Rudin's Proof about Winding Numbers
$begingroup$
This is kind of a softball question, an untied loose end that has always bugged me. It is well-known that if $Gamma_1sim Gamma_2$ are two homotopic closed paths in a region $Omega$, and if $alphanotin Omega$, then $n(Gamma_1;alpha)=n(Gamma_2;alpha).$ I've seen several proofs of this, using approximation by polygonal paths. Rudin's is (surprise!) the slickest, but of course, he leaves some of the details to the reader, and when I do the calculation, I am off by a factor of two at a certain step, which does not affect the proof (one can scale the original hypothesis), but I must be making a mistake, and it has always bugged me. So I'd like to see where my error is.
Let $H:Itimes Ito Omega$ be the homotopy. and choose an integer $n$ such that
$|s-t|+|s'-t'|<1/nRightarrow$
$ |H(s)-H(t)|+|H(s')-H(t')|<epsilon. $
Define the paths ${gamma_0,cdots ,gamma_n}$ by
$gamma_k(s)=H(i/k,k/n)(ns+1-i)+H((i-1)/n,k/n)(i-ns)$
if $i-1le nsle i.$
The claim is then that $|gamma_k(s)-H(s,k/n)|<epsilon.$
Here is what I am getting, after substituting and applying the triangle inequality:
$|H(i/n,k/n)-H((i-1)/n,k/n)|(ns-i)+|H(i/n,k/n)-H(s,k/n)|$
which is easily seen to be $<2epsilon.$ It seems like the only way to avoid the factor of two, would be to arrive at a tractable expression without using the triangle inequality. But I do not see how to do this. Unless at the outset, we should have simply required that
$|s-t|+|s'-t'|<1/nRightarrow$
$|H(s)-H(t)|+|H(s')-H(t')|<epsilon/2. $
complex-analysis analysis analytic-geometry winding-number
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
$endgroup$
add a comment |
$begingroup$
This is kind of a softball question, an untied loose end that has always bugged me. It is well-known that if $Gamma_1sim Gamma_2$ are two homotopic closed paths in a region $Omega$, and if $alphanotin Omega$, then $n(Gamma_1;alpha)=n(Gamma_2;alpha).$ I've seen several proofs of this, using approximation by polygonal paths. Rudin's is (surprise!) the slickest, but of course, he leaves some of the details to the reader, and when I do the calculation, I am off by a factor of two at a certain step, which does not affect the proof (one can scale the original hypothesis), but I must be making a mistake, and it has always bugged me. So I'd like to see where my error is.
Let $H:Itimes Ito Omega$ be the homotopy. and choose an integer $n$ such that
$|s-t|+|s'-t'|<1/nRightarrow$
$ |H(s)-H(t)|+|H(s')-H(t')|<epsilon. $
Define the paths ${gamma_0,cdots ,gamma_n}$ by
$gamma_k(s)=H(i/k,k/n)(ns+1-i)+H((i-1)/n,k/n)(i-ns)$
if $i-1le nsle i.$
The claim is then that $|gamma_k(s)-H(s,k/n)|<epsilon.$
Here is what I am getting, after substituting and applying the triangle inequality:
$|H(i/n,k/n)-H((i-1)/n,k/n)|(ns-i)+|H(i/n,k/n)-H(s,k/n)|$
which is easily seen to be $<2epsilon.$ It seems like the only way to avoid the factor of two, would be to arrive at a tractable expression without using the triangle inequality. But I do not see how to do this. Unless at the outset, we should have simply required that
$|s-t|+|s'-t'|<1/nRightarrow$
$|H(s)-H(t)|+|H(s')-H(t')|<epsilon/2. $
complex-analysis analysis analytic-geometry winding-number
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
$endgroup$
add a comment |
$begingroup$
This is kind of a softball question, an untied loose end that has always bugged me. It is well-known that if $Gamma_1sim Gamma_2$ are two homotopic closed paths in a region $Omega$, and if $alphanotin Omega$, then $n(Gamma_1;alpha)=n(Gamma_2;alpha).$ I've seen several proofs of this, using approximation by polygonal paths. Rudin's is (surprise!) the slickest, but of course, he leaves some of the details to the reader, and when I do the calculation, I am off by a factor of two at a certain step, which does not affect the proof (one can scale the original hypothesis), but I must be making a mistake, and it has always bugged me. So I'd like to see where my error is.
Let $H:Itimes Ito Omega$ be the homotopy. and choose an integer $n$ such that
$|s-t|+|s'-t'|<1/nRightarrow$
$ |H(s)-H(t)|+|H(s')-H(t')|<epsilon. $
Define the paths ${gamma_0,cdots ,gamma_n}$ by
$gamma_k(s)=H(i/k,k/n)(ns+1-i)+H((i-1)/n,k/n)(i-ns)$
if $i-1le nsle i.$
The claim is then that $|gamma_k(s)-H(s,k/n)|<epsilon.$
Here is what I am getting, after substituting and applying the triangle inequality:
$|H(i/n,k/n)-H((i-1)/n,k/n)|(ns-i)+|H(i/n,k/n)-H(s,k/n)|$
which is easily seen to be $<2epsilon.$ It seems like the only way to avoid the factor of two, would be to arrive at a tractable expression without using the triangle inequality. But I do not see how to do this. Unless at the outset, we should have simply required that
$|s-t|+|s'-t'|<1/nRightarrow$
$|H(s)-H(t)|+|H(s')-H(t')|<epsilon/2. $
complex-analysis analysis analytic-geometry winding-number
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
$endgroup$
This is kind of a softball question, an untied loose end that has always bugged me. It is well-known that if $Gamma_1sim Gamma_2$ are two homotopic closed paths in a region $Omega$, and if $alphanotin Omega$, then $n(Gamma_1;alpha)=n(Gamma_2;alpha).$ I've seen several proofs of this, using approximation by polygonal paths. Rudin's is (surprise!) the slickest, but of course, he leaves some of the details to the reader, and when I do the calculation, I am off by a factor of two at a certain step, which does not affect the proof (one can scale the original hypothesis), but I must be making a mistake, and it has always bugged me. So I'd like to see where my error is.
Let $H:Itimes Ito Omega$ be the homotopy. and choose an integer $n$ such that
$|s-t|+|s'-t'|<1/nRightarrow$
$ |H(s)-H(t)|+|H(s')-H(t')|<epsilon. $
Define the paths ${gamma_0,cdots ,gamma_n}$ by
$gamma_k(s)=H(i/k,k/n)(ns+1-i)+H((i-1)/n,k/n)(i-ns)$
if $i-1le nsle i.$
The claim is then that $|gamma_k(s)-H(s,k/n)|<epsilon.$
Here is what I am getting, after substituting and applying the triangle inequality:
$|H(i/n,k/n)-H((i-1)/n,k/n)|(ns-i)+|H(i/n,k/n)-H(s,k/n)|$
which is easily seen to be $<2epsilon.$ It seems like the only way to avoid the factor of two, would be to arrive at a tractable expression without using the triangle inequality. But I do not see how to do this. Unless at the outset, we should have simply required that
$|s-t|+|s'-t'|<1/nRightarrow$
$|H(s)-H(t)|+|H(s')-H(t')|<epsilon/2. $
complex-analysis analysis analytic-geometry winding-number
complex-analysis analysis analytic-geometry winding-number
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
edited Dec 2 '18 at 5:11
Matematleta
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.2k2918
10.2k2918
add a comment |
add a comment |
0
active
oldest
votes
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021562%2frudins-proof-about-winding-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021562%2frudins-proof-about-winding-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
