Supremum of the image of a monotonic function












1












$begingroup$


For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?










      share|cite|improve this question









      $endgroup$




      For a monotonic function $f:Dtomathbb{R}$ where $Dsubsetmathbb{R}$, and two sets $A$ and $B$ such that $sup A=sup B$, is it true that $sup f(A)=sup f(B)$, where $f(A)$ denotes the image of set $A$ under function $f$? If so, how to prove it, assuming $f(sup A)$ and $f(sup B)$ may be undefined?







      real-analysis supremum-and-infimum






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 17 at 13:12









      Yutong ZhangYutong Zhang

      587




      587






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if $f$ is monotonically increasing?
            $endgroup$
            – Yutong Zhang
            Jan 17 at 13:30






          • 1




            $begingroup$
            @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
            $endgroup$
            – user3482749
            Jan 17 at 13:33










          • $begingroup$
            For proving something we need proof. And to disprove things even counter example is sufficient.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:34










          • $begingroup$
            But I'll write for increasing as well....
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:35






          • 1




            $begingroup$
            Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 14:03












          Your Answer








          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%2f3076953%2fsupremum-of-the-image-of-a-monotonic-function%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if $f$ is monotonically increasing?
            $endgroup$
            – Yutong Zhang
            Jan 17 at 13:30






          • 1




            $begingroup$
            @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
            $endgroup$
            – user3482749
            Jan 17 at 13:33










          • $begingroup$
            For proving something we need proof. And to disprove things even counter example is sufficient.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:34










          • $begingroup$
            But I'll write for increasing as well....
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:35






          • 1




            $begingroup$
            Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 14:03
















          1












          $begingroup$

          No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What if $f$ is monotonically increasing?
            $endgroup$
            – Yutong Zhang
            Jan 17 at 13:30






          • 1




            $begingroup$
            @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
            $endgroup$
            – user3482749
            Jan 17 at 13:33










          • $begingroup$
            For proving something we need proof. And to disprove things even counter example is sufficient.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:34










          • $begingroup$
            But I'll write for increasing as well....
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:35






          • 1




            $begingroup$
            Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 14:03














          1












          1








          1





          $begingroup$

          No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?






          share|cite|improve this answer









          $endgroup$



          No. This need not be true. The function may be monotonic but it may be decreasing. e.g. y= -x and choose D= [0,1] A =[1/2, 1] B=[1/3, 1] then Sup(A)=Sup(B)=1 and next thing is reversed. Right?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 13:26









          Ravi SatputeRavi Satpute

          267




          267












          • $begingroup$
            What if $f$ is monotonically increasing?
            $endgroup$
            – Yutong Zhang
            Jan 17 at 13:30






          • 1




            $begingroup$
            @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
            $endgroup$
            – user3482749
            Jan 17 at 13:33










          • $begingroup$
            For proving something we need proof. And to disprove things even counter example is sufficient.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:34










          • $begingroup$
            But I'll write for increasing as well....
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:35






          • 1




            $begingroup$
            Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 14:03


















          • $begingroup$
            What if $f$ is monotonically increasing?
            $endgroup$
            – Yutong Zhang
            Jan 17 at 13:30






          • 1




            $begingroup$
            @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
            $endgroup$
            – user3482749
            Jan 17 at 13:33










          • $begingroup$
            For proving something we need proof. And to disprove things even counter example is sufficient.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:34










          • $begingroup$
            But I'll write for increasing as well....
            $endgroup$
            – Ravi Satpute
            Jan 17 at 13:35






          • 1




            $begingroup$
            Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
            $endgroup$
            – Ravi Satpute
            Jan 17 at 14:03
















          $begingroup$
          What if $f$ is monotonically increasing?
          $endgroup$
          – Yutong Zhang
          Jan 17 at 13:30




          $begingroup$
          What if $f$ is monotonically increasing?
          $endgroup$
          – Yutong Zhang
          Jan 17 at 13:30




          1




          1




          $begingroup$
          @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
          $endgroup$
          – user3482749
          Jan 17 at 13:33




          $begingroup$
          @YutongZhang Still no: take $A = (0,1)$, $B = [0,1]$, and take $$f: x mapsto left{array{x&x< 1\x+1&x geq 1}right..$$ Then $sup f(A) = 1$, but $sup f(B) = 2$.
          $endgroup$
          – user3482749
          Jan 17 at 13:33












          $begingroup$
          For proving something we need proof. And to disprove things even counter example is sufficient.
          $endgroup$
          – Ravi Satpute
          Jan 17 at 13:34




          $begingroup$
          For proving something we need proof. And to disprove things even counter example is sufficient.
          $endgroup$
          – Ravi Satpute
          Jan 17 at 13:34












          $begingroup$
          But I'll write for increasing as well....
          $endgroup$
          – Ravi Satpute
          Jan 17 at 13:35




          $begingroup$
          But I'll write for increasing as well....
          $endgroup$
          – Ravi Satpute
          Jan 17 at 13:35




          1




          1




          $begingroup$
          Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
          $endgroup$
          – Ravi Satpute
          Jan 17 at 14:03




          $begingroup$
          Only continuity won't be sufficient. As my counter example is also continuous. But if $f$ is from [a,b] continuous and increasing then these are sufficient conditions and much stronger. I'm intentionally adding these conditions so that you won't have to worry about boundary points. You can also try with A and B finite sets and keeping Sup included. So that the functions behaviour on in between points will matter and sup f(A) you can try with curves.
          $endgroup$
          – Ravi Satpute
          Jan 17 at 14:03


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076953%2fsupremum-of-the-image-of-a-monotonic-function%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

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