Continuous self-maps on $mathbb{Q}$












8














1) Are $mathbb{Q}$ and $mathbb{Q}setminus{0}$ homeomorphic?



2) If $Ssubseteq mathbb{Q}$ is a non-empty subset, is there a continuous surjection $f:mathbb{Q}to S$?






general-topology







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  • 1




    That's right - will correct!
    – Dominic van der Zypen
    Aug 11 '16 at 6:24






  • 2




    Q is already missing $e $ in a similar way. It will certainly not miss 0.
    – Alephnull
    Aug 11 '16 at 7:28
















8














1) Are $mathbb{Q}$ and $mathbb{Q}setminus{0}$ homeomorphic?



2) If $Ssubseteq mathbb{Q}$ is a non-empty subset, is there a continuous surjection $f:mathbb{Q}to S$?






general-topology







share|cite|improve this question




















  • 1




    That's right - will correct!
    – Dominic van der Zypen
    Aug 11 '16 at 6:24






  • 2




    Q is already missing $e $ in a similar way. It will certainly not miss 0.
    – Alephnull
    Aug 11 '16 at 7:28














8












8








8


3





1) Are $mathbb{Q}$ and $mathbb{Q}setminus{0}$ homeomorphic?



2) If $Ssubseteq mathbb{Q}$ is a non-empty subset, is there a continuous surjection $f:mathbb{Q}to S$?






general-topology







share|cite|improve this question















1) Are $mathbb{Q}$ and $mathbb{Q}setminus{0}$ homeomorphic?



2) If $Ssubseteq mathbb{Q}$ is a non-empty subset, is there a continuous surjection $f:mathbb{Q}to S$?






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 11 '16 at 6:24

























asked Aug 11 '16 at 6:14









Dominic van der Zypen

1,351613




1,351613








  • 1




    That's right - will correct!
    – Dominic van der Zypen
    Aug 11 '16 at 6:24






  • 2




    Q is already missing $e $ in a similar way. It will certainly not miss 0.
    – Alephnull
    Aug 11 '16 at 7:28














  • 1




    That's right - will correct!
    – Dominic van der Zypen
    Aug 11 '16 at 6:24






  • 2




    Q is already missing $e $ in a similar way. It will certainly not miss 0.
    – Alephnull
    Aug 11 '16 at 7:28








1




1




That's right - will correct!
– Dominic van der Zypen
Aug 11 '16 at 6:24




That's right - will correct!
– Dominic van der Zypen
Aug 11 '16 at 6:24




2




2




Q is already missing $e $ in a similar way. It will certainly not miss 0.
– Alephnull
Aug 11 '16 at 7:28




Q is already missing $e $ in a similar way. It will certainly not miss 0.
– Alephnull
Aug 11 '16 at 7:28










3 Answers
3






active

oldest

votes


















6














There exists a monotonic bijection $Bbb QtoBbb Qsetminus{0}$. This is then automatically a homeomorphism.



To see this, let ${q_nmid ninBbb N}$ be an enumeration of $Bbb Q$ and define $f(q_n)$ recursively: Find the smallest $m$ such that $q_mne 0$ and for all $k<n$ we have $q_m {<atop>} f(q_k)iff q_n{<atop>} q_k$. Then define $f(q_n)=q_m$.
The crucial point is that $Bbb Qsetminus{0}$ is still linearly ordered such that there is no largest and no no smallest element and there is another element between any two given elements, thus always allowing us to find such a $q_m$.




The same method helps find a continuous surjection $Bbb Qto S$. Just if there is no matching element in $S$, relax the $<atop>$ condition to allow equality.






share|cite|improve this answer































    5














    The answer to both questions is yes.



    It's a standard result that the rationals are the unique countable metrizable space without isolated points, so $Bbb Qsetminus{0}$ is homeomorphic to $Bbb Q$.



    Let $S$ be a non-empty subset of $Bbb Q$. Then $StimesBbb Q$ is homeomorphic to $Bbb Q$, and the projection $pi:StimesBbb Qto S$ maps it continuously onto $S$.






    share|cite|improve this answer





















    • what about Q^n?
      – Alephnull
      Aug 11 '16 at 7:27










    • @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
      – Brian M. Scott
      Aug 11 '16 at 7:29



















    3














    Let me give yet another argument for the second question. Note that you can partition $mathbb{Q}$ into infinitely many nonempty open sets $U_n$ (for instance, let $alpha_0<alpha_1<dots$ be your favorite unbounded increasing sequence of irrational numbers and let $U_n=(alpha_{n-1},alpha_n)capmathbb{Q}$ where $alpha_{-1}=-infty$). The surjection $f:mathbb{Q}tomathbb{N}$ sending $U_n$ to $n$ is continuous. Since any map with domain $mathbb{N}$ is continuous, $mathbb{Q}$ can continuously surject onto any space that $mathbb{N}$ can surject onto, i.e. any countable nonempty space.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      There exists a monotonic bijection $Bbb QtoBbb Qsetminus{0}$. This is then automatically a homeomorphism.



