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 $k in mathbb Z$
real-analysis
asked Jan 15 at 21:27
MP3129MP3129
910211
$endgroup$
add a comment |
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 $k in mathbb Z$
real-analysis
asked Jan 15 at 21:27
MP3129MP3129
910211
$endgroup$
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
add a comment |
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 $k in mathbb Z$
real-analysis
asked Jan 15 at 21:27
MP3129MP3129
910211
$endgroup$
$g:mathbb R rightarrow mathbb R$, $g(x)= begin{cases} sin(x) & x in mathbb Q\ 0 & x notin mathbb Q end{cases}$.
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
real-analysis
asked Jan 15 at 21:27
MP3129MP3129
910211
asked Jan 15 at 21:27
MP3129MP3129
910211
edited Feb 6 at 14:59
MP3129
asked Jan 15 at 21:27
MP3129MP3129
910211
asked Jan 15 at 21:27
MP3129MP3129
910211
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
add a comment |
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
add a comment |
0
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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