Is the set $ (x,sin(frac1x) ) $ a manifold with and without the originIntersection of a manifold with open setcomplement of a finite subset of a path-connected space is path-connectedShowing a function is a manifoldNot sure about my proof that orthogonal matrices are a manifold in $rm Mat_n times n(mathbbR)$Is an open subset $U subset mathbfR^n$ diffeomorphic to the product $U' times mathbfR$ with $U' subset mathbfR^n - 1$ open?More general than topological/smooth manifoldsWhy is the boundary of an oriented manifold with its (opposite oriented) copy the empty set?2-manifold with involution without fixed points is a boundary of a 3-manifold.$M =(t, vert t vert) text vert t in mathbbR $ isn't a Smooth SubmanifoldDoes the set $ (x,y,z) $ determine a manifold?
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Is the set $ x in (0,1)$ a manifold with and without the origin
Intersection of a manifold with open setcomplement of a finite subset of a path-connected space is path-connectedShowing a function is a manifoldNot sure about my proof that orthogonal matrices are a manifold in $rm Mat_n times n(mathbbR)$Is an open subset $U subset mathbfR^n$ diffeomorphic to the product $U' times mathbfR$ with $U' subset mathbfR^n - 1$ open?More general than topological/smooth manifoldsWhy is the boundary of an oriented manifold with its (opposite oriented) copy the empty set?2-manifold with involution without fixed points is a boundary of a 3-manifold.$M =(t, vert t vert) text vert t in mathbbR $ isn't a Smooth SubmanifoldDoes the set $ (x,y,z) $ determine a manifold?
$begingroup$
Is the set $A = x in (0,1)$ a manifold with and without the origin.
So without the origin $(0,0)$ I'm pretty sure that it is a manifold, since in each point the set is a graph of a smooth function. we can also present $A$ with the implicit function $F(x,y) = y - sin(frac1x)$.
However, with the origin I am pretty sure that $A$ is not a manifold, but I don't know how to prove it formally. The way I know to prove a set is not a manifold, is to show that at any open environment of the origin $(0,0)$, $A$ is not a graph of a function. I haven't learned anything about manifolds and connectedness.
Any help would be appreciated.
calculus multivariable-calculus manifolds smooth-manifolds
asked Mar 21 at 18:45
Gabi GGabi G
500110
$endgroup$
add a comment |
$begingroup$
Is the set $A = x in (0,1)$ a manifold with and without the origin.
So without the origin $(0,0)$ I'm pretty sure that it is a manifold, since in each point the set is a graph of a smooth function. we can also present $A$ with the implicit function $F(x,y) = y - sin(frac1x)$.
However, with the origin I am pretty sure that $A$ is not a manifold, but I don't know how to prove it formally. The way I know to prove a set is not a manifold, is to show that at any open environment of the origin $(0,0)$, $A$ is not a graph of a function. I haven't learned anything about manifolds and connectedness.
Any help would be appreciated.
calculus multivariable-calculus manifolds smooth-manifolds
asked Mar 21 at 18:45
Gabi GGabi G
500110
$endgroup$
1
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46
add a comment |
$begingroup$
Is the set $A = x in (0,1)$ a manifold with and without the origin.
So without the origin $(0,0)$ I'm pretty sure that it is a manifold, since in each point the set is a graph of a smooth function. we can also present $A$ with the implicit function $F(x,y) = y - sin(frac1x)$.
However, with the origin I am pretty sure that $A$ is not a manifold, but I don't know how to prove it formally. The way I know to prove a set is not a manifold, is to show that at any open environment of the origin $(0,0)$, $A$ is not a graph of a function. I haven't learned anything about manifolds and connectedness.
Any help would be appreciated.
calculus multivariable-calculus manifolds smooth-manifolds
asked Mar 21 at 18:45
Gabi GGabi G
500110
$endgroup$
Is the set $A = x in (0,1)$ a manifold with and without the origin.
So without the origin $(0,0)$ I'm pretty sure that it is a manifold, since in each point the set is a graph of a smooth function. we can also present $A$ with the implicit function $F(x,y) = y - sin(frac1x)$.
However, with the origin I am pretty sure that $A$ is not a manifold, but I don't know how to prove it formally. The way I know to prove a set is not a manifold, is to show that at any open environment of the origin $(0,0)$, $A$ is not a graph of a function. I haven't learned anything about manifolds and connectedness.
Any help would be appreciated.
calculus multivariable-calculus manifolds smooth-manifolds
calculus multivariable-calculus manifolds smooth-manifolds
asked Mar 21 at 18:45
Gabi GGabi G
500110
asked Mar 21 at 18:45
Gabi GGabi G
500110
asked Mar 21 at 18:45
Gabi GGabi G
500110
asked Mar 21 at 18:45
Gabi GGabi G
500110
asked Mar 21 at 18:45
Gabi GGabi G
500110
500110
1
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46
add a comment |
1
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46
1
1
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $mathbbR^n$. But open subset of $mathbbR^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
$endgroup$
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $mathbbR^n$. But open subset of $mathbbR^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
$endgroup$
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
add a comment |
$begingroup$
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $mathbbR^n$. But open subset of $mathbbR^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
$endgroup$
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
add a comment |
$begingroup$
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $mathbbR^n$. But open subset of $mathbbR^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
$endgroup$
If $A$ is a manifold then any neighborhood of $(0,0)$ is homeomorphic of open subset of $mathbbR^n$. But open subset of $mathbbR^n$ are path connected while every neighborhood of $(0,0)$ in $A$ are not path-connected. Hence $A$ can not be manifold if $(0,0)$ is included because path-connectedness property is preserved by homeomorphism.
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
answered Mar 21 at 19:05
M. A. SARKARM. A. SARKAR
2,4621820
2,4621820
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
add a comment |
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
Is there a way to show this without using path connectivity?
$endgroup$
– Gabi G
Mar 21 at 22:46
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
@GabiG, It perhaps has.
$endgroup$
– M. A. SARKAR
Mar 22 at 4:33
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
I am pretty sure that you can show that in every neighborhood of $(0,0)$ the set $M$ is not a graph of a function. Since if $x = f(y)$ then we have two values of $x$ for the same value of $y$, and if $y = f(x)$ then i every neighborhood of $x=0$ the function is not smooth. I think this proves it.
$endgroup$
– Gabi G
Mar 23 at 17:48
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
@GabiG, $M$ or $A$ ?
$endgroup$
– M. A. SARKAR
Mar 23 at 19:43
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
$begingroup$
$A$, my mistake
$endgroup$
– Gabi G
Mar 23 at 22:16
add a comment |
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1
$begingroup$
Do you mean Differentiable Manifold by manifold?
$endgroup$
– M. A. SARKAR
Mar 21 at 18:55
$begingroup$
A smooth manifold. There 4 different definitions that are equal
$endgroup$
– Gabi G
Mar 21 at 20:46