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!






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














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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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






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














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























    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012935%2fopenness-of-upper-contour-set%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012935%2fopenness-of-upper-contour-set%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
            Read more

            “minimization” problem in Euclidean space related to orthonormal basis

            Image
            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
            Read more