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 $n$.



$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$



$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$






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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
















2












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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 $n$.



$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$



$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$






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  • $begingroup$
    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










  • $begingroup$
    @Holo I got it from here.
    $endgroup$
    – Le Anh Dung
    Jan 22 at 16:44














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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 $n$.



$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$



$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$






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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 $n$.



$$C_{n+1}=frac{C_n}{3} cupleft(frac{2}{3}+frac{C_n}{3}right)$$



$$=left(frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right]right) cup left(frac{2}{3}+frac{1}{3} bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^m},frac{2k+1}{3^m}right] right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k+2.3^m}{3^{m+1}},frac{2k+2.3^m+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2(k+3^m)}{3^{m+1}},frac{2(k+3^m)+1}{3^{m+1}}right]right)$$



$$=left(bigcap_{m=0}^{n}bigcup_{k=0}^{leftlfloor frac{3^{m}}{2}rightrfloor}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right) cup left(bigcap_{m=0}^{n}bigcup_{k=3^m}^{leftlfloor frac{3^{m}}{2}rightrfloor+3^m}left[frac{2k}{3^{m+1}},frac{2k+1}{3^{m+1}}right]right)$$






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edited Jan 23 at 6:07







Le Anh Dung

















asked Jan 21 at 9:52









Le Anh DungLe Anh Dung

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  • $begingroup$
    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










  • $begingroup$
    @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










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 $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$






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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.






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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 $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        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 $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          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 $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$






          share|cite|improve this answer











          $endgroup$



          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 $C_n cap bigcup_{k=0}^{lfloor 3^{n+1}/2rfloor}left[frac{2k}{3^{n+1}},frac{2k+1}{3^{n+1}}right]subseteq C_{n+1}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 23 at 16:03

























          answered Jan 22 at 15:14









          HoloHolo

          5,75421131




          5,75421131























              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$



                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 9:12

























                  answered Jan 23 at 9:05









                  Le Anh DungLe Anh Dung

                  1,2131621




                  1,2131621






























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