Prove that $Graph (f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}([frac{i}{2^n},frac{i+1}{2^n}))times...












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Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$



but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
is a singleton, but I can't.






measure-theory







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$



    but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
    then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
    is a singleton, but I can't.






    measure-theory







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$



      but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
      then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
      is a singleton, but I can't.






      measure-theory







      share|cite|improve this question









      $endgroup$




      Let $f:mathbb Rlongrightarrow mathbb R$ a funcion. I try to prove that $$text{Graph}(f)=bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$



      but I have difficulty to prove the $supset$ inclusion. So let $$(x,y)in bigcap_{ninmathbb N}bigcup_{iinmathbb Z}f^{-1}left(left[frac{i}{2^n},frac{i+1}{2^n}right)right)times left[frac{i}{2^n},frac{i+1}{2^n}right),$$
      then, for all $n$, there is $i_n$ s.t. $(x,y)in f^{-1}left(left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)right)times left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. In particular, $f(x)in left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$ for all $n$, i.e. $f(x)=bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)$. Now, I would like to prove that $$bigcap_{ninmathbb N}left[frac{i_n}{2^n},frac{i_n+1}{2^n}right)={*},$$
      is a singleton, but I can't.






      measure-theory




      measure-theory






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      asked Dec 10 '18 at 22:23









      NewMathNewMath

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          In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.






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            $begingroup$

            In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.






            share|cite|improve this answer











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              $begingroup$

              In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.






              share|cite|improve this answer











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                $begingroup$

                In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.






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                In one hand, $f(x)in bigcap_{ninmathbb N}I_n$. In the other hand, if $yin bigcap_{ninmathbb N}I_n$, then $|y-f(x)|leq frac{1}{2^n}to 0$ and thus $f(x)=y$. The claim follow.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Dec 10 '18 at 22:51

























                answered Dec 10 '18 at 22:43









                SurbSurb

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