Fdtxjr

Multi tool use

What problems would a superhuman have who's skin is constantly hot?

Can the alpha, lambda values of a glmnet object output determine whether ridge or Lasso?

Can I use a violin G string for D?

For which categories of spectra is there an explicit description of the fibrant objects via lifting properties?

Why aren't there more Gauls like Obelix?

What is the generally accepted pronunciation of “topoi”?

MySQL importing CSV files really slow

How can I manipulate the output of Information?

Recommendation letter by significant other if you worked with them professionally?

How do we create new idioms and use them in a novel?

Has a sovereign Communist government ever run, and conceded loss, on a fair election?

What is Tony Stark injecting into himself in Iron Man 3?

What is better: yes / no radio, or simple checkbox?

Doesn't allowing a user mode program to access kernel space memory and execute the IN and OUT instructions defeat the purpose of having CPU modes?

Is it possible that a question has only two answers?

Trig Subsitution When There's No Square Root

Professor forcing me to attend a conference, I can't afford even with 50% funding

Does a difference of tense count as a difference of meaning in a minimal pair?

What do you call someone who likes to pick fights?

Street obstacles in New Zealand

What's the 'present simple' form of the word "нашла́" in 3rd person singular female?

Help find my computational error for logarithms

Why do we say ‘pairwise disjoint’, rather than ‘disjoint’?

Why is there an extra space when I type "ls" in the Desktop directory?

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

0

$begingroup$

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

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, 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

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

add a comment | 

0

$begingroup$

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

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, 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

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

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

  
  
  
  1. $X subseteq R_1 cup R_2 cup dots$
  
  
  2. 
     $|R_1|+|R_2| + dots < varepsilon$, 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

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

$endgroup$

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

share|cite|improve this question

asked yesterday

ZacharyZachary

1799

asked yesterday

ZacharyZachary

1799

asked yesterday

ZacharyZachary

1799

1 Answer
1

active

oldest

votes

0

$begingroup$

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 $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3141020%2fa-piecewise-smooth-curve-in-mathbbrn-is-a-null-set-in-mathbbrn%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

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 $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

0

$begingroup$

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 $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

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 $mathbb R^n$.

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

$endgroup$

share|cite|improve this answer

answered yesterday

Christian BlatterChristian Blatter

175k8115327

answered yesterday

Christian BlatterChristian Blatter

175k8115327

answered yesterday

Christian BlatterChristian Blatter

175k8115327