Space filling curve












1












$begingroup$


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







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$endgroup$












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


















1












$begingroup$


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











$endgroup$












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
















1












1








1


1



$begingroup$


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











$endgroup$




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






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share|cite|improve this question













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edited Dec 11 '18 at 12:58







Santanu Debnath

















asked Dec 11 '18 at 7:03









Santanu DebnathSantanu Debnath

1579




1579












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




















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


















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






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












1 Answer
1






active

oldest

votes


















3












$begingroup$

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









$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Santanu Debnath
    Dec 11 '18 at 16:28












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

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









$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Santanu Debnath
    Dec 11 '18 at 16:28
















3












$begingroup$

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









$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Santanu Debnath
    Dec 11 '18 at 16:28














3












3








3





$begingroup$

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









$endgroup$



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 FrostPaul Frost

11.7k3935




11.7k3935












  • $begingroup$
    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
    $endgroup$
    – Santanu Debnath
    Dec 11 '18 at 16:28


















  • $begingroup$
    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
    $endgroup$
    – Santanu Debnath
    Dec 11 '18 at 16:28
















$begingroup$
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
$endgroup$
– Santanu Debnath
Dec 11 '18 at 16:28




$begingroup$
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
$endgroup$
– Santanu Debnath
Dec 11 '18 at 16:28


















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