How to prove that every open convex set in $mathbb{R}^{n}$ is homeomorphic to an open ball?












1












$begingroup$



If $A,B$ are compacts,, convex with nonempty interiors, prove that they are homeomorphic




My ideia is to find a homeomorphism between $A$ and $B$ and extended to closure, since in this case $text{int}(overline{A}) = text{int}(A)$ (Can I do this?). My question is: prove that $text{int}(A)$ and $text{int}(B)$ are homeomorphic? I'm trying to prove that a open convex set in $mathbb{R}^{n}$ is homeomorphic to an open ball, but I still cannot. Can someone give me a hint?



I didnt look for other questions because I dont want to see the complete solution. I just wanted a hint of how to prove it.




Lastly, is possible to generalize this result for a complete metric space?






real-analysis metric-spaces convex-analysis







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$endgroup$








  • 1




    $begingroup$
    I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:48






  • 1




    $begingroup$
    Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
    $endgroup$
    – Cheerful Parsnip
    Dec 23 '18 at 19:54
















1












$begingroup$



If $A,B$ are compacts,, convex with nonempty interiors, prove that they are homeomorphic




My ideia is to find a homeomorphism between $A$ and $B$ and extended to closure, since in this case $text{int}(overline{A}) = text{int}(A)$ (Can I do this?). My question is: prove that $text{int}(A)$ and $text{int}(B)$ are homeomorphic? I'm trying to prove that a open convex set in $mathbb{R}^{n}$ is homeomorphic to an open ball, but I still cannot. Can someone give me a hint?



I didnt look for other questions because I dont want to see the complete solution. I just wanted a hint of how to prove it.




Lastly, is possible to generalize this result for a complete metric space?






real-analysis metric-spaces convex-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:48






  • 1




    $begingroup$
    Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
    $endgroup$
    – Cheerful Parsnip
    Dec 23 '18 at 19:54














1












1








1





$begingroup$



If $A,B$ are compacts,, convex with nonempty interiors, prove that they are homeomorphic




My ideia is to find a homeomorphism between $A$ and $B$ and extended to closure, since in this case $text{int}(overline{A}) = text{int}(A)$ (Can I do this?). My question is: prove that $text{int}(A)$ and $text{int}(B)$ are homeomorphic? I'm trying to prove that a open convex set in $mathbb{R}^{n}$ is homeomorphic to an open ball, but I still cannot. Can someone give me a hint?



I didnt look for other questions because I dont want to see the complete solution. I just wanted a hint of how to prove it.




Lastly, is possible to generalize this result for a complete metric space?






real-analysis metric-spaces convex-analysis







share|cite|improve this question









$endgroup$





If $A,B$ are compacts,, convex with nonempty interiors, prove that they are homeomorphic




My ideia is to find a homeomorphism between $A$ and $B$ and extended to closure, since in this case $text{int}(overline{A}) = text{int}(A)$ (Can I do this?). My question is: prove that $text{int}(A)$ and $text{int}(B)$ are homeomorphic? I'm trying to prove that a open convex set in $mathbb{R}^{n}$ is homeomorphic to an open ball, but I still cannot. Can someone give me a hint?



I didnt look for other questions because I dont want to see the complete solution. I just wanted a hint of how to prove it.




Lastly, is possible to generalize this result for a complete metric space?






real-analysis metric-spaces convex-analysis




real-analysis metric-spaces convex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 19:44









Lucas CorrêaLucas Corrêa

1,5711321




1,5711321








  • 1




    $begingroup$
    I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:48






  • 1




    $begingroup$
    Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
    $endgroup$
    – Cheerful Parsnip
    Dec 23 '18 at 19:54














  • 1




    $begingroup$
    I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
    $endgroup$
    – Exodd
    Dec 23 '18 at 19:48






  • 1




    $begingroup$
    Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
    $endgroup$
    – Cheerful Parsnip
    Dec 23 '18 at 19:54








1




1




$begingroup$
I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
$endgroup$
– Exodd
Dec 23 '18 at 19:48




$begingroup$
I would try to fix a point in the interior of each set, $P,Q$ and then map every segment starting from $P$ into a segment starting from $Q$. More or less what you do when you define a seminorm associated to a convex set
$endgroup$
– Exodd
Dec 23 '18 at 19:48




1




1




$begingroup$
Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
$endgroup$
– Cheerful Parsnip
Dec 23 '18 at 19:54




$begingroup$
Google came up with this: math.uni-hamburg.de/home/geschke/papers/ConvexOpen.pdf
$endgroup$
– Cheerful Parsnip
Dec 23 '18 at 19:54










2 Answers
2






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oldest

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2












$begingroup$

Hints:



$(1). $ Let $zin U$, an open and bounded convex set in $mathbb R^n$. Then, each ray from $z$ intersects $partial U$ in exactly one point.



$(2). $ Set $f(x)=x/|x|$, use $(1)$ to show that $f$ restricts to a bijection from $partial U$ to the sphere.



