lines of force converging on three or more points (?)
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From: David Bernier on 25 Jun 2010 00:48 This is inspired by lines of force in magnetic fields. We can take all circles C_r (1<=r<=2) which are in the upper half plane, are tangent to the x-axis at (0, 0) and where C_r has a radius of r. They all pass through origin (0, 0). If we remove the point at the origin, each C_r can be obtained through a continuous map f_r: ]0, 1[ -> R^2 which is injective. f_r's inverse won't be continuous, however. This can, I think, by mapping the open set ]0, 1/2[ in the space ]0, 1[ through f_r to C_r. The result in C_r is not open, I think. It may be likely that bi-continuous maps for each C_r can be obtained by doing some stretching ... The initial figure can be imagined using a "Nested Apollonian Gasket", such as can be found here: < http://www.bogotobogo.com/svg_source/ApolloNested2367.svg > . By separating the two "pincers" that meet (or almost meet) at one point, one can obtain something like a crescent (probably) where the deformed C_r start and finish at one and the other end of the crescent. That way, we have 2^(aleph_0) disjoint copies of ]0, 1[ in the plane such that any two lines have points arbitrarily close in any open neigborhood (in R^2) of either end of the crescent. I don't see how one could have a third point which every curve gets arbitrarily close to while not intersecting. In 3D, with the Lorenz attractor, there might be 2^(aleph_0) disjoint copies of ]0, 1[ such that each curve gets arbitrarily close to points A, B and C specially chosen in the "strange attractor". I'm not sure. Moreover, a line of flow for the time interval (0, oo) might not be homeomorphic to ]0, 1[, because a bijective continuous p: (0, 1) -> "the flow" curve might not have a continuous inverse, i.e. p^{-1} : "the flow" curve --> (0, 1) could be discontinuous. So, can one get all 2^(aleph_0) disjoint curves in R^2 going arbitrarily close to a third point C in R^2 if we drop the requirement that each image of (0, 1) be homeomorphic to (0, 1)? David Bernier
From: David Bernier on 25 Jun 2010 01:08 David Bernier wrote: > This is inspired by lines of force in magnetic fields. > > We can take all circles C_r (1<=r<=2) which are in the upper > half plane, are tangent to the x-axis at (0, 0) and > where C_r has a radius of r. > > They all pass through origin (0, 0). If we remove the > point at the origin, each C_r can be obtained through > a continuous map f_r: ]0, 1[ -> R^2 which is injective. > f_r's inverse won't be continuous, however. This can, I think, > by mapping the open set ]0, 1/2[ in the space ]0, 1[ through > f_r to C_r. The result in C_r is not open, I think. [...] I forgotten some topology. But the sequence 1/10, 9/10, 1/100, 99/100, 1/1000, 999/1000 ... doesn't converge in (0, 1) and neither does it converge if mapping t in (0, 1) to exp(2*pi*i*t) on the unit circle less {1}. So is the unit circle minus a point homeomorphic to (0, 1) as a metric space? David Bernier
From: William Elliot on 25 Jun 2010 01:39 On Fri, 25 Jun 2010, David Bernier wrote: > We can take all circles C_r (1<=r<=2) which are in the upper > half plane, are tangent to the x-axis at (0, 0) and > where C_r has a radius of r. > > They all pass through origin (0, 0). If we remove the > point at the origin, each C_r can be obtained through > a continuous map f_r: ]0, 1[ -> R^2 which is injective. > f_r's inverse won't be continuous, however. This can, I think, > by mapping the open set ]0, 1/2[ in the space ]0, 1[ through > f_r to C_r. The result in C_r is not open, I think. > Incorrect. An open line segement and a circle with a single point remove, are homeomorphic. > It may be likely that bi-continuous maps for each C_r > can be obtained by doing some stretching ... > > The initial figure can be imagined using > a "Nested Apollonian Gasket", such as can be found here: > < http://www.bogotobogo.com/svg_source/ApolloNested2367.svg > . There seems to be some confusion between C_r, C_r\(0,0) and C = { C_r\(0,0) | r in [1,2] } > By separating the two "pincers" that meet (or almost meet) > at one point, one can obtain something like a crescent > (probably) where the deformed C_r start and finish at > one and the other end of the crescent. > Find the point diametrically opposed to the missing point. Let L be a line tangent to that point. Roll by bending each side of the circle missing one point onto L. > That way, we have 2^(aleph_0) disjoint copies of ]0, 1[ in the plane > such that any two lines have points arbitrarily close in any open > neigborhood (in R^2) of either end of the crescent. > For example ]0,1[ x {r} for r in [1,2]. > I don't see how one could have a third point which every curve gets > arbitrarily close to while not intersecting. The third guess point is not in any of the C_r\(0,0). C_r\(0,0) is homeomorphic to (0,1). \/{ C_r | r in [1,2] } - {(0,0)} is homemorphic [0,1]^2 - { (0,0), (1,1) } and { (x,y) | x^2 + y^2 = 1 } - { (-1,0), (1,0) }.
