Proof that interior of a simple closed curve in plane is simply connecte
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Assuming the Jordan Curve Theorem, we can consider the 2 connected components of the complement of the simple closed curve C in the Riemann sphere. I am trying to establish the Jordan-Schoenflies theorem via Caratheodory's mapping theorem. Is there a basic way of establishing the connected components are simply connected so that we can get a conformal mapping from the unit disk to the component and hence use Caratheodory's theorem?
complex-analysis
asked Nov 22 at 21:54
shc
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Assuming the Jordan Curve Theorem, we can consider the 2 connected components of the complement of the simple closed curve C in the Riemann sphere. I am trying to establish the Jordan-Schoenflies theorem via Caratheodory's mapping theorem. Is there a basic way of establishing the connected components are simply connected so that we can get a conformal mapping from the unit disk to the component and hence use Caratheodory's theorem?
complex-analysis
asked Nov 22 at 21:54
shc
62
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assuming the Jordan Curve Theorem, we can consider the 2 connected components of the complement of the simple closed curve C in the Riemann sphere. I am trying to establish the Jordan-Schoenflies theorem via Caratheodory's mapping theorem. Is there a basic way of establishing the connected components are simply connected so that we can get a conformal mapping from the unit disk to the component and hence use Caratheodory's theorem?
complex-analysis
asked Nov 22 at 21:54
shc
62
Assuming the Jordan Curve Theorem, we can consider the 2 connected components of the complement of the simple closed curve C in the Riemann sphere. I am trying to establish the Jordan-Schoenflies theorem via Caratheodory's mapping theorem. Is there a basic way of establishing the connected components are simply connected so that we can get a conformal mapping from the unit disk to the component and hence use Caratheodory's theorem?
complex-analysis
complex-analysis
asked Nov 22 at 21:54
shc
62
asked Nov 22 at 21:54
shc
62
asked Nov 22 at 21:54
shc
62
asked Nov 22 at 21:54
shc
62
asked Nov 22 at 21:54
shc
62
62
add a comment |
add a comment |
1 Answer
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The Jordan Curve Theorem says that any simple closed curve $C$ in the two-dimensional sphere $S^2$ separates $S^2$ into two connected regions (that is, $S^2 setminus C$ has exactly two nonempty connected components).
The Jordan-Schoenflies Theorem strengthens this by stating that these two regions are homeomorphic to an open disk.
Based only on the Jordan Curve Theorem, is there a basic way of establishing the connected components are simply connected? It depends on your understanding of "basic", but it seems to me that the simple connectedness of the regions is not a straightforward corollary of Jordan Curve. In fact, you have to strengthen the proof.
Look at higher dimensions, for example an embedded copy $C$ of $S^2$ in the three-dimensional sphere $S^3$. Then $S^3 setminus C$ also has exactly two nonempty connected components, but they are not necessarily simply connected. An example for this phenomenom is the Alexander horned sphere. See https://en.wikipedia.org/wiki/Alexander_horned_sphere.
answered Nov 22 at 22:42
Paul Frost
8,4571528
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
|
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1 Answer
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The Jordan Curve Theorem says that any simple closed curve $C$ in the two-dimensional sphere $S^2$ separates $S^2$ into two connected regions (that is, $S^2 setminus C$ has exactly two nonempty connected components).
The Jordan-Schoenflies Theorem strengthens this by stating that these two regions are homeomorphic to an open disk.
Based only on the Jordan Curve Theorem, is there a basic way of establishing the connected components are simply connected? It depends on your understanding of "basic", but it seems to me that the simple connectedness of the regions is not a straightforward corollary of Jordan Curve. In fact, you have to strengthen the proof.
Look at higher dimensions, for example an embedded copy $C$ of $S^2$ in the three-dimensional sphere $S^3$. Then $S^3 setminus C$ also has exactly two nonempty connected components, but they are not necessarily simply connected. An example for this phenomenom is the Alexander horned sphere. See https://en.wikipedia.org/wiki/Alexander_horned_sphere.
answered Nov 22 at 22:42
Paul Frost
8,4571528
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
|
show 1 more comment
up vote
0
down vote
The Jordan Curve Theorem says that any simple closed curve $C$ in the two-dimensional sphere $S^2$ separates $S^2$ into two connected regions (that is, $S^2 setminus C$ has exactly two nonempty connected components).
The Jordan-Schoenflies Theorem strengthens this by stating that these two regions are homeomorphic to an open disk.
