Mixing
How to derive closed formulas of Cantor set?
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The Cantor set $mathcal{C}$ is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
From Wikiwand's page, The explicit formulas of Cantor sets are
$$mathcal{C} = bigcap_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right) space (1)$$ and $$mathcal{C} = [0,1] setminus bigcupleft{left(frac {3k+1}{3^n}, frac{3k+2}{3^n}right) ,middlevert, k,nin mathbb Z^+right} space (2)$$
I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.
Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!
cantor-set
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
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show 8 more comments
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The Cantor set $mathcal{C}$ is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
From Wikiwand's page, The explicit formulas of Cantor sets are
$$mathcal{C} = bigcap_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right) space (1)$$ and $$mathcal{C} = [0,1] setminus bigcupleft{left(frac {3k+1}{3^n}, frac{3k+2}{3^n}right) ,middlevert, k,nin mathbb Z^+right} space (2)$$
I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.
Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!
cantor-set
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
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1
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I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
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– Henno Brandsma
Jan 19 at 9:18
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Thank you so much @HennoBrandsma! You are correct.
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– Le Anh Dung
Jan 19 at 9:20
1
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The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
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– Henno Brandsma
Jan 19 at 9:20
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@HennoBrandsma I have fixed it :)
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– Le Anh Dung
Jan 19 at 9:22
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Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
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– J.G.
Jan 19 at 9:35
|
show 8 more comments
$begingroup$
The Cantor set $mathcal{C}$ is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
From Wikiwand's page, The explicit formulas of Cantor sets are
$$mathcal{C} = bigcap_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right) space (1)$$ and $$mathcal{C} = [0,1] setminus bigcupleft{left(frac {3k+1}{3^n}, frac{3k+2}{3^n}right) ,middlevert, k,nin mathbb Z^+right} space (2)$$
I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.
Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!
cantor-set
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
$endgroup$
The Cantor set $mathcal{C}$ is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty}C_n$$ where $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
From Wikiwand's page, The explicit formulas of Cantor sets are
$$mathcal{C} = bigcap_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right) space (1)$$ and $$mathcal{C} = [0,1] setminus bigcupleft{left(frac {3k+1}{3^n}, frac{3k+2}{3^n}right) ,middlevert, k,nin mathbb Z^+right} space (2)$$
I have tried for several days to get $(1)$ and $(2)$ from the definition of Cantor set, but to no avail.
Could you please help me derive formulas $(1)$ and $(2)$? Thank you so much!
cantor-set
cantor-set
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
edited Jan 19 at 9:56
Le Anh Dung
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
asked Jan 19 at 9:09
Le Anh DungLe Anh Dung
1,2911621
1,2911621
1
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I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
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– Henno Brandsma
Jan 19 at 9:18
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Thank you so much @HennoBrandsma! You are correct.
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– Le Anh Dung
Jan 19 at 9:20
1
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The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
$endgroup$
– Henno Brandsma
Jan 19 at 9:20
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@HennoBrandsma I have fixed it :)
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– Le Anh Dung
Jan 19 at 9:22
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Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
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– J.G.
Jan 19 at 9:35
|
show 8 more comments
1
$begingroup$
I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
$endgroup$
– Henno Brandsma
Jan 19 at 9:18
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Thank you so much @HennoBrandsma! You are correct.
$endgroup$
– Le Anh Dung
Jan 19 at 9:20
1
$begingroup$
The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
$endgroup$
– Henno Brandsma
Jan 19 at 9:20
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@HennoBrandsma I have fixed it :)
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– Le Anh Dung
Jan 19 at 9:22
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Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
$endgroup$
– J.G.
Jan 19 at 9:35
1
1
$begingroup$
I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
$endgroup$
– Henno Brandsma
Jan 19 at 9:18
$begingroup$
I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
$endgroup$
– Henno Brandsma
Jan 19 at 9:18
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Thank you so much @HennoBrandsma! You are correct.
$endgroup$
– Le Anh Dung
Jan 19 at 9:20
$begingroup$
Thank you so much @HennoBrandsma! You are correct.
