Cantor set with pairs of points identified

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Consider the middle-thirds Cantor set in $[0,1]$. I want to identify points in the following way.

First, identify the two points $1/3$ and $2/3$.

Then, idenfity $1/9$ with $2/9$, and also identify $7/9$ with $8/9$.

At the third step there will be four pairs of points: $1/27sim 2/27$; $7/27sim 8/27$; $19/27sim 20/27$; $25/27sim 26/27$.

Continue.

Essentially I am squeezing together the consecutive gaps in the Cantor set. I would like to know of the resulting space is homeomorphic to $[0,1]$.

general-topology cantor-set

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edited Jul 22 at 23:31

Eric Wofsey

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asked Jul 22 at 20:14

Forever Mozart

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up vote
11
down vote

favorite

4

Consider the middle-thirds Cantor set in $[0,1]$. I want to identify points in the following way.

First, identify the two points $1/3$ and $2/3$.

Then, idenfity $1/9$ with $2/9$, and also identify $7/9$ with $8/9$.

At the third step there will be four pairs of points: $1/27sim 2/27$; $7/27sim 8/27$; $19/27sim 20/27$; $25/27sim 26/27$.

Continue.

Essentially I am squeezing together the consecutive gaps in the Cantor set. I would like to know of the resulting space is homeomorphic to $[0,1]$.

general-topology cantor-set

share|cite|improve this question

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

asked Jul 22 at 20:14

Forever Mozart

5,37621240

add a comment | 

up vote
11
down vote

favorite

4

up vote
11
down vote

favorite

4

4

Consider the middle-thirds Cantor set in $[0,1]$. I want to identify points in the following way.

First, identify the two points $1/3$ and $2/3$.

Then, idenfity $1/9$ with $2/9$, and also identify $7/9$ with $8/9$.

At the third step there will be four pairs of points: $1/27sim 2/27$; $7/27sim 8/27$; $19/27sim 20/27$; $25/27sim 26/27$.

Continue.

Essentially I am squeezing together the consecutive gaps in the Cantor set. I would like to know of the resulting space is homeomorphic to $[0,1]$.

general-topology cantor-set

share|cite|improve this question

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

asked Jul 22 at 20:14

Forever Mozart

5,37621240

Consider the middle-thirds Cantor set in $[0,1]$. I want to identify points in the following way.

First, identify the two points $1/3$ and $2/3$.

Then, idenfity $1/9$ with $2/9$, and also identify $7/9$ with $8/9$.

At the third step there will be four pairs of points: $1/27sim 2/27$; $7/27sim 8/27$; $19/27sim 20/27$; $25/27sim 26/27$.

Continue.

Essentially I am squeezing together the consecutive gaps in the Cantor set. I would like to know of the resulting space is homeomorphic to $[0,1]$.

general-topology cantor-set

share|cite|improve this question

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

asked Jul 22 at 20:14

Forever Mozart

5,37621240

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

edited Jul 22 at 23:31

Eric Wofsey

162k12189300

162k12189300

asked Jul 22 at 20:14

Forever Mozart

5,37621240

asked Jul 22 at 20:14

Forever Mozart

5,37621240

asked Jul 22 at 20:14

Forever Mozart

5,37621240

5,37621240

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
13
down vote



accepted

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=sum_n=1^inftyfrac2a_n3^nin K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=sum_n=1^inftyfraca_n2^n.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:Kto [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
  – DanielWainfleet
  Jul 23 at 4:24
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
13
down vote



accepted

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=sum_n=1^inftyfrac2a_n3^nin K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=sum_n=1^inftyfraca_n2^n.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:Kto [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
  – DanielWainfleet
  Jul 23 at 4:24
  
  

  
  
  
  

add a comment | 

up vote
13
down vote



accepted

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=sum_n=1^inftyfrac2a_n3^nin K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=sum_n=1^inftyfraca_n2^n.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:Kto [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
  – DanielWainfleet
  Jul 23 at 4:24
  
  

  
  
  
  

add a comment | 

up vote
13
down vote



accepted

up vote
13
down vote



accepted

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=sum_n=1^inftyfrac2a_n3^nin K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=sum_n=1^inftyfraca_n2^n.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:Kto [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=sum_n=1^inftyfrac2a_n3^nin K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=sum_n=1^inftyfraca_n2^n.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:Kto [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

answered Jul 22 at 20:30

Eric Wofsey

162k12189300

162k12189300

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
  – DanielWainfleet
  Jul 23 at 4:24
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
  – DanielWainfleet
  Jul 23 at 4:24
  
  

  
  
  
  

You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
– DanielWainfleet
Jul 23 at 4:24

You could also say that the process of identifying the end-points of the open intervals results in an ordered space which has 2 endpoints, is connected, and is separable, which, by a variant of the Souslin question, is therefore homeomorphic to $[0,1]$.
– DanielWainfleet
Jul 23 at 4:24

add a comment | 

 

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