Non-Borel set in arbitrary metric space












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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







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    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.






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







    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.






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







      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




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






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      asked 4 hours ago









      Daniel LiDaniel Li

      752414




      752414






















          2 Answers
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          $begingroup$

          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











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            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.






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              2 Answers
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              active

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              2 Answers
              2






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              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






























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