How to prove this closed formula for Cantor set?
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Let $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
Theorem: $$C_n=bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!
My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$
$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$
cantor-set
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
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$begingroup$
Let $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
Theorem: $$C_n=bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!
My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$
$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$
cantor-set
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
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Btw, where did you saw this form? This is the first time I see it
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– Holo
Jan 22 at 15:20
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@Holo I got it from here.
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– Le Anh Dung
Jan 22 at 16:44
add a comment |
$begingroup$
Let $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
Theorem: $$C_n=bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!
My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$
$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$
cantor-set
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
$endgroup$
Let $C_0=[0,1]$ and $C_{n+1} = dfrac{C_n}{3} bigcupleft(dfrac{2}{3}+dfrac{C_n}{3}right)$.
Theorem: $$C_n=bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]$$
I have tried to prove this assertion by induction on $n$, but to no avail. I am stuck at inductive step.
Please shed me some light to accomplish the proof. Thank you so much!
My attempt:
The formula is trivially true for $n=0$. Let it hold for $n$.
$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$
$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$
$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$
cantor-set
cantor-set
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
edited Jan 23 at 6:07
Le Anh Dung
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
asked Jan 21 at 9:52
Le Anh DungLe Anh Dung
1,2131621
1,2131621
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Btw, where did you saw this form? This is the first time I see it
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– Holo
Jan 22 at 15:20
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@Holo I got it from here.
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– Le Anh Dung
Jan 22 at 16:44
add a comment |
$begingroup$
Btw, where did you saw this form? This is the first time I see it
$endgroup$
– Holo
Jan 22 at 15:20
$begingroup$
@Holo I got it from here.
$endgroup$
– Le Anh Dung
Jan 22 at 16:44
$begingroup$
Btw, where did you saw this form? This is the first time I see it
$endgroup$
– Holo
Jan 22 at 15:20
$begingroup$
Btw, where did you saw this form? This is the first time I see it
$endgroup$
– Holo
Jan 22 at 15:20
$begingroup$
@Holo I got it from here.
$endgroup$
– Le Anh Dung
Jan 22 at 16:44
$begingroup$
@Holo I got it from here.
$endgroup$
– Le Anh Dung
Jan 22 at 16:44
add a comment |
2 Answers
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Notice that
$$bigcap_{m=0}^{n+1}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]=left(bigcap_{m=0}^{n}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right)cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]\=C_ncap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,frac{1}{3^{n+1}}],ldots,[frac{3^{n+1}-1}{3^{n+1}},1]$.
$bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$ is just taking the even parts of the list together, notice that $C_{n+1}subseteq bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $xin C_n$, it had to be in the form $0.d_1d_2ldots_3$ where $iin {1,2,ldots, n-1}$ implies $d_iin{0,2}$, or in other words $xin[0.d_1ldots d_{n-1}0_3,0.d_1ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$
answered Jan 22 at 15:14
HoloHolo
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After several hours of thinking, I have figured out a proof and posted it here as an answer.
Lemma: $$2 . leftlfloor frac{3^{m}}{2}rightrfloor = 3^m-1$$
Proof:
We prove this assertion by induction on $n$. The statement is trivially true for $n=0$. Let it hold for $n$.
$2 . leftlfloor dfrac{3^{m+1}}{2}rightrfloor = 2 . leftlfloor dfrac{2.3^{m}+3^m}{2}rightrfloor=2 . leftlfloor 3^m+ dfrac{3^m}{2}rightrfloor=2left(3^m+leftlfloor dfrac{3^{m}}{2}rightrfloorright)=$ $2.3^m+2 . leftlfloor dfrac{3^{m}}{2}rightrfloor$ $=2.3^m+(3^m-1)=3^{m+1}-1$. This completes the proof.
The formula is trivially true for $n=0$. Let it hold for $n$.
$begin{align}C_{n+1} &=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)\ &=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)\ &=left(bigcap_{t=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{t-1}}{2}rightrfloor}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right) cup left(bigcap_{t=1}^{n+1}bigcup_{k=3^{t-1}}^{leftlfloor frac{3^{t-1}}{2}rightrfloor+3^{t-1}}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right)
text{ Let }t=m+1\ &=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) text{ Let }m=tend{align}$
We have some observations.
1.
$bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=left[0,frac{1}{3}right]$.
2.
