Mixing
Example of a simply connected Lipschitz domain non-homeomorphic to unit ball?
1
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What is example of a simply connected Lipschitz domain which is not homeomorphic to unit ball? In $R^2$, such a domain is necessarily unit disc (Are simply connected open sets in $mathbb{R}^2$ homeomorphic to an open ball?).
Even, I am curious to just see the example without being "Lipschitz".
real-analysis general-topology differential-geometry algebraic-topology
asked Jan 6 at 19:55
ershersh
320112
$endgroup$
add a comment |
1
$begingroup$
What is example of a simply connected Lipschitz domain which is not homeomorphic to unit ball? In $R^2$, such a domain is necessarily unit disc (Are simply connected open sets in $mathbb{R}^2$ homeomorphic to an open ball?).
Even, I am curious to just see the example without being "Lipschitz".
real-analysis general-topology differential-geometry algebraic-topology
asked Jan 6 at 19:55
ershersh
320112
$endgroup$
2
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
2
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32
add a comment |
1
1
1
0
$begingroup$
What is example of a simply connected Lipschitz domain which is not homeomorphic to unit ball? In $R^2$, such a domain is necessarily unit disc (Are simply connected open sets in $mathbb{R}^2$ homeomorphic to an open ball?).
Even, I am curious to just see the example without being "Lipschitz".
real-analysis general-topology differential-geometry algebraic-topology
asked Jan 6 at 19:55
ershersh
320112
$endgroup$
What is example of a simply connected Lipschitz domain which is not homeomorphic to unit ball? In $R^2$, such a domain is necessarily unit disc (Are simply connected open sets in $mathbb{R}^2$ homeomorphic to an open ball?).
Even, I am curious to just see the example without being "Lipschitz".
real-analysis general-topology differential-geometry algebraic-topology
real-analysis general-topology differential-geometry algebraic-topology
asked Jan 6 at 19:55
ershersh
320112
asked Jan 6 at 19:55
ershersh
320112
asked Jan 6 at 19:55
ershersh
320112
asked Jan 6 at 19:55
ershersh
320112
asked Jan 6 at 19:55
ershersh
320112
320112
2
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
2
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32
add a comment |
2
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
2
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32
2
2
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
2
2
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32
add a comment |
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2
$begingroup$
Maybe $mathbb{R}^3 backslash {0}$?
$endgroup$
– Mindlack
Jan 6 at 20:11
$begingroup$
Right, that is the example of a simply connected domain! But this doesn't have a Lipschitz boundary? Now, if I consider a three dimensional open-annulus which is simply connected having boundry as the union of two spheres, I think this would constitute the Lipschitz domain?
$endgroup$
– ersh
Jan 6 at 20:29
2
$begingroup$
I think so, indeed, since the boundary is going to be a smooth $2$-dimensional manifold.
$endgroup$
– Mindlack
Jan 6 at 20:32