      To see this, let ${q_nmid ninBbb N}$ be an enumeration of $Bbb Q$ and define $f(q_n)$ recursively: Find the smallest $m$ such that $q_mne 0$ and for all $k<n$ we have $q_m {<atop>} f(q_k)iff q_n{<atop>} q_k$. Then define $f(q_n)=q_m$.
      The crucial point is that $Bbb Qsetminus{0}$ is still linearly ordered such that there is no largest and no no smallest element and there is another element between any two given elements, thus always allowing us to find such a $q_m$.




      The same method helps find a continuous surjection $Bbb Qto S$. Just if there is no matching element in $S$, relax the $<atop>$ condition to allow equality.






      share|cite|improve this answer




























        6














        There exists a monotonic bijection $Bbb QtoBbb Qsetminus{0}$. This is then automatically a homeomorphism.



        To see this, let ${q_nmid ninBbb N}$ be an enumeration of $Bbb Q$ and define $f(q_n)$ recursively: Find the smallest $m$ such that $q_mne 0$ and for all $k<n$ we have $q_m {<atop>} f(q_k)iff q_n{<atop>} q_k$. Then define $f(q_n)=q_m$.
        The crucial point is that $Bbb Qsetminus{0}$ is still linearly ordered such that there is no largest and no no smallest element and there is another element between any two given elements, thus always allowing us to find such a $q_m$.




        The same method helps find a continuous surjection $Bbb Qto S$. Just if there is no matching element in $S$, relax the $<atop>$ condition to allow equality.






        share|cite|improve this answer


























          6












          6








          6






          There exists a monotonic bijection $Bbb QtoBbb Qsetminus{0}$. This is then automatically a homeomorphism.



          To see this, let ${q_nmid ninBbb N}$ be an enumeration of $Bbb Q$ and define $f(q_n)$ recursively: Find the smallest $m$ such that $q_mne 0$ and for all $k<n$ we have $q_m {<atop>} f(q_k)iff q_n{<atop>} q_k$. Then define $f(q_n)=q_m$.
          The crucial point is that $Bbb Qsetminus{0}$ is still linearly ordered such that there is no largest and no no smallest element and there is another element between any two given elements, thus always allowing us to find such a $q_m$.




          The same method helps find a continuous surjection $Bbb Qto S$. Just if there is no matching element in $S$, relax the $<atop>$ condition to allow equality.






          share|cite|improve this answer














          There exists a monotonic bijection $Bbb QtoBbb Qsetminus{0}$. This is then automatically a homeomorphism.



          To see this, let ${q_nmid ninBbb N}$ be an enumeration of $Bbb Q$ and define $f(q_n)$ recursively: Find the smallest $m$ such that $q_mne 0$ and for all $k<n$ we have $q_m {<atop>} f(q_k)iff q_n{<atop>} q_k$. Then define $f(q_n)=q_m$.
          The crucial point is that $Bbb Qsetminus{0}$ is still linearly ordered such that there is no largest and no no smallest element and there is another element between any two given elements, thus always allowing us to find such a $q_m$.




          The same method helps find a continuous surjection $Bbb Qto S$. Just if there is no matching element in $S$, relax the $<atop>$ condition to allow equality.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 13:19









          Martin Sleziak

          44.6k7115270




          44.6k7115270










          answered Aug 11 '16 at 6:48









          Hagen von Eitzen

          275k21268495




          275k21268495























              5














              The answer to both questions is yes.



              It's a standard result that the rationals are the unique countable metrizable space without isolated points, so $Bbb Qsetminus{0}$ is homeomorphic to $Bbb Q$.



              Let $S$ be a non-empty subset of $Bbb Q$. Then $StimesBbb Q$ is homeomorphic to $Bbb Q$, and the projection $pi:StimesBbb Qto S$ maps it continuously onto $S$.






              share|cite|improve this answer





















              • what about Q^n?
                – Alephnull
                Aug 11 '16 at 7:27










              • @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
                – Brian M. Scott
                Aug 11 '16 at 7:29
















              5














              The answer to both questions is yes.



              It's a standard result that the rationals are the unique countable metrizable space without isolated points, so $Bbb Qsetminus{0}$ is homeomorphic to $Bbb Q$.



              Let $S$ be a non-empty subset of $Bbb Q$. Then $StimesBbb Q$ is homeomorphic to $Bbb Q$, and the projection $pi:StimesBbb Qto S$ maps it continuously onto $S$.






              share|cite|improve this answer





















              • what about Q^n?
                – Alephnull
                Aug 11 '16 at 7:27










              • @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
                – Brian M. Scott
                Aug 11 '16 at 7:29














              5












              5








              5






              The answer to both questions is yes.