$frac{1}{3}<frac{2k}{3^{m}}$ and $frac{2k+1}{3^{m}}<frac{2}{3}$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. Then $left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. It follows that $bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ and thus $left(bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) =emptyset$.
3.
From 1. and 2., we get
$begin{align}bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] &= bigcap_{m=1}^{n+1} left[left( bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor+1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) right]\&=bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
4.
$2 left(leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1} right)=2 left(leftlfloor frac{3^{m-1}}{2}rightrfloorright)+2.3^{m-1}=(3^{m-1}-1)+2.3^{m-1}=3.3^{m-1}-1=$ $3^m-1=2 . leftlfloor frac{3^{m}}{2}rightrfloor$. Hence $leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}=leftlfloor frac{3^{m}}{2}rightrfloor$.
As a result,
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right)$$
Let $I_m^k=left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$.
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1} I_m^k right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor} I_m^k right)$$
$frac{2k+1}{3^{m}}<frac{2}{3}$ for all $k le 3^{m-1}-1$ and $frac{2}{3} le frac{2k}{3^{m}}$ for all $k ge 3^{m-1}$ $implies$ $left(bigcup_{k=0}^{3^{m_1-1}-1} I_m^kright) cap left( bigcup_{k=3^{m_2-1}}^{leftlfloor frac{3^{m_2}}{2}rightrfloor} I_m^k right) =emptyset$ for all $m_1,m_2 le n+1$.
Lemma: Let $I_n= {iinBbb N mid 0 le i le n}$ and $(A_i mid iin I_n)$, $(B_i mid iin I_n)$ be collections of nonempty sets such that $A_i cap B_j =emptyset$ for all $i,jin I_n$. Then $$left(bigcap_{iin I_n} A_i right) cup left(bigcap_{iin I_n} B_iright)= bigcap_{iin I_n} (A_icup B_i)$$
Proof: It is easy to verify this lemma.
We apply this lemma for $C_{n+1}$ and get $$C_{n+1}=bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$$
First, $C_{n+1} subseteq [0,1]$ and thus $C_{n+1}=C_{n+1} cap [0,1]$.
Second, $bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] =[0,1]$.
As a result,
$begin{align}C_{n+1}&=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap [0,1]\&= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right)\&=bigcap_{m=0}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
This completes the proof.
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
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$begingroup$
Notice that
$$bigcap_{m=0}^{n+1}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]=left(bigcap_{m=0}^{n}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right)cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]\=C_ncap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,frac{1}{3^{n+1}}],ldots,[frac{3^{n+1}-1}{3^{n+1}},1]$.
$bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$ is just taking the even parts of the list together, notice that $C_{n+1}subseteq bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $xin C_n$, it had to be in the form $0.d_1d_2ldots_3$ where $iin {1,2,ldots, n-1}$ implies $d_iin{0,2}$, or in other words $xin[0.d_1ldots d_{n-1}0_3,0.d_1ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$
answered Jan 22 at 15:14
HoloHolo
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$begingroup$
Notice that
$$bigcap_{m=0}^{n+1}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]=left(bigcap_{m=0}^{n}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right)cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]\=C_ncap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,frac{1}{3^{n+1}}],ldots,[frac{3^{n+1}-1}{3^{n+1}},1]$.
$bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$ is just taking the even parts of the list together, notice that $C_{n+1}subseteq bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $xin C_n$, it had to be in the form $0.d_1d_2ldots_3$ where $iin {1,2,ldots, n-1}$ implies $d_iin{0,2}$, or in other words $xin[0.d_1ldots d_{n-1}0_3,0.d_1ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$
answered Jan 22 at 15:14
HoloHolo
5,75421131
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$begingroup$
Notice that
$$bigcap_{m=0}^{n+1}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]=left(bigcap_{m=0}^{n}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right)cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]\=C_ncap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,frac{1}{3^{n+1}}],ldots,[frac{3^{n+1}-1}{3^{n+1}},1]$.
$bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$ is just taking the even parts of the list together, notice that $C_{n+1}subseteq bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $xin C_n$, it had to be in the form $0.d_1d_2ldots_3$ where $iin {1,2,ldots, n-1}$ implies $d_iin{0,2}$, or in other words $xin[0.d_1ldots d_{n-1}0_3,0.d_1ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$
answered Jan 22 at 15:14
HoloHolo
5,75421131
$endgroup$
Notice that
$$bigcap_{m=0}^{n+1}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]=left(bigcap_{m=0}^{n}bigcup_{k=0}^{lfloor 3^m/2rfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right)cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]\=C_ncap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$$
We may notice that if we divide the interval $[0,1]$ into $3^{n+1}$ parts we will get $[0,frac{1}{3^{n+1}}],ldots,[frac{3^{n+1}-1}{3^{n+1}},1]$.
$bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]$ is just taking the even parts of the list together, notice that $C_{n+1}subseteq bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]cap C_n$, now notice that the even intervals of this stage are always the first and last thirds of the even intervals of the previous stages, so if $xin C_n$, it had to be in the form $0.d_1d_2ldots_3$ where $iin {1,2,ldots, n-1}$ implies $d_iin{0,2}$, or in other words $xin[0.d_1ldots d_{n-1}0_3,0.d_1ldots d_{n-1}2_3]$, $d_n$ will be $0$ on the first third and $2$ on the last third, so $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$
answered Jan 22 at 15:14
HoloHolo
5,75421131
edited Jan 23 at 16:03
answered Jan 22 at 15:14
HoloHolo
5,75421131
answered Jan 22 at 15:14
HoloHolo
5,75421131
answered Jan 22 at 15:14
HoloHolo
5,75421131
5,75421131
add a comment |
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After several hours of thinking, I have figured out a proof and posted it here as an answer.
Lemma: $$2 . leftlfloor frac{3^{m}}{2}rightrfloor = 3^m-1$$
Proof:
We prove this assertion by induction on $n$. The statement is trivially true for $n=0$. Let it hold for $n$.
$2 . leftlfloor dfrac{3^{m+1}}{2}rightrfloor = 2 . leftlfloor dfrac{2.3^{m}+3^m}{2}rightrfloor=2 . leftlfloor 3^m+ dfrac{3^m}{2}rightrfloor=2left(3^m+leftlfloor dfrac{3^{m}}{2}rightrfloorright)=$ $2.3^m+2 . leftlfloor dfrac{3^{m}}{2}rightrfloor$ $=2.3^m+(3^m-1)=3^{m+1}-1$. This completes the proof.
The formula is trivially true for $n=0$. Let it hold for $n$.
$begin{align}C_{n+1} &=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)\ &=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)\ &=left(bigcap_{t=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{t-1}}{2}rightrfloor}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right) cup left(bigcap_{t=1}^{n+1}bigcup_{k=3^{t-1}}^{leftlfloor frac{3^{t-1}}{2}rightrfloor+3^{t-1}}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right)
text{ Let }t=m+1\ &=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) text{ Let }m=tend{align}$
We have some observations.
1.
$bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=left[0,frac{1}{3}right]$.
2.
$frac{1}{3}<frac{2k}{3^{m}}$ and $frac{2k+1}{3^{m}}<frac{2}{3}$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. Then $left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. It follows that $bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ and thus $left(bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) =emptyset$.
3.
From 1. and 2., we get
$begin{align}bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] &= bigcap_{m=1}^{n+1} left[left( bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor+1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) right]\&=bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
4.
$2 left(leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1} right)=2 left(leftlfloor frac{3^{m-1}}{2}rightrfloorright)+2.3^{m-1}=(3^{m-1}-1)+2.3^{m-1}=3.3^{m-1}-1=$ $3^m-1=2 . leftlfloor frac{3^{m}}{2}rightrfloor$. Hence $leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}=leftlfloor frac{3^{m}}{2}rightrfloor$.
As a result,
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right)$$
Let $I_m^k=left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$.
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1} I_m^k right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor} I_m^k right)$$
$frac{2k+1}{3^{m}}<frac{2}{3}$ for all $k le 3^{m-1}-1$ and $frac{2}{3} le frac{2k}{3^{m}}$ for all $k ge 3^{m-1}$ $implies$ $left(bigcup_{k=0}^{3^{m_1-1}-1} I_m^kright) cap left( bigcup_{k=3^{m_2-1}}^{leftlfloor frac{3^{m_2}}{2}rightrfloor} I_m^k right) =emptyset$ for all $m_1,m_2 le n+1$.
Lemma: Let $I_n= {iinBbb N mid 0 le i le n}$ and $(A_i mid iin I_n)$, $(B_i mid iin I_n)$ be collections of nonempty sets such that $A_i cap B_j =emptyset$ for all $i,jin I_n$. Then $$left(bigcap_{iin I_n} A_i right) cup left(bigcap_{iin I_n} B_iright)= bigcap_{iin I_n} (A_icup B_i)$$
Proof: It is easy to verify this lemma.
