Openness of Upper Contour Set












1














Let $f: {mathbb{R}^n}$ to $mathbb{R} $ is a continuous function. For any $x$ belonging to $mathbb{R}^n$ define
$(U(x)={y|f(y)>f(x)})$ is an open set.



I tried proving that the complement of the set is closed. The property of continuity of function could be invoked. I cannot seem to properly form the proof. Please help!






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  • I edited your question (the equation which defines the open set). Please make sure I did it properly.
    – Yanko
    Nov 25 at 15:13










  • Thank You for the correction
    – Shinjini Rana
    Nov 25 at 15:51
















1














Let $f: {mathbb{R}^n}$ to $mathbb{R} $ is a continuous function. For any $x$ belonging to $mathbb{R}^n$ define
$(U(x)={y|f(y)>f(x)})$ is an open set.



I tried proving that the complement of the set is closed. The property of continuity of function could be invoked. I cannot seem to properly form the proof. Please help!






real-analysis functions self-learning







share|cite|improve this question
























  • I edited your question (the equation which defines the open set). Please make sure I did it properly.
    – Yanko
    Nov 25 at 15:13










  • Thank You for the correction
    – Shinjini Rana
    Nov 25 at 15:51














1












1








1







Let $f: {mathbb{R}^n}$ to $mathbb{R} $ is a continuous function. For any $x$ belonging to $mathbb{R}^n$ define
$(U(x)={y|f(y)>f(x)})$ is an open set.



I tried proving that the complement of the set is closed. The property of continuity of function could be invoked. I cannot seem to properly form the proof. Please help!






real-analysis functions self-learning







share|cite|improve this question















Let $f: {mathbb{R}^n}$ to $mathbb{R} $ is a continuous function. For any $x$ belonging to $mathbb{R}^n$ define
$(U(x)={y|f(y)>f(x)})$ is an open set.



I tried proving that the complement of the set is closed. The property of continuity of function could be invoked. I cannot seem to properly form the proof. Please help!






real-analysis functions self-learning




real-analysis functions self-learning






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share|cite|improve this question













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edited Nov 25 at 19:02









Andrés E. Caicedo

64.7k8158246




64.7k8158246










asked Nov 25 at 14:56









Shinjini Rana

7915




7915












  • I edited your question (the equation which defines the open set). Please make sure I did it properly.
    – Yanko
    Nov 25 at 15:13










  • Thank You for the correction
    – Shinjini Rana
    Nov 25 at 15:51


















  • I edited your question (the equation which defines the open set). Please make sure I did it properly.
    – Yanko
    Nov 25 at 15:13










  • Thank You for the correction
    – Shinjini Rana
    Nov 25 at 15:51
















I edited your question (the equation which defines the open set). Please make sure I did it properly.
– Yanko
Nov 25 at 15:13




I edited your question (the equation which defines the open set). Please make sure I did it properly.
– Yanko
Nov 25 at 15:13












Thank You for the correction
– Shinjini Rana
Nov 25 at 15:51




Thank You for the correction
– Shinjini Rana
Nov 25 at 15:51










2 Answers
2






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1














$U(x)=f^{-1}((f(x),infty)$ and $((f(x),infty)$ is open in $mathbb R$ so $U(x)$ is open.






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  • Short and clever. Fantastic proof.
    – Dog_69
    Dec 7 at 9:39



















1














I will prove that $U$ is sequentially open. In $mathbb R^n$ it implies that the set is also open:



Definition (Reminder). A set $Usubseteq mathbb R^n$ is sequentially open if, for every sequence ${x_n}$ converging to a point $xin U$, there exists $n_0inmathbb N$ such that $x_nin U$ whenever $ngeq n_0$.



Now, consider a point $yin U$ and a sequence ${y_n}$ such that $y_nto y$. Since $f$ is continuous, the sequence ${f(y_n)}$ converges to $f(y)$ on $mathbb R$. By definition of convergence, there exists $n_0inmathbb N$ such that



$$f(y_n) in (f(y)-r,f(y)+r), $$



(here $0<r<f(y)-f(x)$) whenever $ngeq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $ngeq n_0$, i.e. $y_nin U$ whenever $ngeq n_0$, as we wanted.






