Mixing
What is the shortest non-trivial logical deduction about overlapping circles?
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1
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For line segments in a 1D infinite space the shortest non-trivial statement I can think of is:
"For line segments $A$, $B$ and $C$. If there are regions where (at least) $A$ and $B$ overlap and regions where $B$ and $C$ overlap and regions where $A$ and $C$ overlap there must be a region where $A$, $B$ and $C$ overlap."
Now in a 2D infinite plane with a number of filled circles $A,B,C,D...$ , you are told some incomplete information about if there are areas where the circles overlap and which circles these are. What it the simplest non-trivial logical deduction that one could make from this information about overlaps in the same sort of way as was done with line segments?
geometry circle
asked 2 days ago
zooby
946616
add a comment |
up vote
1
down vote
favorite
For line segments in a 1D infinite space the shortest non-trivial statement I can think of is:
"For line segments $A$, $B$ and $C$. If there are regions where (at least) $A$ and $B$ overlap and regions where $B$ and $C$ overlap and regions where $A$ and $C$ overlap there must be a region where $A$, $B$ and $C$ overlap."
Now in a 2D infinite plane with a number of filled circles $A,B,C,D...$ , you are told some incomplete information about if there are areas where the circles overlap and which circles these are. What it the simplest non-trivial logical deduction that one could make from this information about overlaps in the same sort of way as was done with line segments?
geometry circle
asked 2 days ago
zooby
946616
If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For line segments in a 1D infinite space the shortest non-trivial statement I can think of is:
"For line segments $A$, $B$ and $C$. If there are regions where (at least) $A$ and $B$ overlap and regions where $B$ and $C$ overlap and regions where $A$ and $C$ overlap there must be a region where $A$, $B$ and $C$ overlap."
Now in a 2D infinite plane with a number of filled circles $A,B,C,D...$ , you are told some incomplete information about if there are areas where the circles overlap and which circles these are. What it the simplest non-trivial logical deduction that one could make from this information about overlaps in the same sort of way as was done with line segments?
geometry circle
asked 2 days ago
zooby
946616
For line segments in a 1D infinite space the shortest non-trivial statement I can think of is:
"For line segments $A$, $B$ and $C$. If there are regions where (at least) $A$ and $B$ overlap and regions where $B$ and $C$ overlap and regions where $A$ and $C$ overlap there must be a region where $A$, $B$ and $C$ overlap."
Now in a 2D infinite plane with a number of filled circles $A,B,C,D...$ , you are told some incomplete information about if there are areas where the circles overlap and which circles these are. What it the simplest non-trivial logical deduction that one could make from this information about overlaps in the same sort of way as was done with line segments?
geometry circle
geometry circle
asked 2 days ago
zooby
946616
asked 2 days ago
zooby
946616
edited 2 days ago
asked 2 days ago
zooby
946616
asked 2 days ago
zooby
946616
asked 2 days ago
zooby
946616
946616
If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago
add a comment |
If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago
If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If three or more (but only finitely-many) circular regions in the plane are such that any three have a point in common, then all of them have a point in common.
(The case of three regions is, of course, tautological, but including it makes for the most-complete statement of the result.) This, and OP's segment example, are special cases of Helly's Theorem, which can be expressed as:
If $d+1$ or more (but only finitely-many) convex subsets of $mathbb{R}^d$ are such that any $d+1$ of them have a point in common, then all of them have a point in common.
(Again, the case of $d+1$ subsets is tautological.) As the Wikipedia article notes, the version of the theorem for infinitely-many regions requires the regions to be compact as well as convex.
It's worth noting that the topology of $mathbb{R}^d$ is important here. If OP's example were not about segments on the line but arcs on a circle, pairwise intersections do not imply a common intersection. Nor does the statement I mentioned if "circular regions in the plane" is replaced by "caps on the sphere".
answered 2 days ago
Blue
46.3k869146
1
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If three or more (but only finitely-many) circular regions in the plane are such that any three have a point in common, then all of them have a point in common.
(The case of three regions is, of course, tautological, but including it makes for the most-complete statement of the result.) This, and OP's segment example, are special cases of Helly's Theorem, which can be expressed as:
If $d+1$ or more (but only finitely-many) convex subsets of $mathbb{R}^d$ are such that any $d+1$ of them have a point in common, then all of them have a point in common.
