What is the shortest non-trivial logical deduction about overlapping circles?











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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







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  • If you really just want the "simplest" deduction then this is surely opinion-based.
    – Eric Wofsey
    Nov 19 at 21:56










  • Well OK, by simplest I mean fewest circles and shortest statement.
    – zooby
    Nov 19 at 21:57










  • 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
    Nov 19 at 22:03










  • @NazimJ Not true! You could have a hole in the middle!
    – zooby
    Nov 19 at 22:08















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







share|cite|improve this question
























  • If you really just want the "simplest" deduction then this is surely opinion-based.
    – Eric Wofsey
    Nov 19 at 21:56










  • Well OK, by simplest I mean fewest circles and shortest statement.
    – zooby
    Nov 19 at 21:57










  • 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
    Nov 19 at 22:03










  • @NazimJ Not true! You could have a hole in the middle!
    – zooby
    Nov 19 at 22:08













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 21:58

























asked Nov 19 at 21:48









zooby

966616




966616












  • If you really just want the "simplest" deduction then this is surely opinion-based.
    – Eric Wofsey
    Nov 19 at 21:56










  • Well OK, by simplest I mean fewest circles and shortest statement.
    – zooby
    Nov 19 at 21:57










  • 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
    Nov 19 at 22:03










  • @NazimJ Not true! You could have a hole in the middle!
    – zooby
    Nov 19 at 22:08


















  • If you really just want the "simplest" deduction then this is surely opinion-based.
    – Eric Wofsey
    Nov 19 at 21:56










  • Well OK, by simplest I mean fewest circles and shortest statement.
    – zooby
    Nov 19 at 21:57










  • 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
    Nov 19 at 22:03










  • @NazimJ Not true! You could have a hole in the middle!
    – zooby
    Nov 19 at 22:08
















If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
Nov 19 at 21:56




If you really just want the "simplest" deduction then this is surely opinion-based.
– Eric Wofsey
Nov 19 at 21:56












Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
Nov 19 at 21:57




Well OK, by simplest I mean fewest circles and shortest statement.
– zooby
Nov 19 at 21:57












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
Nov 19 at 22:03




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
Nov 19 at 22:03












@NazimJ Not true! You could have a hole in the middle!
– zooby
Nov 19 at 22:08




@NazimJ Not true! You could have a hole in the middle!
– zooby
Nov 19 at 22:08










1 Answer
1






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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".






share|cite|improve this answer



















  • 1




    Thanks. That answers the question.
    – zooby
    Nov 20 at 0:41










  • BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
    – zooby
    Nov 21 at 13:25










  • A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
    – Blue
    Nov 21 at 17:09











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1 Answer
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active

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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".






share|cite|improve this answer



















  • 1




    Thanks. That answers the question.
    – zooby
    Nov 20 at 0:41










  • BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
    – zooby
    Nov 21 at 13:25










  • A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
    – Blue
    Nov 21 at 17:09















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".






share|cite|improve this answer



















  • 1




    Thanks. That answers the question.
    – zooby
    Nov 20 at 0:41










  • BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
    – zooby
    Nov 21 at 13:25










  • A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
    – Blue
    Nov 21 at 17:09













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".






share|cite|improve this answer















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".







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 at 1:30

























answered Nov 19 at 23:36









Blue

47.3k870149




47.3k870149








  • 1




    Thanks. That answers the question.
    – zooby
    Nov 20 at 0:41










  • BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
    – zooby
    Nov 21 at 13:25










  • A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
    – Blue
    Nov 21 at 17:09














  • 1




    Thanks. That answers the question.
    – zooby
    Nov 20 at 0:41










  • BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
    – zooby
    Nov 21 at 13:25










  • A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
    – Blue
    Nov 21 at 17:09








1




1




Thanks. That answers the question.
– zooby
Nov 20 at 0:41




Thanks. That answers the question.
– zooby
Nov 20 at 0:41












BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
Nov 21 at 13:25




BTW Is there a version of Helly's theoream where the N-spheres are on a the surface of an (N+1)-sphere?
– zooby
Nov 21 at 13:25












A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
Nov 21 at 17:09




A web search for "Helly theorem on sphere" turns up a number of references to "Helly-type" results.
– Blue
Nov 21 at 17:09


















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