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?






real-analysis supremum-and-infimum







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?






    real-analysis supremum-and-infimum







    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?






      real-analysis supremum-and-infimum







      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




      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

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
          Read more

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

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
          Read more

          File:DeusFollowingSea.jpg

          Image
          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
          Read more