$endgroup$
– Le Anh Dung
Jan 19 at 9:20
1
1
$begingroup$
The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
$endgroup$
– Henno Brandsma
Jan 19 at 9:20
$begingroup$
The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
$endgroup$
– Henno Brandsma
Jan 19 at 9:20
$begingroup$
@HennoBrandsma I have fixed it :)
$endgroup$
– Le Anh Dung
Jan 19 at 9:22
$begingroup$
@HennoBrandsma I have fixed it :)
$endgroup$
– Le Anh Dung
Jan 19 at 9:22
$begingroup$
Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
$endgroup$
– J.G.
Jan 19 at 9:35
$begingroup$
Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
$endgroup$
– J.G.
Jan 19 at 9:35
|
show 8 more comments
2 Answers
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The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
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Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
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– Le Anh Dung
Jan 19 at 12:11
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@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
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– Henno Brandsma
Jan 19 at 12:24
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@LeAnhDung also see my answer here for a more accurate description.
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– Henno Brandsma
Jan 19 at 12:30
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Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
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– Le Anh Dung
Jan 19 at 12:33
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I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
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– Le Anh Dung
Jan 19 at 12:34
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I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!
Let $mathbf{L}(x)=dfrac{x}{3}$ and $mathbf{R}(x)=dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=mathbf{L}(I_n) cup mathbf{R}(I_n)$. Cantor set is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty} I_n$$
Theorem: $$mathcal{C} = [0,1] - bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left( frac{3k+1}{3^{n+1}} ,frac{3k+2}{3^{n+1}}right)$$
Let $T_n=I_0-I_n$ for all $ninBbb N$.
Lemma 1: $$T_{n+1}=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)$$
Proof:
Notice that $mathbf{R}(T_n)=mathbf{R}(I_0-I_n)=mathbf{R}(I_0)- mathbf{R}(I_n) subseteq mathbf{R}(I_0) subseteq I_0-mathbf{L}(I_0)$. Thus $(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)=mathbf{R}(T_n)$. Similarly, $mathbf{L}(T_n) subseteq mathbf{L}(I_0) subseteq I_0 -mathbf{R}(I_0)$ and thus $(I_0-mathbf{R}(I_0)) capmathbf{L}(T_n)=mathbf{L}(T_n)$. Moreover, $mathbf{L}(T_n)capmathbf{R}(T_n) subseteq mathbf{L}(I_0)capmathbf{R}(I_0)=emptyset$.
By the definition of $T_{n+1}$:
$begin{align}T_{n+1} &=I_0-I_{n+1}\ &=I_0-(mathbf{L}(I_n) cup mathbf{R}(I_n))\ &=(I_0-mathbf{L}(I_n)) cap (I_0 - mathbf{R}(I_n))\ &=[I_0-mathbf{L}(I_0-T_n)] cap [I_0 - mathbf{R}(I_0-T_n)]\ &=[I_0-(mathbf{L}(I_0)-mathbf{L}(T_n))] cap [I_0 - (mathbf{R}(I_0)-mathbf{R}(T_n))]\ &=[(I_0-mathbf{L}(I_0)) cup mathbf{L}(T_n)] cap [(I_0-mathbf{R}(I_0)) cup mathbf{R}(T_n)]\ &= [(I_0-mathbf{L}(I_0)) cap (I_0-mathbf{R}(I_0))] cup [(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)] cup [mathbf{L}(T_n)cap(I_0-mathbf{R}(I_0))] cup [mathbf{L}(T_n)capmathbf{R}(T_n)]\ &=[I_0 -mathbf{(L}(I_0)cupmathbf{R}(I_0))]cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=(I_0-I_1)cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &= T_1 cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)end{align}$
Lemma 2: $$bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) = left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$$
Proof:
Let $t=m+1$. Then $RHS=left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]$.
Moreover, $bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^{3^0-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^0 left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)=$ $left(frac{1}{3},frac{2}{3} right)$.
It follows that $RHS = left [ bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) right] cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]=$ $bigcup_{t=0}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) =LHS$.
Lemma 3: $$T_{n+1}=bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$$
Proof:
We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.
