Question

Series question

Answer 1

The set F is a fractal constructed as follows:

i) Begin with the line segment \[I _{0} = \left[ 0,1 \right]\]
ii) Remove the open middle third \[\left( 1/3 \right) , \left( 2/3 \right)\] to get \[I _{1} = \left[ 0, \frac{ 1 }{ 3 } \right] \cup \left[ \frac{ 2 }{ 3 }, 1 \right]\]
iii) Remove the open middle third from every remaining line segment, to get \[I _{2} = \left[ 0, \frac{ 1 }{ 9 } \right] \cup \left[ \frac{ 2 }{ 9 } , \frac{ 1 }{ 3 } \right]\] \[\cup \left[ \frac{ 2 }{ 3 }, \frac{ 7 }{ 9 } \right]\] \[\cup \left[ \frac{ 8 }{ 9 }, 1 \right]\]
iv) Repeat the process ad infintum (forever)

(a) By considering a suitable series, show that the total length of all of the intervals removed is equal to 1.
(b) If the total length removed is 1, have we removed all of the points from the interval [0, 1]? Justify your answer.

Answer 2

For a clearer representation: https://i.imgsafe.org/856536cef1.png

Answer 3

remove \(1/3\) and you have \(1-1/3\)

remove \(1/3\) of that and you have \((1/3)(1-1/3)\)
remove \(1/3\) of that and you have \((1/3)(1-1/3)(1-1/3)\)

keep doing this 'a bunch' and add them up

\((1/3)+(1/3)(1-1/3)+(1/3)(1-1/3)(1-1/3)+...\\=\sum_1^\infty \frac{1}{3}(1-\frac{1}{3})^{n-1}=\frac{1}{3}\sum_1^\infty(\frac{2}{3})^{n-1}=\frac{1}{3}\sum_0^\infty(\frac{2}{3})^n=\frac{1}{3}(\frac{1}{1-\frac{2}{3}})=\frac{1}{3}(\dfrac{1}{\frac{1}{3}})=1\)

Answer 4

https://en.wikipedia.org/wiki/Cantor_set

Answer 5

Very weird indeed. When I learned about the Cantor set in analysis, our teacher started out by saying "today we go to the dark side"