We apply this lemma for $C_{n+1}$ and get $$C_{n+1}=bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$$
First, $C_{n+1} subseteq [0,1]$ and thus $C_{n+1}=C_{n+1} cap [0,1]$.
Second, $bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] =[0,1]$.
As a result,
$begin{align}C_{n+1}&=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap [0,1]\&= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right)\&=bigcap_{m=0}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
This completes the proof.
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
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After several hours of thinking, I have figured out a proof and posted it here as an answer.
Lemma: $$2 . leftlfloor frac{3^{m}}{2}rightrfloor = 3^m-1$$
Proof:
We prove this assertion by induction on $n$. The statement is trivially true for $n=0$. Let it hold for $n$.
$2 . leftlfloor dfrac{3^{m+1}}{2}rightrfloor = 2 . leftlfloor dfrac{2.3^{m}+3^m}{2}rightrfloor=2 . leftlfloor 3^m+ dfrac{3^m}{2}rightrfloor=2left(3^m+leftlfloor dfrac{3^{m}}{2}rightrfloorright)=$ $2.3^m+2 . leftlfloor dfrac{3^{m}}{2}rightrfloor$ $=2.3^m+(3^m-1)=3^{m+1}-1$. This completes the proof.
The formula is trivially true for $n=0$. Let it hold for $n$.
$begin{align}C_{n+1} &=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)\ &=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)\ &=left(bigcap_{t=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{t-1}}{2}rightrfloor}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right) cup left(bigcap_{t=1}^{n+1}bigcup_{k=3^{t-1}}^{leftlfloor frac{3^{t-1}}{2}rightrfloor+3^{t-1}}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right)
text{ Let }t=m+1\ &=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) text{ Let }m=tend{align}$
We have some observations.
1.
$bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=left[0,frac{1}{3}right]$.
2.
$frac{1}{3}<frac{2k}{3^{m}}$ and $frac{2k+1}{3^{m}}<frac{2}{3}$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. Then $left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. It follows that $bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ and thus $left(bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) =emptyset$.
3.
From 1. and 2., we get
$begin{align}bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] &= bigcap_{m=1}^{n+1} left[left( bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor+1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) right]\&=bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
4.
$2 left(leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1} right)=2 left(leftlfloor frac{3^{m-1}}{2}rightrfloorright)+2.3^{m-1}=(3^{m-1}-1)+2.3^{m-1}=3.3^{m-1}-1=$ $3^m-1=2 . leftlfloor frac{3^{m}}{2}rightrfloor$. Hence $leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}=leftlfloor frac{3^{m}}{2}rightrfloor$.
As a result,
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right)$$
Let $I_m^k=left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$.
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1} I_m^k right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor} I_m^k right)$$
$frac{2k+1}{3^{m}}<frac{2}{3}$ for all $k le 3^{m-1}-1$ and $frac{2}{3} le frac{2k}{3^{m}}$ for all $k ge 3^{m-1}$ $implies$ $left(bigcup_{k=0}^{3^{m_1-1}-1} I_m^kright) cap left( bigcup_{k=3^{m_2-1}}^{leftlfloor frac{3^{m_2}}{2}rightrfloor} I_m^k right) =emptyset$ for all $m_1,m_2 le n+1$.
Lemma: Let $I_n= {iinBbb N mid 0 le i le n}$ and $(A_i mid iin I_n)$, $(B_i mid iin I_n)$ be collections of nonempty sets such that $A_i cap B_j =emptyset$ for all $i,jin I_n$. Then $$left(bigcap_{iin I_n} A_i right) cup left(bigcap_{iin I_n} B_iright)= bigcap_{iin I_n} (A_icup B_i)$$
Proof: It is easy to verify this lemma.
We apply this lemma for $C_{n+1}$ and get $$C_{n+1}=bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$$
First, $C_{n+1} subseteq [0,1]$ and thus $C_{n+1}=C_{n+1} cap [0,1]$.
Second, $bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] =[0,1]$.
As a result,
$begin{align}C_{n+1}&=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap [0,1]\&= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right)\&=bigcap_{m=0}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
This completes the proof.
