Problem with simply connected 3D domains
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I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is aÃÂ path-connected domain where one can continuously shrink anyÃÂ simpleÃÂ closedÃÂ curve into a point while remaining in the domain "
I thought I understood it and my understanding went well with the bellow 2D domains:
I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)
But now when I see the bellow 3D domains, I get confused.
I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?
Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.
complex-analysis
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up vote
1
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favorite
I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is aÃÂ path-connected domain where one can continuously shrink anyÃÂ simpleÃÂ closedÃÂ curve into a point while remaining in the domain "
I thought I understood it and my understanding went well with the bellow 2D domains:
I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)
But now when I see the bellow 3D domains, I get confused.
I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?
Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.
complex-analysis
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is aÃÂ path-connected domain where one can continuously shrink anyÃÂ simpleÃÂ closedÃÂ curve into a point while remaining in the domain "
I thought I understood it and my understanding went well with the bellow 2D domains:
I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)
But now when I see the bellow 3D domains, I get confused.
I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?
Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.
complex-analysis
I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is aÃÂ path-connected domain where one can continuously shrink anyÃÂ simpleÃÂ closedÃÂ curve into a point while remaining in the domain "
I thought I understood it and my understanding went well with the bellow 2D domains:
I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)
But now when I see the bellow 3D domains, I get confused.
I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?
Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.
complex-analysis
asked Aug 21 at 14:08
paulplusx
15812
15812
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1 Answer
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I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
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
I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
add a comment |Â
up vote
1
down vote
accepted
I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.
I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.
answered Aug 21 at 14:16
Kusma
2,373213
2,373213
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
add a comment |Â
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
Wow! I didn't know that. This makes perfect sense. Thanks.
â paulplusx
Aug 21 at 15:15
add a comment |Â
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