Non-Borel set in arbitrary metric space












1












$begingroup$


Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.










      share|cite|improve this question









      $endgroup$




      Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.







      real-analysis general-topology functional-analysis measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Daniel LiDaniel Li

      752414




      752414






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.



          This result can be found in:
          Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.



          In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $({x,y},d)$ equipped with the discrete metric $d:{x,y}times {x,y} to {0,1}$ given by
          $$
          d(x,y)=1, quad d(x,x)=d(y,y)=0.
          $$

          The Borel sigma algebra on this metric space is given by
          $$
          {{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
          $$

          where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:



            In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.






            share|cite|improve this answer









            $endgroup$













              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%2f3154781%2fnon-borel-set-in-arbitrary-metric-space%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









              4












              $begingroup$

              Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.



              This result can be found in:
              Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.



              In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $({x,y},d)$ equipped with the discrete metric $d:{x,y}times {x,y} to {0,1}$ given by
              $$
              d(x,y)=1, quad d(x,x)=d(y,y)=0.
              $$

              The Borel sigma algebra on this metric space is given by
              $$
              {{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
              $$

              where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.



                This result can be found in:
                Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.



                In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $({x,y},d)$ equipped with the discrete metric $d:{x,y}times {x,y} to {0,1}$ given by
                $$
                d(x,y)=1, quad d(x,x)=d(y,y)=0.
                $$

                The Borel sigma algebra on this metric space is given by
                $$
                {{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
                $$

                where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.



                  This result can be found in:
                  Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.



                  In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $({x,y},d)$ equipped with the discrete metric $d:{x,y}times {x,y} to {0,1}$ given by
                  $$
                  d(x,y)=1, quad d(x,x)=d(y,y)=0.
                  $$

                  The Borel sigma algebra on this metric space is given by
                  $$
                  {{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
                  $$

                  where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.






                  share|cite|improve this answer











                  $endgroup$



                  Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.



                  This result can be found in:
                  Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.



                  In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $({x,y},d)$ equipped with the discrete metric $d:{x,y}times {x,y} to {0,1}$ given by
                  $$
                  d(x,y)=1, quad d(x,x)=d(y,y)=0.
                  $$

                  The Borel sigma algebra on this metric space is given by
                  $$
                  {{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
                  $$

                  where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  MartinMartin

                  1,096917




                  1,096917























                      2












                      $begingroup$

                      Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:



                      In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:



                        In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:



                          In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.






                          share|cite|improve this answer









                          $endgroup$



                          Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:



                          In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Noah SchweberNoah Schweber

                          127k10151290




                          127k10151290






























                              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%2f3154781%2fnon-borel-set-in-arbitrary-metric-space%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

                              Liste der Baudenkmale in Friedland (Mecklenburg)

                              Single-Malt-Whisky

                              Czorneboh