Notice that $frac{1}{3} < frac{3k+1}{3^{m+2}} < frac{3k+2}{3^{m+2}} < frac{2}{3}$ for all $3^m le k le 2.3^m-1$. It follows that $bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) subseteq left(frac{1}{3},frac{2}{3} right)$ and thus $left(frac{1}{3},frac{2}{3} right)= left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ]$.
By Lemma 1,
$T_{(n+1)+1}=T_{n+2}=mathbf{L}(T_{n+1}) cup T_1 cup mathbf{R}(T_{n+1})$
$=mathbf{L}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right) cup left(frac{1}{3},frac{2}{3} right) cup mathbf{R}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right)$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1+2.3^{m+1}}{3^{m+2}}, frac{3k+2+2.3^{m+1}}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3(k+2.3^m)+1}{3^{m+2}}, frac{3(k+2.3^m)+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left [bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ] cup left[ bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$= left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$ by Lemma 2
We proceed to prove our main theorem.
$mathcal{C}:=bigcap_{n=0}^{infty} I_n=bigcap_{n=0}^{infty} (I_0-T_n)=I_0-bigcup_{n=0}^{infty} T_n=I_0-bigcup_{n=1}^{infty}T_n$ [since $T_0=emptyset$]
$=I_0-bigcup_{n=0}^{infty}T_{n+1}=I_0-bigcup_{n=0}^{infty}bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{m=0}^{infty} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(frac{3k+1}{3^{n+1}}, frac{3k+2}{3^{n+1}}right)$.
This completes the proof.
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
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The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
|
show 3 more comments
$begingroup$
The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
|
show 3 more comments
$begingroup$
The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
The finite union in (1) is just the expression for $C_n$; you can prove its truth by induction. And $C = bigcap_n C_n$ by definition.
Then (2) follows by de Morgan: the complement of $C_n$ in $[0,1]$ is just a finite union of open intervals and the complement of $C$ is just the union of the complements of the $C_n$. You then take the complement of that to get $C$ back.
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:26
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
|
show 3 more comments
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
Hi @Henno. Let $T_n=bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$. As a result, $T_0=[0,frac{1}{3}]cup [frac{2}{3},1]$ and $T_1=[0,frac{1}{9}]cup [frac{2}{9},frac{3}{9}]cup [frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}] cup[frac{6}{9},frac{7}{9}]cup [frac{8}{9},1]$. As a result, $T_0=C_1$, but $T_1neq C_2$. I am unable to verify that the formulas given in Wikipedia is correct or not. Please help me check the formula!
$endgroup$
– Le Anh Dung
Jan 19 at 12:11
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung every $C_n$ is a disjoint union of $2^n$ many intervals of length $frac{1}{3^n}$, so the formulae on the wikiwand page are wrong.
$endgroup$
– Henno Brandsma
Jan 19 at 12:24
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
@LeAnhDung also see my answer here for a more accurate description.
$endgroup$
– Henno Brandsma
Jan 19 at 12:30
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
Hi @Henno, I think that formula is correct. Although $T_n=:bigcup_{k=0}^{3^n-1} left(left[frac{3k+0}{3^{n+1}},,, frac{3k+1}{3^{n+1}}right] cup left[frac{3k+2}{3^{n+1}},,,frac{3k+3}{3^{n+1}}right] right)$ is not equal to $C_n$. But the redundant intervals in $T_{n+1}$ and $T_n$ are actually disjoint. As you can see in my previous comment, $[frac{3}{9},frac{4}{9}]cup [frac{5}{9},frac{6}{9}]$ from $T_1$ are disjoint with $T_0$. From this observation, I guess $T_1cap T_0=C_2$ and $bigcap {T_0,cdots,T_n}=C_{n+1}$.
$endgroup$
– Le Anh Dung
Jan 19 at 12:33
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
$begingroup$
I am mainly concerned with the explicit formulas of Cantor set, while your answer in that link does not address that :)
$endgroup$
– Le Anh Dung
Jan 19 at 12:34
|
show 3 more comments
$begingroup$
I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!
