Are $C$ and $Ctimes C$ homeomorphic?












4












$begingroup$



Are $C$ and $Ctimes C$ homeomorphic? (Here, $C$ denotes the "first middle third" Cantor set.)




Seems that they are but I can't come up with an idea how to show it.






general-topology cantor-set







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
    $endgroup$
    – Asaf Karagila♦
    Jan 23 '17 at 8:43








  • 3




    $begingroup$
    Also, if you view $C$ as $2^ω$, it is almost obvious.
    $endgroup$
    – user87690
    Jan 23 '17 at 10:06










  • $begingroup$
    See also math.stackexchange.com/q/3034989.
    $endgroup$
    – Paul Frost
    Dec 11 '18 at 15:03
















4












$begingroup$



Are $C$ and $Ctimes C$ homeomorphic? (Here, $C$ denotes the "first middle third" Cantor set.)




Seems that they are but I can't come up with an idea how to show it.






general-topology cantor-set







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
    $endgroup$
    – Asaf Karagila♦
    Jan 23 '17 at 8:43








  • 3




    $begingroup$
    Also, if you view $C$ as $2^ω$, it is almost obvious.
    $endgroup$
    – user87690
    Jan 23 '17 at 10:06










  • $begingroup$
    See also math.stackexchange.com/q/3034989.
    $endgroup$
    – Paul Frost
    Dec 11 '18 at 15:03














4












4








4


0



$begingroup$



Are $C$ and $Ctimes C$ homeomorphic? (Here, $C$ denotes the "first middle third" Cantor set.)




Seems that they are but I can't come up with an idea how to show it.






general-topology cantor-set







share|cite|improve this question











$endgroup$





Are $C$ and $Ctimes C$ homeomorphic? (Here, $C$ denotes the "first middle third" Cantor set.)




Seems that they are but I can't come up with an idea how to show it.






general-topology cantor-set




general-topology 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 13:20









Batominovski

33k33293




33k33293










asked Jan 23 '17 at 8:40









Gleb ChiliGleb Chili

44428




44428








  • 2




    $begingroup$
    Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
    $endgroup$
    – Asaf Karagila♦
    Jan 23 '17 at 8:43








  • 3




    $begingroup$
    Also, if you view $C$ as $2^ω$, it is almost obvious.
    $endgroup$
    – user87690
    Jan 23 '17 at 10:06










  • $begingroup$
    See also math.stackexchange.com/q/3034989.
    $endgroup$
    – Paul Frost
    Dec 11 '18 at 15:03














  • 2




    $begingroup$
    Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
    $endgroup$
    – Asaf Karagila♦
    Jan 23 '17 at 8:43








  • 3




    $begingroup$
    Also, if you view $C$ as $2^ω$, it is almost obvious.
    $endgroup$
    – user87690
    Jan 23 '17 at 10:06










  • $begingroup$
    See also math.stackexchange.com/q/3034989.
    $endgroup$
    – Paul Frost
    Dec 11 '18 at 15:03








2




2




$begingroup$
Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
$endgroup$
– Asaf Karagila♦
Jan 23 '17 at 8:43






$begingroup$
Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points.
$endgroup$
– Asaf Karagila♦
Jan 23 '17 at 8:43






3




3




$begingroup$
Also, if you view $C$ as $2^ω$, it is almost obvious.
$endgroup$
– user87690
Jan 23 '17 at 10:06




$begingroup$
Also, if you view $C$ as $2^ω$, it is almost obvious.
$endgroup$
– user87690
Jan 23 '17 at 10:06












$begingroup$
See also math.stackexchange.com/q/3034989.
$endgroup$
– Paul Frost
Dec 11 '18 at 15:03




$begingroup$
See also math.stackexchange.com/q/3034989.
$endgroup$
– Paul Frost
Dec 11 '18 at 15:03










1 Answer
1






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oldest

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$begingroup$

Yes the Cantor set $C$ is homeomorphic to $C times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:




[A] topological space is a Cantor space if it is homeomorphic to the Cantor set.



. . .



[T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space ${0, 1}$. This is usually written as $2^mathbb{N}$ or $2^omega$ (where $2$ denotes the $2$-element set ${0,1}$ with the discrete topology).



. . .



[M]any properties of Cantor spaces can be established using $2^omega$, because its construction as a product makes it amenable to analysis.



Cantor spaces have the following properties:




  • . . .

  • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.







share|cite|improve this answer











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    0












    $begingroup$

    Yes the Cantor set $C$ is homeomorphic to $C times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:




    [A] topological space is a Cantor space if it is homeomorphic to the Cantor set.



    . . .



    [T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space ${0, 1}$. This is usually written as $2^mathbb{N}$ or $2^omega$ (where $2$ denotes the $2$-element set ${0,1}$ with the discrete topology).



    . . .



    [M]any properties of Cantor spaces can be established using $2^omega$, because its construction as a product makes it amenable to analysis.



    Cantor spaces have the following properties:




    • . . .

    • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Yes the Cantor set $C$ is homeomorphic to $C times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:




      [A] topological space is a Cantor space if it is homeomorphic to the Cantor set.



      . . .



      [T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space ${0, 1}$. This is usually written as $2^mathbb{N}$ or $2^omega$ (where $2$ denotes the $2$-element set ${0,1}$ with the discrete topology).



      . . .



      [M]any properties of Cantor spaces can be established using $2^omega$, because its construction as a product makes it amenable to analysis.



      Cantor spaces have the following properties:




      • . . .

      • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Yes the Cantor set $C$ is homeomorphic to $C times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:




        [A] topological space is a Cantor space if it is homeomorphic to the Cantor set.



        . . .



        [T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space ${0, 1}$. This is usually written as $2^mathbb{N}$ or $2^omega$ (where $2$ denotes the $2$-element set ${0,1}$ with the discrete topology).



        . . .



        [M]any properties of Cantor spaces can be established using $2^omega$, because its construction as a product makes it amenable to analysis.



        Cantor spaces have the following properties:




        • . . .

        • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.







        share|cite|improve this answer











        $endgroup$



        Yes the Cantor set $C$ is homeomorphic to $C times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:




        [A] topological space is a Cantor space if it is homeomorphic to the Cantor set.



        . . .



        [T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space ${0, 1}$. This is usually written as $2^mathbb{N}$ or $2^omega$ (where $2$ denotes the $2$-element set ${0,1}$ with the discrete topology).



        . . .



        [M]any properties of Cantor spaces can be established using $2^omega$, because its construction as a product makes it amenable to analysis.



        Cantor spaces have the following properties:




        • . . .

        • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 8:47


























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