The oscillation of a bounded function at a point
$begingroup$
Enumerate the rationals in $[0,1]$ (ie. $mathbb{Q}cap[0,1]$) by $q_n$. Define $f:[0,1]tomathbb{R}$ by
$$
f(x)=
begin{cases}
1/n & text{if } x=q_n text{ for some }n\
0 & text{otherwise}
end{cases}
$$
Also, define for $cin[0,1]$ and $r>0$
$$text{Osc}(f,c,r)=sup{ |f(x)-f(y)|:x,yin[0,1]cap(c-r,c+r)}$$
Since $text{Osc}(f,c,r)$ is non-increasing as $rto0^+$, we can define oscillation at $c$ as $text{Osc}(f,c):= lim_{rto0^+}text{Osc}(f,c,r)$.
The question is then what is the value of $text{Osc}(f,x)$ when $x=q_n$ or when $x$ is irrational?
real-analysis limits continuity rational-numbers discontinuous-functions
asked Dec 6 '18 at 3:59
JohnJohn
1067
$endgroup$
add a comment |
$begingroup$
Enumerate the rationals in $[0,1]$ (ie. $mathbb{Q}cap[0,1]$) by $q_n$. Define $f:[0,1]tomathbb{R}$ by
$$
f(x)=
begin{cases}
1/n & text{if } x=q_n text{ for some }n\
0 & text{otherwise}
end{cases}
$$
Also, define for $cin[0,1]$ and $r>0$
$$text{Osc}(f,c,r)=sup{ |f(x)-f(y)|:x,yin[0,1]cap(c-r,c+r)}$$
Since $text{Osc}(f,c,r)$ is non-increasing as $rto0^+$, we can define oscillation at $c$ as $text{Osc}(f,c):= lim_{rto0^+}text{Osc}(f,c,r)$.
The question is then what is the value of $text{Osc}(f,x)$ when $x=q_n$ or when $x$ is irrational?
real-analysis limits continuity rational-numbers discontinuous-functions
asked Dec 6 '18 at 3:59
JohnJohn
1067
$endgroup$
add a comment |
$begingroup$
Enumerate the rationals in $[0,1]$ (ie. $mathbb{Q}cap[0,1]$) by $q_n$. Define $f:[0,1]tomathbb{R}$ by
$$
f(x)=
begin{cases}
1/n & text{if } x=q_n text{ for some }n\
0 & text{otherwise}
end{cases}
$$
Also, define for $cin[0,1]$ and $r>0$
$$text{Osc}(f,c,r)=sup{ |f(x)-f(y)|:x,yin[0,1]cap(c-r,c+r)}$$
Since $text{Osc}(f,c,r)$ is non-increasing as $rto0^+$, we can define oscillation at $c$ as $text{Osc}(f,c):= lim_{rto0^+}text{Osc}(f,c,r)$.
The question is then what is the value of $text{Osc}(f,x)$ when $x=q_n$ or when $x$ is irrational?
real-analysis limits continuity rational-numbers discontinuous-functions
asked Dec 6 '18 at 3:59
JohnJohn
1067
$endgroup$
Enumerate the rationals in $[0,1]$ (ie. $mathbb{Q}cap[0,1]$) by $q_n$. Define $f:[0,1]tomathbb{R}$ by
$$
f(x)=
begin{cases}
1/n & text{if } x=q_n text{ for some }n\
0 & text{otherwise}
end{cases}
$$
Also, define for $cin[0,1]$ and $r>0$
$$text{Osc}(f,c,r)=sup{ |f(x)-f(y)|:x,yin[0,1]cap(c-r,c+r)}$$
Since $text{Osc}(f,c,r)$ is non-increasing as $rto0^+$, we can define oscillation at $c$ as $text{Osc}(f,c):= lim_{rto0^+}text{Osc}(f,c,r)$.
The question is then what is the value of $text{Osc}(f,x)$ when $x=q_n$ or when $x$ is irrational?
real-analysis limits continuity rational-numbers discontinuous-functions
real-analysis limits continuity rational-numbers discontinuous-functions
asked Dec 6 '18 at 3:59
JohnJohn
1067
asked Dec 6 '18 at 3:59
JohnJohn
1067
asked Dec 6 '18 at 3:59
JohnJohn
1067
asked Dec 6 '18 at 3:59
JohnJohn
1067
asked Dec 6 '18 at 3:59
JohnJohn
1067
1067
add a comment |
add a comment |
1 Answer
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$begingroup$
When $x=q_n$ the oscillation is $frac 1 n$ and when $x$ is irrational it is $0$. It is a well known result that the oscillation of $f$ at $x$ is $0$ iff $f$ is continuous at $x$. (You can try to prove this). In this case you can verify that $f$ is continuous at each irrational number.
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
When $x=q_n$ the oscillation is $frac 1 n$ and when $x$ is irrational it is $0$. It is a well known result that the oscillation of $f$ at $x$ is $0$ iff $f$ is continuous at $x$. (You can try to prove this). In this case you can verify that $f$ is continuous at each irrational number.
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
When $x=q_n$ the oscillation is $frac 1 n$ and when $x$ is irrational it is $0$. It is a well known result that the oscillation of $f$ at $x$ is $0$ iff $f$ is continuous at $x$. (You can try to prove this). In this case you can verify that $f$ is continuous at each irrational number.
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
When $x=q_n$ the oscillation is $frac 1 n$ and when $x$ is irrational it is $0$. It is a well known result that the oscillation of $f$ at $x$ is $0$ iff $f$ is continuous at $x$. (You can try to prove this). In this case you can verify that $f$ is continuous at each irrational number.
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
When $x=q_n$ the oscillation is $frac 1 n$ and when $x$ is irrational it is $0$. It is a well known result that the oscillation of $f$ at $x$ is $0$ iff $f$ is continuous at $x$. (You can try to prove this). In this case you can verify that $f$ is continuous at each irrational number.
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Dec 6 '18 at 5:23
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
53.8k32055
add a comment |
add a comment |
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