Let $mathbf{L}(x)=dfrac{x}{3}$ and $mathbf{R}(x)=dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=mathbf{L}(I_n) cup mathbf{R}(I_n)$. Cantor set is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty} I_n$$
Theorem: $$mathcal{C} = [0,1] - bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left( frac{3k+1}{3^{n+1}} ,frac{3k+2}{3^{n+1}}right)$$
Let $T_n=I_0-I_n$ for all $ninBbb N$.
Lemma 1: $$T_{n+1}=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)$$
Proof:
Notice that $mathbf{R}(T_n)=mathbf{R}(I_0-I_n)=mathbf{R}(I_0)- mathbf{R}(I_n) subseteq mathbf{R}(I_0) subseteq I_0-mathbf{L}(I_0)$. Thus $(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)=mathbf{R}(T_n)$. Similarly, $mathbf{L}(T_n) subseteq mathbf{L}(I_0) subseteq I_0 -mathbf{R}(I_0)$ and thus $(I_0-mathbf{R}(I_0)) capmathbf{L}(T_n)=mathbf{L}(T_n)$. Moreover, $mathbf{L}(T_n)capmathbf{R}(T_n) subseteq mathbf{L}(I_0)capmathbf{R}(I_0)=emptyset$.
By the definition of $T_{n+1}$:
$begin{align}T_{n+1} &=I_0-I_{n+1}\ &=I_0-(mathbf{L}(I_n) cup mathbf{R}(I_n))\ &=(I_0-mathbf{L}(I_n)) cap (I_0 - mathbf{R}(I_n))\ &=[I_0-mathbf{L}(I_0-T_n)] cap [I_0 - mathbf{R}(I_0-T_n)]\ &=[I_0-(mathbf{L}(I_0)-mathbf{L}(T_n))] cap [I_0 - (mathbf{R}(I_0)-mathbf{R}(T_n))]\ &=[(I_0-mathbf{L}(I_0)) cup mathbf{L}(T_n)] cap [(I_0-mathbf{R}(I_0)) cup mathbf{R}(T_n)]\ &= [(I_0-mathbf{L}(I_0)) cap (I_0-mathbf{R}(I_0))] cup [(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)] cup [mathbf{L}(T_n)cap(I_0-mathbf{R}(I_0))] cup [mathbf{L}(T_n)capmathbf{R}(T_n)]\ &=[I_0 -mathbf{(L}(I_0)cupmathbf{R}(I_0))]cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=(I_0-I_1)cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &= T_1 cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)end{align}$
Lemma 2: $$bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) = left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$$
Proof:
Let $t=m+1$. Then $RHS=left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]$.
Moreover, $bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^{3^0-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^0 left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)=$ $left(frac{1}{3},frac{2}{3} right)$.
It follows that $RHS = left [ bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) right] cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]=$ $bigcup_{t=0}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) =LHS$.
Lemma 3: $$T_{n+1}=bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$$
Proof:
We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.
Notice that $frac{1}{3} < frac{3k+1}{3^{m+2}} < frac{3k+2}{3^{m+2}} < frac{2}{3}$ for all $3^m le k le 2.3^m-1$. It follows that $bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) subseteq left(frac{1}{3},frac{2}{3} right)$ and thus $left(frac{1}{3},frac{2}{3} right)= left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ]$.
By Lemma 1,
$T_{(n+1)+1}=T_{n+2}=mathbf{L}(T_{n+1}) cup T_1 cup mathbf{R}(T_{n+1})$
$=mathbf{L}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right) cup left(frac{1}{3},frac{2}{3} right) cup mathbf{R}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right)$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1+2.3^{m+1}}{3^{m+2}}, frac{3k+2+2.3^{m+1}}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3(k+2.3^m)+1}{3^{m+2}}, frac{3(k+2.3^m)+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left [bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ] cup left[ bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$= left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$ by Lemma 2
We proceed to prove our main theorem.
$mathcal{C}:=bigcap_{n=0}^{infty} I_n=bigcap_{n=0}^{infty} (I_0-T_n)=I_0-bigcup_{n=0}^{infty} T_n=I_0-bigcup_{n=1}^{infty}T_n$ [since $T_0=emptyset$]
$=I_0-bigcup_{n=0}^{infty}T_{n+1}=I_0-bigcup_{n=0}^{infty}bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{m=0}^{infty} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(frac{3k+1}{3^{n+1}}, frac{3k+2}{3^{n+1}}right)$.
