Countable discontinuities and pointwise convergence

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If $f_n$ converges pointwise to $f$ and each $f_n$ has a countable number of discontinuities, what is a counterexample to $f$ also having a countable number of discontinuities?






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    If $f_n$ converges pointwise to $f$ and each $f_n$ has a countable number of discontinuities, what is a counterexample to $f$ also having a countable number of discontinuities?






    real-analysis analysis







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      If $f_n$ converges pointwise to $f$ and each $f_n$ has a countable number of discontinuities, what is a counterexample to $f$ also having a countable number of discontinuities?






      real-analysis analysis







      share|cite|improve this question















      If $f_n$ converges pointwise to $f$ and each $f_n$ has a countable number of discontinuities, what is a counterexample to $f$ also having a countable number of discontinuities?






      real-analysis analysis




      real-analysis analysis






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      edited Aug 30 at 7:00









      Santosh Linkha

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      asked Sep 21 '17 at 19:51









      Salamander

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          Let $a_i$ be an enumeration of $[0,1] cap mathbbQ$ and let $f_n$ be defined on $[0,1]$ so that
          $$f_n(x) = textbf1_a_i : i<n$$ which converges pointwise to $textbf1_[0,1] cap mathbbQ(x)$ and is discontinuous on all of $[0,1]$.






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          • I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
            – Salamander
            Sep 21 '17 at 20:22







          • 1




            @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
            – MathematicsStudent1122
            Sep 21 '17 at 20:25










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          up vote
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          accepted










          Let $a_i$ be an enumeration of $[0,1] cap mathbbQ$ and let $f_n$ be defined on $[0,1]$ so that
          $$f_n(x) = textbf1_a_i : i<n$$ which converges pointwise to $textbf1_[0,1] cap mathbbQ(x)$ and is discontinuous on all of $[0,1]$.






          share|cite|improve this answer






















          • I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
            – Salamander
            Sep 21 '17 at 20:22







          • 1




            @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
            – MathematicsStudent1122
            Sep 21 '17 at 20:25














          up vote
          4
          down vote



          accepted










          Let $a_i$ be an enumeration of $[0,1] cap mathbbQ$ and let $f_n$ be defined on $[0,1]$ so that
          $$f_n(x) = textbf1_a_i : i<n$$ which converges pointwise to $textbf1_[0,1] cap mathbbQ(x)$ and is discontinuous on all of $[0,1]$.






          share|cite|improve this answer






















          • I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
            – Salamander
            Sep 21 '17 at 20:22







          • 1




            @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
            – MathematicsStudent1122
            Sep 21 '17 at 20:25












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Let $a_i$ be an enumeration of $[0,1] cap mathbbQ$ and let $f_n$ be defined on $[0,1]$ so that
          $$f_n(x) = textbf1_a_i : i<n$$ which converges pointwise to $textbf1_[0,1] cap mathbbQ(x)$ and is discontinuous on all of $[0,1]$.






          share|cite|improve this answer














          Let $a_i$ be an enumeration of $[0,1] cap mathbbQ$ and let $f_n$ be defined on $[0,1]$ so that
          $$f_n(x) = textbf1_a_i : i<n$$ which converges pointwise to $textbf1_[0,1] cap mathbbQ(x)$ and is discontinuous on all of $[0,1]$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 30 at 7:02









          Santosh Linkha

          9,46922751




          9,46922751










          answered Sep 21 '17 at 20:03









          MathematicsStudent1122

          7,66622159




          7,66622159











          • I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
            – Salamander
            Sep 21 '17 at 20:22







          • 1




            @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
            – MathematicsStudent1122
            Sep 21 '17 at 20:25
















          • I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
            – Salamander
            Sep 21 '17 at 20:22







          • 1




            @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
            – MathematicsStudent1122
            Sep 21 '17 at 20:25















          I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
          – Salamander
          Sep 21 '17 at 20:22





          I would think that function has $ mathbbQ $ discontinuities which is countable, why does it instead have a discontinuity at every real in $[0,1]$ ?
          – Salamander
          Sep 21 '17 at 20:22





          1




          1




          @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
          – MathematicsStudent1122
          Sep 21 '17 at 20:25




          @Salamander No, it's discontinuous at every real in $[0,1]$. Since $[0,1] cap mathbbQ$ and $[0,1] setminus mathbbQ$ are dense in $[0,1]$,for any $x_0 in [0,1]$ and any neighbourhood $(x_0 - delta, x_0 + delta)$ we can see that $f$ is jumping between $1$ and $0$, hence no continuity.
          – MathematicsStudent1122
          Sep 21 '17 at 20:25

















           

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