Properties of discontinuity of the second kind












1












$begingroup$


Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either $f(x^+)$ or $f(x^-)$ does not exist.



Supposing that $f$ has a discontinuity of the second kind on an interval $(a, b)$. I want to show that there is subinterval $(c, d)subseteq (a, b)$ such that $f$ is not monotonic on any interval $(e, f)subseteq (c, d)$.



We already know that if $f$ is monotonic on any open interval, then it cannot have any discontinuities of the second kind.



I can't figure out how to construct the interval around the discontinuity. My attempt so far is



Assume without loss of generality that $f$ has a discontinuity of the second kind at $xin (a, b)$ and that $f(x^+)$ does not exist. Then there must exist a sequence ${t_n}in(x, b)$ such that $t_nrightarrow x$ but $f(t_n)$ does not converge.



I draw a blank once I get here, this could very well be false.






real-analysis monotone-functions discontinuous-functions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either $f(x^+)$ or $f(x^-)$ does not exist.



    Supposing that $f$ has a discontinuity of the second kind on an interval $(a, b)$. I want to show that there is subinterval $(c, d)subseteq (a, b)$ such that $f$ is not monotonic on any interval $(e, f)subseteq (c, d)$.



    We already know that if $f$ is monotonic on any open interval, then it cannot have any discontinuities of the second kind.



    I can't figure out how to construct the interval around the discontinuity. My attempt so far is



    Assume without loss of generality that $f$ has a discontinuity of the second kind at $xin (a, b)$ and that $f(x^+)$ does not exist. Then there must exist a sequence ${t_n}in(x, b)$ such that $t_nrightarrow x$ but $f(t_n)$ does not converge.



    I draw a blank once I get here, this could very well be false.






    real-analysis monotone-functions discontinuous-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either $f(x^+)$ or $f(x^-)$ does not exist.



      Supposing that $f$ has a discontinuity of the second kind on an interval $(a, b)$. I want to show that there is subinterval $(c, d)subseteq (a, b)$ such that $f$ is not monotonic on any interval $(e, f)subseteq (c, d)$.



      We already know that if $f$ is monotonic on any open interval, then it cannot have any discontinuities of the second kind.



      I can't figure out how to construct the interval around the discontinuity. My attempt so far is



      Assume without loss of generality that $f$ has a discontinuity of the second kind at $xin (a, b)$ and that $f(x^+)$ does not exist. Then there must exist a sequence ${t_n}in(x, b)$ such that $t_nrightarrow x$ but $f(t_n)$ does not converge.



      I draw a blank once I get here, this could very well be false.






      real-analysis monotone-functions discontinuous-functions







      share|cite|improve this question









      $endgroup$




      Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either $f(x^+)$ or $f(x^-)$ does not exist.



      Supposing that $f$ has a discontinuity of the second kind on an interval $(a, b)$. I want to show that there is subinterval $(c, d)subseteq (a, b)$ such that $f$ is not monotonic on any interval $(e, f)subseteq (c, d)$.



      We already know that if $f$ is monotonic on any open interval, then it cannot have any discontinuities of the second kind.



      I can't figure out how to construct the interval around the discontinuity. My attempt so far is



      Assume without loss of generality that $f$ has a discontinuity of the second kind at $xin (a, b)$ and that $f(x^+)$ does not exist. Then there must exist a sequence ${t_n}in(x, b)$ such that $t_nrightarrow x$ but $f(t_n)$ does not converge.



      I draw a blank once I get here, this could very well be false.






      real-analysis monotone-functions discontinuous-functions




      real-analysis monotone-functions discontinuous-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 19 at 21:17









      Andrew ShedlockAndrew Shedlock

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      1688






















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          $begingroup$

          You are on the right track. Let $m=liminf_{ntoinfty}f(t_n)$ and $M=limsup_{ntoinfty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find ${a_n}$ and ${b_b}$ subsequences of ${t_n}$ and $delta>0$ such that $lim_{ntoinfty}a_n=lim_{ntoinfty}b_n=x$ and
          $$
          f(a_n)-f(b_m)>deltaquadforall m,n.
          $$
          Then $f$ is not monotonic on $(x,z)$ for some $z>x$.






          share|cite|improve this answer









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            $begingroup$

            You are on the right track. Let $m=liminf_{ntoinfty}f(t_n)$ and $M=limsup_{ntoinfty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find ${a_n}$ and ${b_b}$ subsequences of ${t_n}$ and $delta>0$ such that $lim_{ntoinfty}a_n=lim_{ntoinfty}b_n=x$ and
            $$
            f(a_n)-f(b_m)>deltaquadforall m,n.
            $$
            Then $f$ is not monotonic on $(x,z)$ for some $z>x$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You are on the right track. Let $m=liminf_{ntoinfty}f(t_n)$ and $M=limsup_{ntoinfty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find ${a_n}$ and ${b_b}$ subsequences of ${t_n}$ and $delta>0$ such that $lim_{ntoinfty}a_n=lim_{ntoinfty}b_n=x$ and
              $$
              f(a_n)-f(b_m)>deltaquadforall m,n.
              $$
              Then $f$ is not monotonic on $(x,z)$ for some $z>x$.






              share|cite|improve this answer









              $endgroup$
















                1












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                1





                $begingroup$

                You are on the right track. Let $m=liminf_{ntoinfty}f(t_n)$ and $M=limsup_{ntoinfty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find ${a_n}$ and ${b_b}$ subsequences of ${t_n}$ and $delta>0$ such that $lim_{ntoinfty}a_n=lim_{ntoinfty}b_n=x$ and
                $$
                f(a_n)-f(b_m)>deltaquadforall m,n.
                $$
                Then $f$ is not monotonic on $(x,z)$ for some $z>x$.






                share|cite|improve this answer









                $endgroup$



                You are on the right track. Let $m=liminf_{ntoinfty}f(t_n)$ and $M=limsup_{ntoinfty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find ${a_n}$ and ${b_b}$ subsequences of ${t_n}$ and $delta>0$ such that $lim_{ntoinfty}a_n=lim_{ntoinfty}b_n=x$ and
                $$
                f(a_n)-f(b_m)>deltaquadforall m,n.
                $$
                Then $f$ is not monotonic on $(x,z)$ for some $z>x$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 23:13









                Julián AguirreJulián Aguirre

                69.1k24096




                69.1k24096






























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