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Formula for canonical basis of the Cantor set

1

$begingroup$

Is there a natural formula for the canonical clopen subsets of the middle-thirds Cantor set $C$ which is based on elements $sigmain 2^<omega$:

$B(varnothing)=C$

$B(langle 0rangle)=[0,1/3]$; $B(langle 1rangle)=[2/3,1]$

$B(langle 0,0rangle)=[0,1/9]$; $B(langle 0,1rangle)=[2/9,1/3]$; $B(langle 1,0rangle)=[2/3,7/9]$; $B(langle 1,1rangle)=[8/9,1]$

...

Is there an explicit formula for the interval $B(sigma)$?

general-topology cantor-set

share|cite|improve this question

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

$endgroup$

add a comment | 

1

$begingroup$

Is there a natural formula for the canonical clopen subsets of the middle-thirds Cantor set $C$ which is based on elements $sigmain 2^<omega$:

$B(varnothing)=C$

$B(langle 0rangle)=[0,1/3]$; $B(langle 1rangle)=[2/3,1]$

$B(langle 0,0rangle)=[0,1/9]$; $B(langle 0,1rangle)=[2/9,1/3]$; $B(langle 1,0rangle)=[2/3,7/9]$; $B(langle 1,1rangle)=[8/9,1]$

...

Is there an explicit formula for the interval $B(sigma)$?

general-topology cantor-set

share|cite|improve this question

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

$endgroup$

add a comment | 

$begingroup$

Is there a natural formula for the canonical clopen subsets of the middle-thirds Cantor set $C$ which is based on elements $sigmain 2^<omega$:

$B(varnothing)=C$

$B(langle 0rangle)=[0,1/3]$; $B(langle 1rangle)=[2/3,1]$

$B(langle 0,0rangle)=[0,1/9]$; $B(langle 0,1rangle)=[2/9,1/3]$; $B(langle 1,0rangle)=[2/3,7/9]$; $B(langle 1,1rangle)=[8/9,1]$

...

Is there an explicit formula for the interval $B(sigma)$?

general-topology cantor-set

share|cite|improve this question

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

$endgroup$

share|cite|improve this question

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

share|cite|improve this question

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

edited Mar 25 at 21:04

Eric Wofsey

193k14221352

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

asked Mar 25 at 20:25

D.S. LiphamD.S. Lipham

1083

1 Answer
1

active

oldest

votes

1

$begingroup$

If $sigma$ is a sequence of length $n$ (which I will consider as a function $1,dots,nto0,1)$, then $B(sigma)=[a,b]cap C$ where $$a=sum_k=1^n frac2sigma(k)3^k$$ and $b=a+1/3^n$.

To verify that this formula is correct, you can check that it satisfies the recurrence corresponding to the "middle thirds" construction of $C$. Namely, if $B(sigma)=[a,b]cap C$, then $B(sigma0)=[a,c]cap C$ and $B(sigma 1)=[d,b] cap C$ where $c$ is $1/3$ of the way from $a$ to $b$ and $d$ is $2/3$ of the way from $a$ to $d$.

share|cite|improve this answer

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

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1

$begingroup$

If $sigma$ is a sequence of length $n$ (which I will consider as a function $1,dots,nto0,1)$, then $B(sigma)=[a,b]cap C$ where $$a=sum_k=1^n frac2sigma(k)3^k$$ and $b=a+1/3^n$.

To verify that this formula is correct, you can check that it satisfies the recurrence corresponding to the "middle thirds" construction of $C$. Namely, if $B(sigma)=[a,b]cap C$, then $B(sigma0)=[a,c]cap C$ and $B(sigma 1)=[d,b] cap C$ where $c$ is $1/3$ of the way from $a$ to $b$ and $d$ is $2/3$ of the way from $a$ to $d$.

share|cite|improve this answer

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

1

$begingroup$

If $sigma$ is a sequence of length $n$ (which I will consider as a function $1,dots,nto0,1)$, then $B(sigma)=[a,b]cap C$ where $$a=sum_k=1^n frac2sigma(k)3^k$$ and $b=a+1/3^n$.

To verify that this formula is correct, you can check that it satisfies the recurrence corresponding to the "middle thirds" construction of $C$. Namely, if $B(sigma)=[a,b]cap C$, then $B(sigma0)=[a,c]cap C$ and $B(sigma 1)=[d,b] cap C$ where $c$ is $1/3$ of the way from $a$ to $b$ and $d$ is $2/3$ of the way from $a$ to $d$.

share|cite|improve this answer

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

$begingroup$

If $sigma$ is a sequence of length $n$ (which I will consider as a function $1,dots,nto0,1)$, then $B(sigma)=[a,b]cap C$ where $$a=sum_k=1^n frac2sigma(k)3^k$$ and $b=a+1/3^n$.

To verify that this formula is correct, you can check that it satisfies the recurrence corresponding to the "middle thirds" construction of $C$. Namely, if $B(sigma)=[a,b]cap C$, then $B(sigma0)=[a,c]cap C$ and $B(sigma 1)=[d,b] cap C$ where $c$ is $1/3$ of the way from $a$ to $b$ and $d$ is $2/3$ of the way from $a$ to $d$.

share|cite|improve this answer

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

$endgroup$

share|cite|improve this answer

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352

answered Mar 25 at 21:00

Eric WofseyEric Wofsey

193k14221352