(a)

#### To determine

**To show:** The total length of all the intervals of cantor set are removed is 1: Despite that, the cantor set contains the infinitely many numbers and give examples of some numbers in the cantor set.

#### Answer

The few numbers of cantor set are
13,23,19,29,79 and 89
.

#### Explanation

**Result used:**

The sum of the infinite geometric series
∑n=1∞arn−1
is
a1−r if |r|<1
.

**Calculation:**

Consider the closed interval
[0,1]

*Level 0*:

Remove the open interval
(13,23)
from the interval
[0,1]
.

This results in two intervals
[0,13] and [23,1]

The remaining interval from
[0,1]
is
[0,13]∪[13,1]
.

The length of the interval removed is
1⋅13=13

*Level 1*:

The remaining interval from
[0,1]
is
[0,13]∪[13,1]

Remove the open interval
(19,29)
from the interval
[0,13]
.

That is,
(19,29)
leaves two intervals
[0,19] and [29,39]
,

Remove the open interval
(79,89)
from the interval
[23,1]

That is, the open interval
(79,89)
leaves two intervals
[69,79] and [89,1]
.

The lengths of the intervals removed are
2⋅19=29

The remaining interval from
[0,13]∪[13,1]
is
[0,19]∪[29,39]∪[69,79]∪[89,1]

*Level 3*:

Remove the open interval
(127,227)
from the interval
[0,19]
, remove the open interval
(727,827)
from the interval
[29,39]
, remove the open interval
(1927,2027)
from the interval
[69,79]
and remove the open interval
(2527,2627)
from the interval
[89,1]
.

That is,
(127,227)
leaves two intervals
[0,127] and [227,327]
, the open interval
(727,827)
leaves two intervals
[627,727] and [827,927]
,
(1927,2027)
leaves two intervals
[1827,1927] and [2027,2127]
, the open interval
(2527,2627)
leaves two intervals
[2427,2527] and [2627,1]
.

The lengths of the intervals removed are
4⋅127=22.127

The remaining interval from
[0,19]∪[29,39]∪[69,79]∪[89,1]
is
[0,127]∪[227,327]∪[627,727]∪[627,727]∪[1827,1927]∪[2027,2127][2427,2527]∪[2627,1]

Proceed in this similar way, at level *n*,

The lengths of the intervals removed are
2n⋅13n+1

That is,

2n⋅13n+1=2n3n+1=2n3n⋅3=13(23)n

Therefore, the sum of the removed interval lengths
∑n=0∞13(23)n

In the geometric series
∑n=0∞13(23)n
, where
r=23 and a=13

∑n=0∞13(23)n=131−23=1313=1

Therefore, the sum of the removed interval lengths is 1.

Note that, the left most open intervals
((13)n,(23)n)
are removed and right most intervals
(1−(23)n,1−(13)n)
are removed in the interval
[0,1]
. The numbers 0 and 1 is not removed.

Therefore, the numbers in the cantor set are the boundary numbers.

The some of the numbers in cantor set are
13,23,19,29,79 and 89,

(b)

#### To determine

**To show:** The sum of the areas of the removed squares is 1.

#### Explanation

**Proof:**

*Step 1:*

The area of the square removed is 1 center square out of 9 squares.

That is,
A0=1⋅19

*Step 2:*

The area of the square in this step removed by 9 out of remaining 8 squares is
A1=8⋅181

That is,
A1=8⋅(19)2

Proceed in this similar way, at the step *n* is
An=8n⋅19n+1

An=8n⋅19n⋅9=19(89)n

The sum of the removed squares is
∑n=0∞19(89)n

Here, the geometric series
∑n=0∞19(89)n
where
a=19 and r=89

∑n=0∞19(89)n=191−89=1919=99=1

Hence, the sum of the removed squares is 1.