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A piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$
Prove that the image of a curve has zero contentNull velocity and piecewise smooth pathDefinitions of multiple Riemann integrals and boundednessPartitions of unity from PMA RudinUsing Green's Theorem to Express the Integral $I=int_C (Pdx+Qdy)$ as an expression of $I_i=int _C_i (Pdx+Qdy)$Question about Manifolds (without boundaries)$partial U$ is a smooth curve (i.e., the image of a smooth path $mathbfc$ with $mathbfc' not= 0$)?Smooth map with null differential at each point are constant on the connected component of the domainAbout change of coordinates of vector fields in smooth manifold theorySmooth curve segments and smooth charts
$begingroup$
Definitions:
A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.
A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that
- $X subseteq R_1 cup R_2 cup dots$
$|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?
Thanks in advance.
multivariable-calculus
$endgroup$
add a comment |
$begingroup$
Definitions:
A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.
A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that
- $X subseteq R_1 cup R_2 cup dots$
$|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?
Thanks in advance.
multivariable-calculus
$endgroup$
add a comment |
$begingroup$
Definitions:
A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.
A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that
- $X subseteq R_1 cup R_2 cup dots$
$|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?
Thanks in advance.
multivariable-calculus
$endgroup$
Definitions:
A piecewise smooth curve in $mathbbR^2$ is a subset of $mathbbR^2$ that can be written as $boldsymbolvarphi_1[a_1,b_1] cup boldsymbolvarphi_2[a_2,b_2]cup dotscup boldsymbolvarphi_N [a_N,b_N]$, with each $boldsymbolvarphi_i:[a_i,b_i] to mathbbR^2$ a parameterization of a smooth curve, and where each end point $boldsymbolvarphi_i(b_i)$ corresponds with the next starting point $boldsymbolvarphi_i+1(a_i+1)$.
A subset $X subseteq mathbbR^2$ is a null set in $mathbbR^2$ if for each $varepsilon > 0$ there is a sequence $R_1,R_2,dots$ of rectangles with sides parallel to the coordinate axes such that
- $X subseteq R_1 cup R_2 cup dots$
$|R_1|+|R_2| + dots < varepsilon$, with $|R_i|$ the area of rectangle $R_i$.
My notes proceed with stating that a piecewise smooth curve in $mathbbR^n$ is a null set in $mathbbR^n$. I would like to understand to intuiton behind this theorem. Is there an easily understandable explanation behind it?
Thanks in advance.
multivariable-calculus
multivariable-calculus
asked yesterday
ZacharyZachary
1799
1799
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1 Answer
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$begingroup$
The following is not a proof; it just helps the intuition.
It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.
A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.
One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.
$endgroup$
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$begingroup$
The following is not a proof; it just helps the intuition.
It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.
A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.
One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.
$endgroup$
add a comment |
$begingroup$
The following is not a proof; it just helps the intuition.
It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.
A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.
One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.
$endgroup$
add a comment |
$begingroup$
The following is not a proof; it just helps the intuition.
It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.
A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.
One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.
$endgroup$
The following is not a proof; it just helps the intuition.
It is sufficient to consider a smooth curve, since a finite union of null sets is again a null set.
A smooth curve $phi:>[a,b]tomathbb R^2$ has finite length $L>0$. Given an $Ngg1$ you should be able to cover it with $N$ squares of side length $4Lover N$, say. These squares have a total area $$Ncdotleft(4Lover Nright)^2=16L^2over N .$$
Given an $epsilon>0$ we can make the right hand side $<epsilon$ by choosing $N$ large enough.
One therefore can say that the basic reason for the intended theorem is the fact that the dimension of $mathbb R^2$ is $1$ greater than the dimension of $mathbb R$, and similarly when $mathbb R^2$ is replaced by $mathbb R^n$.
answered yesterday
Christian BlatterChristian Blatter
175k8115327
175k8115327
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