   Chapter 11.2, Problem 89E

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Textbook Problem

# The Cantor set, named after the German mathematician Georg Cantor (1845–1918), 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. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0. (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.

Explanation

Result used:

The sum of the infinite geometric series n=1arn1 is a1r 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 113=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 219=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]

(b)

To determine

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

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