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

192k28225439




192k28225439










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

        50.7k457120




        50.7k457120






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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