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Non-Borel set in arbitrary metric space


Derived Sets in arbitrary metric space$A subseteq (X,d)$ is compact. Which metric $p$ makes $(A times A,p)$ also compact and $d: (A times A,p) rightarrow [0,infty)$ continuous?Borel sets and measurabilityapproximate a Borel set by a continuousAn example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example.A Borel subset of a topological spaceseparability of a metric spacetotally disconnected and non Borel set.What do metric spaces look like?How do we get the notion “Borel regular” measures?













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$begingroup$


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.










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$endgroup$
















    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






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      asked 4 hours ago









      Daniel LiDaniel Li

      752414




      752414




















          2 Answers
          2






          active

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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 $mathbbR$ 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 $d:x,ytimes x,y to 0,1$ 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
          $$
          x,y,x,y,emptyset = mathcalP(x,y)
          $$

          where $mathcalP(x,y)$ 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









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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 $mathbbR$ 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 $d:x,ytimes x,y to 0,1$ 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
              $$
              x,y,x,y,emptyset = mathcalP(x,y)
              $$

              where $mathcalP(x,y)$ 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 $mathbbR$ 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 $d:x,ytimes x,y to 0,1$ 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
                $$
                x,y,x,y,emptyset = mathcalP(x,y)
                $$

                where $mathcalP(x,y)$ 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 $mathbbR$ 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 $d:x,ytimes x,y to 0,1$ 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
                  $$
                  x,y,x,y,emptyset = mathcalP(x,y)
                  $$

                  where $mathcalP(x,y)$ 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 $mathbbR$ 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 $d:x,ytimes x,y to 0,1$ 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
                  $$
                  x,y,x,y,emptyset = mathcalP(x,y)
                  $$

                  where $mathcalP(x,y)$ 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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