Continuous self-maps on $mathbb{Q}$











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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






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















up vote
8
down vote

favorite
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




    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













up vote
8
down vote

favorite
3









up vote
8
down vote

favorite
3






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

















up vote
6
down vote



accepted










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






























    up vote
    5
    down vote













    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


















    up vote
    3
    down vote













    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








      up vote
      6
      down vote



      accepted










      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



























        up vote
        6
        down vote



        accepted










        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

























          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          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

          44.6k7115269




          44.6k7115269










          answered Aug 11 '16 at 6:48









          Hagen von Eitzen

          275k21268495




          275k21268495






















              up vote
              5
              down vote













              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















              up vote
              5
              down vote













              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













              up vote
              5
              down vote










              up vote
              5
              down vote









              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

              454k38505906




              454k38505906












              • 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










              up vote
              3
              down vote













              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



























                up vote
                3
                down vote













                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

























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  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

                  63.2k6101183




                  63.2k6101183










                  answered Aug 11 '16 at 7:20









                  Eric Wofsey

                  178k12202328




                  178k12202328






























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