Fdtxjr

Multi tool use

Complexity of many constant time steps with occasional logarithmic steps

Mortgage adviser recommends a longer term than necessary combined with overpayments

How many things? AとBがふたつ

Strange behaviour of Check

How to set letter above or below the symbol?

Did the new image of black hole confirm the general theory of relativity?

Jazz greats knew nothing of modes. Why are they used to improvise on standards?

Why does tar appear to skip file contents when output file is /dev/null?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

Can a monk deflect thrown melee weapons?

No baking right

Unable to start mainnet node docker container

Biased dice probability question

Estimate capacitor parameters

Area of a 2D convex hull

Single author papers against my advisor's will?

What do you call the holes in a flute?

Working around an AWS network ACL rule limit

Why does this iterative way of solving of equation work?

Stars Make Stars

Can a zero nonce be safely used with AES-GCM if the key is random and never used again?

Fishing simulator

Is 1 ppb equal to 1 μg/kg?

When communicating altitude with a '9' in it, should it be pronounced "nine hundred" or "niner hundred"?

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 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162277%2fformula-for-canonical-basis-of-the-cantor-set%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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