If a compact set's boundary is the finite union of smooth Jordan curves, is it a domain?
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Let $V subset mathbbR^2$ be a compact set such that it's boundary is a finite union of piecewise smooth Jordan curves.
For example:
Does this imply that $V setminus partial V$ is a domain?
(If yes, does this also apply to higher finite dimensions, namely $mathbbR^n$?)
EDIT: Thanks to @UmbertoP it is clear that the answer is no, and a simple counter-example is the union of two disjoint closed disks.
Now, if I add the following conditions:
- The jordan curves mentioned are pairwise disjoint (Thanks @LeeMosher)
- One of the jordan curves is the boundary of a set that contains all other jordan curves
Do I have a domain now?
general-topology
asked May 31 '16 at 13:37
Pedro A
1,6811723
add a comment |Â
up vote
0
down vote
favorite
Let $V subset mathbbR^2$ be a compact set such that it's boundary is a finite union of piecewise smooth Jordan curves.
For example:
Does this imply that $V setminus partial V$ is a domain?
(If yes, does this also apply to higher finite dimensions, namely $mathbbR^n$?)
EDIT: Thanks to @UmbertoP it is clear that the answer is no, and a simple counter-example is the union of two disjoint closed disks.
Now, if I add the following conditions:
- The jordan curves mentioned are pairwise disjoint (Thanks @LeeMosher)
- One of the jordan curves is the boundary of a set that contains all other jordan curves
Do I have a domain now?
general-topology
asked May 31 '16 at 13:37
Pedro A
1,6811723
2
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
2
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V subset mathbbR^2$ be a compact set such that it's boundary is a finite union of piecewise smooth Jordan curves.
For example:
Does this imply that $V setminus partial V$ is a domain?
(If yes, does this also apply to higher finite dimensions, namely $mathbbR^n$?)
EDIT: Thanks to @UmbertoP it is clear that the answer is no, and a simple counter-example is the union of two disjoint closed disks.
Now, if I add the following conditions:
- The jordan curves mentioned are pairwise disjoint (Thanks @LeeMosher)
- One of the jordan curves is the boundary of a set that contains all other jordan curves
Do I have a domain now?
general-topology
asked May 31 '16 at 13:37
Pedro A
1,6811723
Let $V subset mathbbR^2$ be a compact set such that it's boundary is a finite union of piecewise smooth Jordan curves.
For example:
Does this imply that $V setminus partial V$ is a domain?
(If yes, does this also apply to higher finite dimensions, namely $mathbbR^n$?)
EDIT: Thanks to @UmbertoP it is clear that the answer is no, and a simple counter-example is the union of two disjoint closed disks.
Now, if I add the following conditions:
- The jordan curves mentioned are pairwise disjoint (Thanks @LeeMosher)
- One of the jordan curves is the boundary of a set that contains all other jordan curves
Do I have a domain now?
general-topology
asked May 31 '16 at 13:37
Pedro A
1,6811723
edited Jun 1 '16 at 0:07
asked May 31 '16 at 13:37
Pedro A
1,6811723
asked May 31 '16 at 13:37
Pedro A
1,6811723
asked May 31 '16 at 13:37
Pedro A
1,6811723
1,6811723
2
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
2
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00
add a comment |Â
2
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
2
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00
2
2
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
2
2
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.
Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V backslash partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.
This shows that we may focus to connected $V$ having an interior point. In this case $V backslash partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).
answered Aug 23 at 23:01
Paul Frost
4,693424
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
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
accepted
Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.
Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V backslash partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.
This shows that we may focus to connected $V$ having an interior point. In this case $V backslash partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).
answered Aug 23 at 23:01
Paul Frost
4,693424
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
add a comment |Â
up vote
1
down vote
accepted
Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.
Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V backslash partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.
This shows that we may focus to connected $V$ having an interior point. In this case $V backslash partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).
answered Aug 23 at 23:01
Paul Frost
4,693424
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.
Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V backslash partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.
This shows that we may focus to connected $V$ having an interior point. In this case $V backslash partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).
answered Aug 23 at 23:01
Paul Frost
4,693424
Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.
Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V backslash partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.
This shows that we may focus to connected $V$ having an interior point. In this case $V backslash partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).
answered Aug 23 at 23:01
Paul Frost
4,693424
answered Aug 23 at 23:01
Paul Frost
4,693424
answered Aug 23 at 23:01
Paul Frost
4,693424
answered Aug 23 at 23:01
Paul Frost
4,693424
4,693424
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
add a comment |Â
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
Thank you very much, especially for taking the time to answer a question that is over two years old! :)
– Pedro A
Aug 24 at 0:36
add a comment |Â
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2
Do you intend, as your picture suggests, that the Jordan curves forming this union are pairwise disjoint?
– Lee Mosher
May 31 '16 at 14:03
2
What happens if $V$ is a union of two disjoint closed disks?
– Umberto P.
May 31 '16 at 14:13
@LeeMosher thanks for pointing that out, yes I want that. I was indeed feeling that some more conditions would be necessary.
– Pedro A
May 31 '16 at 23:59
@UmbertoP. Thank you!! I don't know how I missed that. If I add the condition that one of the jordan curves fully encloses all the others, is it sufficient now, or still lacking other conditions?
– Pedro A
Jun 1 '16 at 0:00