Ex3-2

# b
3

{\textstyle b=3}

 in Problem 3.1, the expansion is called the triadic or ternary expansion of

x

{\textstyle x}

.

(a) Show that the Cantor set

C

{\textstyle {\mathcal {C}}}

 consists of all

x

{\textstyle x}

 such that some triadic expansion for which every

c

k

{\textstyle c\_{k}}

is either 0 or 2.

# D

[

0
,
1

]

−

C

=

⋂

# k
1

∞

⋃

# ℓ
0

3

k
−
1

−
1

(

3
ℓ
1

3

k

,

3
ℓ
2

3

k

)

{\displaystyle D=\left[0,1\right]-{\mathcal {C}}=\bigcap \_{k=1}^{\infty }\bigcup \_{\ell =0}^{3^{k-1}-1}\left({\dfrac {3\ell +1}{3^{k}}},{\dfrac {3\ell +2}{3^{k}}}\right)}

1,

# E

{

x
∈

[

0
,
1

]

x

 has a digit 1 in its ternary expansion

}

{\textstyle E=\left\{x\in \left[0,1\right]\ {\biggr |}\ x{\text{ has a digit 1 in its ternary expansion}}\right\}}

 and

# S

{

∑

# n
1

∞

a

n

3

n

a

n

∈
{
0
,
2
}

}

{\displaystyle S=\left\{\sum \_{n=1}^{\infty }{\dfrac {a\_{n}}{3^{n}}}\ {\biggr |}\ a\_{n}\in \{0,2\}\right\}}

.

We suffice to show that

x
∈

C

{\textstyle x\in {\mathcal {C}}}

 if and only if

x
∈
S

{\displaystyle x\in S}

. Suppose that

x
∈
S

{\displaystyle x\in S}

, then

# n
1

∞

a

n

3

n

{\displaystyle x=\sum \_{n=1}^{\infty }{\dfrac {a\_{n}}{3^{n}}}}

 for some

a

n

∈

{

0
,
2

}

{\displaystyle a\_{n}\in \left\{0,2\right\}}

.

*Claim 1.* The partial sum

x

k

=

∑

# n
1

k

a

n

3

n

{\displaystyle x\_{k}=\sum \_{n=1}^{k}{\dfrac {a\_{n}}{3^{n}}}}

 is the left endpoint of a closed subinterval of

C

k

{\displaystyle {\mathcal {C}}\_{k}}

.

*Proof.* We prove it using induction. If

# k
1

{\displaystyle k = 1}

, then

x
1

{\displaystyle x\_1}

 is either

0

{\displaystyle 0}

 or

2
3

{\displaystyle {\dfrac {2}{3}}}

, either of these is the left endpoint of

C

1

=

[

0
,

1
3

]

∪

[

2
3

,
1

]

{\displaystyle {\mathcal {C}}\_{1}=\left[0,{\dfrac {1}{3}}\right]\cup \left[{\dfrac {2}{3}},1\right]}

. Assume that

x

k
−
1

{\displaystyle x\_{k-1}}

 is a left endpoint of

C

k
−
1

{\displaystyle {\mathcal {C}}\_{k-1}}

, then

[

x

k
−
1

,

x

k
−
1

1

3

k
−
1

]

⊂

C

k
−
1

{\displaystyle \left[x\_{k-1},x\_{k-1}+{\dfrac {1}{3^{k-1}}}\right]\subset {\mathcal {C}}\_{k-1}}

. By the construction of the Cantor set

C

{\displaystyle \mathcal{C}}

, we trisect the interval

[

x

k
−
1

,

x

k
−
1

1

3

k
−
1

]

{\displaystyle \left[x\_{k-1},x\_{k-1}+{\dfrac {1}{3^{k-1}}}\right]}

 and remove the middle closed interval to obtain the closed subinterval of

C

k

{\displaystyle {\mathcal {C}}\_{k}}

:

[

x

k
−
1

,

x

k
−
1

1

3

k

]

∪

[

x

k
−
1

2

3

k

,

x

k
−
1

1

3

k
−
1

]

⊂

C

k

{\displaystyle \left[x\_{k-1},x\_{k-1}+{\dfrac {1}{3^{k}}}\right]\cup \left[x\_{k-1}+{\dfrac {2}{3^{k}}},x\_{k-1}+{\dfrac {1}{3^{k-1}}}\right]\subset {\mathcal {C}}\_{k}}

