Mixing
How are “handles†described in point set topology?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I recently began learning about topology in terms of neighborhoods, open and closed sets, boundary and limit points. I'm trying to reconcile this with the view of topology as "rubber sheet geometry".
In particular, I'm picturing the set of points comprising the interior and boundary of a circle on the plane. If I understand correctly, this would be a closed set, and also an open set because the rest of the plane is closed.
From the rubber sheet view, I understand that the plane is pretty much like the surface of a sphere, and that if you add a "handle" to the sphere, you get a surface that's equivalent to a torus.
How do we express this "handle" concept in the language of point set topology?
general-topology
asked Jul 26 at 22:44
tangentstorm
36617
add a comment |Â
up vote
3
down vote
favorite
I recently began learning about topology in terms of neighborhoods, open and closed sets, boundary and limit points. I'm trying to reconcile this with the view of topology as "rubber sheet geometry".
In particular, I'm picturing the set of points comprising the interior and boundary of a circle on the plane. If I understand correctly, this would be a closed set, and also an open set because the rest of the plane is closed.
From the rubber sheet view, I understand that the plane is pretty much like the surface of a sphere, and that if you add a "handle" to the sphere, you get a surface that's equivalent to a torus.
How do we express this "handle" concept in the language of point set topology?
general-topology
asked Jul 26 at 22:44
tangentstorm
36617
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I recently began learning about topology in terms of neighborhoods, open and closed sets, boundary and limit points. I'm trying to reconcile this with the view of topology as "rubber sheet geometry".
In particular, I'm picturing the set of points comprising the interior and boundary of a circle on the plane. If I understand correctly, this would be a closed set, and also an open set because the rest of the plane is closed.
From the rubber sheet view, I understand that the plane is pretty much like the surface of a sphere, and that if you add a "handle" to the sphere, you get a surface that's equivalent to a torus.
How do we express this "handle" concept in the language of point set topology?
general-topology
asked Jul 26 at 22:44
tangentstorm
36617
I recently began learning about topology in terms of neighborhoods, open and closed sets, boundary and limit points. I'm trying to reconcile this with the view of topology as "rubber sheet geometry".
In particular, I'm picturing the set of points comprising the interior and boundary of a circle on the plane. If I understand correctly, this would be a closed set, and also an open set because the rest of the plane is closed.
From the rubber sheet view, I understand that the plane is pretty much like the surface of a sphere, and that if you add a "handle" to the sphere, you get a surface that's equivalent to a torus.
How do we express this "handle" concept in the language of point set topology?
general-topology
asked Jul 26 at 22:44
tangentstorm
36617
edited Jul 27 at 3:28
asked Jul 26 at 22:44
tangentstorm
36617
asked Jul 26 at 22:44
tangentstorm
36617
asked Jul 26 at 22:44
tangentstorm
36617
36617
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46
add a comment |Â
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
The interior of the circle together with its boundary is a closed set. It's not also open, because the "rest of the plane" isn't closed - it's open.
The plane isn't quite like the surface of a sphere. It is near any point, but on the plane you can find an infinite sequence of points that never cluster anywhere. On the surface of the sphere that's impossible. (Of course this depends on a precise definition of "cluster").
To talk rigorously about the handle concept you have to do a fair amount of work. You define paths as continuous functions from the circle (just the boundary) into the space. Then you make precise what it means to "shrink a path to a single point". Then you have a (kind of) handle when you have a path that can't shrink to a point.
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
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 interior of the circle together with its boundary is a closed set. It's not also open, because the "rest of the plane" isn't closed - it's open.
The plane isn't quite like the surface of a sphere. It is near any point, but on the plane you can find an infinite sequence of points that never cluster anywhere. On the surface of the sphere that's impossible. (Of course this depends on a precise definition of "cluster").
To talk rigorously about the handle concept you have to do a fair amount of work. You define paths as continuous functions from the circle (just the boundary) into the space. Then you make precise what it means to "shrink a path to a single point". Then you have a (kind of) handle when you have a path that can't shrink to a point.
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
add a comment |Â
up vote
1
down vote
The interior of the circle together with its boundary is a closed set. It's not also open, because the "rest of the plane" isn't closed - it's open.
