Mixing
Topology - proving a point in the image is in the interior
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Suppose I have a bounded region in the complex plane, M. I want to show that a point P is in the interior of this region.
At first I thought that if through P, for any arbitrary direction, I could find a continuous set of points within M whose tangent at P was parallel to that direction, then that would be sufficient to prove P is an interior point.
But this isn't true. Take for example two filled circles tangentially intersecting at a single point P. Through P I can move infinitesimally in any direction, but P is a boundary point.
But what if I could prove that the same property of being able to move infinitesimally in any direction and remain in the region applies to all points "near" P. Would that be sufficient to prove P is in the interior?
Is there any applicable theorem in topology here I can use?
general-topology
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
$endgroup$
add a comment |
$begingroup$
Suppose I have a bounded region in the complex plane, M. I want to show that a point P is in the interior of this region.
At first I thought that if through P, for any arbitrary direction, I could find a continuous set of points within M whose tangent at P was parallel to that direction, then that would be sufficient to prove P is an interior point.
But this isn't true. Take for example two filled circles tangentially intersecting at a single point P. Through P I can move infinitesimally in any direction, but P is a boundary point.
But what if I could prove that the same property of being able to move infinitesimally in any direction and remain in the region applies to all points "near" P. Would that be sufficient to prove P is in the interior?
Is there any applicable theorem in topology here I can use?
general-topology
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
$endgroup$
2
$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38
add a comment |
$begingroup$
Suppose I have a bounded region in the complex plane, M. I want to show that a point P is in the interior of this region.
At first I thought that if through P, for any arbitrary direction, I could find a continuous set of points within M whose tangent at P was parallel to that direction, then that would be sufficient to prove P is an interior point.
But this isn't true. Take for example two filled circles tangentially intersecting at a single point P. Through P I can move infinitesimally in any direction, but P is a boundary point.
But what if I could prove that the same property of being able to move infinitesimally in any direction and remain in the region applies to all points "near" P. Would that be sufficient to prove P is in the interior?
Is there any applicable theorem in topology here I can use?
general-topology
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
$endgroup$
Suppose I have a bounded region in the complex plane, M. I want to show that a point P is in the interior of this region.
At first I thought that if through P, for any arbitrary direction, I could find a continuous set of points within M whose tangent at P was parallel to that direction, then that would be sufficient to prove P is an interior point.
But this isn't true. Take for example two filled circles tangentially intersecting at a single point P. Through P I can move infinitesimally in any direction, but P is a boundary point.
But what if I could prove that the same property of being able to move infinitesimally in any direction and remain in the region applies to all points "near" P. Would that be sufficient to prove P is in the interior?
Is there any applicable theorem in topology here I can use?
general-topology
general-topology
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
asked Jan 24 at 1:19
Ameet SharmaAmeet Sharma
666517
666517
2
$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38
add a comment |
2
$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38
2
2
$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38
$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38
add a comment |
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$begingroup$
If you refine it as you suggest, then it's trivially true: if there is an open neighborhood of $P$ so that if $Q$ is in the neighborhood, you can move some tiny amount in any direction from $Q$ and end up in $M$, then don't move at all from those points, and you have yourself a neighborhood of $P$ contained in $M$. What if you rule out "not moving at all"? Then it's false again: take the unit disk minus ${ 1/n : nin mathbb{N}}$ and $P=0$.
$endgroup$
– csprun
Jan 24 at 1:38