Mixing
Formal construction of the Cantor set $K$ and determination of $[0,1]setminus K.$
$begingroup$
Let $J_{0,1}:=[0,1]$.
Step 1. We remove the central open interval $I_{0,1}=big(frac{1}{3},frac{2}{3}big)$. We denote with $J_{1,1}:=big[0,frac{1}{3}big]$ and with $J_{1,2}:=big[frac{2}{3},1big]$.
We define $K_1:=J_{1,1}cup J_{1,2}$ then $complement{K_1}=I_{0,1}.$
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}cup J_{2,2}cup J_{2,3}cup J_{2,4}$, where $J_{2,1}=big[0,frac{1}{9}big]$, $J_{2,2}=big[frac{2}{9},frac{1}{3}big]$, $J_{2,3}=big[frac{2}{3},frac{7}{9}big]$, $J_{2,4}=big[frac{8}{9}, 1big]$.
At this point we define with $I_{1,1}=big(frac{1}{9}, frac{2}{9}big)$, $I_{1,2}=big(frac{7}{9},frac{8}{9}big)$ the two open interval just removed. Therefore, $complement{K_2}=complement{K_1}cup big(I_{1,1}cup I_{1,2}big)$
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,cdots, 2^n$ of length $1/3^n.$
For all $ninmathbb{N}$ we define $$K_{n}:=bigcup_{k=1}^{2^n} J_{n,k}$$
We define the Cantor set $K$ in the follow way $$K:=bigcap_{n=1}^{+infty} K_n.$$
In general, as discussed above, $$complement{K_n}= complement{K_{n-1}}cupbigcup_{k=1}^{2^n/2} I_{n-1,k}tag1$$
Question 1. It's correct (1)?
We determine,
$$[0,1]setminus K=[0,1]cap bigg[bigcup_{n=1}^{+infty}complement{K_{n-1}}cup bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}bigg]tag2$$
Question 2. How can I proceed in (2)? I would like to prove that $$[0,1]setminus K =bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n}I_{n,k}.$$
My answer at question 2.
begin{equation}
begin{split}
[0,1]setminus K &= [0,1]capbigg[bigcup_{n=1}^{+infty}complement{K_n}bigg]\
=&bigcup_{n=1}^{+infty} complement{K_n}\
=&bigcup_{n=1}^{+infty}complement{K_{n-1}}cupbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k}\
color{BLUE}{=}& bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}\
color{RED}{=}&bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n} I_{n,k}.
end{split}
end{equation}
For the blue equality I used the fact that $$bigcup_{n=1}^{+infty}complement{K_{n-1}}subseteqbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k} $$
and the red equality it is a simple rewriting of the indexes.
Question 3. It's correct my answer?
Thanks!
real-analysis measure-theory lebesgue-measure cantor-set
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
$endgroup$
add a comment |
$begingroup$
Let $J_{0,1}:=[0,1]$.
Step 1. We remove the central open interval $I_{0,1}=big(frac{1}{3},frac{2}{3}big)$. We denote with $J_{1,1}:=big[0,frac{1}{3}big]$ and with $J_{1,2}:=big[frac{2}{3},1big]$.
We define $K_1:=J_{1,1}cup J_{1,2}$ then $complement{K_1}=I_{0,1}.$
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}cup J_{2,2}cup J_{2,3}cup J_{2,4}$, where $J_{2,1}=big[0,frac{1}{9}big]$, $J_{2,2}=big[frac{2}{9},frac{1}{3}big]$, $J_{2,3}=big[frac{2}{3},frac{7}{9}big]$, $J_{2,4}=big[frac{8}{9}, 1big]$.
At this point we define with $I_{1,1}=big(frac{1}{9}, frac{2}{9}big)$, $I_{1,2}=big(frac{7}{9},frac{8}{9}big)$ the two open interval just removed. Therefore, $complement{K_2}=complement{K_1}cup big(I_{1,1}cup I_{1,2}big)$
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,cdots, 2^n$ of length $1/3^n.$
For all $ninmathbb{N}$ we define $$K_{n}:=bigcup_{k=1}^{2^n} J_{n,k}$$
We define the Cantor set $K$ in the follow way $$K:=bigcap_{n=1}^{+infty} K_n.$$
In general, as discussed above, $$complement{K_n}= complement{K_{n-1}}cupbigcup_{k=1}^{2^n/2} I_{n-1,k}tag1$$
Question 1. It's correct (1)?
We determine,
$$[0,1]setminus K=[0,1]cap bigg[bigcup_{n=1}^{+infty}complement{K_{n-1}}cup bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}bigg]tag2$$
Question 2. How can I proceed in (2)? I would like to prove that $$[0,1]setminus K =bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n}I_{n,k}.$$
My answer at question 2.
begin{equation}
begin{split}
[0,1]setminus K &= [0,1]capbigg[bigcup_{n=1}^{+infty}complement{K_n}bigg]\
=&bigcup_{n=1}^{+infty} complement{K_n}\
=&bigcup_{n=1}^{+infty}complement{K_{n-1}}cupbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k}\
color{BLUE}{=}& bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}\
color{RED}{=}&bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n} I_{n,k}.
end{split}
end{equation}
For the blue equality I used the fact that $$bigcup_{n=1}^{+infty}complement{K_{n-1}}subseteqbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k} $$
and the red equality it is a simple rewriting of the indexes.
Question 3. It's correct my answer?
Thanks!
real-analysis measure-theory lebesgue-measure cantor-set
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
$endgroup$
add a comment |
$begingroup$
Let $J_{0,1}:=[0,1]$.
Step 1. We remove the central open interval $I_{0,1}=big(frac{1}{3},frac{2}{3}big)$. We denote with $J_{1,1}:=big[0,frac{1}{3}big]$ and with $J_{1,2}:=big[frac{2}{3},1big]$.