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
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After several hours of thinking, I have figured out a proof and posted it here as an answer.
Lemma: $$2 . leftlfloor frac{3^{m}}{2}rightrfloor = 3^m-1$$
Proof:
We prove this assertion by induction on $n$. The statement is trivially true for $n=0$. Let it hold for $n$.
$2 . leftlfloor dfrac{3^{m+1}}{2}rightrfloor = 2 . leftlfloor dfrac{2.3^{m}+3^m}{2}rightrfloor=2 . leftlfloor 3^m+ dfrac{3^m}{2}rightrfloor=2left(3^m+leftlfloor dfrac{3^{m}}{2}rightrfloorright)=$ $2.3^m+2 . leftlfloor dfrac{3^{m}}{2}rightrfloor$ $=2.3^m+(3^m-1)=3^{m+1}-1$. This completes the proof.
The formula is trivially true for $n=0$. Let it hold for $n$.
$begin{align}C_{n+1} &=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)\ &=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)\ &=left(bigcap_{t=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{t-1}}{2}rightrfloor}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right) cup left(bigcap_{t=1}^{n+1}bigcup_{k=3^{t-1}}^{leftlfloor frac{3^{t-1}}{2}rightrfloor+3^{t-1}}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right)
text{ Let }t=m+1\ &=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) text{ Let }m=tend{align}$
We have some observations.
1.
$bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=left[0,frac{1}{3}right]$.
2.
$frac{1}{3}<frac{2k}{3^{m}}$ and $frac{2k+1}{3^{m}}<frac{2}{3}$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. Then $left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. It follows that $bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ and thus $left(bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) =emptyset$.
3.
From 1. and 2., we get
$begin{align}bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] &= bigcap_{m=1}^{n+1} left[left( bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor+1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) right]\&=bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
4.
$2 left(leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1} right)=2 left(leftlfloor frac{3^{m-1}}{2}rightrfloorright)+2.3^{m-1}=(3^{m-1}-1)+2.3^{m-1}=3.3^{m-1}-1=$ $3^m-1=2 . leftlfloor frac{3^{m}}{2}rightrfloor$. Hence $leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}=leftlfloor frac{3^{m}}{2}rightrfloor$.
As a result,
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right)$$
Let $I_m^k=left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$.
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1} I_m^k right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor} I_m^k right)$$
$frac{2k+1}{3^{m}}<frac{2}{3}$ for all $k le 3^{m-1}-1$ and $frac{2}{3} le frac{2k}{3^{m}}$ for all $k ge 3^{m-1}$ $implies$ $left(bigcup_{k=0}^{3^{m_1-1}-1} I_m^kright) cap left( bigcup_{k=3^{m_2-1}}^{leftlfloor frac{3^{m_2}}{2}rightrfloor} I_m^k right) =emptyset$ for all $m_1,m_2 le n+1$.
Lemma: Let $I_n= {iinBbb N mid 0 le i le n}$ and $(A_i mid iin I_n)$, $(B_i mid iin I_n)$ be collections of nonempty sets such that $A_i cap B_j =emptyset$ for all $i,jin I_n$. Then $$left(bigcap_{iin I_n} A_i right) cup left(bigcap_{iin I_n} B_iright)= bigcap_{iin I_n} (A_icup B_i)$$
Proof: It is easy to verify this lemma.
We apply this lemma for $C_{n+1}$ and get $$C_{n+1}=bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$$
First, $C_{n+1} subseteq [0,1]$ and thus $C_{n+1}=C_{n+1} cap [0,1]$.
Second, $bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] =[0,1]$.
As a result,
$begin{align}C_{n+1}&=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap [0,1]\&= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right)\&=bigcap_{m=0}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
This completes the proof.
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
$endgroup$
After several hours of thinking, I have figured out a proof and posted it here as an answer.
Lemma: $$2 . leftlfloor frac{3^{m}}{2}rightrfloor = 3^m-1$$
Proof:
We prove this assertion by induction on $n$. The statement is trivially true for $n=0$. Let it hold for $n$.
$2 . leftlfloor dfrac{3^{m+1}}{2}rightrfloor = 2 . leftlfloor dfrac{2.3^{m}+3^m}{2}rightrfloor=2 . leftlfloor 3^m+ dfrac{3^m}{2}rightrfloor=2left(3^m+leftlfloor dfrac{3^{m}}{2}rightrfloorright)=$ $2.3^m+2 . leftlfloor dfrac{3^{m}}{2}rightrfloor$ $=2.3^m+(3^m-1)=3^{m+1}-1$. This completes the proof.