This completes the proof.
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
$endgroup$
add a comment |
$begingroup$
I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!
Let $mathbf{L}(x)=dfrac{x}{3}$ and $mathbf{R}(x)=dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=mathbf{L}(I_n) cup mathbf{R}(I_n)$. Cantor set is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty} I_n$$
Theorem: $$mathcal{C} = [0,1] - bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left( frac{3k+1}{3^{n+1}} ,frac{3k+2}{3^{n+1}}right)$$
Let $T_n=I_0-I_n$ for all $ninBbb N$.
Lemma 1: $$T_{n+1}=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)$$
Proof:
Notice that $mathbf{R}(T_n)=mathbf{R}(I_0-I_n)=mathbf{R}(I_0)- mathbf{R}(I_n) subseteq mathbf{R}(I_0) subseteq I_0-mathbf{L}(I_0)$. Thus $(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)=mathbf{R}(T_n)$. Similarly, $mathbf{L}(T_n) subseteq mathbf{L}(I_0) subseteq I_0 -mathbf{R}(I_0)$ and thus $(I_0-mathbf{R}(I_0)) capmathbf{L}(T_n)=mathbf{L}(T_n)$. Moreover, $mathbf{L}(T_n)capmathbf{R}(T_n) subseteq mathbf{L}(I_0)capmathbf{R}(I_0)=emptyset$.
By the definition of $T_{n+1}$:
$begin{align}T_{n+1} &=I_0-I_{n+1}\ &=I_0-(mathbf{L}(I_n) cup mathbf{R}(I_n))\ &=(I_0-mathbf{L}(I_n)) cap (I_0 - mathbf{R}(I_n))\ &=[I_0-mathbf{L}(I_0-T_n)] cap [I_0 - mathbf{R}(I_0-T_n)]\ &=[I_0-(mathbf{L}(I_0)-mathbf{L}(T_n))] cap [I_0 - (mathbf{R}(I_0)-mathbf{R}(T_n))]\ &=[(I_0-mathbf{L}(I_0)) cup mathbf{L}(T_n)] cap [(I_0-mathbf{R}(I_0)) cup mathbf{R}(T_n)]\ &= [(I_0-mathbf{L}(I_0)) cap (I_0-mathbf{R}(I_0))] cup [(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)] cup [mathbf{L}(T_n)cap(I_0-mathbf{R}(I_0))] cup [mathbf{L}(T_n)capmathbf{R}(T_n)]\ &=[I_0 -mathbf{(L}(I_0)cupmathbf{R}(I_0))]cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=(I_0-I_1)cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &= T_1 cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)end{align}$
Lemma 2: $$bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) = left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$$
Proof:
Let $t=m+1$. Then $RHS=left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]$.
Moreover, $bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^{3^0-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^0 left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)=$ $left(frac{1}{3},frac{2}{3} right)$.
It follows that $RHS = left [ bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) right] cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]=$ $bigcup_{t=0}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) =LHS$.
Lemma 3: $$T_{n+1}=bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$$
Proof:
We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.
Notice that $frac{1}{3} < frac{3k+1}{3^{m+2}} < frac{3k+2}{3^{m+2}} < frac{2}{3}$ for all $3^m le k le 2.3^m-1$. It follows that $bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) subseteq left(frac{1}{3},frac{2}{3} right)$ and thus $left(frac{1}{3},frac{2}{3} right)= left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ]$.
By Lemma 1,
$T_{(n+1)+1}=T_{n+2}=mathbf{L}(T_{n+1}) cup T_1 cup mathbf{R}(T_{n+1})$
$=mathbf{L}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right) cup left(frac{1}{3},frac{2}{3} right) cup mathbf{R}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right)$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1+2.3^{m+1}}{3^{m+2}}, frac{3k+2+2.3^{m+1}}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3(k+2.3^m)+1}{3^{m+2}}, frac{3(k+2.3^m)+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left [bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ] cup left[ bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$= left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$ by Lemma 2
We proceed to prove our main theorem.