We define $K_1:=J_{1,1}cup J_{1,2}$ then $complement{K_1}=I_{0,1}.$
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}cup J_{2,2}cup J_{2,3}cup J_{2,4}$, where $J_{2,1}=big[0,frac{1}{9}big]$, $J_{2,2}=big[frac{2}{9},frac{1}{3}big]$, $J_{2,3}=big[frac{2}{3},frac{7}{9}big]$, $J_{2,4}=big[frac{8}{9}, 1big]$.
At this point we define with $I_{1,1}=big(frac{1}{9}, frac{2}{9}big)$, $I_{1,2}=big(frac{7}{9},frac{8}{9}big)$ the two open interval just removed. Therefore, $complement{K_2}=complement{K_1}cup big(I_{1,1}cup I_{1,2}big)$
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,cdots, 2^n$ of length $1/3^n.$
For all $ninmathbb{N}$ we define $$K_{n}:=bigcup_{k=1}^{2^n} J_{n,k}$$
We define the Cantor set $K$ in the follow way $$K:=bigcap_{n=1}^{+infty} K_n.$$
In general, as discussed above, $$complement{K_n}= complement{K_{n-1}}cupbigcup_{k=1}^{2^n/2} I_{n-1,k}tag1$$
Question 1. It's correct (1)?
We determine,
$$[0,1]setminus K=[0,1]cap bigg[bigcup_{n=1}^{+infty}complement{K_{n-1}}cup bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}bigg]tag2$$
Question 2. How can I proceed in (2)? I would like to prove that $$[0,1]setminus K =bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n}I_{n,k}.$$
My answer at question 2.
begin{equation}
begin{split}
[0,1]setminus K &= [0,1]capbigg[bigcup_{n=1}^{+infty}complement{K_n}bigg]\
=&bigcup_{n=1}^{+infty} complement{K_n}\
=&bigcup_{n=1}^{+infty}complement{K_{n-1}}cupbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k}\
color{BLUE}{=}& bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}\
color{RED}{=}&bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n} I_{n,k}.
end{split}
end{equation}
For the blue equality I used the fact that $$bigcup_{n=1}^{+infty}complement{K_{n-1}}subseteqbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k} $$
and the red equality it is a simple rewriting of the indexes.
Question 3. It's correct my answer?
Thanks!
real-analysis measure-theory lebesgue-measure cantor-set
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
$endgroup$
Let $J_{0,1}:=[0,1]$.
Step 1. We remove the central open interval $I_{0,1}=big(frac{1}{3},frac{2}{3}big)$. We denote with $J_{1,1}:=big[0,frac{1}{3}big]$ and with $J_{1,2}:=big[frac{2}{3},1big]$.
We define $K_1:=J_{1,1}cup J_{1,2}$ then $complement{K_1}=I_{0,1}.$
Step 2. We remove from $J_{1,1}$ and $J_{1,2}$ the central open interval of length $frac{1}{9}$. Then we define what remains with $K_2:=J_{2,1}cup J_{2,2}cup J_{2,3}cup J_{2,4}$, where $J_{2,1}=big[0,frac{1}{9}big]$, $J_{2,2}=big[frac{2}{9},frac{1}{3}big]$, $J_{2,3}=big[frac{2}{3},frac{7}{9}big]$, $J_{2,4}=big[frac{8}{9}, 1big]$.
At this point we define with $I_{1,1}=big(frac{1}{9}, frac{2}{9}big)$, $I_{1,2}=big(frac{7}{9},frac{8}{9}big)$ the two open interval just removed. Therefore, $complement{K_2}=complement{K_1}cup big(I_{1,1}cup I_{1,2}big)$
At the end of step $n$ we will have $2^n$ close interval $J_{n,k}$ for $k=1,cdots, 2^n$ of length $1/3^n.$
For all $ninmathbb{N}$ we define $$K_{n}:=bigcup_{k=1}^{2^n} J_{n,k}$$
We define the Cantor set $K$ in the follow way $$K:=bigcap_{n=1}^{+infty} K_n.$$
In general, as discussed above, $$complement{K_n}= complement{K_{n-1}}cupbigcup_{k=1}^{2^n/2} I_{n-1,k}tag1$$
Question 1. It's correct (1)?
We determine,
$$[0,1]setminus K=[0,1]cap bigg[bigcup_{n=1}^{+infty}complement{K_{n-1}}cup bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}bigg]tag2$$
Question 2. How can I proceed in (2)? I would like to prove that $$[0,1]setminus K =bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n}I_{n,k}.$$
My answer at question 2.
begin{equation}
begin{split}
[0,1]setminus K &= [0,1]capbigg[bigcup_{n=1}^{+infty}complement{K_n}bigg]\
=&bigcup_{n=1}^{+infty} complement{K_n}\
=&bigcup_{n=1}^{+infty}complement{K_{n-1}}cupbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k}\
color{BLUE}{=}& bigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2} I_{n-1,k}\
color{RED}{=}&bigcup_{n=0}^{+infty}bigcup_{k=1}^{2^n} I_{n,k}.
end{split}
end{equation}
For the blue equality I used the fact that $$bigcup_{n=1}^{+infty}complement{K_{n-1}}subseteqbigcup_{n=1}^{+infty}bigcup_{k=1}^{2^n/2}I_{n-1,k} $$
and the red equality it is a simple rewriting of the indexes.
Question 3. It's correct my answer?
Thanks!
real-analysis measure-theory lebesgue-measure cantor-set
real-analysis measure-theory lebesgue-measure cantor-set
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
edited Jan 22 at 21:38
Jack J.
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
asked Jan 22 at 16:47
Jack J.Jack J.
3492419
3492419
add a comment |
add a comment |
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