Chapter 8.2, Problem 65E

(a)

To determine

To Find: The example of the some numbers in the cantor set.

Answer to Problem 65E

The example of the some numbers in the cantor set is obtained.

Explanation of Solution

Given:

The total length of all the intervals that are removed is 1

The closed interval $\left[0,1\right]$ , remove the open interval $\left(\frac{1}{3},\frac{2}{3}\right)$ .

The two interval is left as $\left[0,\frac{1}{3}\right]$ and $\left[\frac{2}{3},1\right]$ . Then, open the middle third of each of them.

Calculation:

To measure the size of the cantor set sue the infinite series by removing the middle terms 0, 1 and segment size is,

  $\left|\frac{2}{3}-\frac{1}{3}\right|=\frac{1}{3}$

The first term is $\frac{1}{3}$ , the middle term is $\frac{1}{9}$ and the length of the middle term is $\frac{1}{9}$ .

Then,

  $\begin{array}{c}{a}_2=2\left(\frac{1}{9}\right)\\ =\frac{2}{9}\end{array}$

Then, the series is,

  $\begin{array}{l}{a}_n=\frac{2^{n-1}}{3^n}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{2^{n-1}}{3^n}}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{3}{\left(\frac{2}{3}\right)}^{n-1}}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{2^{n-1}}{3^n}}=\frac{\frac{1}{3}}{1-\frac{2}{3}}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{2^{n-1}}{3^n}}=1\end{array}$

Consider for the second part 0 and 1 are not removed and the other numbers still in the cantor are $\frac{1}{3},\frac{2}{3},\frac{1}{9},\frac{2}{9},\frac{7}{9},\frac{8}{9}$ and in general any end point of the subintervals that is done it is removed. Oddly how the only points that survive all of the cutting is performed. Take for instances $\frac{1}{4}$ is still the cantor set even though $\frac{1}{4}$ is never the one of the endpoints of one of the subintervals is removed.

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }2{\left(\frac{1}{9}\right)}^m}=\frac{\frac{2}{9}}{1-\frac{1}{9}}\\ =\frac{1}{4}\end{array}$

This, implies that $\frac{1}{4}=\frac{2}{9}+\frac{2}{81}+\frac{2}{729}+\mathrm{...}$ is the infinite sum of the points that is the end points of the intervals and are to be removed in creating the cantor set.

(b)

To determine

To Prove: The sum of the areas of the removed squares is 1 and this shows that implies that the sierpinski carpet has the area 0.

Explanation of Solution

Given:

The centre of the square of side 1 then removing the centres of the eight smaller squares and so on.

The given diagram is shown in Figure 1

  

Figure 1

The sum of the areas of the removed square is 1

Calculation:

Consider the area of the square is,

  $\begin{array}{l}{a}_1=\frac{1}{3^2}\\ a_2=8\times \frac{1}{{\left(3^2\right)}^2}\\ a_3=8^2\times {\left(\frac{1}{{\left(3^2\right)}^2}\right)}^2\\ a_n=8^{n-1}{\left(\frac{1}{3^n}\right)}^2\end{array}$

Then, the area is,

  $\begin{array}{c}{a}_1=\frac{1}{3^2}+8\times \frac{1}{{\left(3^2\right)}^2}+8^2\times {\left(\frac{1}{{\left(3^2\right)}^2}\right)}^2+8^{n-1}{\left(\frac{1}{3^n}\right)}^2\\ =\frac{{\left(\frac{1}{3}\right)}^2}{1-\frac{8}{9}}\\ =1\end{array}$

Hence, proved.