Based only on the Jordan Curve Theorem, is there a basic way of establishing the connected components are simply connected? It depends on your understanding of "basic", but it seems to me that the simple connectedness of the regions is not a straightforward corollary of Jordan Curve. In fact, you have to strengthen the proof.
Look at higher dimensions, for example an embedded copy $C$ of $S^2$ in the three-dimensional sphere $S^3$. Then $S^3 setminus C$ also has exactly two nonempty connected components, but they are not necessarily simply connected. An example for this phenomenom is the Alexander horned sphere. See https://en.wikipedia.org/wiki/Alexander_horned_sphere.
answered Nov 22 at 22:42
Paul Frost
8,4571528
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
The Jordan Curve Theorem says that any simple closed curve $C$ in the two-dimensional sphere $S^2$ separates $S^2$ into two connected regions (that is, $S^2 setminus C$ has exactly two nonempty connected components).
The Jordan-Schoenflies Theorem strengthens this by stating that these two regions are homeomorphic to an open disk.
Based only on the Jordan Curve Theorem, is there a basic way of establishing the connected components are simply connected? It depends on your understanding of "basic", but it seems to me that the simple connectedness of the regions is not a straightforward corollary of Jordan Curve. In fact, you have to strengthen the proof.
Look at higher dimensions, for example an embedded copy $C$ of $S^2$ in the three-dimensional sphere $S^3$. Then $S^3 setminus C$ also has exactly two nonempty connected components, but they are not necessarily simply connected. An example for this phenomenom is the Alexander horned sphere. See https://en.wikipedia.org/wiki/Alexander_horned_sphere.
answered Nov 22 at 22:42
Paul Frost
8,4571528
The Jordan Curve Theorem says that any simple closed curve $C$ in the two-dimensional sphere $S^2$ separates $S^2$ into two connected regions (that is, $S^2 setminus C$ has exactly two nonempty connected components).
The Jordan-Schoenflies Theorem strengthens this by stating that these two regions are homeomorphic to an open disk.
Based only on the Jordan Curve Theorem, is there a basic way of establishing the connected components are simply connected? It depends on your understanding of "basic", but it seems to me that the simple connectedness of the regions is not a straightforward corollary of Jordan Curve. In fact, you have to strengthen the proof.
Look at higher dimensions, for example an embedded copy $C$ of $S^2$ in the three-dimensional sphere $S^3$. Then $S^3 setminus C$ also has exactly two nonempty connected components, but they are not necessarily simply connected. An example for this phenomenom is the Alexander horned sphere. See https://en.wikipedia.org/wiki/Alexander_horned_sphere.
answered Nov 22 at 22:42
Paul Frost
8,4571528
edited Nov 23 at 8:22
answered Nov 22 at 22:42
Paul Frost
8,4571528
answered Nov 22 at 22:42
Paul Frost
8,4571528
answered Nov 22 at 22:42
Paul Frost
8,4571528
8,4571528
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
|
show 1 more comment
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
Thanks for the answer. I am aware of the example of the Alexander horned sphere and the extension to higher dimensions via Jordan-Brouwer Theorem. What I wanted to know is how to prove the Schoenflies Theorem via Caratheodory's mapping theorem. The proof as outlined here (en.wikipedia.org/wiki/Schoenflies_problem#cite_ref-6) is starightforward but only if one can apply Riemann mapping which needs the components to be simply connected. Hence my question.
– shc
Nov 22 at 22:52
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
The horned sphere is an example that the components are not necessarily simply connected. This is a special feature in dimension two, and it is not trivial. The proof goes far deeper than just invoking Jordan Curve. That is, there is a gap between Jordan curve and the applicability of Caratheodory.
– Paul Frost
Nov 22 at 23:01
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
Certainly. However I am only interested in the dimension 2 case. If a proof without invoklng Schoenflies can be given then the RMT/Caratheodory proof will be complete. Else there seems to be some circular reasoning here, atleast in my head.
– shc
Nov 22 at 23:04
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
As I said: To fill the gap you have to go back to the origin and prove more than Jordan Curve. Simple connectedness is not an addition you will get almost for free.
– Paul Frost
Nov 22 at 23:12
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
@shc If you want, I shall delete my answer. This gives you a better chance that somebody else has a look to you question (if a question has an answer, then this is visible at first glance in the list of questions, and many people will not have a closer look in this case).
– Paul Frost
Nov 23 at 8:20
|
show 1 more comment
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