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
10.3k2918
$endgroup$
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
10.3k2918
$endgroup$
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
10.3k2918
$endgroup$
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
10.3k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.3k2918
edited Dec 2 '18 at 5:11
Matematleta
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.3k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.3k2918
asked Dec 1 '18 at 16:56
MatematletaMatematleta
10.3k2918
10.3k2918
add a comment |
add a comment |
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