Measure of $C^1$ path in $mathbb{R}^2$
$begingroup$
I started studying multivariable integration and still trying to grasp the conecpt of the measure.
I`m doing excersices and I keep getting the feeling im doing something wrong so I hope one of you could help me :
Definition: $E subset mathbb{R}^n$ is neglectible if for every $epsilon$ >0 there exists a series of open cubes $Q_1,Q_2,...$ such that $Esubset cup Q_i$ and $sum_1^infty V(Q_i)<epsilon$
Now, I`m trying to show that if $phi:[0,1]to mathbb{R}^2$ a continously differentiable path, then $phi([0,1])$ is neglectible.
My idea was to say that $phi$ is continously differentiable on compact set and hence lipchitz, so we have : $d(phi(x),phi(y))leq Ld(x,y)$.
Now, [0,1] is compact so we can cover it by finitely many open sets of diameter $sqrt{ epsilon/L}$,
Excplicitly: $[0,1]subset cup_1^NI_i$.
Now we can define a cover for $phi([0,1])$ as folllowing :
For each $i$, take $Q_i$ coverage of $phi(I_i)$. Its sides are bounded from the inequality : $d(phi(x),phi(y))leq Ld(x,y)<sqrt{ epsilon/L}$, so overall $V(Q_i)<epsilon$
To finish off the proof - and this is where my problem is :
$phi([0,1])subset cup_1^Nphi(I_i)$
$implies$
$V(phi([0,1]))leq sum_1^NV(Q_i)=Nepsilon$
However- N is the number of open sets required to cover $[0,1]$, so it obviously depends on $epsilon$ and most likely to get large as $epsilon$ gets smaller.
I can't understand what I'm missing here. I'll be glad if someone can help me.
real-analysis multivariable-calculus proof-verification
asked Dec 10 '18 at 9:01
SarSar
513113
$endgroup$
add a comment |
$begingroup$
I started studying multivariable integration and still trying to grasp the conecpt of the measure.
I`m doing excersices and I keep getting the feeling im doing something wrong so I hope one of you could help me :
Definition: $E subset mathbb{R}^n$ is neglectible if for every $epsilon$ >0 there exists a series of open cubes $Q_1,Q_2,...$ such that $Esubset cup Q_i$ and $sum_1^infty V(Q_i)<epsilon$
Now, I`m trying to show that if $phi:[0,1]to mathbb{R}^2$ a continously differentiable path, then $phi([0,1])$ is neglectible.
My idea was to say that $phi$ is continously differentiable on compact set and hence lipchitz, so we have : $d(phi(x),phi(y))leq Ld(x,y)$.
Now, [0,1] is compact so we can cover it by finitely many open sets of diameter $sqrt{ epsilon/L}$,
Excplicitly: $[0,1]subset cup_1^NI_i$.
Now we can define a cover for $phi([0,1])$ as folllowing :
For each $i$, take $Q_i$ coverage of $phi(I_i)$. Its sides are bounded from the inequality : $d(phi(x),phi(y))leq Ld(x,y)<sqrt{ epsilon/L}$, so overall $V(Q_i)<epsilon$
To finish off the proof - and this is where my problem is :
$phi([0,1])subset cup_1^Nphi(I_i)$
$implies$
$V(phi([0,1]))leq sum_1^NV(Q_i)=Nepsilon$
However- N is the number of open sets required to cover $[0,1]$, so it obviously depends on $epsilon$ and most likely to get large as $epsilon$ gets smaller.
I can't understand what I'm missing here. I'll be glad if someone can help me.
real-analysis multivariable-calculus proof-verification
asked Dec 10 '18 at 9:01
SarSar
513113
$endgroup$
$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
1
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
1
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41
add a comment |
$begingroup$
I started studying multivariable integration and still trying to grasp the conecpt of the measure.
I`m doing excersices and I keep getting the feeling im doing something wrong so I hope one of you could help me :
Definition: $E subset mathbb{R}^n$ is neglectible if for every $epsilon$ >0 there exists a series of open cubes $Q_1,Q_2,...$ such that $Esubset cup Q_i$ and $sum_1^infty V(Q_i)<epsilon$
Now, I`m trying to show that if $phi:[0,1]to mathbb{R}^2$ a continously differentiable path, then $phi([0,1])$ is neglectible.
