Mixing
Local Euclidean Spaces
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I'm confused about some properties of local Euclidean spaces. I'm working with the definition:
A topological space, $X$, is locally Euclidean of dimension $n$, iff
$forall xin X. exists U$ a neighborhood of $x$ such that $U$ is
homeomorphic to an open ball of $mathbb{R}^n$.
- If we're looking at an open disk in $mathbb{R}^3$, i.e. ${(x,y,z)in mathbb{R}^3|x^2+y^2< 1, z=0}$, then this has to be homeomorphic to an open ball in $mathbb{R}^2$, right? How about a non-connected union of say a circle ($S^1$) and a disk as before in $mathbb{R}^3$. Is this locally Euclidean but just for varying $n$ in this case? Which means my definition doesn't hold for the space $X$ but for each connected part.
- Is locally euclidean a property of open sets? Or are we looking at the subset topology in each case? If we have a closed rectangle in $mathbb{R}^2$, how could we map a neighborhood to an open set in $mathbb{R}^2$. Or am I missing something here?
general-topology euclidean-geometry
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
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add a comment |
$begingroup$
I'm confused about some properties of local Euclidean spaces. I'm working with the definition:
A topological space, $X$, is locally Euclidean of dimension $n$, iff
$forall xin X. exists U$ a neighborhood of $x$ such that $U$ is
homeomorphic to an open ball of $mathbb{R}^n$.
- If we're looking at an open disk in $mathbb{R}^3$, i.e. ${(x,y,z)in mathbb{R}^3|x^2+y^2< 1, z=0}$, then this has to be homeomorphic to an open ball in $mathbb{R}^2$, right? How about a non-connected union of say a circle ($S^1$) and a disk as before in $mathbb{R}^3$. Is this locally Euclidean but just for varying $n$ in this case? Which means my definition doesn't hold for the space $X$ but for each connected part.
- Is locally euclidean a property of open sets? Or are we looking at the subset topology in each case? If we have a closed rectangle in $mathbb{R}^2$, how could we map a neighborhood to an open set in $mathbb{R}^2$. Or am I missing something here?
general-topology euclidean-geometry
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
$endgroup$
$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04
add a comment |
$begingroup$
I'm confused about some properties of local Euclidean spaces. I'm working with the definition:
A topological space, $X$, is locally Euclidean of dimension $n$, iff
$forall xin X. exists U$ a neighborhood of $x$ such that $U$ is
homeomorphic to an open ball of $mathbb{R}^n$.
- If we're looking at an open disk in $mathbb{R}^3$, i.e. ${(x,y,z)in mathbb{R}^3|x^2+y^2< 1, z=0}$, then this has to be homeomorphic to an open ball in $mathbb{R}^2$, right? How about a non-connected union of say a circle ($S^1$) and a disk as before in $mathbb{R}^3$. Is this locally Euclidean but just for varying $n$ in this case? Which means my definition doesn't hold for the space $X$ but for each connected part.
- Is locally euclidean a property of open sets? Or are we looking at the subset topology in each case? If we have a closed rectangle in $mathbb{R}^2$, how could we map a neighborhood to an open set in $mathbb{R}^2$. Or am I missing something here?
general-topology euclidean-geometry
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
$endgroup$
I'm confused about some properties of local Euclidean spaces. I'm working with the definition:
A topological space, $X$, is locally Euclidean of dimension $n$, iff
$forall xin X. exists U$ a neighborhood of $x$ such that $U$ is
homeomorphic to an open ball of $mathbb{R}^n$.
- If we're looking at an open disk in $mathbb{R}^3$, i.e. ${(x,y,z)in mathbb{R}^3|x^2+y^2< 1, z=0}$, then this has to be homeomorphic to an open ball in $mathbb{R}^2$, right? How about a non-connected union of say a circle ($S^1$) and a disk as before in $mathbb{R}^3$. Is this locally Euclidean but just for varying $n$ in this case? Which means my definition doesn't hold for the space $X$ but for each connected part.
- Is locally euclidean a property of open sets? Or are we looking at the subset topology in each case? If we have a closed rectangle in $mathbb{R}^2$, how could we map a neighborhood to an open set in $mathbb{R}^2$. Or am I missing something here?
general-topology euclidean-geometry
general-topology euclidean-geometry
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
edited Jan 29 at 20:45
GottlobtFrege
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
asked Jan 29 at 19:03
GottlobtFregeGottlobtFrege
778
778
$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04
add a comment |
$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04
$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04
add a comment |
1 Answer
1
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votes
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"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.
The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
add a comment |
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1 Answer
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$begingroup$
"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.
The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
add a comment |
$begingroup$
"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.
The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
add a comment |
$begingroup$
"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.
The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
"Locally Euclidean of dimension $n$" implies that $n$ is invariant for all points of the space.
The closed rectangle is not locally Euclidean in that definition because the boundary points do not have a neighbourhood (in the subspace topology always) that is homeomorphic to an open set of a Euclidean space. That is why the notion of "manifolds with boundary" was introduced. Open subsets of Euclidean spaces are trivially locally Euclidean of the same dimension.
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
answered Jan 29 at 22:58
Henno BrandsmaHenno Brandsma
114k348124
114k348124
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
add a comment |
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
$begingroup$
Therefore the set I described in point 1 (non-connected union of line and disk) is not locally euclidean? But the connected parts themselves are?
$endgroup$
– GottlobtFrege
Jan 30 at 11:09
1
1
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
$begingroup$
@GottlobtFrege indeed. Not locally Euclidean of 1 or 2 dimensions that is.
$endgroup$
– Henno Brandsma
Jan 30 at 12:33
add a comment |
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$begingroup$
What you called an "open disk in $Bbb R^3$" isn't homeomorphic to an open ball in $Bbb R^2$.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:05
$begingroup$
How is that? (Maybe the name "open disk in $mathbb{R}^3$" is wrong. I mean the set I wrote above.)
$endgroup$
– GottlobtFrege
Jan 29 at 19:50
$begingroup$
What you wrote is homeomorphic to a closed disc in $Bbb R^2$, which is compact, and so not homeomorphic to an open disc.
$endgroup$
– Lord Shark the Unknown
Jan 29 at 19:56
$begingroup$
Ah, yes. That was a typo. Edited.
$endgroup$
– GottlobtFrege
Jan 29 at 20:03
$begingroup$
The trick here is that "$n$" is specified before and independently of $x$, so you should read this as "$exists nforall x exists U ldots$" instead of "$forall xexists Uexists n ldots$"
$endgroup$
– Neal
Jan 29 at 23:04