Space filling curve












0














Let $C$ be the Cantor set. Then the Cantor function $f:C to [0,1]$ can be extended to $F:[0,1]to [0,1]$ linearly as the end points of an removed interval takes the same value. For example $f(1/3)=f(2/3)$.



But in order to get the space filling curve, the map $g:C to [0,1]×[0,1]$ can't be extended linearly to [0,1] as $g(1/3)≠g(2/3)$. Then how to extend the map $g$ from $C$ to all over [0,1] to get a continuous space filling curve?



Edit: For any $t in C$, $t$ can be written uniquely as $t=0.(2a_1)(2a_2)(2a_3) dots $(base 3), where $a_i$ is either 0 or 1.



$f(t)=0.a_1a_2a_3dots$(base 2)



$g(t)=(x(t),y(t))$, where $x(t)=0.a_1a_3a_5dots$(base 2) and $y(t)=0.a_2a_4a_6dots$(base 2)






continuity curves cantor-set







share|cite|improve this question
























  • You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
    – Paul Frost
    Dec 11 '18 at 9:24


















0














Let $C$ be the Cantor set. Then the Cantor function $f:C to [0,1]$ can be extended to $F:[0,1]to [0,1]$ linearly as the end points of an removed interval takes the same value. For example $f(1/3)=f(2/3)$.



But in order to get the space filling curve, the map $g:C to [0,1]×[0,1]$ can't be extended linearly to [0,1] as $g(1/3)≠g(2/3)$. Then how to extend the map $g$ from $C$ to all over [0,1] to get a continuous space filling curve?



Edit: For any $t in C$, $t$ can be written uniquely as $t=0.(2a_1)(2a_2)(2a_3) dots $(base 3), where $a_i$ is either 0 or 1.



$f(t)=0.a_1a_2a_3dots$(base 2)



$g(t)=(x(t),y(t))$, where $x(t)=0.a_1a_3a_5dots$(base 2) and $y(t)=0.a_2a_4a_6dots$(base 2)






continuity curves cantor-set







share|cite|improve this question
























  • You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
    – Paul Frost
    Dec 11 '18 at 9:24
















0












0








0







Let $C$ be the Cantor set. Then the Cantor function $f:C to [0,1]$ can be extended to $F:[0,1]to [0,1]$ linearly as the end points of an removed interval takes the same value. For example $f(1/3)=f(2/3)$.



But in order to get the space filling curve, the map $g:C to [0,1]×[0,1]$ can't be extended linearly to [0,1] as $g(1/3)≠g(2/3)$. Then how to extend the map $g$ from $C$ to all over [0,1] to get a continuous space filling curve?



Edit: For any $t in C$, $t$ can be written uniquely as $t=0.(2a_1)(2a_2)(2a_3) dots $(base 3), where $a_i$ is either 0 or 1.



$f(t)=0.a_1a_2a_3dots$(base 2)



$g(t)=(x(t),y(t))$, where $x(t)=0.a_1a_3a_5dots$(base 2) and $y(t)=0.a_2a_4a_6dots$(base 2)






continuity curves cantor-set







share|cite|improve this question















Let $C$ be the Cantor set. Then the Cantor function $f:C to [0,1]$ can be extended to $F:[0,1]to [0,1]$ linearly as the end points of an removed interval takes the same value. For example $f(1/3)=f(2/3)$.



But in order to get the space filling curve, the map $g:C to [0,1]×[0,1]$ can't be extended linearly to [0,1] as $g(1/3)≠g(2/3)$. Then how to extend the map $g$ from $C$ to all over [0,1] to get a continuous space filling curve?



Edit: For any $t in C$, $t$ can be written uniquely as $t=0.(2a_1)(2a_2)(2a_3) dots $(base 3), where $a_i$ is either 0 or 1.



$f(t)=0.a_1a_2a_3dots$(base 2)



$g(t)=(x(t),y(t))$, where $x(t)=0.a_1a_3a_5dots$(base 2) and $y(t)=0.a_2a_4a_6dots$(base 2)






continuity curves cantor-set




continuity curves cantor-set






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 12:58

























asked Dec 11 '18 at 7:03









Santanu Debnath

1529




1529












  • You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
    – Paul Frost
    Dec 11 '18 at 9:24




















  • You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
    – Paul Frost
    Dec 11 '18 at 9:24


















You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
– Paul Frost
Dec 11 '18 at 9:24






You should perhaps explain what the Cantor function is. The relevant property is that it is continuous surjection. What is the map $g$?
– Paul Frost
Dec 11 '18 at 9:24












1 Answer
1






active

oldest

votes


















2














Let us collect some well-known facts about the Cantor set $C$. Usually it is defined by iteratively deleting the open middle third from each closed line segment in a finite disjoint union of such segments. Start by deleting the open middle third from the interval $[0, 1]$ which leaves $[0,1/3] cup [2/3,1]$ and continue as described.