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









    1














    $U(x)=f^{-1}((f(x),infty)$ and $((f(x),infty)$ is open in $mathbb R$ so $U(x)$ is open.






    share|cite|improve this answer





















    • Short and clever. Fantastic proof.
      – Dog_69
      Dec 7 at 9:39
















    1














    $U(x)=f^{-1}((f(x),infty)$ and $((f(x),infty)$ is open in $mathbb R$ so $U(x)$ is open.






    share|cite|improve this answer





















    • Short and clever. Fantastic proof.
      – Dog_69
      Dec 7 at 9:39














    1












    1








    1






    $U(x)=f^{-1}((f(x),infty)$ and $((f(x),infty)$ is open in $mathbb R$ so $U(x)$ is open.






    share|cite|improve this answer












    $U(x)=f^{-1}((f(x),infty)$ and $((f(x),infty)$ is open in $mathbb R$ so $U(x)$ is open.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 at 23:24









    Kavi Rama Murthy

    48.6k31854




    48.6k31854












    • Short and clever. Fantastic proof.
      – Dog_69
      Dec 7 at 9:39


















    • Short and clever. Fantastic proof.
      – Dog_69
      Dec 7 at 9:39
















    Short and clever. Fantastic proof.
    – Dog_69
    Dec 7 at 9:39




    Short and clever. Fantastic proof.
    – Dog_69
    Dec 7 at 9:39











    1














    I will prove that $U$ is sequentially open. In $mathbb R^n$ it implies that the set is also open:



    Definition (Reminder). A set $Usubseteq mathbb R^n$ is sequentially open if, for every sequence ${x_n}$ converging to a point $xin U$, there exists $n_0inmathbb N$ such that $x_nin U$ whenever $ngeq n_0$.



    Now, consider a point $yin U$ and a sequence ${y_n}$ such that $y_nto y$. Since $f$ is continuous, the sequence ${f(y_n)}$ converges to $f(y)$ on $mathbb R$. By definition of convergence, there exists $n_0inmathbb N$ such that



    $$f(y_n) in (f(y)-r,f(y)+r), $$



    (here $0<r<f(y)-f(x)$) whenever $ngeq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $ngeq n_0$, i.e. $y_nin U$ whenever $ngeq n_0$, as we wanted.






    share|cite|improve this answer




























      1














      I will prove that $U$ is sequentially open. In $mathbb R^n$ it implies that the set is also open:



      Definition (Reminder). A set $Usubseteq mathbb R^n$ is sequentially open if, for every sequence ${x_n}$ converging to a point $xin U$, there exists $n_0inmathbb N$ such that $x_nin U$ whenever $ngeq n_0$.



      Now, consider a point $yin U$ and a sequence ${y_n}$ such that $y_nto y$. Since $f$ is continuous, the sequence ${f(y_n)}$ converges to $f(y)$ on $mathbb R$. By definition of convergence, there exists $n_0inmathbb N$ such that



      $$f(y_n) in (f(y)-r,f(y)+r), $$



      (here $0<r<f(y)-f(x)$) whenever $ngeq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $ngeq n_0$, i.e. $y_nin U$ whenever $ngeq n_0$, as we wanted.






      share|cite|improve this answer


























        1












        1








        1






        I will prove that $U$ is sequentially open. In $mathbb R^n$ it implies that the set is also open:



        Definition (Reminder). A set $Usubseteq mathbb R^n$ is sequentially open if, for every sequence ${x_n}$ converging to a point $xin U$, there exists $n_0inmathbb N$ such that $x_nin U$ whenever $ngeq n_0$.



        Now, consider a point $yin U$ and a sequence ${y_n}$ such that $y_nto y$. Since $f$ is continuous, the sequence ${f(y_n)}$ converges to $f(y)$ on $mathbb R$. By definition of convergence, there exists $n_0inmathbb N$ such that



        $$f(y_n) in (f(y)-r,f(y)+r), $$



        (here $0<r<f(y)-f(x)$) whenever $ngeq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $ngeq n_0$, i.e. $y_nin U$ whenever $ngeq n_0$, as we wanted.






        share|cite|improve this answer














        I will prove that $U$ is sequentially open. In $mathbb R^n$ it implies that the set is also open:



        Definition (Reminder). A set $Usubseteq mathbb R^n$ is sequentially open if, for every sequence ${x_n}$ converging to a point $xin U$, there exists $n_0inmathbb N$ such that $x_nin U$ whenever $ngeq n_0$.



        Now, consider a point $yin U$ and a sequence ${y_n}$ such that $y_nto y$. Since $f$ is continuous, the sequence ${f(y_n)}$ converges to $f(y)$ on $mathbb R$. By definition of convergence, there exists $n_0inmathbb N$ such that



        $$f(y_n) in (f(y)-r,f(y)+r), $$



        (here $0<r<f(y)-f(x)$) whenever $ngeq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $ngeq n_0$, i.e. $y_nin U$ whenever $ngeq n_0$, as we wanted.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 at 23:05

























        answered Nov 25 at 16:53









        Dog_69

        5401422




        5401422






























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