$mathcal{C}:=bigcap_{n=0}^{infty} I_n=bigcap_{n=0}^{infty} (I_0-T_n)=I_0-bigcup_{n=0}^{infty} T_n=I_0-bigcup_{n=1}^{infty}T_n$ [since $T_0=emptyset$]
$=I_0-bigcup_{n=0}^{infty}T_{n+1}=I_0-bigcup_{n=0}^{infty}bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{m=0}^{infty} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(frac{3k+1}{3^{n+1}}, frac{3k+2}{3^{n+1}}right)$.
This completes the proof.
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
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I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!
Let $mathbf{L}(x)=dfrac{x}{3}$ and $mathbf{R}(x)=dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=mathbf{L}(I_n) cup mathbf{R}(I_n)$. Cantor set is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty} I_n$$
Theorem: $$mathcal{C} = [0,1] - bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left( frac{3k+1}{3^{n+1}} ,frac{3k+2}{3^{n+1}}right)$$
Let $T_n=I_0-I_n$ for all $ninBbb N$.
Lemma 1: $$T_{n+1}=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)$$
Proof:
Notice that $mathbf{R}(T_n)=mathbf{R}(I_0-I_n)=mathbf{R}(I_0)- mathbf{R}(I_n) subseteq mathbf{R}(I_0) subseteq I_0-mathbf{L}(I_0)$. Thus $(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)=mathbf{R}(T_n)$. Similarly, $mathbf{L}(T_n) subseteq mathbf{L}(I_0) subseteq I_0 -mathbf{R}(I_0)$ and thus $(I_0-mathbf{R}(I_0)) capmathbf{L}(T_n)=mathbf{L}(T_n)$. Moreover, $mathbf{L}(T_n)capmathbf{R}(T_n) subseteq mathbf{L}(I_0)capmathbf{R}(I_0)=emptyset$.
By the definition of $T_{n+1}$:
$begin{align}T_{n+1} &=I_0-I_{n+1}\ &=I_0-(mathbf{L}(I_n) cup mathbf{R}(I_n))\ &=(I_0-mathbf{L}(I_n)) cap (I_0 - mathbf{R}(I_n))\ &=[I_0-mathbf{L}(I_0-T_n)] cap [I_0 - mathbf{R}(I_0-T_n)]\ &=[I_0-(mathbf{L}(I_0)-mathbf{L}(T_n))] cap [I_0 - (mathbf{R}(I_0)-mathbf{R}(T_n))]\ &=[(I_0-mathbf{L}(I_0)) cup mathbf{L}(T_n)] cap [(I_0-mathbf{R}(I_0)) cup mathbf{R}(T_n)]\ &= [(I_0-mathbf{L}(I_0)) cap (I_0-mathbf{R}(I_0))] cup [(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)] cup [mathbf{L}(T_n)cap(I_0-mathbf{R}(I_0))] cup [mathbf{L}(T_n)capmathbf{R}(T_n)]\ &=[I_0 -mathbf{(L}(I_0)cupmathbf{R}(I_0))]cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=(I_0-I_1)cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &= T_1 cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)end{align}$
Lemma 2: $$bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) = left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$$
Proof:
Let $t=m+1$. Then $RHS=left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]$.
Moreover, $bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^{3^0-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^0 left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)=$ $left(frac{1}{3},frac{2}{3} right)$.
It follows that $RHS = left [ bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) right] cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]=$ $bigcup_{t=0}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) =LHS$.
Lemma 3: $$T_{n+1}=bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$$
Proof:
We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.
Notice that $frac{1}{3} < frac{3k+1}{3^{m+2}} < frac{3k+2}{3^{m+2}} < frac{2}{3}$ for all $3^m le k le 2.3^m-1$. It follows that $bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) subseteq left(frac{1}{3},frac{2}{3} right)$ and thus $left(frac{1}{3},frac{2}{3} right)= left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ]$.
