Extension of order-preserving bijection from rationals to reals.












2












$begingroup$


If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.



Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.



Next we would need to prove such an extension is continuous and continuous inverse?






real-analysis order-theory







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  • 2




    $begingroup$
    Dedekind cuts${}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 18:07
















2












$begingroup$


If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.



Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.



Next we would need to prove such an extension is continuous and continuous inverse?






real-analysis order-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Dedekind cuts${}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 18:07














2












2








2





$begingroup$


If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.



Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.



Next we would need to prove such an extension is continuous and continuous inverse?






real-analysis order-theory







share|cite|improve this question











$endgroup$




If $f:mathbb{Q}rightarrowmathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:mathbb{R}rightarrowmathbb{R}$.



Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.



Next we would need to prove such an extension is continuous and continuous inverse?






real-analysis order-theory




real-analysis order-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 19:52









Andrés E. Caicedo

65.8k8160252




65.8k8160252










asked Dec 19 '18 at 18:02









ershersh

438113




438113








  • 2




    $begingroup$
    Dedekind cuts${}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 18:07














  • 2




    $begingroup$
    Dedekind cuts${}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 18:07








2




2




$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07




$begingroup$
Dedekind cuts${}$?
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 18:07










1 Answer
1






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oldest

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1












$begingroup$

My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.



Here is another approach using Lord Shark the Unknown natural idea.



Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
$$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
Then





  • $Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.


  • $F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$


  • $F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.


Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
and prove the three properties above.



As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.






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    1 Answer
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    $begingroup$

    My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.



    Here is another approach using Lord Shark the Unknown natural idea.



    Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
    $$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
    Then





    • $Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.


    • $F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$


    • $F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.


    Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
    and prove the three properties above.



    As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.



      Here is another approach using Lord Shark the Unknown natural idea.



      Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
      $$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
      Then





      • $Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.


      • $F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$


      • $F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.


      Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
      and prove the three properties above.



      As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.



        Here is another approach using Lord Shark the Unknown natural idea.



        Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
        $$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
        Then





        • $Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.


        • $F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$


        • $F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.


        Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
        and prove the three properties above.



        As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.






        share|cite|improve this answer











        $endgroup$



        My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.



        Here is another approach using Lord Shark the Unknown natural idea.



        Given $xin mathbb{R}$ define $F:mathbb{R}rightarrow mathbb{R}$ by
        $$F(x):=sup left ( f((-infty,x) cap mathbb{Q})right )$$
        Then





        • $Fmid_mathbb{Q}=f$: This is just the identity $f(-infty,a)=(-infty,f(a))$ for $ain mathbb{Q}$.


        • $F$ is monotone: This came from $x<y implies f((-infty,x) cap mathbb{Q})subseteq f((-infty,y) cap mathbb{Q})$


        • $F$ is continuous: Take $xin mathbb{R}$ and $varepsilon>0$. Take $ain (f(x),f(x)+varepsilon)cap mathbb{Q}$ an define $b=f^{-1}(a)$. Then $delta=b-x$ is such that $f(x,x+delta)subseteq (f(x),f(x)+varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.


        Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=sup left ( g((-infty,x) cap mathbb{Q})right )$$
        and prove the three properties above.



        As $Fcirc Gmid_mathbb{Q}=Gcirc Fmid_mathbb{Q}=text{Id}_mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 19:33

























        answered Dec 19 '18 at 19:15









        yamete kudasaiyamete kudasai

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