The formula is trivially true for $n=0$. Let it hold for $n$.
$begin{align}C_{n+1} &=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)\ &=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)\ &=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)\ &=left(bigcap_{t=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{t-1}}{2}rightrfloor}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right) cup left(bigcap_{t=1}^{n+1}bigcup_{k=3^{t-1}}^{leftlfloor frac{3^{t-1}}{2}rightrfloor+3^{t-1}}left[frac{2k}{3^{t}},frac{2k+1}{3^{t}}right]right)
text{ Let }t=m+1\ &=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) text{ Let }m=tend{align}$
We have some observations.
1.
$bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=left[0,frac{1}{3}right]$.
2.
$frac{1}{3}<frac{2k}{3^{m}}$ and $frac{2k+1}{3^{m}}<frac{2}{3}$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. Then $left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ for all $leftlfloor frac{3^{m-1}}{2}rightrfloor +1le k le 3^{m-1}-1$. It follows that $bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] subsetneq left(frac{1}{3},frac{2}{3}right)$ and thus $left(bigcap_{m=1}^{1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor +1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) =emptyset$.
3.
From 1. and 2., we get
$begin{align}bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] &= bigcap_{m=1}^{n+1} left[left( bigcup_{k=0}^{leftlfloor frac{3^{m-1}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcup_{k=leftlfloor frac{3^{m-1}}{2}rightrfloor+1}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) right]\&=bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
4.
$2 left(leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1} right)=2 left(leftlfloor frac{3^{m-1}}{2}rightrfloorright)+2.3^{m-1}=(3^{m-1}-1)+2.3^{m-1}=3.3^{m-1}-1=$ $3^m-1=2 . leftlfloor frac{3^{m}}{2}rightrfloor$. Hence $leftlfloor frac{3^{m-1}}{2}rightrfloor+3^{m-1}=leftlfloor frac{3^{m}}{2}rightrfloor$.
As a result,
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right)$$
Let $I_m^k=left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$.
$$C_{n+1}= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{3^{m-1}-1} I_m^k right) cup left(bigcap_{m=1}^{n+1}bigcup_{k=3^{m-1}}^{leftlfloor frac{3^{m}}{2}rightrfloor} I_m^k right)$$
$frac{2k+1}{3^{m}}<frac{2}{3}$ for all $k le 3^{m-1}-1$ and $frac{2}{3} le frac{2k}{3^{m}}$ for all $k ge 3^{m-1}$ $implies$ $left(bigcup_{k=0}^{3^{m_1-1}-1} I_m^kright) cap left( bigcup_{k=3^{m_2-1}}^{leftlfloor frac{3^{m_2}}{2}rightrfloor} I_m^k right) =emptyset$ for all $m_1,m_2 le n+1$.
Lemma: Let $I_n= {iinBbb N mid 0 le i le n}$ and $(A_i mid iin I_n)$, $(B_i mid iin I_n)$ be collections of nonempty sets such that $A_i cap B_j =emptyset$ for all $i,jin I_n$. Then $$left(bigcap_{iin I_n} A_i right) cup left(bigcap_{iin I_n} B_iright)= bigcap_{iin I_n} (A_icup B_i)$$
Proof: It is easy to verify this lemma.
We apply this lemma for $C_{n+1}$ and get $$C_{n+1}=bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]$$
First, $C_{n+1} subseteq [0,1]$ and thus $C_{n+1}=C_{n+1} cap [0,1]$.
Second, $bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]=bigcup_{k=0}^{0}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] =[0,1]$.
As a result,
$begin{align}C_{n+1}&=left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap [0,1]\&= left(bigcap_{m=1}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]right) cap left( bigcap_{m=0}^{0}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right] right)\&=bigcap_{m=0}^{n+1}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m}},frac{2k+1}{3^{m}}right]end{align}$
This completes the proof.
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
edited Jan 23 at 9:12
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
answered Jan 23 at 9:05
Le Anh DungLe Anh Dung
1,2131621
1,2131621
add a comment |
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Btw, where did you saw this form? This is the first time I see it
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– Holo
Jan 22 at 15:20
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@Holo I got it from here.
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– Le Anh Dung
Jan 22 at 16:44