$(3).$ Extend the inverse of the restricted map in $(2)$ to the ball.



$(4)$. Prove that the map and its inverse are continuous bijections.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: Without lost of generality assume $0 in int A $. Then consider $f: A to B$ with



    $$ f(x)= frac{x }{| x |} quad forall x in bd A $$



    and $f(0) = 0 .$ For the case $x in int A setminus {0}$, note that $f^{-1} {frac{x }{| x |} } in bd A $ is singleton so lets call it by $eta (x)$ then define $$ f(x) = frac{x}{| eta (x) |} $$



    Here $B$ is the unit closed ball.



    It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $eta : bd B to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.






    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












      $begingroup$

      Hints:



      $(1). $ Let $zin U$, an open and bounded convex set in $mathbb R^n$. Then, each ray from $z$ intersects $partial U$ in exactly one point.



      $(2). $ Set $f(x)=x/|x|$, use $(1)$ to show that $f$ restricts to a bijection from $partial U$ to the sphere.



      $(3).$ Extend the inverse of the restricted map in $(2)$ to the ball.



      $(4)$. Prove that the map and its inverse are continuous bijections.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hints:



        $(1). $ Let $zin U$, an open and bounded convex set in $mathbb R^n$. Then, each ray from $z$ intersects $partial U$ in exactly one point.



        $(2). $ Set $f(x)=x/|x|$, use $(1)$ to show that $f$ restricts to a bijection from $partial U$ to the sphere.



        $(3).$ Extend the inverse of the restricted map in $(2)$ to the ball.



        $(4)$. Prove that the map and its inverse are continuous bijections.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hints:



          $(1). $ Let $zin U$, an open and bounded convex set in $mathbb R^n$. Then, each ray from $z$ intersects $partial U$ in exactly one point.



          $(2). $ Set $f(x)=x/|x|$, use $(1)$ to show that $f$ restricts to a bijection from $partial U$ to the sphere.



          $(3).$ Extend the inverse of the restricted map in $(2)$ to the ball.



          $(4)$. Prove that the map and its inverse are continuous bijections.






          share|cite|improve this answer









          $endgroup$



          Hints:



          $(1). $ Let $zin U$, an open and bounded convex set in $mathbb R^n$. Then, each ray from $z$ intersects $partial U$ in exactly one point.



          $(2). $ Set $f(x)=x/|x|$, use $(1)$ to show that $f$ restricts to a bijection from $partial U$ to the sphere.



          $(3).$ Extend the inverse of the restricted map in $(2)$ to the ball.



          $(4)$. Prove that the map and its inverse are continuous bijections.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 20:02









          MatematletaMatematleta

          11.1k2918




          11.1k2918























              1












              $begingroup$

              Hint: Without lost of generality assume $0 in int A $. Then consider $f: A to B$ with



              $$ f(x)= frac{x }{| x |} quad forall x in bd A $$



              and $f(0) = 0 .$ For the case $x in int A setminus {0}$, note that $f^{-1} {frac{x }{| x |} } in bd A $ is singleton so lets call it by $eta (x)$ then define $$ f(x) = frac{x}{| eta (x) |} $$



              Here $B$ is the unit closed ball.



              It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $eta : bd B to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Hint: Without lost of generality assume $0 in int A $. Then consider $f: A to B$ with



                $$ f(x)= frac{x }{| x |} quad forall x in bd A $$



                and $f(0) = 0 .$ For the case $x in int A setminus {0}$, note that $f^{-1} {frac{x }{| x |} } in bd A $ is singleton so lets call it by $eta (x)$ then define $$ f(x) = frac{x}{| eta (x) |} $$



                Here $B$ is the unit closed ball.



                It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $eta : bd B to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint: Without lost of generality assume $0 in int A $. Then consider $f: A to B$ with



                  $$ f(x)= frac{x }{| x |} quad forall x in bd A $$



                  and $f(0) = 0 .$ For the case $x in int A setminus {0}$, note that $f^{-1} {frac{x }{| x |} } in bd A $ is singleton so lets call it by $eta (x)$ then define $$ f(x) = frac{x}{| eta (x) |} $$



                  Here $B$ is the unit closed ball.



                  It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $eta : bd B to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.






                  share|cite|improve this answer











                  $endgroup$



                  Hint: Without lost of generality assume $0 in int A $. Then consider $f: A to B$ with



                  $$ f(x)= frac{x }{| x |} quad forall x in bd A $$



                  and $f(0) = 0 .$ For the case $x in int A setminus {0}$, note that $f^{-1} {frac{x }{| x |} } in bd A $ is singleton so lets call it by $eta (x)$ then define $$ f(x) = frac{x}{| eta (x) |} $$



                  Here $B$ is the unit closed ball.



                  It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $eta : bd B to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 23 '18 at 20:24

























                  answered Dec 23 '18 at 19:52









                  Red shoesRed shoes

                  4,761621




                  4,761621






























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