Hello, TreeMoney!

Good ol' Cantor and Sierpinski . . . with mind-boggling concepts.

The Cantor set is constructed as follows.

We start with the closed interval $\displaystyle [0,\,1]$ and remove the open interval $\displaystyle \left(\frac{1}{3},\,\frac{2}{3}\right)$

That leaves the two intervals $\displaystyle \left[0,\,\frac{1}{3}\right]$ and $\displaystyle \left[\frac{2}{3},\,1\right]$ and we remove the open middle third of each.

Four intervals remain and again we remove the open middle third of each of them.

We continue this procedure indefinitely, at each step removing the open middle third

of every interval that remains from the preceding step.

The Cantor set consists of the numbers that remain in $\displaystyle [0,\,1]$ after all those intervals have been removed.

a) Show that the total length of all the intervals that are removed is 1.

Despite that, the Cantor set contains infinitely many numbers.

Give examples of some numbers in the Cantor set.

. . $\displaystyle * -=- * === * -=- * = = = = = = = = = * -=- * === * -=- * $

. . $\displaystyle 0\qquad\;\;\;\frac{1}{9}\qquad\;\;\,\frac{2}{9}\qq uad\;\;\,\frac{1}{3}\qquad\qquad\qquad\quad\;\frac {2}{3}\qquad\;\;\;\frac{7}{9}\qquad\;\;\,\frac{8}{ 9}\qquad\;\;\;1$

There is 1 segment of length $\displaystyle \frac{1}{3}$

. . . . . . 2 segments of length $\displaystyle \frac{1}{9}$

. . . . . . 4 segments of length $\displaystyle \frac{1}{27}$

. . . . . . 8 segments of length $\displaystyle \frac{1}{81}$ . . . and so on.

The total length is: .$\displaystyle S \;= \;\frac{1}{3} + 2\cdot\frac{1}{9} + 4\cdot\frac{1}{27} + 8\cdot\frac{1}{81} + \hdots$

Hence: .$\displaystyle S \;= \;\frac{1}{3} + \frac{2}{3^2} + \frac{2^2}{3^3} + \frac{2^3}{3^4} + \hdots$

We have a Geometric Series with first term $\displaystyle a = \frac{1}{3}$ and common ratio $\displaystyle r = \frac{2}{3}$

Its sum is: .$\displaystyle S \;= \;\frac{\frac{1}{3}}{1 - \frac{2}{3}} \;= \;\frac{\frac{1}{3}}{\frac{1}{3}} \;= \;1$

Evidently, *all* of the interval has been removed.

Yet there are infinitely many numbers remaining in the Cantor set.

I'll let you figure out which ones are left.

b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set.

It is constructed by removing the center one-ninth of a square of side 1,

then removing the centers of the eight smaller remaining squares, and so on.

Show that the sum of the areas of the removed squares is 1.

This implies that the Sierpinski carpet has area 0. Code:

*-----------------------------------*
| ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ |
| *---* *---* *---* |
| ϧ |:::| ϧ ϧ |:::| ϧ ϧ |:::| ϧ |
| *---* *---* *---* |
| ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ |
| *-----------* |
| ϧ ϧ ϧ |:::::::::::| ϧ ϧ ϧ |
| *---* |:::::::::::| *---* |
| ϧ |:::| ϧ |:::::::::::| ϧ |:::| ϧ |
| *---* |:::::::::::| *---* |
| ϧ ϧ ϧ |:::::::::::| ϧ ϧ ϧ |
| *-----------* |
| ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ |
| *---* *---* *---* |
| ϧ |:::| ϧ ϧ |:::| ϧ ϧ |:::| ϧ |
| *---* *---* *---* |
| ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ ϧ |
*-----------------------------------*

There is: 1 square with area $\displaystyle \left(\frac{1}{3}\right)^2$

. . . . . . . 8 squares with area$\displaystyle \left(\frac{1}{9}\right)^2$

. . . . . . . 64 squares with area $\displaystyle \left(\frac{1}{27}\right)^2$

. . . . . . . 512 squares with area $\displaystyle \left(\frac{1}{81}\right)^2$ . . . and so on.

The total area is: .$\displaystyle A \;= \;1\left(\frac{1}{3}\right)^2 + 8\left(\frac{1}{9}\right)^2 + 64\left(\frac{1}{27}\right)^2 + 512\left(\frac{1}{81}\right)^2 + \hdots$

Hence: .$\displaystyle A \;= \;\frac{1}{3^2} + \frac{2^3}{3^4} + \frac{2^6}{3^6} + \frac{2^9}{3^8} + \hdots$

We have a Geometric Series with first term $\displaystyle a = \frac{1}{9}$ and common ratio $\displaystyle r = \frac{2^3}{3^2} = \frac{8}{9}$

Its sum is: .$\displaystyle A \;= \;\frac{\frac{1}{9}}{1 - \frac{8}{9}} \;=\;\frac{\frac{1}{9}}{\frac{1}{9}} \;= \;1$