From: David Bernier on 25 Jun 2010 03:46 David Bernier wrote: > David Bernier wrote: >> This is inspired by lines of force in magnetic fields. >> >> We can take all circles C_r (1<=r<=2) which are in the upper >> half plane, are tangent to the x-axis at (0, 0) and >> where C_r has a radius of r. >> >> They all pass through origin (0, 0). If we remove the >> point at the origin, each C_r can be obtained through >> a continuous map f_r: ]0, 1[ -> R^2 which is injective. >> f_r's inverse won't be continuous, however. This can, I think, >> by mapping the open set ]0, 1/2[ in the space ]0, 1[ through >> f_r to C_r. The result in C_r is not open, I think. > [...] > > I forgotten some topology. But the sequence > 1/10, 9/10, 1/100, 99/100, 1/1000, 999/1000 ... > doesn't converge in (0, 1) and neither > does it converge if mapping t in (0, 1) to exp(2*pi*i*t) > on the unit circle less {1}. > > So is the unit circle minus a point homeomorphic to > (0, 1) as a metric space? [...] Thanks to William Elliot. I looked through my 500 filters and changed one, so I hope I can now see your posts. David
From: David Bernier on 25 Jun 2010 21:15
David Bernier wrote: > This is inspired by lines of force in magnetic fields. > > We can take all circles C_r (1<=r<=2) which are in the upper > half plane, are tangent to the x-axis at (0, 0) and > where C_r has a radius of r. > > They all pass through origin (0, 0). If we remove the > point at the origin, each C_r can be obtained through > a continuous map f_r: ]0, 1[ -> R^2 which is injective. > f_r's inverse won't be continuous, however. This can, I think, > by mapping the open set ]0, 1/2[ in the space ]0, 1[ through > f_r to C_r. The result in C_r is not open, I think. In a subsequent post, William Elliot wrote: > Incorrect. An open line segement and a circle with a single > point remove, are homeomorphic. > >> It may be likely that bi-continuous maps for each C_r >> can be obtained by doing some stretching ... > >> The initial figure can be imagined using >> a "Nested Apollonian Gasket", such as can be found here: >> < http://www.bogotobogo.com/svg_source/ApolloNested2367.svg > . > > There seems to be some confusion between C_r, C_r\(0,0) and > C = { C_r\(0,0) | r in [1,2] } The idea of removing one point on the largest circle is to get 2^(aleph_0) homeomorphs of the open segment (0, 1) such that no two lines intersect. Also, for any line, the infimum of distances from a point on one of the lines to the point (0,0) is zero. >> By separating the two "pincers" that meet (or almost meet) >> at one point, one can obtain something like a crescent >> (probably) where the deformed C_r start and finish at >> one and the other end of the crescent. > > Find the point diametrically opposed to the missing point. > Let L be a line tangent to that point. > > Roll by bending each side of the circle missing one point onto L. Ok. I can follow that. Then if U is the rolled-out solid figure containing the 2^(aleph_0) homeomorphs of (0, 1), then we can write the closure of U as U \/ {A, B} where A, B are distinct points not in U. For any of the (0, 1) homeomorphs X, the infimum of distances from a point in X to A (respectively B) is zero. >> That way, we have 2^(aleph_0) disjoint copies of ]0, 1[ in the plane >> such that any two lines have points arbitrarily close in any open >> neigborhood (in R^2) of either end of the crescent. > > For example ]0,1[ x {r} for r in [1,2]. > >> I don't see how one could have a third point which every curve gets >> arbitrarily close to while not intersecting. > > The third guess point is not in any of the C_r\(0,0). Sorry for the bad formulation of my question. Is it possible to trace 2^(aleph_0) lines in the plane such that no two lines intersect and such that there are three distinct points A_i (i=1, 2, 3) each with the property that any given line gets arbitrarily close to A_i ? Also, let's assume that the A_i's aren't on any of the lines. The lines must be the image of ]0, 1[ by continuous injective mappings q_{j}: ]0, 1[ --> R^2 , j belonging to J, where J is merely a set of indices, each j in J being an index value. Here, we drop the requirement that the q_{j}'s be open maps. Then for R^2, I don't know the answer. If we still want three special points A_i (i=1, 2, 3) but change the ambient space to R^3, I wouldn't be too surprised if mappings q_{j}: ]0, 1[ --> R^3 exist, but can't visualize that. David Bernier > C_r\(0,0) is homeomorphic to (0,1). > > \/{ C_r | r in [1,2] } - {(0,0)} > is homemorphic > [0,1]^2 - { (0,0), (1,1) } > and > { (x,y) | x^2 + y^2 = 1 } - { (-1,0), (1,0) }. |