My idea was to say that $phi$ is continously differentiable on compact set and hence lipchitz, so we have : $d(phi(x),phi(y))leq Ld(x,y)$.
Now, [0,1] is compact so we can cover it by finitely many open sets of diameter $sqrt{ epsilon/L}$,
Excplicitly: $[0,1]subset cup_1^NI_i$.
Now we can define a cover for $phi([0,1])$ as folllowing :
For each $i$, take $Q_i$ coverage of $phi(I_i)$. Its sides are bounded from the inequality : $d(phi(x),phi(y))leq Ld(x,y)<sqrt{ epsilon/L}$, so overall $V(Q_i)<epsilon$
To finish off the proof - and this is where my problem is :
$phi([0,1])subset cup_1^Nphi(I_i)$
$implies$
$V(phi([0,1]))leq sum_1^NV(Q_i)=Nepsilon$
However- N is the number of open sets required to cover $[0,1]$, so it obviously depends on $epsilon$ and most likely to get large as $epsilon$ gets smaller.
I can't understand what I'm missing here. I'll be glad if someone can help me.
real-analysis multivariable-calculus proof-verification
asked Dec 10 '18 at 9:01
SarSar
513113
$endgroup$
I started studying multivariable integration and still trying to grasp the conecpt of the measure.
I`m doing excersices and I keep getting the feeling im doing something wrong so I hope one of you could help me :
Definition: $E subset mathbb{R}^n$ is neglectible if for every $epsilon$ >0 there exists a series of open cubes $Q_1,Q_2,...$ such that $Esubset cup Q_i$ and $sum_1^infty V(Q_i)<epsilon$
Now, I`m trying to show that if $phi:[0,1]to mathbb{R}^2$ a continously differentiable path, then $phi([0,1])$ is neglectible.
My idea was to say that $phi$ is continously differentiable on compact set and hence lipchitz, so we have : $d(phi(x),phi(y))leq Ld(x,y)$.
Now, [0,1] is compact so we can cover it by finitely many open sets of diameter $sqrt{ epsilon/L}$,
Excplicitly: $[0,1]subset cup_1^NI_i$.
Now we can define a cover for $phi([0,1])$ as folllowing :
For each $i$, take $Q_i$ coverage of $phi(I_i)$. Its sides are bounded from the inequality : $d(phi(x),phi(y))leq Ld(x,y)<sqrt{ epsilon/L}$, so overall $V(Q_i)<epsilon$
To finish off the proof - and this is where my problem is :
$phi([0,1])subset cup_1^Nphi(I_i)$
$implies$
$V(phi([0,1]))leq sum_1^NV(Q_i)=Nepsilon$
However- N is the number of open sets required to cover $[0,1]$, so it obviously depends on $epsilon$ and most likely to get large as $epsilon$ gets smaller.
I can't understand what I'm missing here. I'll be glad if someone can help me.
real-analysis multivariable-calculus proof-verification
real-analysis multivariable-calculus proof-verification
asked Dec 10 '18 at 9:01
SarSar
513113
asked Dec 10 '18 at 9:01
SarSar
513113
asked Dec 10 '18 at 9:01
SarSar
513113
asked Dec 10 '18 at 9:01
SarSar
513113
asked Dec 10 '18 at 9:01
SarSar
513113
513113
$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
1
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
1
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41
add a comment |
$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
1
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
1
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41
$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
1
1
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
1
1
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41
add a comment |
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$begingroup$
Can you write $N$ in terms of $epsilon$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:44
$begingroup$
Youre right, its about $1/ epsilon$ , Which still isn`t good
$endgroup$
– Sar
Dec 10 '18 at 9:48
1
$begingroup$
Isn't it $frac{sqrt{L}}{sqrt{epsilon}}$?
$endgroup$
– Uskebasi
Dec 10 '18 at 9:50
1
$begingroup$
Aha! True. Thanks alot .
$endgroup$
– Sar
Dec 10 '18 at 9:52
$begingroup$
@Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments.
$endgroup$
– Brahadeesh
Dec 10 '18 at 18:41