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I'm looking at the proof of Cauchy's theorem $oint_Cf(z)dz = 0$ in Titchmarsh's Theory of Functions (second edition p.75). This assumes $f$ is holomorphic and $C$ is a piece-wise smooth rectifiable closed simple curve.
He puts a grid over the curve and its interior to divide it into "a large number of small parts" and then separately shows that the integral round each of the interior squares is zero and that the sum of the integrals round the squares intersecting the curve is zero.
The proof seems to assume that for squares smaller than some size the curve will always intersect each square with a single connected arc in order to justify forming a closed contour with with the sides of the square interior to the curve. Is this assumption true and if so how is it proved ? If true does it still hold without the smoothness assumption ?
I have tried several approaches to proving this without any success. Considering $C$ as a rectifiable Jordan curve didn't get me anywhere, nor did knowing that disjoint compact sets have a minimum separation.
general-topology complex-analysis
asked 2 days ago
Tom Collinge
4,331932
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I'm looking at the proof of Cauchy's theorem $oint_Cf(z)dz = 0$ in Titchmarsh's Theory of Functions (second edition p.75). This assumes $f$ is holomorphic and $C$ is a piece-wise smooth rectifiable closed simple curve.
He puts a grid over the curve and its interior to divide it into "a large number of small parts" and then separately shows that the integral round each of the interior squares is zero and that the sum of the integrals round the squares intersecting the curve is zero.
The proof seems to assume that for squares smaller than some size the curve will always intersect each square with a single connected arc in order to justify forming a closed contour with with the sides of the square interior to the curve. Is this assumption true and if so how is it proved ? If true does it still hold without the smoothness assumption ?
I have tried several approaches to proving this without any success. Considering $C$ as a rectifiable Jordan curve didn't get me anywhere, nor did knowing that disjoint compact sets have a minimum separation.
general-topology complex-analysis
asked 2 days ago
Tom Collinge
4,331932
It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm looking at the proof of Cauchy's theorem $oint_Cf(z)dz = 0$ in Titchmarsh's Theory of Functions (second edition p.75). This assumes $f$ is holomorphic and $C$ is a piece-wise smooth rectifiable closed simple curve.
He puts a grid over the curve and its interior to divide it into "a large number of small parts" and then separately shows that the integral round each of the interior squares is zero and that the sum of the integrals round the squares intersecting the curve is zero.
The proof seems to assume that for squares smaller than some size the curve will always intersect each square with a single connected arc in order to justify forming a closed contour with with the sides of the square interior to the curve. Is this assumption true and if so how is it proved ? If true does it still hold without the smoothness assumption ?
I have tried several approaches to proving this without any success. Considering $C$ as a rectifiable Jordan curve didn't get me anywhere, nor did knowing that disjoint compact sets have a minimum separation.
general-topology complex-analysis
asked 2 days ago
Tom Collinge
4,331932
I'm looking at the proof of Cauchy's theorem $oint_Cf(z)dz = 0$ in Titchmarsh's Theory of Functions (second edition p.75). This assumes $f$ is holomorphic and $C$ is a piece-wise smooth rectifiable closed simple curve.
He puts a grid over the curve and its interior to divide it into "a large number of small parts" and then separately shows that the integral round each of the interior squares is zero and that the sum of the integrals round the squares intersecting the curve is zero.
The proof seems to assume that for squares smaller than some size the curve will always intersect each square with a single connected arc in order to justify forming a closed contour with with the sides of the square interior to the curve. Is this assumption true and if so how is it proved ? If true does it still hold without the smoothness assumption ?
I have tried several approaches to proving this without any success. Considering $C$ as a rectifiable Jordan curve didn't get me anywhere, nor did knowing that disjoint compact sets have a minimum separation.
general-topology complex-analysis
asked 2 days ago
Tom Collinge
4,331932
asked 2 days ago
Tom Collinge
4,331932
asked 2 days ago
Tom Collinge
4,331932
asked 2 days ago
Tom Collinge
4,331932
4,331932
It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago
add a comment |Â
It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago
It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago
add a comment |Â
1 Answer
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The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve. Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:
Suppose $f:D subset mathbbC to mathbbC$ is continuous and $C
= phi(t): a leqslant t leqslant b$ is any rectifiable curve in $D$ defined by a continuous function $phi:[a,b] to mathbbC$.
