Proving two path connected subsets of $BbbR^2$ are not homeomorphic when the number of cut-points/pairs are the same. The Next CEO of Stack OverflowCut points of topology?Are any two open connected subsets of $BbbR^n$ homeomorphic?Are space of paths between two different points and space of pointed loops only homotopy equivalent? What about smooth case?How do I show using cut points that these two subsets of the plane are not homeomorphic?Connected open subsets in $mathbbR^2$ are path connected.Are the complements of two homeomorphic compact, connected subsets of $mathbbR^2$ homeomorphic?Proving two spaces are homeomorphicAn Almost Homeomorphsim Between Discs is a Homeomorphism.Is the Product of two path connected space path connected?Proving two sequences are not homeomorphic
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Proving two path connected subsets of $BbbR^2$ are not homeomorphic when the number of cut-points/pairs are the same.
The Next CEO of Stack OverflowCut points of topology?Are any two open connected subsets of $BbbR^n$ homeomorphic?Are space of paths between two different points and space of pointed loops only homotopy equivalent? What about smooth case?How do I show using cut points that these two subsets of the plane are not homeomorphic?Connected open subsets in $mathbbR^2$ are path connected.Are the complements of two homeomorphic compact, connected subsets of $mathbbR^2$ homeomorphic?Proving two spaces are homeomorphicAn Almost Homeomorphsim Between Discs is a Homeomorphism.Is the Product of two path connected space path connected?Proving two sequences are not homeomorphic
$begingroup$
Given the two path connected subsets of $BbbR^2$ depicted in the image attached, where the end of the line segment on the left is included, how do I go about proving they are not homeomorphic?
This question comes from a larger set of questions in which the lack of topological equivalence was proven with simple arguments based on the number of cut-points and pairs. However, as far as I can tell these have the same number of n-type cuts (they have infinite 1 points and 2 points, infinite 1 pairs, 2 pairs, and 3 pairs etc).
They do seem intuitively different to me as the 1-pairs on the left are based on selecting the end point and any point on the circle, where as the choice of points for the pairs is not as restrictive on the right. But I can't figure any argument from this.
general-topology
edited Mar 18 at 19:47
Clayton
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
$endgroup$
add a comment |
$begingroup$
Given the two path connected subsets of $BbbR^2$ depicted in the image attached, where the end of the line segment on the left is included, how do I go about proving they are not homeomorphic?
This question comes from a larger set of questions in which the lack of topological equivalence was proven with simple arguments based on the number of cut-points and pairs. However, as far as I can tell these have the same number of n-type cuts (they have infinite 1 points and 2 points, infinite 1 pairs, 2 pairs, and 3 pairs etc).
They do seem intuitively different to me as the 1-pairs on the left are based on selecting the end point and any point on the circle, where as the choice of points for the pairs is not as restrictive on the right. But I can't figure any argument from this.
general-topology
edited Mar 18 at 19:47
Clayton
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
$endgroup$
1
$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03
add a comment |
$begingroup$
Given the two path connected subsets of $BbbR^2$ depicted in the image attached, where the end of the line segment on the left is included, how do I go about proving they are not homeomorphic?
This question comes from a larger set of questions in which the lack of topological equivalence was proven with simple arguments based on the number of cut-points and pairs. However, as far as I can tell these have the same number of n-type cuts (they have infinite 1 points and 2 points, infinite 1 pairs, 2 pairs, and 3 pairs etc).
They do seem intuitively different to me as the 1-pairs on the left are based on selecting the end point and any point on the circle, where as the choice of points for the pairs is not as restrictive on the right. But I can't figure any argument from this.
general-topology
edited Mar 18 at 19:47
Clayton
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
$endgroup$
Given the two path connected subsets of $BbbR^2$ depicted in the image attached, where the end of the line segment on the left is included, how do I go about proving they are not homeomorphic?
This question comes from a larger set of questions in which the lack of topological equivalence was proven with simple arguments based on the number of cut-points and pairs. However, as far as I can tell these have the same number of n-type cuts (they have infinite 1 points and 2 points, infinite 1 pairs, 2 pairs, and 3 pairs etc).
They do seem intuitively different to me as the 1-pairs on the left are based on selecting the end point and any point on the circle, where as the choice of points for the pairs is not as restrictive on the right. But I can't figure any argument from this.
general-topology
general-topology
edited Mar 18 at 19:47
Clayton
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
edited Mar 18 at 19:47
Clayton
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
edited Mar 18 at 19:47
Clayton
19.5k33288
edited Mar 18 at 19:47
Clayton
19.5k33288
edited Mar 18 at 19:47
Clayton
19.5k33288
19.5k33288
asked Mar 18 at 19:40
GoetheGoethe
61
asked Mar 18 at 19:40
GoetheGoethe
61
asked Mar 18 at 19:40
GoetheGoethe
61
61
1
$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03
add a comment |
1
$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03
1
1
$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
add a comment |
$begingroup$
Here are two approaches which use cut-points:
The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.
The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
answered Mar 18 at 21:10
triitrii
80817
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
add a comment |
$begingroup$
In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
add a comment |
$begingroup$
In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
In the left hand picture if we have a cutpoint, one of its induced components is contractible, while this is not the case for the cutpoints of the right hand picture (these all lie on the middle segment). This is a topological property that shows they're not homeomorphic.
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
answered Mar 18 at 21:28
Henno BrandsmaHenno Brandsma
114k348124
114k348124
add a comment |
add a comment |
$begingroup$
Here are two approaches which use cut-points:
The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.
The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
answered Mar 18 at 21:10
triitrii
80817
$endgroup$
add a comment |
$begingroup$
Here are two approaches which use cut-points:
The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.
The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
answered Mar 18 at 21:10
triitrii
80817
$endgroup$
add a comment |
$begingroup$
Here are two approaches which use cut-points:
The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.
The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
answered Mar 18 at 21:10
triitrii
80817
$endgroup$
Here are two approaches which use cut-points:
The space on the left has the property that one can remove a non-cut point (namely any point on the left on the circle) such that the remaining space has exactly one non-cut point. For the space on the right this is not true since after removing a non-cut point almost any point on the remaining full circle is a non-cut point.
The space on the right has the property that one can remove one point (namely the point in the middle of the line segment) and get two components such that each component has infinitely many non-cut points. For the space on the left this is not true as after removing a point and getting two components one of them is a half open intervall which has only one non-cut point.
answered Mar 18 at 21:10
triitrii
80817
edited Mar 19 at 7:32
answered Mar 18 at 21:10
triitrii
80817
answered Mar 18 at 21:10
triitrii
80817
answered Mar 18 at 21:10
triitrii
80817
80817
add a comment |
add a comment |
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$begingroup$
@Clayton edit: Surely all points along the line segment bar the end disconnect the sets though?
$endgroup$
– Goethe
Mar 18 at 19:58
$begingroup$
Path-components and connected components are the same here, since the space is locally path-connected. The idea described by @Clayton comes from the fact that homeomorphisms induce bijections on path-components (or connected components, but again, for this space there is no difference).
$endgroup$
– o.h.
Mar 18 at 20:03