1. ## Cantor Set

Does anyone have any idea how to do this problem? Because I sure don't!!!!

The Cantor set is constructed as follows. We start with the closed interval [0, 1] and remove the open interval ( 1/3 , 2/3 ). That leaves the two intervals [0, 1/3 ] and [ 2/3 , 1] 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 [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.
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.

2. Wow, the Cantor set (I am afraid of it).

You removed $\displaystyle 1/3$ then from the the remaining you removed $\displaystyle 2/9$ and from that $\displaystyle 4/27$...
The total length is,
$\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+...=1$ Infinite geometric series.

I think what Cantor wanted to show is that there cannot exist an interval of length non-zero. For that would imply that there length of the removed is not one. Hence the Cantor set is just a collection of discontinous points. Note, it is not non-empty because $\displaystyle 0,1$ are its elements.

3. For the Cantor set, the middle third is successively removed from the remaining pieces.

Start with the line interval $\displaystyle I \subset R$ from 0 to 1.
Remove one third from the middle. What remains is two disconnected pieces a third in length.
Now remove two middle thirds from each; that is, remove one ninth the original length from each, which is two removals of one ninth.
Now four even smaller pieces are left. Remove the middle third of each of these four pieces, which is 1/27th the original length removed four times.
This goes on forever.

So the total amount removed is $\displaystyle \sum_{n=1}^{\infty} 2^{n-1}\left(\frac{1}{3^n}\right) = 1$

However, they are open segments removed, so for one thing, the end points at any iteration will never be removed. So 0 and 1 are definitely in the Cantor set, as are 1/3 and 2/3.

For the carpet, I've never worked with that before, but it seems to be the same idea just extrapolated in 2 dimensions.
For each iteration, you take out 8 times more holes than before, and the holes are one ninth in size.

My initial analysis (which might be wrong) is that the sum of removed pieces adds up as follows:

$\displaystyle \sum_{n=1}^{\infty} 8^{n-1}\left(\frac{1}{9^n}\right) = 1$.

Hope that helps.

4. 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: .$\displaystyle A \;= \;\frac{\frac{1}{9}}{1 - \frac{8}{9}} \;=\;\frac{\frac{1}{9}}{\frac{1}{9}} \;= \;1$

5. Not only does it has infintely many number it has uncountable many.

6. Originally Posted by ThePerfectHacker
Not only does it has infintely many number it has uncountable many.
Hello, THP,

I've a question with your statement:

In my opinion the set $\displaystyle \mathbb{N}$ is mapped to the summands of the series if you use such an equation: $\displaystyle \sum_{n=1}^{\infty} 2^{n-1}\left(\frac{1}{3^n}\right) = 1$.

Doesn't that mean that the set has countable infinite numbers?

(Probably I miss the meaning of "uncountable")

EB

7. Originally Posted by earboth
Hello, THP,

I've a question with your statement:

In my opinion the set $\displaystyle \mathbb{N}$ is mapped to the summands of the series if you use such an equation: $\displaystyle \sum_{n=1}^{\infty} 2^{n-1}\left(\frac{1}{3^n}\right) = 1$.

Doesn't that mean that the set has countable infinite numbers?

(Probably I miss the meaning of "uncountable")

EB
Contable is the cardinality of any finite set or $\displaystyle \mathbb{Z}$ (or $\displaystyle \mathbb{N}$ both the same). Uncountable is anything larger than this for example, $\displaystyle \mathbb{R}$ is larger. The rule is usually that rational and algebraic numbers are countable while transendental are not. So I am saything that this set is so large that you cannot count them (so it contains transendental elements). But the nice feature is that all intervals are discontinous because any interval $\displaystyle [a,b]$ always have length. This is why 1 length was removed. Thing of it like that a line segment and a point is removed the length is still 1. So if you remove points that do not form an interval you still have the full length. It reminds me of the Dirichlet function because it too is discontinous everywhere uncountable.

,

# the sierpinski carpet is a two-dimensional counterpart of the cantor set. it is constructed

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