(Again, the case of $d+1$ subsets is tautological.) As the Wikipedia article notes, the version of the theorem for infinitely-many regions requires the regions to be compact as well as convex.
It's worth noting that the topology of $mathbb{R}^d$ is important here. If OP's example were not about segments on the line but arcs on a circle, pairwise intersections do not imply a common intersection. Nor does the statement I mentioned if "circular regions in the plane" is replaced by "caps on the sphere".
answered 2 days ago
Blue
46.3k869146
1
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
add a comment |
up vote
2
down vote
accepted
If three or more (but only finitely-many) circular regions in the plane are such that any three have a point in common, then all of them have a point in common.
(The case of three regions is, of course, tautological, but including it makes for the most-complete statement of the result.) This, and OP's segment example, are special cases of Helly's Theorem, which can be expressed as:
If $d+1$ or more (but only finitely-many) convex subsets of $mathbb{R}^d$ are such that any $d+1$ of them have a point in common, then all of them have a point in common.
(Again, the case of $d+1$ subsets is tautological.) As the Wikipedia article notes, the version of the theorem for infinitely-many regions requires the regions to be compact as well as convex.
It's worth noting that the topology of $mathbb{R}^d$ is important here. If OP's example were not about segments on the line but arcs on a circle, pairwise intersections do not imply a common intersection. Nor does the statement I mentioned if "circular regions in the plane" is replaced by "caps on the sphere".
answered 2 days ago
Blue
46.3k869146
1
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If three or more (but only finitely-many) circular regions in the plane are such that any three have a point in common, then all of them have a point in common.
(The case of three regions is, of course, tautological, but including it makes for the most-complete statement of the result.) This, and OP's segment example, are special cases of Helly's Theorem, which can be expressed as:
If $d+1$ or more (but only finitely-many) convex subsets of $mathbb{R}^d$ are such that any $d+1$ of them have a point in common, then all of them have a point in common.
(Again, the case of $d+1$ subsets is tautological.) As the Wikipedia article notes, the version of the theorem for infinitely-many regions requires the regions to be compact as well as convex.
It's worth noting that the topology of $mathbb{R}^d$ is important here. If OP's example were not about segments on the line but arcs on a circle, pairwise intersections do not imply a common intersection. Nor does the statement I mentioned if "circular regions in the plane" is replaced by "caps on the sphere".
answered 2 days ago
Blue
46.3k869146
If three or more (but only finitely-many) circular regions in the plane are such that any three have a point in common, then all of them have a point in common.
(The case of three regions is, of course, tautological, but including it makes for the most-complete statement of the result.) This, and OP's segment example, are special cases of Helly's Theorem, which can be expressed as:
If $d+1$ or more (but only finitely-many) convex subsets of $mathbb{R}^d$ are such that any $d+1$ of them have a point in common, then all of them have a point in common.
(Again, the case of $d+1$ subsets is tautological.) As the Wikipedia article notes, the version of the theorem for infinitely-many regions requires the regions to be compact as well as convex.
It's worth noting that the topology of $mathbb{R}^d$ is important here. If OP's example were not about segments on the line but arcs on a circle, pairwise intersections do not imply a common intersection. Nor does the statement I mentioned if "circular regions in the plane" is replaced by "caps on the sphere".
answered 2 days ago
Blue
46.3k869146
edited 2 days ago
answered 2 days ago
Blue
46.3k869146
answered 2 days ago
Blue
46.3k869146
answered 2 days ago
Blue
46.3k869146
46.3k869146
1
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
add a comment |
1
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
1
1
Thanks. That answers the question.
– zooby
2 days ago
Thanks. That answers the question.
– zooby
2 days ago
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
yesterday
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
21 hours ago
add a comment |
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If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
2 days ago
Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
2 days ago
Same as the line segments. If A and B intersect, B and C intersect, C and A intersect, then there is a region (possibly a point) where all 3 intersect
– NazimJ
2 days ago
@NazimJ Not true! You could have a hole in the middle!
– zooby
2 days ago