Given $epsilon > 0$, there exists $delta = delta(epsilon)> 0$ such
that for any partition $P= a< t_0 < t_1 < ldots < t_n = b$ with
$|P| < delta$, the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $ is contained in $D$ and
$$left|int_C f(z) , dz - int_Gamma f(z) , dz right| <
epsilon.$$
Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E subset D$ and $rho > 0$ such that the closed disks $barD(z;rho) subset E$ for all $z in C$.
(2) Using uniform continuity of $f$ on $E$ find $eta > 0$ such that $|f(z) - f(z')| < epsilon/(2L)$ when $|z - z'| < eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $phi$ on $[a,b]$, find $delta_2 > 0$ such that $|phi(t) - phi(t')| < min(rho,eta)$ when $|t-t'| < delta_2$.
(4) There exists $delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < ldots < t_n = b)$ with $|P| < delta_1$, we have (since $f$ is integrable),
$$tag*left| int_C f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
(5) Finally, by requiring further that $|P| leqslant min(delta_1,delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $,
$$tag**left| int_Gamma f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
In view of (*) and (**), the lemma is proved.
answered yesterday
RRL
43.4k42160
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve. Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:
Suppose $f:D subset mathbbC to mathbbC$ is continuous and $C
= phi(t): a leqslant t leqslant b$ is any rectifiable curve in $D$ defined by a continuous function $phi:[a,b] to mathbbC$.
Given $epsilon > 0$, there exists $delta = delta(epsilon)> 0$ such
that for any partition $P= a< t_0 < t_1 < ldots < t_n = b$ with
$|P| < delta$, the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $ is contained in $D$ and
$$left|int_C f(z) , dz - int_Gamma f(z) , dz right| <
epsilon.$$
Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E subset D$ and $rho > 0$ such that the closed disks $barD(z;rho) subset E$ for all $z in C$.
(2) Using uniform continuity of $f$ on $E$ find $eta > 0$ such that $|f(z) - f(z')| < epsilon/(2L)$ when $|z - z'| < eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $phi$ on $[a,b]$, find $delta_2 > 0$ such that $|phi(t) - phi(t')| < min(rho,eta)$ when $|t-t'| < delta_2$.
(4) There exists $delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < ldots < t_n = b)$ with $|P| < delta_1$, we have (since $f$ is integrable),
$$tag*left| int_C f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
(5) Finally, by requiring further that $|P| leqslant min(delta_1,delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $,
$$tag**left| int_Gamma f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
In view of (*) and (**), the lemma is proved.
answered yesterday
RRL
43.4k42160
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
add a comment |Â
up vote
1
down vote
The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve. Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:
Suppose $f:D subset mathbbC to mathbbC$ is continuous and $C
= phi(t): a leqslant t leqslant b$ is any rectifiable curve in $D$ defined by a continuous function $phi:[a,b] to mathbbC$.
Given $epsilon > 0$, there exists $delta = delta(epsilon)> 0$ such
that for any partition $P= a< t_0 < t_1 < ldots < t_n = b$ with
$|P| < delta$, the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $ is contained in $D$ and
$$left|int_C f(z) , dz - int_Gamma f(z) , dz right| <
epsilon.$$
Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E subset D$ and $rho > 0$ such that the closed disks $barD(z;rho) subset E$ for all $z in C$.
(2) Using uniform continuity of $f$ on $E$ find $eta > 0$ such that $|f(z) - f(z')| < epsilon/(2L)$ when $|z - z'| < eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $phi$ on $[a,b]$, find $delta_2 > 0$ such that $|phi(t) - phi(t')| < min(rho,eta)$ when $|t-t'| < delta_2$.
(4) There exists $delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < ldots < t_n = b)$ with $|P| < delta_1$, we have (since $f$ is integrable),
$$tag*left| int_C f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
(5) Finally, by requiring further that $|P| leqslant min(delta_1,delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $,
$$tag**left| int_Gamma f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
In view of (*) and (**), the lemma is proved.
answered yesterday
RRL
43.4k42160
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve. Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:
Suppose $f:D subset mathbbC to mathbbC$ is continuous and $C
= phi(t): a leqslant t leqslant b$ is any rectifiable curve in $D$ defined by a continuous function $phi:[a,b] to mathbbC$.