              It's a standard result that the rationals are the unique countable metrizable space without isolated points, so $Bbb Qsetminus{0}$ is homeomorphic to $Bbb Q$.



              Let $S$ be a non-empty subset of $Bbb Q$. Then $StimesBbb Q$ is homeomorphic to $Bbb Q$, and the projection $pi:StimesBbb Qto S$ maps it continuously onto $S$.






              share|cite|improve this answer












              The answer to both questions is yes.



              It's a standard result that the rationals are the unique countable metrizable space without isolated points, so $Bbb Qsetminus{0}$ is homeomorphic to $Bbb Q$.



              Let $S$ be a non-empty subset of $Bbb Q$. Then $StimesBbb Q$ is homeomorphic to $Bbb Q$, and the projection $pi:StimesBbb Qto S$ maps it continuously onto $S$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 11 '16 at 6:59









              Brian M. Scott

              455k38505907




              455k38505907












              • what about Q^n?
                – Alephnull
                Aug 11 '16 at 7:27










              • @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
                – Brian M. Scott
                Aug 11 '16 at 7:29


















              • what about Q^n?
                – Alephnull
                Aug 11 '16 at 7:27










              • @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
                – Brian M. Scott
                Aug 11 '16 at 7:29
















              what about Q^n?
              – Alephnull
              Aug 11 '16 at 7:27




              what about Q^n?
              – Alephnull
              Aug 11 '16 at 7:27












              @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
              – Brian M. Scott
              Aug 11 '16 at 7:29




              @Alephnull: It's homeomorphic to $Bbb Q$, if that's what you're asking.
              – Brian M. Scott
              Aug 11 '16 at 7:29











              3














              Let me give yet another argument for the second question. Note that you can partition $mathbb{Q}$ into infinitely many nonempty open sets $U_n$ (for instance, let $alpha_0<alpha_1<dots$ be your favorite unbounded increasing sequence of irrational numbers and let $U_n=(alpha_{n-1},alpha_n)capmathbb{Q}$ where $alpha_{-1}=-infty$). The surjection $f:mathbb{Q}tomathbb{N}$ sending $U_n$ to $n$ is continuous. Since any map with domain $mathbb{N}$ is continuous, $mathbb{Q}$ can continuously surject onto any space that $mathbb{N}$ can surject onto, i.e. any countable nonempty space.






              share|cite|improve this answer




























                3














                Let me give yet another argument for the second question. Note that you can partition $mathbb{Q}$ into infinitely many nonempty open sets $U_n$ (for instance, let $alpha_0<alpha_1<dots$ be your favorite unbounded increasing sequence of irrational numbers and let $U_n=(alpha_{n-1},alpha_n)capmathbb{Q}$ where $alpha_{-1}=-infty$). The surjection $f:mathbb{Q}tomathbb{N}$ sending $U_n$ to $n$ is continuous. Since any map with domain $mathbb{N}$ is continuous, $mathbb{Q}$ can continuously surject onto any space that $mathbb{N}$ can surject onto, i.e. any countable nonempty space.






                share|cite|improve this answer


























                  3












                  3








                  3






                  Let me give yet another argument for the second question. Note that you can partition $mathbb{Q}$ into infinitely many nonempty open sets $U_n$ (for instance, let $alpha_0<alpha_1<dots$ be your favorite unbounded increasing sequence of irrational numbers and let $U_n=(alpha_{n-1},alpha_n)capmathbb{Q}$ where $alpha_{-1}=-infty$). The surjection $f:mathbb{Q}tomathbb{N}$ sending $U_n$ to $n$ is continuous. Since any map with domain $mathbb{N}$ is continuous, $mathbb{Q}$ can continuously surject onto any space that $mathbb{N}$ can surject onto, i.e. any countable nonempty space.






                  share|cite|improve this answer














                  Let me give yet another argument for the second question. Note that you can partition $mathbb{Q}$ into infinitely many nonempty open sets $U_n$ (for instance, let $alpha_0<alpha_1<dots$ be your favorite unbounded increasing sequence of irrational numbers and let $U_n=(alpha_{n-1},alpha_n)capmathbb{Q}$ where $alpha_{-1}=-infty$). The surjection $f:mathbb{Q}tomathbb{N}$ sending $U_n$ to $n$ is continuous. Since any map with domain $mathbb{N}$ is continuous, $mathbb{Q}$ can continuously surject onto any space that $mathbb{N}$ can surject onto, i.e. any countable nonempty space.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 11 '16 at 7:28









                  Ittay Weiss

                  63.4k6101183




                  63.4k6101183










                  answered Aug 11 '16 at 7:20









                  Eric Wofsey

                  179k12204331




                  179k12204331






























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