This implies that

x

k
−
1

a

k

3

k

{\displaystyle x\_{k-1}+{\dfrac {a\_{k}}{3^{k}}}}

 is a left endpoint of the closed subinterval of

C

k

{\displaystyle {\mathcal {C}}\_{k}}

 for each

a

k

∈

{

0
,
2

}

{\displaystyle a\_{k}\in \left\{0,2\right\}}

. Since each

x
n

{\displaystyle x\_n}

 is now the left endpoint of

C

n

{\displaystyle \mathcal{C}\_n}

, we have

x

n

∈

C

n

{\displaystyle x\_{n}\in {\mathcal {C}}\_{n}}

. As the Cantor set

C

{\displaystyle \mathcal{C}}

 is the arbitrary intersection of closed intervals

C

n

{\displaystyle \mathcal{C}\_n}

, that is,

# n
1

∞

C

n

{\displaystyle C=\bigcap \_{n=1}^{\infty }{\mathcal {C}}\_{n}}

, so

C

{\displaystyle \mathcal{C}}

 is closed and it contains all limit points of

S

{\displaystyle S}

, and finally we have

x
:=

lim

n
→
∞

x

n

∈

C

{\displaystyle x:=\lim \limits \_{n\to \infty }x\_{n}\in {\mathcal {C}}}

 as desired. This shows

S
⊆

C

{\displaystyle S\subseteq {\mathcal {C}}}

.

◻

{\displaystyle \square}

On the other hands, suppose that

x
∉
S

{\displaystyle x\not \in S}

 for the sake of contrapositive, which implies that

x
∈
E

{\displaystyle x \in E}

 instead, then

x

{\displaystyle x}

 appears to have the least number

M

{\displaystyle M}

 such that

a

M

=
1

{\displaystyle a\_{M}=1}

 for its ternary representation. Note that there cannot exists

N
∈

N

{\displaystyle N \in \N}

 such that

x
n

=
0

{\displaystyle x\_n = 0}

 or

2

{\displaystyle 2}

 for every

n
>
N

{\displaystyle n > N}

, otherwise

x

{\displaystyle x}

 would have another ternary expansion without the digit

1

{\displaystyle 1}

, as

1.000

⋯

10

=
0.222

⋯

3

{\displaystyle 1.000\cdots \_{10}=0.222\cdots \_{3}}

 holds by geometric series. Write

x

{\displaystyle x}

 as

# n
1

M
−
1

a

n

3

n

1

3

M

∑

# n
M
1

∞

a

n

3

n

{\displaystyle x=\sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n}}}+{\dfrac {1}{3^{M}}}+\sum \_{n=M+1}^{\infty }{\dfrac {a\_{n}}{3^{n}}}}

Now we rewrite

∑

# n
1

M
−
1

a

n

3

n

=

3

3

M

∑

# n
1

M
−
1

a

n

3

n
−
M
1

{\displaystyle \sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n}}}={\dfrac {3}{3^{M}}}\sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n-M+1}}}}

 and since

0
<

∑

# n
M
1

∞

a

n

3

n

<

1

3

M

{\displaystyle 0<\sum \_{n=M+1}^{\infty }{\dfrac {a\_{n}}{3^{n}}}<{\dfrac {1}{3^{M}}}}

 holds2 for

a

n

∈

{

0
,
2

}

{\displaystyle a\_{n}\in \left\{0,2\right\}}

 and any

M
∈

N

{\displaystyle M\in \mathbb {N} }

, it follows that

0
<

∑

# n
1

∞

a

n

3

n

−

∑

# n
1

M

a

n

3

n

<

1

3

M

{\displaystyle 0<\sum \_{n=1}^{\infty }{\dfrac {a\_{n}}{3^{n}}}-\sum \_{n=1}^{M}{\dfrac {a\_{n}}{3^{n}}}<{\dfrac {1}{3^{M}}}}

0
<
x
−

∑

# n
1

M
−
1

a

n

3

n

−

1

3

M

<

1

3

M

{\displaystyle 0<x-\sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n}}}-{\dfrac {1}{3^{M}}}<{\dfrac {1}{3^{M}}}}

1

3

M

(

3

∑

# n
1

M
−
1

a

n

3

n
−
M
1

1

)

<
x
<

1

3

M

(

3

∑

# n
1

M
−
1

a

n

3

n
−
M
1

2

)

{\displaystyle {\dfrac {1}{3^{M}}}\left(3\sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n-M+1}}}+1\right)<x<{\dfrac {1}{3^{M}}}\left(3\sum \_{n=1}^{M-1}{\dfrac {a\_{n}}{3^{n-M+1}}}+2\right)}

This shows that

x
∈

⋂

# k
1

∞

⋃

# ℓ
0

3

k
−
1

−
1

(

3
ℓ
1

3

k

,

3
ℓ
2

3

k

)

{\displaystyle x\in \bigcap \_{k=1}^{\infty }\bigcup \_{\ell =0}^{3^{k-1}-1}\left({\dfrac {3\ell +1}{3^{k}}},{\dfrac {3\ell +2}{3^{k}}}\right)}

 for suitable

ℓ

{\displaystyle \ell}

 and

k

{\displaystyle k}

, hence

x
∉

C

{\displaystyle x\not \in {\mathcal {C}}}

. This completes the proof.

◼

{\displaystyle \blacksquare }

1 One can show the equality using induction and some topology facts. This is left as an exercise to the reader.

2 This can be proved by comparison test. Note that we do not allow

a
n

=
0

{\displaystyle a\_n = 0}

 or

2

{\displaystyle 2}

 for all

n
∈

N

{\displaystyle n \in \N}

.