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
asked Nov 19 at 21:48
zooby
966616
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 Nov 19 at 21:48
zooby
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
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 Nov 19 at 21:48
zooby
966616
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 Nov 19 at 21:48
zooby
966616
asked Nov 19 at 21:48
zooby
966616
edited Nov 19 at 21:58
asked Nov 19 at 21:48
zooby
966616
asked Nov 19 at 21:48
zooby
966616
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
add a comment |
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
add a comment |
1 Answer
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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 Nov 19 at 23:36
Blue
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
add a comment |
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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 Nov 19 at 23:36
Blue
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
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 Nov 19 at 23:36
Blue
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
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 Nov 19 at 23:36
Blue
47.3k870149
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 Nov 19 at 23:36
Blue
47.3k870149
edited Nov 20 at 1:30
answered Nov 19 at 23:36
Blue
47.3k870149
answered Nov 19 at 23:36
Blue
47.3k870149
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
add a comment |
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
add a comment |
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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