By Lemma 1,
$T_{(n+1)+1}=T_{n+2}=mathbf{L}(T_{n+1}) cup T_1 cup mathbf{R}(T_{n+1})$
$=mathbf{L}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right) cup left(frac{1}{3},frac{2}{3} right) cup mathbf{R}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right)$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1+2.3^{m+1}}{3^{m+2}}, frac{3k+2+2.3^{m+1}}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3(k+2.3^m)+1}{3^{m+2}}, frac{3(k+2.3^m)+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left [bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ] cup left[ bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$= left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$ by Lemma 2
We proceed to prove our main theorem.
$mathcal{C}:=bigcap_{n=0}^{infty} I_n=bigcap_{n=0}^{infty} (I_0-T_n)=I_0-bigcup_{n=0}^{infty} T_n=I_0-bigcup_{n=1}^{infty}T_n$ [since $T_0=emptyset$]
$=I_0-bigcup_{n=0}^{infty}T_{n+1}=I_0-bigcup_{n=0}^{infty}bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{m=0}^{infty} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(frac{3k+1}{3^{n+1}}, frac{3k+2}{3^{n+1}}right)$.
This completes the proof.
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
$endgroup$
I have figured out the proof and posted it as an answer here. It will be great and beneficial if someone helps me verify my attempt. Thank you so much!
Let $mathbf{L}(x)=dfrac{x}{3}$ and $mathbf{R}(x)=dfrac{x+2}{3}$. Let $I_0=[0,1]$ and $I_{n+1}=mathbf{L}(I_n) cup mathbf{R}(I_n)$. Cantor set is defined as follows: $$mathcal{C}:=bigcap_{n=0}^{infty} I_n$$
Theorem: $$mathcal{C} = [0,1] - bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left( frac{3k+1}{3^{n+1}} ,frac{3k+2}{3^{n+1}}right)$$
Let $T_n=I_0-I_n$ for all $ninBbb N$.
Lemma 1: $$T_{n+1}=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)$$
Proof:
Notice that $mathbf{R}(T_n)=mathbf{R}(I_0-I_n)=mathbf{R}(I_0)- mathbf{R}(I_n) subseteq mathbf{R}(I_0) subseteq I_0-mathbf{L}(I_0)$. Thus $(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)=mathbf{R}(T_n)$. Similarly, $mathbf{L}(T_n) subseteq mathbf{L}(I_0) subseteq I_0 -mathbf{R}(I_0)$ and thus $(I_0-mathbf{R}(I_0)) capmathbf{L}(T_n)=mathbf{L}(T_n)$. Moreover, $mathbf{L}(T_n)capmathbf{R}(T_n) subseteq mathbf{L}(I_0)capmathbf{R}(I_0)=emptyset$.
By the definition of $T_{n+1}$:
$begin{align}T_{n+1} &=I_0-I_{n+1}\ &=I_0-(mathbf{L}(I_n) cup mathbf{R}(I_n))\ &=(I_0-mathbf{L}(I_n)) cap (I_0 - mathbf{R}(I_n))\ &=[I_0-mathbf{L}(I_0-T_n)] cap [I_0 - mathbf{R}(I_0-T_n)]\ &=[I_0-(mathbf{L}(I_0)-mathbf{L}(T_n))] cap [I_0 - (mathbf{R}(I_0)-mathbf{R}(T_n))]\ &=[(I_0-mathbf{L}(I_0)) cup mathbf{L}(T_n)] cap [(I_0-mathbf{R}(I_0)) cup mathbf{R}(T_n)]\ &= [(I_0-mathbf{L}(I_0)) cap (I_0-mathbf{R}(I_0))] cup [(I_0-mathbf{L}(I_0)) capmathbf{R}(T_n)] cup [mathbf{L}(T_n)cap(I_0-mathbf{R}(I_0))] cup [mathbf{L}(T_n)capmathbf{R}(T_n)]\ &=[I_0 -mathbf{(L}(I_0)cupmathbf{R}(I_0))]cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=(I_0-I_1)cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &= T_1 cup mathbf{R}(T_n) cup mathbf{L}(T_n)\ &=mathbf{L}(T_n) cup T_1 cup mathbf{R}(T_n)end{align}$
Lemma 2: $$bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) = left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$$
Proof:
Let $t=m+1$. Then $RHS=left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]$.
Moreover, $bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^{3^0-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{k=0}^0 left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)=$ $left(frac{1}{3},frac{2}{3} right)$.
