Can a Homeomorphism exist between two discontinuous spaces.
1
$begingroup$
In trying to solve a complex problem, I encountered a sub problem about homeomorphisms between level sets of functions. To explain the problem I have created a specific example pictured in a sketch below:
Additional information about the example:
- The black lines are level sets of some function which is defined for some space $mathbb{R}^n$ (for convenience the sketch is for $mathbb{R}^2$). The red line is to indicate there is a discontinuity between the level sets between the two regions $V_{1,1}$ and $V_{2,1}$ (and the other two regions $V_{1,2}$ and $V_{2,2}$).
- Points on the red boundary belong to region $V_{2,1}$ (and $V_{2,2}$).
- There exists a homeomorphism between the level sets for regions $V_{1,1}$ and $V_{1,2}$, and a homeomorphism between the level sets for regions $V_{2,1}$ and $V_{2,2}$.
My question is: With this discontinuity between $V_{1,1}$ and $V_{2,1}$ (or the other two regions $V_{1,2}$ and $V_{2,2}$) can there exist a homeomorphism between the two different level sets.
My concern is at the red boundary line and the open set definition of continuity, as I do not believe that open sets maps to open sets along this red boundary. On the other hand the existence of homeomorphisms for the different regions suggests that I could combine them to form a single homeomorphism.
Additional Notes
- If you have any additional questions I can provide needed clarification.
general-topology continuity
asked Jan 9 at 23:06
AzJAzJ
290218
$endgroup$
add a comment |
1
$begingroup$
In trying to solve a complex problem, I encountered a sub problem about homeomorphisms between level sets of functions. To explain the problem I have created a specific example pictured in a sketch below:
Additional information about the example:
- The black lines are level sets of some function which is defined for some space $mathbb{R}^n$ (for convenience the sketch is for $mathbb{R}^2$). The red line is to indicate there is a discontinuity between the level sets between the two regions $V_{1,1}$ and $V_{2,1}$ (and the other two regions $V_{1,2}$ and $V_{2,2}$).
- Points on the red boundary belong to region $V_{2,1}$ (and $V_{2,2}$).
- There exists a homeomorphism between the level sets for regions $V_{1,1}$ and $V_{1,2}$, and a homeomorphism between the level sets for regions $V_{2,1}$ and $V_{2,2}$.
My question is: With this discontinuity between $V_{1,1}$ and $V_{2,1}$ (or the other two regions $V_{1,2}$ and $V_{2,2}$) can there exist a homeomorphism between the two different level sets.
My concern is at the red boundary line and the open set definition of continuity, as I do not believe that open sets maps to open sets along this red boundary. On the other hand the existence of homeomorphisms for the different regions suggests that I could combine them to form a single homeomorphism.
Additional Notes
- If you have any additional questions I can provide needed clarification.
general-topology continuity
asked Jan 9 at 23:06
AzJAzJ
290218
$endgroup$
1
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34
add a comment |
1
1
1
$begingroup$
In trying to solve a complex problem, I encountered a sub problem about homeomorphisms between level sets of functions. To explain the problem I have created a specific example pictured in a sketch below:
Additional information about the example:
- The black lines are level sets of some function which is defined for some space $mathbb{R}^n$ (for convenience the sketch is for $mathbb{R}^2$). The red line is to indicate there is a discontinuity between the level sets between the two regions $V_{1,1}$ and $V_{2,1}$ (and the other two regions $V_{1,2}$ and $V_{2,2}$).
- Points on the red boundary belong to region $V_{2,1}$ (and $V_{2,2}$).
- There exists a homeomorphism between the level sets for regions $V_{1,1}$ and $V_{1,2}$, and a homeomorphism between the level sets for regions $V_{2,1}$ and $V_{2,2}$.
My question is: With this discontinuity between $V_{1,1}$ and $V_{2,1}$ (or the other two regions $V_{1,2}$ and $V_{2,2}$) can there exist a homeomorphism between the two different level sets.
My concern is at the red boundary line and the open set definition of continuity, as I do not believe that open sets maps to open sets along this red boundary. On the other hand the existence of homeomorphisms for the different regions suggests that I could combine them to form a single homeomorphism.
Additional Notes
- If you have any additional questions I can provide needed clarification.
general-topology continuity
asked Jan 9 at 23:06
AzJAzJ
290218
$endgroup$
In trying to solve a complex problem, I encountered a sub problem about homeomorphisms between level sets of functions. To explain the problem I have created a specific example pictured in a sketch below:
Additional information about the example:
- The black lines are level sets of some function which is defined for some space $mathbb{R}^n$ (for convenience the sketch is for $mathbb{R}^2$). The red line is to indicate there is a discontinuity between the level sets between the two regions $V_{1,1}$ and $V_{2,1}$ (and the other two regions $V_{1,2}$ and $V_{2,2}$).
- Points on the red boundary belong to region $V_{2,1}$ (and $V_{2,2}$).
- There exists a homeomorphism between the level sets for regions $V_{1,1}$ and $V_{1,2}$, and a homeomorphism between the level sets for regions $V_{2,1}$ and $V_{2,2}$.