The plane isn't quite like the surface of a sphere. It is near any point, but on the plane you can find an infinite sequence of points that never cluster anywhere. On the surface of the sphere that's impossible. (Of course this depends on a precise definition of "cluster").
To talk rigorously about the handle concept you have to do a fair amount of work. You define paths as continuous functions from the circle (just the boundary) into the space. Then you make precise what it means to "shrink a path to a single point". Then you have a (kind of) handle when you have a path that can't shrink to a point.
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The interior of the circle together with its boundary is a closed set. It's not also open, because the "rest of the plane" isn't closed - it's open.
The plane isn't quite like the surface of a sphere. It is near any point, but on the plane you can find an infinite sequence of points that never cluster anywhere. On the surface of the sphere that's impossible. (Of course this depends on a precise definition of "cluster").
To talk rigorously about the handle concept you have to do a fair amount of work. You define paths as continuous functions from the circle (just the boundary) into the space. Then you make precise what it means to "shrink a path to a single point". Then you have a (kind of) handle when you have a path that can't shrink to a point.
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
The interior of the circle together with its boundary is a closed set. It's not also open, because the "rest of the plane" isn't closed - it's open.
The plane isn't quite like the surface of a sphere. It is near any point, but on the plane you can find an infinite sequence of points that never cluster anywhere. On the surface of the sphere that's impossible. (Of course this depends on a precise definition of "cluster").
To talk rigorously about the handle concept you have to do a fair amount of work. You define paths as continuous functions from the circle (just the boundary) into the space. Then you make precise what it means to "shrink a path to a single point". Then you have a (kind of) handle when you have a path that can't shrink to a point.
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
answered Jul 26 at 22:50
Ethan Bolker
35.7k54199
35.7k54199
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
add a comment |Â
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
Hrm, thanks for clearing up my mistakes. I can't tell if this answers my question or not. Like: I don't even see how the idea of a boundary or a closed set can apply to a circle on the plane if there's a handle sticking out of the middle and connecting it to the rest of the plane. It seems like there's no longer really an inside or an outside to it, so how can you even say it's bounded?
– tangentstorm
Jul 27 at 3:37
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
"Bounded" and "boundary" have the same root, but they are not related concepts topologically. In the plane, any set such that lies completely inside some circle is bounded. That has nothing to do with the boundary of a set. As my answer suggests, to understand point set topology takes some time - learning and getting used to the definitions and the theorems.
– Ethan Bolker
Jul 27 at 11:40
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
I think I'm starting to understand. "Inside a circle" makes sense on the plane or on the surface of a sphere, but "inside" no longer makes sense if the circle has a handle leading out over the boundary. So instead, you define a path around the circle, and define shrinking the path to point. Am I getting it now?
– tangentstorm
Jul 27 at 17:32
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
We are both still confused. "Inside a circle" does not make sense on a sphere. A handle "leading out over the boundary" doesn't make any sense to me. I think you just have to learn more formal topology - further back and forth here isn't going to help you.
– Ethan Bolker
Jul 27 at 17:36
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
If you have a sphere, you can draw a circle on it, and a point on the sphere is either "inside" or "outside" the circle. (It might not be clear which side is "inside", but a point is either on one side of the boundary or the other, so you can just pick one direction and call it "inward".) If you have a sphere with a handle on it, you can draw a circle around the handle, but it's no longer possible to say whether a given point is "inside" or "outside" the circle, because you can draw a path along the surface to the point in either the "inward" or "outward" direction.
– tangentstorm
Jul 27 at 18:20
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863890%2fhow-are-handles-described-in-point-set-topology%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
You mean something like: 1) remove two discs from the sphere; 2) take the disjoint union with a cylinder; 3) take the quotient by the equivalence relation identifying boundaries of the cylinder with boundaries of the removed discs; 4) prove the resulting topological space is homeomorphic to the torus?
– Daniel Schepler
Jul 26 at 22:53
I'm talking about handles: encyclopediaofmath.org/index.php/Handle_theory ... What you're saying sounds right, but I'm not looking for a proof, I'm just looking for what concepts or words in point set topology correspond to this idea...
– tangentstorm
Jul 27 at 3:27
For connected $2$-manifolds $X$, adding a "handle" means taking the connected sum with a torus. Since $X$ is connected, this will give the same result as above
– leibnewtz
Jul 27 at 7:46