3 dense uncountable pairwise disjoint subsets of real line












3












$begingroup$


Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?



I feel like it's not possible.



I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.






real-analysis general-topology real-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 16:40












  • $begingroup$
    @NotMike can you explain a bit more.. some precise example would be appreciated
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 16:43










  • $begingroup$
    You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 17:22










  • $begingroup$
    Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
    $endgroup$
    – timtfj
    Dec 4 '18 at 18:07












  • $begingroup$
    @timtfj partition property is not needed as long as it serves the other conditions
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:16
















3












$begingroup$


Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?



I feel like it's not possible.



I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.






real-analysis general-topology real-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 16:40












  • $begingroup$
    @NotMike can you explain a bit more.. some precise example would be appreciated
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 16:43










  • $begingroup$
    You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 17:22










  • $begingroup$
    Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
    $endgroup$
    – timtfj
    Dec 4 '18 at 18:07












  • $begingroup$
    @timtfj partition property is not needed as long as it serves the other conditions
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:16














3












3








3


1



$begingroup$


Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?



I feel like it's not possible.



I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.






real-analysis general-topology real-numbers







share|cite|improve this question











$endgroup$




Can we find three dense uncountable pairwise disjoint subsets of $mathbb{R}$? If so, what are these three sets?



I feel like it's not possible.



I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.






real-analysis general-topology real-numbers




real-analysis general-topology real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 16:36









amWhy

1




1










asked Dec 4 '18 at 16:29









ChakSayantanChakSayantan

1426




1426








  • 1




    $begingroup$
    Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 16:40












  • $begingroup$
    @NotMike can you explain a bit more.. some precise example would be appreciated
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 16:43










  • $begingroup$
    You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 17:22










  • $begingroup$
    Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
    $endgroup$
    – timtfj
    Dec 4 '18 at 18:07












  • $begingroup$
    @timtfj partition property is not needed as long as it serves the other conditions
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:16














  • 1




    $begingroup$
    Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 16:40












  • $begingroup$
    @NotMike can you explain a bit more.. some precise example would be appreciated
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 16:43










  • $begingroup$
    You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 17:22










  • $begingroup$
    Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
    $endgroup$
    – timtfj
    Dec 4 '18 at 18:07












  • $begingroup$
    @timtfj partition property is not needed as long as it serves the other conditions
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:16








1




1




$begingroup$
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
$endgroup$
– Not Mike
Dec 4 '18 at 16:40






$begingroup$
Yes, begin with three arbitrary uncountable disjoint subsets of the irrationals and find three disjoint dense subsets of the rationals. To each of the sets of irrationals adjoin one of the countable dense subsets of the rationals.
$endgroup$
– Not Mike
Dec 4 '18 at 16:40














$begingroup$
@NotMike can you explain a bit more.. some precise example would be appreciated
$endgroup$
– ChakSayantan
Dec 4 '18 at 16:43




$begingroup$
@NotMike can you explain a bit more.. some precise example would be appreciated
$endgroup$
– ChakSayantan
Dec 4 '18 at 16:43












$begingroup$
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:22




$begingroup$
You can find $2^{aleph_0}$ pairwise disjoint uncountable subsets of $Bbb R$.
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 17:22












$begingroup$
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
$endgroup$
– timtfj
Dec 4 '18 at 18:07






$begingroup$
Do you want to partition $mathbb{R}$ into the three subsetsor just find three pairwise disjoint subsets?
$endgroup$
– timtfj
Dec 4 '18 at 18:07














$begingroup$
@timtfj partition property is not needed as long as it serves the other conditions
$endgroup$
– ChakSayantan
Dec 4 '18 at 18:16




$begingroup$
@timtfj partition property is not needed as long as it serves the other conditions
$endgroup$
– ChakSayantan
Dec 4 '18 at 18:16










2 Answers
2






active

oldest

votes


















2












$begingroup$

Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.



Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$



Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 17:13












  • $begingroup$
    You take the union
    $endgroup$
    – Not Mike
    Dec 4 '18 at 17:26










  • $begingroup$
    then is it dence in R? I can't view it properly
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:17












  • $begingroup$
    Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 19:02










  • $begingroup$
    ah! Got it. Thanks.
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 19:05



















1












$begingroup$

Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.



Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.



Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.



Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.



    Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$



    Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 17:13












    • $begingroup$
      You take the union
      $endgroup$
      – Not Mike
      Dec 4 '18 at 17:26










    • $begingroup$
      then is it dence in R? I can't view it properly
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 18:17












    • $begingroup$
      Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
      $endgroup$
      – Not Mike
      Dec 4 '18 at 19:02










    • $begingroup$
      ah! Got it. Thanks.
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 19:05
















    2












    $begingroup$

    Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.



    Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$



    Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 17:13












    • $begingroup$
      You take the union
      $endgroup$
      – Not Mike
      Dec 4 '18 at 17:26










    • $begingroup$
      then is it dence in R? I can't view it properly
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 18:17












    • $begingroup$
      Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
      $endgroup$
      – Not Mike
      Dec 4 '18 at 19:02










    • $begingroup$
      ah! Got it. Thanks.
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 19:05














    2












    2








    2





    $begingroup$

    Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.



    Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$



    Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.






    share|cite|improve this answer









    $endgroup$



    Fix three disjoint dense subsets of $mathbb{Q}$, call them $D_0, D_1$ and $D_2$.



    Let $I_0=(0,1)backslash mathbb{Q}$, $I_1=(1,2)backslash mathbb{Q}$, $I_2=(2,3)backslash mathbb{Q}$ and set $J_k = I_k cup D_k.$



    Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 '18 at 16:53









    Not MikeNot Mike

    36329




    36329












    • $begingroup$
      So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 17:13












    • $begingroup$
      You take the union
      $endgroup$
      – Not Mike
      Dec 4 '18 at 17:26










    • $begingroup$
      then is it dence in R? I can't view it properly
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 18:17












    • $begingroup$
      Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
      $endgroup$
      – Not Mike
      Dec 4 '18 at 19:02










    • $begingroup$
      ah! Got it. Thanks.
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 19:05


















    • $begingroup$
      So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 17:13












    • $begingroup$
      You take the union
      $endgroup$
      – Not Mike
      Dec 4 '18 at 17:26










    • $begingroup$
      then is it dence in R? I can't view it properly
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 18:17












    • $begingroup$
      Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
      $endgroup$
      – Not Mike
      Dec 4 '18 at 19:02










    • $begingroup$
      ah! Got it. Thanks.
      $endgroup$
      – ChakSayantan
      Dec 4 '18 at 19:05
















    $begingroup$
    So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 17:13






    $begingroup$
    So I have taken Q_p ={i/p^j: i,j integers}Z for three different primes. Now, J_k would be union or plus?
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 17:13














    $begingroup$
    You take the union
    $endgroup$
    – Not Mike
    Dec 4 '18 at 17:26




    $begingroup$
    You take the union
    $endgroup$
    – Not Mike
    Dec 4 '18 at 17:26












    $begingroup$
    then is it dence in R? I can't view it properly
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:17






    $begingroup$
    then is it dence in R? I can't view it properly
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 18:17














    $begingroup$
    Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 19:02




    $begingroup$
    Yes. Let $Usubset mathbb{R}$ be open then $ emptyset neq Ucap D_k subset Ucap J_k$ so we have $Ucap J_kneq emptyset $. Which sense $U$ was arbitrary, entials each $J_k$ is dense.
    $endgroup$
    – Not Mike
    Dec 4 '18 at 19:02












    $begingroup$
    ah! Got it. Thanks.
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 19:05




    $begingroup$
    ah! Got it. Thanks.
    $endgroup$
    – ChakSayantan
    Dec 4 '18 at 19:05











    1












    $begingroup$

    Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.



    Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.



    Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.



    Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.



      Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.



      Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.



      Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.



        Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.



        Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.



        Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.






        share|cite|improve this answer











        $endgroup$



        Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${1,2,3}$.



        Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${4,5,6}$.



        Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in ${7,8,9}$.



        Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $mathbb R$, and $D=mathbb Rsetminus(Acup Bcup C)$ is a fourth one.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 23:15

























        answered Dec 4 '18 at 22:33









        bofbof

        51.4k557120




        51.4k557120






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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