“A manifold with boundary has dimension at least 1” if it has a dimension and if it has nonempty boundary?Can a topological manifold be non-connected and each component with different dimension?Why do connected oriented manifolds have compactly supported forms with integral one but with support contained in a given open proper subset?Understanding topological and manifold boundaries on the real lineIs $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.Is $x^3-6xy+y^2=-108$ a regular submanifold but not a regular $k$-submanifold?Since not all compact subspaces of $mathbb R^n$ are manifolds with boundaries, how can we this apply index of a vector field formula?Submanifold with boundary of a manifold with boundaryNumber of Differentiable Structures on a Smooth ManifoldIs there a condition for the closure of an open subset to be a manifold with boundary?Can we extend smooth maps when the target manifold has boundary?Can a topological manifold be non-connected and each component with different dimension?Confusion over notion of compact manifold with or without boundaryManifolds with Boundary and Maximal AtlasIs $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.Since not all compact subspaces of $mathbb R^n$ are manifolds with boundaries, how can we this apply index of a vector field formula?Is $x^3-6xy+y^2=-108$ a regular submanifold but not a regular $k$-submanifold?
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“A manifold with boundary has dimension at least 1” if it has a dimension and if it has nonempty boundary?
Can a topological manifold be non-connected and each component with different dimension?Why do connected oriented manifolds have compactly supported forms with integral one but with support contained in a given open proper subset?Understanding topological and manifold boundaries on the real lineIs $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.Is $x^3-6xy+y^2=-108$ a regular submanifold but not a regular $k$-submanifold?Since not all compact subspaces of $mathbb R^n$ are manifolds with boundaries, how can we this apply index of a vector field formula?Submanifold with boundary of a manifold with boundaryNumber of Differentiable Structures on a Smooth ManifoldIs there a condition for the closure of an open subset to be a manifold with boundary?Can we extend smooth maps when the target manifold has boundary?Can a topological manifold be non-connected and each component with different dimension?Confusion over notion of compact manifold with or without boundaryManifolds with Boundary and Maximal AtlasIs $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.Since not all compact subspaces of $mathbb R^n$ are manifolds with boundaries, how can we this apply index of a vector field formula?Is $x^3-6xy+y^2=-108$ a regular submanifold but not a regular $k$-submanifold?
$begingroup$
My book is An Introduction to Manifolds by Loring W. Tu.
As can be found in the following bullet points
Can a topological manifold be non-connected and each component with different dimension?
Is $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.
Understanding topological and manifold boundaries on the real line
we have that
Tu's manifolds with or without boundaries do not necessarily have dimensions.
Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
Question: For Definition 22.6 (see here and here), Tu says that "A manifold with boundary has dimension at least 1". Should this instead be "A manifold with boundary has dimension at least 1 if it has a dimension and if it has nonempty boundary" or "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" (Notice that the prefix "$n-$" precisely gives the manifold with boundary a dimension)?
Embedding photos:
general-topology geometry differential-geometry manifolds differential-topology
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
$endgroup$
|
show 2 more comments
$begingroup$
My book is An Introduction to Manifolds by Loring W. Tu.
As can be found in the following bullet points
Can a topological manifold be non-connected and each component with different dimension?
Is $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.
Understanding topological and manifold boundaries on the real line
we have that
Tu's manifolds with or without boundaries do not necessarily have dimensions.
Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
Question: For Definition 22.6 (see here and here), Tu says that "A manifold with boundary has dimension at least 1". Should this instead be "A manifold with boundary has dimension at least 1 if it has a dimension and if it has nonempty boundary" or "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" (Notice that the prefix "$n-$" precisely gives the manifold with boundary a dimension)?
Embedding photos:
general-topology geometry differential-geometry manifolds differential-topology
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
$endgroup$
1
$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
1
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
2
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35
|
show 2 more comments
$begingroup$
My book is An Introduction to Manifolds by Loring W. Tu.
As can be found in the following bullet points
Can a topological manifold be non-connected and each component with different dimension?
Is $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.
Understanding topological and manifold boundaries on the real line
we have that
Tu's manifolds with or without boundaries do not necessarily have dimensions.
Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
Question: For Definition 22.6 (see here and here), Tu says that "A manifold with boundary has dimension at least 1". Should this instead be "A manifold with boundary has dimension at least 1 if it has a dimension and if it has nonempty boundary" or "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" (Notice that the prefix "$n-$" precisely gives the manifold with boundary a dimension)?
Embedding photos:
general-topology geometry differential-geometry manifolds differential-topology
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
$endgroup$
My book is An Introduction to Manifolds by Loring W. Tu.
As can be found in the following bullet points
Can a topological manifold be non-connected and each component with different dimension?
Is $[0,1) cup 2$ a manifold with boundary? My issue is the $2$.
Understanding topological and manifold boundaries on the real line
we have that
Tu's manifolds with or without boundaries do not necessarily have dimensions.
Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
Question: For Definition 22.6 (see here and here), Tu says that "A manifold with boundary has dimension at least 1". Should this instead be "A manifold with boundary has dimension at least 1 if it has a dimension and if it has nonempty boundary" or "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" (Notice that the prefix "$n-$" precisely gives the manifold with boundary a dimension)?
Embedding photos:
general-topology geometry differential-geometry manifolds differential-topology
general-topology geometry differential-geometry manifolds differential-topology
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
edited May 10 at 13:40
Selene Auckland
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
asked May 10 at 12:20
Selene AucklandSelene Auckland
31414
31414
1
$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
1
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
2
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35
|
show 2 more comments
1
$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
1
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
2
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35
1
1
$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
1
1
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
2
2
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be
If an $n$-dimensional manifold has nonempty boundary, then $nge 1$.