Fact 1. The map $varphi : P = prod_{n=1}^infty {0, 1 } to C, varphi((x_n)) = sum_{n=1}^infty frac{2x_n}{3^n}$, is a homeomorphism. Here the infinite product $P$ of copies of the discrete space ${0, 1 }$ is of course endowed with the product topology.



Fact 2. There exists a continuous surjection $f : C to [0,1]$. In fact, define $psi : P to [0,1], psi((x_n)) = sum_{n=1}^infty frac{x_n}{2^n}$. Then $f = psi varphi^{-1}$ will do.



Fact 3. There exists a homeomorphism $h : C to C times C$. In fact, we have the obvious homeomorphism $H : P to P times P, H((x_n)) = ((x_1,x_3,x_5,dots), (x_2,x_4,x_6,dots))$.



Now define
$$g = (f times f) h : C to [0,1] times [0,1] .$$
This is a continuous surjection. The Tietze extension theorem gives us a continuous extension $G : [0,1] to [0,1] times [0,1]$ which is again a continuous surjection. See for example https://en.wikipedia.org/wiki/Tietze_extension_theorem.



You can also directly define an extension $G$ without applying Tietze's theorem. On each of the removed open intervals $(a,b)$ let $G(t) = (1-frac{t-a}{b-a})g(a) + frac{t-a}{b-a}g(b)$ (note that $[0,1] times [0,1]$ is a convex subset of $mathbb{R}^2$). It can easily be shown that $G$ is continuous.






share|cite|improve this answer





















  • Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
    – Santanu Debnath
    Dec 11 '18 at 16:28











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034989%2fspace-filling-curve%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Let us collect some well-known facts about the Cantor set $C$. Usually it is defined by iteratively deleting the open middle third from each closed line segment in a finite disjoint union of such segments. Start by deleting the open middle third from the interval $[0, 1]$ which leaves $[0,1/3] cup [2/3,1]$ and continue as described.



Fact 1. The map $varphi : P = prod_{n=1}^infty {0, 1 } to C, varphi((x_n)) = sum_{n=1}^infty frac{2x_n}{3^n}$, is a homeomorphism. Here the infinite product $P$ of copies of the discrete space ${0, 1 }$ is of course endowed with the product topology.



Fact 2. There exists a continuous surjection $f : C to [0,1]$. In fact, define $psi : P to [0,1], psi((x_n)) = sum_{n=1}^infty frac{x_n}{2^n}$. Then $f = psi varphi^{-1}$ will do.



Fact 3. There exists a homeomorphism $h : C to C times C$. In fact, we have the obvious homeomorphism $H : P to P times P, H((x_n)) = ((x_1,x_3,x_5,dots), (x_2,x_4,x_6,dots))$.



Now define
$$g = (f times f) h : C to [0,1] times [0,1] .$$
This is a continuous surjection. The Tietze extension theorem gives us a continuous extension $G : [0,1] to [0,1] times [0,1]$ which is again a continuous surjection. See for example https://en.wikipedia.org/wiki/Tietze_extension_theorem.



You can also directly define an extension $G$ without applying Tietze's theorem. On each of the removed open intervals $(a,b)$ let $G(t) = (1-frac{t-a}{b-a})g(a) + frac{t-a}{b-a}g(b)$ (note that $[0,1] times [0,1]$ is a convex subset of $mathbb{R}^2$). It can easily be shown that $G$ is continuous.






share|cite|improve this answer





















  • Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
    – Santanu Debnath
    Dec 11 '18 at 16:28
















2














Let us collect some well-known facts about the Cantor set $C$. Usually it is defined by iteratively deleting the open middle third from each closed line segment in a finite disjoint union of such segments. Start by deleting the open middle third from the interval $[0, 1]$ which leaves $[0,1/3] cup [2/3,1]$ and continue as described.



Fact 1. The map $varphi : P = prod_{n=1}^infty {0, 1 } to C, varphi((x_n)) = sum_{n=1}^infty frac{2x_n}{3^n}$, is a homeomorphism. Here the infinite product $P$ of copies of the discrete space ${0, 1 }$ is of course endowed with the product topology.



Fact 2. There exists a continuous surjection $f : C to [0,1]$. In fact, define $psi : P to [0,1], psi((x_n)) = sum_{n=1}^infty frac{x_n}{2^n}$. Then $f = psi varphi^{-1}$ will do.



Fact 3. There exists a homeomorphism $h : C to C times C$. In fact, we have the obvious homeomorphism $H : P to P times P, H((x_n)) = ((x_1,x_3,x_5,dots), (x_2,x_4,x_6,dots))$.



