Properties of Cantor set












0












$begingroup$


$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.



I know both the proof. I am asking which property of $C$ is the reason of this absurdity!






general-topology cantor-set







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 15:56












  • $begingroup$
    Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 15:59












  • $begingroup$
    I hope that is not true for finite set
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 16:01










  • $begingroup$
    No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:15










  • $begingroup$
    @JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:22
















0












$begingroup$


$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.



I know both the proof. I am asking which property of $C$ is the reason of this absurdity!






general-topology cantor-set







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 15:56












  • $begingroup$
    Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 15:59












  • $begingroup$
    I hope that is not true for finite set
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 16:01










  • $begingroup$
    No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:15










  • $begingroup$
    @JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:22














0












0








0





$begingroup$


$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.



I know both the proof. I am asking which property of $C$ is the reason of this absurdity!






general-topology cantor-set







share|cite|improve this question











$endgroup$




$[0,1]$ is not homeomorphic to $[0,1]×[0,1]$ but $C$ is homeomorphic to $C times C$ where $C$ is the Cantor set.



I know both the proof. I am asking which property of $C$ is the reason of this absurdity!






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 9 '18 at 17:11









Martin Sleziak

44.7k10118272




44.7k10118272










asked Dec 9 '18 at 15:53









Santanu DebnathSantanu Debnath

1529




1529












  • $begingroup$
    If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 15:56












  • $begingroup$
    Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 15:59












  • $begingroup$
    I hope that is not true for finite set
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 16:01










  • $begingroup$
    No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:15










  • $begingroup$
    @JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:22


















  • $begingroup$
    If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 15:56












  • $begingroup$
    Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 15:59












  • $begingroup$
    I hope that is not true for finite set
    $endgroup$
    – Santanu Debnath
    Dec 9 '18 at 16:01










  • $begingroup$
    No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:15










  • $begingroup$
    @JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
    $endgroup$
    – Henno Brandsma
    Dec 9 '18 at 16:22
















$begingroup$
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 15:56






$begingroup$
If you know both proofs, you know the reason: the Cantor set is totally disconnected, hence it is not really surprising that $Ktimes K$ is homeomorphic to $K$, especially if you think to the elements of $K$ as infinite strings over a ternary alphabet. On the other hand if we remove a point from $[0,1]$ we disconnect it, while $[0,1]times[0,1]setminus{(a,b)}$ still is connected.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 15:56














$begingroup$
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
$endgroup$
– Santanu Debnath
Dec 9 '18 at 15:59






$begingroup$
Is it true that if $K subset mathbb{R} $ is totally disconnected then $K$ is homeomorphic to $K×K$?
$endgroup$
– Santanu Debnath
Dec 9 '18 at 15:59














$begingroup$
I hope that is not true for finite set
$endgroup$
– Santanu Debnath
Dec 9 '18 at 16:01




$begingroup$
I hope that is not true for finite set
$endgroup$
– Santanu Debnath
Dec 9 '18 at 16:01












$begingroup$
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
$endgroup$
– Henno Brandsma
Dec 9 '18 at 16:15




$begingroup$
No, it's not true for all totally disconnected subsets, e.g. $K={0} cup {frac1n: n ge 1}$ is an exception.
$endgroup$
– Henno Brandsma
Dec 9 '18 at 16:15












$begingroup$
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
$endgroup$
– Henno Brandsma
Dec 9 '18 at 16:22




$begingroup$
@JackD'Aurizio the rationals certainly is homeomorphic to its square! Even any finite power.
$endgroup$
– Henno Brandsma
Dec 9 '18 at 16:22










1 Answer
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$begingroup$

$C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.



If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.



Other spaces with unique charaterisations also have such preservations by finite products:





  • $mathbb{Q}$: the unique countable metric space without isolated points.


  • $mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).


  • $Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.


And some more exist too. All of the above are homeomorphic to their squares.






share|cite|improve this answer











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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.



    If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.



    Other spaces with unique charaterisations also have such preservations by finite products:





    • $mathbb{Q}$: the unique countable metric space without isolated points.


    • $mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).


    • $Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.


    And some more exist too. All of the above are homeomorphic to their squares.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.



      If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.



      Other spaces with unique charaterisations also have such preservations by finite products:





      • $mathbb{Q}$: the unique countable metric space without isolated points.


      • $mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).


      • $Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.


      And some more exist too. All of the above are homeomorphic to their squares.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.



        If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.



        Other spaces with unique charaterisations also have such preservations by finite products:





        • $mathbb{Q}$: the unique countable metric space without isolated points.


        • $mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).


        • $Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.


        And some more exist too. All of the above are homeomorphic to their squares.






        share|cite|improve this answer











        $endgroup$



        $C$ is the (up to homeomorphism) unique zero-dimensional (or totally disconnected) compact metric space without isolated points.



        If $X$ is such a space, so is $X^n$ for any $n$: still compact, totally disconnected, metric and no isolated points so it's homeomorphic to $X$.



        Other spaces with unique charaterisations also have such preservations by finite products:





        • $mathbb{Q}$: the unique countable metric space without isolated points.


        • $mathbb{P}$ (the irrationals in the reals): the unique completely metrisable zero-dimensional separable metric space that is nowhere locally compact (i.e. the interior of any compact subset is empty).


        • $Csetminus {0}$ (the Cantor set minus a point), the unique locally compact non-compact separable metric space that is totally disconnected.


        And some more exist too. All of the above are homeomorphic to their squares.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 16:28

























        answered Dec 9 '18 at 16:21









        Henno BrandsmaHenno Brandsma

        110k347116




        110k347116






























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