lines of force converging on three or more points (?)
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From: William Elliot on 26 Jun 2010 01:37 >> 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: > >> 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. > The largest circle is C_2 ? > Also, for any line, the infimum of distances > from a point on one of the lines to the point > (0,0) is zero. > x in cl A iff inf{ d(x,a) | a in A } = 0. >>> 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. > > 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. I don't think so. > 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. I don't think the answer is dependent upon the q's nor the dimension of R^n. I think it's still no, and relies upon the fact that a line segement has two endpoints.
From: David Bernier on 26 Jun 2010 04:40
William Elliot 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. >> >> In a subsequent post, William Elliot wrote: >> >>> 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. >> > The largest circle is C_2 ? Yes. >> Also, for any line, the infimum of distances >> from a point on one of the lines to the point >> (0,0) is zero. >> > x in cl A iff inf{ d(x,a) | a in A } = 0. > >>>> 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. >> >> 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. > > I don't think so. Thanks again for pointing out that ]0, 1[ and the unit circle in the plane with one point removed are homeomorphic. >> 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. Possible clue: spirals. I append this comment in rot13; I haven't thought it through thoroughly, however... ----------------- Va cbyne pbbeqvangrf, gur nycun'gu pheir, sbe nycun orgjrra mreb naq bar enqvna, vf ng n enqvhf e rdhny gb bar cyhf gur rkcbaragvny bs gur fhz bs nycun naq gurgn, jurer gurgn sebz zvahf vasvavgl gb cyhf vasvavgl vf nppbeqvat gb phfgbz jura hfvat cynar cbyne pbbeqvangrf. Qbrf gur nycun'gu pheir trg neovgenevyl pybfr gb nal cbvag ba gur havg pvepyr?? ----------------- David Bernier >> 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. > > I don't think the answer is dependent upon the q's nor the > dimension of R^n. I think it's still no, and relies upon > the fact that a line segement has two endpoints. > > |