It follows that $RHS = left [ bigcup_{t=0}^0 bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) right] cup left[bigcup_{t=1}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right)right]=$ $bigcup_{t=0}^{n+1} bigcup_{k=0}^{3^t-1} left(frac{3k+1}{3^{t+1}}, frac{3k+2}{3^{t+1}}right) = bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right) =LHS$.
Lemma 3: $$T_{n+1}=bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$$
Proof:
We prove the assertion by induction on $n$. It is clear that it trivially holds for $n=0$. Let it hold for a positive integer $n$.
Notice that $frac{1}{3} < frac{3k+1}{3^{m+2}} < frac{3k+2}{3^{m+2}} < frac{2}{3}$ for all $3^m le k le 2.3^m-1$. It follows that $bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) subseteq left(frac{1}{3},frac{2}{3} right)$ and thus $left(frac{1}{3},frac{2}{3} right)= left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ]$.
By Lemma 1,
$T_{(n+1)+1}=T_{n+2}=mathbf{L}(T_{n+1}) cup T_1 cup mathbf{R}(T_{n+1})$
$=mathbf{L}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right) cup left(frac{1}{3},frac{2}{3} right) cup mathbf{R}left(bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)right)$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1+2.3^{m+1}}{3^{m+2}}, frac{3k+2+2.3^{m+1}}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3(k+2.3^m)+1}{3^{m+2}}, frac{3(k+2.3^m)+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left(frac{1}{3},frac{2}{3} right) cup left [bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=left[bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right] cup left [ left(frac{1}{3},frac{2}{3} right) cup left( bigcup_{m=0}^n bigcup_{k=3^{m}}^{2.3^m-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right) right)right ] cup left[ bigcup_{m=0}^n bigcup_{k=2.3^{m}}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$= left(frac{1}{3},frac{2}{3} right) cup left[bigcup_{m=0}^n bigcup_{k=0}^{3^{m+1}-1} left(frac{3k+1}{3^{m+2}}, frac{3k+2}{3^{m+2}}right)right]$
$=bigcup_{m=0}^{n+1} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$ by Lemma 2
We proceed to prove our main theorem.
$mathcal{C}:=bigcap_{n=0}^{infty} I_n=bigcap_{n=0}^{infty} (I_0-T_n)=I_0-bigcup_{n=0}^{infty} T_n=I_0-bigcup_{n=1}^{infty}T_n$ [since $T_0=emptyset$]
$=I_0-bigcup_{n=0}^{infty}T_{n+1}=I_0-bigcup_{n=0}^{infty}bigcup_{m=0}^n bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{m=0}^{infty} bigcup_{k=0}^{3^m-1} left(frac{3k+1}{3^{m+1}}, frac{3k+2}{3^{m+1}}right)$
$=I_0-bigcup_{n=0}^{infty} bigcup_{k=0}^{3^n-1} left(frac{3k+1}{3^{n+1}}, frac{3k+2}{3^{n+1}}right)$.
This completes the proof.
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
answered Jan 20 at 15:06
Le Anh DungLe Anh Dung
1,2911621
1,2911621
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1
$begingroup$
I changed the finite intersection to a finite union, as per the page. BTW It's not wikipedia's page but wikiwand...? A sort of commercial (ads (!)) rip-off it seems.
$endgroup$
– Henno Brandsma
Jan 19 at 9:18
$begingroup$
Thank you so much @HennoBrandsma! You are correct.
$endgroup$
– Le Anh Dung
Jan 19 at 9:20
1
$begingroup$
The second formula is wrong semantically: you cannot substract a set of intervals (!) from $[0,1]$. There are unions missing..
$endgroup$
– Henno Brandsma
Jan 19 at 9:20
$begingroup$
@HennoBrandsma I have fixed it :)
$endgroup$
– Le Anh Dung
Jan 19 at 9:22
$begingroup$
Since $mathcal{C}=bigcup_{nge 0}C_n$, to avoid confusion your notation in (1) should probably start the dummy variable $n$ at $0$, not at $1$.
$endgroup$
– J.G.
Jan 19 at 9:35