Now define
$$g = (f times f) h : C to [0,1] times [0,1] .$$
This is a continuous surjection. The Tietze extension theorem gives us a continuous extension $G : [0,1] to [0,1] times [0,1]$ which is again a continuous surjection. See for example https://en.wikipedia.org/wiki/Tietze_extension_theorem.



You can also directly define an extension $G$ without applying Tietze's theorem. On each of the removed open intervals $(a,b)$ let $G(t) = (1-frac{t-a}{b-a})g(a) + frac{t-a}{b-a}g(b)$ (note that $[0,1] times [0,1]$ is a convex subset of $mathbb{R}^2$). It can easily be shown that $G$ is continuous.






share|cite|improve this answer





















  • Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
    – Santanu Debnath
    Dec 11 '18 at 16:28














2












2








2






Let us collect some well-known facts about the Cantor set $C$. Usually it is defined by iteratively deleting the open middle third from each closed line segment in a finite disjoint union of such segments. Start by deleting the open middle third from the interval $[0, 1]$ which leaves $[0,1/3] cup [2/3,1]$ and continue as described.



Fact 1. The map $varphi : P = prod_{n=1}^infty {0, 1 } to C, varphi((x_n)) = sum_{n=1}^infty frac{2x_n}{3^n}$, is a homeomorphism. Here the infinite product $P$ of copies of the discrete space ${0, 1 }$ is of course endowed with the product topology.



Fact 2. There exists a continuous surjection $f : C to [0,1]$. In fact, define $psi : P to [0,1], psi((x_n)) = sum_{n=1}^infty frac{x_n}{2^n}$. Then $f = psi varphi^{-1}$ will do.



Fact 3. There exists a homeomorphism $h : C to C times C$. In fact, we have the obvious homeomorphism $H : P to P times P, H((x_n)) = ((x_1,x_3,x_5,dots), (x_2,x_4,x_6,dots))$.



Now define
$$g = (f times f) h : C to [0,1] times [0,1] .$$
This is a continuous surjection. The Tietze extension theorem gives us a continuous extension $G : [0,1] to [0,1] times [0,1]$ which is again a continuous surjection. See for example https://en.wikipedia.org/wiki/Tietze_extension_theorem.



You can also directly define an extension $G$ without applying Tietze's theorem. On each of the removed open intervals $(a,b)$ let $G(t) = (1-frac{t-a}{b-a})g(a) + frac{t-a}{b-a}g(b)$ (note that $[0,1] times [0,1]$ is a convex subset of $mathbb{R}^2$). It can easily be shown that $G$ is continuous.






share|cite|improve this answer












Let us collect some well-known facts about the Cantor set $C$. Usually it is defined by iteratively deleting the open middle third from each closed line segment in a finite disjoint union of such segments. Start by deleting the open middle third from the interval $[0, 1]$ which leaves $[0,1/3] cup [2/3,1]$ and continue as described.



Fact 1. The map $varphi : P = prod_{n=1}^infty {0, 1 } to C, varphi((x_n)) = sum_{n=1}^infty frac{2x_n}{3^n}$, is a homeomorphism. Here the infinite product $P$ of copies of the discrete space ${0, 1 }$ is of course endowed with the product topology.



Fact 2. There exists a continuous surjection $f : C to [0,1]$. In fact, define $psi : P to [0,1], psi((x_n)) = sum_{n=1}^infty frac{x_n}{2^n}$. Then $f = psi varphi^{-1}$ will do.



Fact 3. There exists a homeomorphism $h : C to C times C$. In fact, we have the obvious homeomorphism $H : P to P times P, H((x_n)) = ((x_1,x_3,x_5,dots), (x_2,x_4,x_6,dots))$.



Now define
$$g = (f times f) h : C to [0,1] times [0,1] .$$
This is a continuous surjection. The Tietze extension theorem gives us a continuous extension $G : [0,1] to [0,1] times [0,1]$ which is again a continuous surjection. See for example https://en.wikipedia.org/wiki/Tietze_extension_theorem.



You can also directly define an extension $G$ without applying Tietze's theorem. On each of the removed open intervals $(a,b)$ let $G(t) = (1-frac{t-a}{b-a})g(a) + frac{t-a}{b-a}g(b)$ (note that $[0,1] times [0,1]$ is a convex subset of $mathbb{R}^2$). It can easily be shown that $G$ is continuous.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 13:54









Paul Frost

9,2762631




9,2762631












  • Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
    – Santanu Debnath
    Dec 11 '18 at 16:28


















  • Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
    – Santanu Debnath
    Dec 11 '18 at 16:28
















Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
– Santanu Debnath
Dec 11 '18 at 16:28




Thanks mate. I was looking for $G$ you defined as I didn't want to apply Tietze's extention theorem. This $G$ is a space filling curve defined by Lebesgue
– Santanu Debnath
Dec 11 '18 at 16:28


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034989%2fspace-filling-curve%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more