

{\displaystyle (x,y)\in D\_{n}}

 for each

n

{\displaystyle n}

, proving

C
×
C
⊂
D
.

{\displaystyle C\times C\subset D.}

Now we observe some facts: when we perform each subdivision of squares from

D

n

{\displaystyle D\_{n}}

 to

D

n
1

{\displaystyle D\_{n+1}}

, it will make each square become 4 smaller squares, hence by noting

D

1

{\displaystyle D\_{1}}

 has 4 disjoint squares, we deduce

D

n

{\displaystyle D\_{n}}

 has

4

n

{\displaystyle 4^{n}}

 disjoint squares. Moreover, each square in

D

n

{\displaystyle D\_{n}}

has area 9 times the area of each square in

D

n
1

{\displaystyle D\_{n+1}}

, hence by noting the area of each square in

D

1

{\displaystyle D\_{1}}

is

1

/

9

{\displaystyle 1/9}

, we deduce that each square in

D

n

{\displaystyle D\_{n}}

has area

1

/

9

n

{\displaystyle 1/9^{n}}

. Therefore, the total area, or, the outer measure of each

D

n

{\displaystyle D\_{n}}

, is exactly

(
4

/

9

)

n

{\displaystyle (4/9)^{n}}

. We also have

D

e

≤

D

n

e

=
(
4

/

9

)

n

 for all

n

{\displaystyle n}

, thus

D

{\displaystyle D}

 is measurable and has measure zero, by Example 2 above Theorem 3.12.

Finally, we prove

D

{\displaystyle D}

 is perfect by proving that each point

(
x
,
y
)

{\displaystyle (x,y)}

in

D

{\displaystyle D}

 is a limit point of

D

{\displaystyle D}

. For each square in

D

n

{\displaystyle D\_{n}}

has side length

1

/

3

n

{\displaystyle 1/3^{n}}

, so two points in a same square have a distance at most

2

/

3

n

{\displaystyle {\sqrt {2}}/3^{n}}

between them. Next, we observe that the four vertices of every square in

D

n

{\displaystyle D\_{n}}

 are also points of

D

{\displaystyle D}

, since the subdivision operation never canceled out the four vertices of each square. Therefore, given

# u
(
x
,
y
)
∈
D

{\displaystyle u=(x,y)\in D}

, we find a square vertex

u

1

=
(

x

1

,

y

1

)

{\displaystyle u\_{1}=(x\_{1},y\_{1})}

 from

D

1

{\displaystyle D\_{1}}

such that

u
,

u

1

{\displaystyle u,u\_{1}}

 lie in a same square in

D

1

{\displaystyle D\_{1}}

. Then for each

n

{\displaystyle n}

 we find a square vertex

u

n

=
(

x

n

,

y

n

)

{\displaystyle u\_{n}=(x\_{n},y\_{n})}

 such that

u
,

u

n

{\displaystyle u,u\_{n}}

 lie in a same square in

D

n

{\displaystyle D\_{n}}

. We then conclude

‖
u
−

u

n

‖
≤

2

3

n

 for all

# n
1
,
2
,
…
,

{\displaystyle \lVert u-u\_{n}\rVert \leq {\dfrac {\sqrt {2}}{3^{n}}}\quad {\text{ for all }}n=1,2,\ldots ,}

which means

{

u

n

}

{\displaystyle \{u\_{n}\}}

 is a sequence in

D

{\displaystyle D}

 converges to

u

{\displaystyle u}

.