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.






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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.






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







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      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







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      $endgroup$




      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

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      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











          $endgroup$













          • $begingroup$
            +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
            $endgroup$
            – DanielWainfleet
            2 mins ago





















          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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            2 Answers
            2






            active

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            active

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            active

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            5












            $begingroup$

            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











            $endgroup$













            • $begingroup$
              +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
              $endgroup$
              – DanielWainfleet
              2 mins ago


















            5












            $begingroup$

            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











            $endgroup$













            • $begingroup$
              +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
              $endgroup$
              – DanielWainfleet
              2 mins ago
















            5












            5








            5





            $begingroup$

            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 4 hours ago









            MartinMartin

            1,106917




            1,106917












            • $begingroup$
              +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
              $endgroup$
              – DanielWainfleet
              2 mins ago




















            • $begingroup$
              +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
              $endgroup$
              – DanielWainfleet
              2 mins ago


















            $begingroup$
            +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
            $endgroup$
            – DanielWainfleet
            2 mins ago






            $begingroup$
            +1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup {{y}:yin Y},$ which is a countable union of closed sets
            $endgroup$
            – DanielWainfleet
            2 mins ago













            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 3 hours ago









                Noah SchweberNoah Schweber

                127k10151290




                127k10151290






























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