Galileo's Paradox
in [Math]
|
From: Virgil on 1 Dec 2006 16:12 In article <45705bc9(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Bob Kolker wrote: > > Tony Orlow wrote: > > > >> > >> > >> To establish an explicit bijection between infinite sets does require > >> an ordering on the sets, at least in general. > > > > Not true. One can map the disk of radius one one onto the disk of radius > > two without ordering points in either disk. Hint: Use a cone. Or if you > > like vectors map the vector V of unit length into 2*V which has length > > 2. No ordering in sight. > > > > So in genaral one does not require an ordering. > > > > Bob Kolker > > Are you saying you can explicitly state the mapping mathematically > without any reference to the ordered coordinates which identify each > point? Not that you have to explicitly address each point individually, > but you do need to make reference to the coordinates in your mapping > formula, no, and those coordinates are each elements of an ordered set, no? That one uses ordered sets to construct unordered but structured sets and constructs bijections between different sets of the latter sort does not mean that the latter sort need be ordered sets. Furthermore, any bijection between sets induces a bijection on their power sets, and those power sets do not necessarily have any ordering induced by the order in infinite originals
From: Virgil on 1 Dec 2006 16:15 In article <45705c2a(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Bob Kolker wrote: > > Bob Kolker wrote: > > > >> Tony Orlow wrote: > >> > >>> > >>> > >>> To establish an explicit bijection between infinite sets does require > >>> an ordering on the sets, at least in general. > >> > >> > >> Not true. One can map the disk of radius one one onto the disk of > >> radius two without ordering points in either disk. Hint: Use a cone. > >> Or if you like vectors map the vector V of unit length into 2*V which > >> has length 2. No ordering in sight. > > > > > > Oops. Map the vector V of length <= 1 to vector 2*V which has length <= 2. > > > > Sorry about that. > > > > Bob Kolker > > > > How do you distinguish the points within each vector, if they are not > ordered. You do not have "points within each vector". It is the vectors that ARE the points. > If you claim you're not mapping points within the vector, then > the vector isn't really an infinite set, is it? A vector is one object, it is the vector spaces that are the sets of those objects. Learn some math, TO. > > TOny
From: Virgil on 1 Dec 2006 16:21 In article <45705cd3(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <456f334d$1(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <456e4621(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > > > > > >>>> Where standard measure is the same, there still may be an infinitesimal > >>>> difference, such as between (0,1) and [0,1], if that's what you mean. > >>> The outer measure of those two sets is exactly the same. > >> Right, and yet, the second is missing two elements, and is therefore > >> infinitesimally smaller in measure. > > > > Except that in outer measure there are no infinitesimals, and the outer > > measure of the difference set, {0,1} is precisely and exactly zero. > > So, you're saying infinitesimals cannot be considered? You're saying one > is not ALLOWED to consider the removal of a finite set from an ifninite > set to make any difference in measure? I say you're wrong. I am not wrong about outer measure of bounded sets in the set of reals when defined as the LUB of the sum of lengths of the intervals of a finite cover by open intervals. One could also use various other definitions of outer measure to get a similar result. > > Tony
From: Virgil on 1 Dec 2006 16:26 In article <45705dcc$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <456f3385(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <456e46b7(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>>> Any uncountable > >>>> set with element indexes expressed as digital numbers will require > >>>> infinitely long indexes for most elements, such as is the case for the > >>>> reals in any nonzero interval. > >>> > >>> There is nothing in being a set, including being a set of reals, that > >>> requires its members to be indexed at all. > >> To establish an explicit bijection between infinite sets does require an > >> ordering on the sets, at least in general. > > > > It requires a function between the sets, but neither set need be ordered. > > > > For example, let P = R^2 be the Cartesian plane, then for any reals a > > and b with a^2 + b^2 > 0, (x,y) |--> (a*x + b*y, -b*x + a*y) bijects > > P to itself, but P is not an ordered set. > > Are x and y ordered? The Cartesian plane is ordered in two dimensions, > not a linear order, but a 2D ordered plane with origin. How about mappings of projective spaces? A "projective line" is not an ordered set.
From: Virgil on 1 Dec 2006 16:28
In article <457069A0.7060208(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > Cantor did know that his fancy was rejected from all important figures > even those hundreds or even thousands of years ago. Is this supposed to mean something? |