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
$endgroup$
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
|
show 1 more comment
$begingroup$
Assuming sensible definitions, an alternative solution is to change the statement to the following:
A connected manifold with non-empty boundary has dimension at least 1
Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
$endgroup$
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be
If an $n$-dimensional manifold has nonempty boundary, then $nge 1$.
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
$endgroup$
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
|
show 1 more comment
$begingroup$
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be
If an $n$-dimensional manifold has nonempty boundary, then $nge 1$.
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
$endgroup$
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
|
show 1 more comment
$begingroup$
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be
If an $n$-dimensional manifold has nonempty boundary, then $nge 1$.
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
$endgroup$
I think Tu’s statement is fine:
A manifold, by definition, always has a dimension. Where are the charts going?
Usually when we say that a manifold “has boundary,” we mean that it has nonempty boundary.
After looking at some of Tu’s (non-standard!) definitions, I think you’re correct. An accurate statement might be
If an $n$-dimensional manifold has nonempty boundary, then $nge 1$.
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
edited May 10 at 15:39
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
answered May 10 at 12:25
Santana AftonSantana Afton
3,5462832
3,5462832
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
|
show 1 more comment
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
I'll add these to the post to make the post hopefully self-contained: 1. Tu's manifolds do not necessarily have dimension. 2. Tu has considered manifolds to be manifolds with boundaries (with empty boundaries).
$endgroup$
– Selene Auckland
May 10 at 12:27
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
$begingroup$
Thanks Santana Afton! About non-standard, Tu said himself on stackexchange that these must be allowed: here. Your response? Haha
$endgroup$
– Selene Auckland
May 10 at 13:11
2
2
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
@SeleneAuckland When learning about these objects, the important part is understanding the core of what they are and how they interact with one another. These edge cases can be important — keep them in mind — but don’t dwell on them at the moment.
$endgroup$
– Santana Afton
May 10 at 13:19
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
$begingroup$
Good advice. Thanks!
$endgroup$
– Selene Auckland
May 10 at 13:20
1
1
$begingroup$
@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
$endgroup$
– Santana Afton
May 10 at 13:31
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@SeleneAuckland I think it’s a tricky issue. It depends on what you think the purpose of an introductory text is, what lessons an author wants a reader to learn from the text, etc. There’s a much larger conversation here about standard definitions vs. “proper” definitions, how mathematics “ought” to behave, teaching students, and so on. I personally disagree, but I’m not sure how much weight that has.
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– Santana Afton
May 10 at 13:31
|
show 1 more comment
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Assuming sensible definitions, an alternative solution is to change the statement to the following:
A connected manifold with non-empty boundary has dimension at least 1
Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
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Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
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– Selene Auckland
May 10 at 12:57
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Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
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– Selene Auckland
May 10 at 12:58
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
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– AnonymousCoward
May 10 at 13:07
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The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
add a comment |
$begingroup$
Assuming sensible definitions, an alternative solution is to change the statement to the following:
A connected manifold with non-empty boundary has dimension at least 1
Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
$endgroup$
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
add a comment |
$begingroup$
Assuming sensible definitions, an alternative solution is to change the statement to the following:
A connected manifold with non-empty boundary has dimension at least 1
Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
$endgroup$
Assuming sensible definitions, an alternative solution is to change the statement to the following:
A connected manifold with non-empty boundary has dimension at least 1
Edit: I rejected the suggested edit to change "manifold with non-empty boundary" to "manifold with boundary with non-empty boundary" because it does not add new information. A manifold with non-empty boundary must be a manifold with boundary, or your definitions are nonsense.
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
edited May 10 at 13:33
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
answered May 10 at 12:55
AnonymousCowardAnonymousCoward
3,2772436
3,2772436
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
add a comment |
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Oh because connected manifolds with or without boundary, under Tu's dimension-less definitions, necessarily have dimensions?
$endgroup$
– Selene Auckland
May 10 at 12:57
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
$begingroup$
Also, the rest of the sentence talks about "a discrete set of points". I don't think connected is what was intended. Why don't we just say "An $n-$manifold with boundary with non-empty boundary has $n ge 1$" ?
$endgroup$
– Selene Auckland
May 10 at 12:58
2
2
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
I don't know what Tu's definition is, but for any sensible definition of manifold, a connected manifold will have a well defined dimension. I also cannot load imgur posts and cant see anything about discrete sets of points directly in your post.
$endgroup$
– AnonymousCoward
May 10 at 13:07
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
$begingroup$
The statement in Section 22: "A manifold with boundary has dimension at least 1 since a manifold of dimension 0, being a discrete set of points, necessarily has empty boundary." The definition of manifolds (without boundary) in Section 5: here
$endgroup$
– Selene Auckland
May 10 at 13:10
add a comment |
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$begingroup$
Of course, this entirely depends on how you defined things and, once you have fixed definitions, the answer will be trivial.
$endgroup$
– AnonymousCoward
May 10 at 12:37
$begingroup$
Also note that imgur is not accessible in every country. It is better to embed images directly into your post.
$endgroup$
– AnonymousCoward
May 10 at 12:52
$begingroup$
@AnonymousCoward You mean one might not access imgur directly but can be embedded imgur images?
$endgroup$
– Selene Auckland
May 10 at 13:11
1
$begingroup$
Oh, they embed from imgur now? this is a shame.
$endgroup$
– AnonymousCoward
May 10 at 13:34
2
$begingroup$
If you are having so many issues with the definitions in a given textbook, it might be a good time to pick up another book. There are so many introductory smooth manifold texts to choose from. The world is your oyster.
$endgroup$
– AnonymousCoward
May 10 at 13:35