My question is: With this discontinuity between $V_{1,1}$ and $V_{2,1}$ (or the other two regions $V_{1,2}$ and $V_{2,2}$) can there exist a homeomorphism between the two different level sets.
My concern is at the red boundary line and the open set definition of continuity, as I do not believe that open sets maps to open sets along this red boundary. On the other hand the existence of homeomorphisms for the different regions suggests that I could combine them to form a single homeomorphism.
Additional Notes
- If you have any additional questions I can provide needed clarification.
general-topology continuity
general-topology continuity
asked Jan 9 at 23:06
AzJAzJ
290218
asked Jan 9 at 23:06
AzJAzJ
290218
edited Jan 10 at 0:34
AzJ
asked Jan 9 at 23:06
AzJAzJ
290218
asked Jan 9 at 23:06
AzJAzJ
290218
asked Jan 9 at 23:06
AzJAzJ
290218
290218
1
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34
add a comment |
1
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34
1
1
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Yes: in particular, if $V_{1,1} = V_{1,2}$ and $V_{2,1} = V_{2,2}$, then the identity is such a map, and is always a homeomorphism, without any assumptions on the nature of the space involved.
answered Jan 9 at 23:39
user3482749user3482749
4,286919
$endgroup$
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
|
show 2 more comments
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1 Answer
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1 Answer
1
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oldest
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1
$begingroup$
Yes: in particular, if $V_{1,1} = V_{1,2}$ and $V_{2,1} = V_{2,2}$, then the identity is such a map, and is always a homeomorphism, without any assumptions on the nature of the space involved.
answered Jan 9 at 23:39
user3482749user3482749
4,286919
$endgroup$
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
|
show 2 more comments
1
$begingroup$
Yes: in particular, if $V_{1,1} = V_{1,2}$ and $V_{2,1} = V_{2,2}$, then the identity is such a map, and is always a homeomorphism, without any assumptions on the nature of the space involved.
answered Jan 9 at 23:39
user3482749user3482749
4,286919
$endgroup$
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
|
show 2 more comments
1
1
1
$begingroup$
Yes: in particular, if $V_{1,1} = V_{1,2}$ and $V_{2,1} = V_{2,2}$, then the identity is such a map, and is always a homeomorphism, without any assumptions on the nature of the space involved.
answered Jan 9 at 23:39
user3482749user3482749
4,286919
$endgroup$
Yes: in particular, if $V_{1,1} = V_{1,2}$ and $V_{2,1} = V_{2,2}$, then the identity is such a map, and is always a homeomorphism, without any assumptions on the nature of the space involved.
answered Jan 9 at 23:39
user3482749user3482749
4,286919
answered Jan 9 at 23:39
user3482749user3482749
4,286919
answered Jan 9 at 23:39
user3482749user3482749
4,286919
answered Jan 9 at 23:39
user3482749user3482749
4,286919
4,286919
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
|
show 2 more comments
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
$begingroup$
Thank you for your answer. For clarification $V_{1,1}=V_{1,2}$ means that these regions are topologically equivalent and not that they are identical. My sketch may have suggested they are identical so I have changed the sketch.
$endgroup$
– AzJ
Jan 10 at 0:34
1
1
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
I just gave an example where it works, so for the purpose of my answer, they are literally the same set.
$endgroup$
– user3482749
Jan 10 at 13:40
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
$begingroup$
Okay does your example work if instead of $V_{1,1}$ being identical to $V_{1,2}$ and $V_{2,1}$ being identical to $V_{2,2}$ they are topologically equivalent to each other?
$endgroup$
– AzJ
Jan 10 at 17:38
1
1
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
$begingroup$
Almost certainly, but there's some more work to be done to prove it: here, I'm using that the two homeomorphisms have a common extension to a homeomorphism $mathbb{R}^2tomathbb{R}^2$.
$endgroup$
– user3482749
Jan 10 at 18:49
2
2
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
$begingroup$
@AzJ No, but the problem is essentially that you've chosen to put the boundary in your sets. If you take your sets to not include the boundary, then that issue goes away: their union is then not connected, and you can map either side separately without issue. Is there a particular reason that you need the boundary included?
$endgroup$
– user3482749
Jan 10 at 21:26
|
show 2 more comments
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1
$begingroup$
I think you should be a little more precise about exactly what data you're working with here. Not necessarily the details of the functions, but say we're given the functions as black boxes, what do you mean by homeomorphisms between the level sets? Do the two homeomorphisms agree on the boundary?
$endgroup$
– jgon
Jan 10 at 0:41
$begingroup$
@jgon To answer your questions: The functions are Lyapunov functions, A homeomorphism between level sets of two Lyapunov function means the Lyapunov functions are topologically equivalent. I do not know the form of the Lyapunov function only their level sets (this is a common problem in dynamical systems). The two homeomorphisms are not defined at the boundary, as the example is written only the hoeomorphism between the level sets in regions $V_{2,1}$ to $V_{2,2}$ includes the red boundary.
$endgroup$
– AzJ
Jan 10 at 17:34