Find all $y in mathbb R$ for which g is continuous in point $y$












1












$begingroup$



$g:mathbb R rightarrow mathbb R$, $g(x)= begin{cases} sin(x) & x in mathbb Q\ 0 & x notin mathbb Q end{cases}$.
Find all $y in mathbb R$ for which function $g$ is continuous at point $y$.




It is my idea but I'm not sure if it's correct so please check:




  • For $y in mathbb Q$:
    $g(y)=sin(y)$
    $lim_{x to y}g(x)=0$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ but $y in mathbb Q$ so it is contradiction

  • For $y notin mathbb Q$:
    $g(y)=0$
    $lim_{x to y}g(x)=sin(y)$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ and $y notin mathbb Q$ so it is truth



Final answer: $y=kpi$ for $k in mathbb Z$










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$endgroup$








  • 3




    $begingroup$
    I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
    $endgroup$
    – ShawSa
    Jan 15 at 21:42


















1












$begingroup$



$g:mathbb R rightarrow mathbb R$, $g(x)= begin{cases} sin(x) & x in mathbb Q\ 0 & x notin mathbb Q end{cases}$.
Find all $y in mathbb R$ for which function $g$ is continuous at point $y$.




It is my idea but I'm not sure if it's correct so please check:




  • For $y in mathbb Q$:
    $g(y)=sin(y)$
    $lim_{x to y}g(x)=0$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ but $y in mathbb Q$ so it is contradiction

  • For $y notin mathbb Q$:
    $g(y)=0$
    $lim_{x to y}g(x)=sin(y)$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ and $y notin mathbb Q$ so it is truth



Final answer: $y=kpi$ for $k in mathbb Z$










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
    $endgroup$
    – ShawSa
    Jan 15 at 21:42
















1












1








1





$begingroup$



$g:mathbb R rightarrow mathbb R$, $g(x)= begin{cases} sin(x) & x in mathbb Q\ 0 & x notin mathbb Q end{cases}$.
Find all $y in mathbb R$ for which function $g$ is continuous at point $y$.




It is my idea but I'm not sure if it's correct so please check:




  • For $y in mathbb Q$:
    $g(y)=sin(y)$
    $lim_{x to y}g(x)=0$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ but $y in mathbb Q$ so it is contradiction

  • For $y notin mathbb Q$:
    $g(y)=0$
    $lim_{x to y}g(x)=sin(y)$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ and $y notin mathbb Q$ so it is truth



Final answer: $y=kpi$ for $k in mathbb Z$










share|cite|improve this question











$endgroup$





$g:mathbb R rightarrow mathbb R$, $g(x)= begin{cases} sin(x) & x in mathbb Q\ 0 & x notin mathbb Q end{cases}$.
Find all $y in mathbb R$ for which function $g$ is continuous at point $y$.




It is my idea but I'm not sure if it's correct so please check:




  • For $y in mathbb Q$:
    $g(y)=sin(y)$
    $lim_{x to y}g(x)=0$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ but $y in mathbb Q$ so it is contradiction

  • For $y notin mathbb Q$:
    $g(y)=0$
    $lim_{x to y}g(x)=sin(y)$

    So I have $sin(y)=0$ so $y=kpi$ for $k in mathbb Z$ and $y notin mathbb Q$ so it is truth



Final answer: $y=kpi$ for $k in mathbb Z$







real-analysis






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share|cite|improve this question













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edited Feb 6 at 14:59







MP3129

















asked Jan 15 at 21:27









MP3129MP3129

910211




910211








  • 3




    $begingroup$
    I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
    $endgroup$
    – ShawSa
    Jan 15 at 21:42
















  • 3




    $begingroup$
    I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
    $endgroup$
    – ShawSa
    Jan 15 at 21:42










3




3




$begingroup$
I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
$endgroup$
– ShawSa
Jan 15 at 21:42






$begingroup$
I agree with your conclusion but I think the arguments are flawed. In particular $lim_{x to y} g(x)$ doesn't exist if $y$ is not a multiple of $pi$. I don't think you should break it up into the cases $y in mathbb{Q}$ and $y notin mathbb{Q}$. Instead try $y=kpi$ for $k in mathbb{Z}$ and $y neq kpi$ for $k in mathbb{Z}$.
$endgroup$
– ShawSa
Jan 15 at 21:42












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