Given $epsilon > 0$, there exists $delta = delta(epsilon)> 0$ such
that for any partition $P= a< t_0 < t_1 < ldots < t_n = b$ with
$|P| < delta$, the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $ is contained in $D$ and
$$left|int_C f(z) , dz - int_Gamma f(z) , dz right| <
epsilon.$$
Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E subset D$ and $rho > 0$ such that the closed disks $barD(z;rho) subset E$ for all $z in C$.
(2) Using uniform continuity of $f$ on $E$ find $eta > 0$ such that $|f(z) - f(z')| < epsilon/(2L)$ when $|z - z'| < eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $phi$ on $[a,b]$, find $delta_2 > 0$ such that $|phi(t) - phi(t')| < min(rho,eta)$ when $|t-t'| < delta_2$.
(4) There exists $delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < ldots < t_n = b)$ with $|P| < delta_1$, we have (since $f$ is integrable),
$$tag*left| int_C f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
(5) Finally, by requiring further that $|P| leqslant min(delta_1,delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $,
$$tag**left| int_Gamma f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
In view of (*) and (**), the lemma is proved.
answered yesterday
RRL
43.4k42160
The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve. Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:
Suppose $f:D subset mathbbC to mathbbC$ is continuous and $C
= phi(t): a leqslant t leqslant b$ is any rectifiable curve in $D$ defined by a continuous function $phi:[a,b] to mathbbC$.
Given $epsilon > 0$, there exists $delta = delta(epsilon)> 0$ such
that for any partition $P= a< t_0 < t_1 < ldots < t_n = b$ with
$|P| < delta$, the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $ is contained in $D$ and
$$left|int_C f(z) , dz - int_Gamma f(z) , dz right| <
epsilon.$$
Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E subset D$ and $rho > 0$ such that the closed disks $barD(z;rho) subset E$ for all $z in C$.
(2) Using uniform continuity of $f$ on $E$ find $eta > 0$ such that $|f(z) - f(z')| < epsilon/(2L)$ when $|z - z'| < eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $phi$ on $[a,b]$, find $delta_2 > 0$ such that $|phi(t) - phi(t')| < min(rho,eta)$ when $|t-t'| < delta_2$.
(4) There exists $delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < ldots < t_n = b)$ with $|P| < delta_1$, we have (since $f$ is integrable),
$$tag*left| int_C f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
(5) Finally, by requiring further that $|P| leqslant min(delta_1,delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $Gamma
=overlinephi(t_0)phi t_1) + ldots + overlinephi(t_n-1)phi(t_n) $,
$$tag**left| int_Gamma f(z) , dz - sum_k=1^n f(phi(t_k))(phi(t_k) - phi(t_k-1))right| < fracepsilon2$$
In view of (*) and (**), the lemma is proved.
answered yesterday
RRL
43.4k42160
edited yesterday
answered yesterday
RRL
43.4k42160
answered yesterday
RRL
43.4k42160
answered yesterday
RRL
43.4k42160
43.4k42160
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
add a comment |Â
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
Many thanks, I will work through this later in detail. One point in the earlier steps I don't understand is why it is necessary to go through convex polygons. I have the proof that any polygon can be triangulated (I think - math.stackexchange.com/q/2871963) so why not go straight from triangles to general polygons ?
– Tom Collinge
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
@TomCollinge: You are correct in that the theorem is extended directly to any region that can be triangulated. There are probably several ways to go straight to polygons and it seems you found one. I became interested in understanding how to prove for the most general curve a while back -- having been frustrated by the lack of rigor (perhaps like you) in most of the standard books. So my focus here was on that last step.
– RRL
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
Thanks again. I follow your proof for this last step, but now find extreme difficulty moving from simple to arbitrary polygons. An arbitrary closed polygonal curve must have a finite number of crossing points and so Intuitively should resolve to the union of a finite number of simple polygons, but that I can't formally prove and haven't managed to google a reference..
– Tom Collinge
yesterday
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It's hand-waving. For a rigorous proof you'll have to consult more modern texts.
– Lord Shark the Unknown
2 days ago
@LordSharktheUnknown, Thanks. Can you recommend a decent proof of Cauchy's theorem for rectifiable Jordan curves (if necessary piece-wise smooth) that doesn't involve homotopy (which I haven't studied) ?
– Tom Collinge
2 days ago