Galileo's Paradox
in [Math]
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From: Virgil on 3 Dec 2006 00:07 In article <45725238$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > 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? NO! Equivalence classes of reals mod 2*pi, as works for bijecting one circle to another, do not form an ordered set unless one is willing to claim order without trichotomy. > > > > 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 > > Oh. Then, "structured". Whatever. What sort of "structures" are required, or allowed?
From: Virgil on 3 Dec 2006 00:15 In article <4572530a(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > 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. > > > > > > Wow, that sounds eerily like "segment sequence topology". Each vectoir > is relative to what, the last endpoint? :) I have no idea what visualization TO uses in trying to understand what a vector in a vector space is like, but "point" works nicely for mathematical vector spaces. > > But anyway, my point is that mapping a vector to another is not a set > bijection unless the vectors are set, and if they are, they are sets of > something, and if not points, then of what? Infinitesimal segments? I > can't argue with that.... One model of a vector space is as the set of functions from an often finite index set to the set of scalars for that space. > > >> 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 > > I'm not the one who gave mapping a vector as an example of an infinite > bijection.... TO missed the point again. It is mappings between vector spaces, which are sets of vectors, which are the bijections, and the vectors are the things being paired up by the bijection.
From: Tony Orlow on 3 Dec 2006 00:16 cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: > > <snip> > >> Anyway, ala Leibniz, each object IS the set of properties which it >> possesses, so any two objects with the exact same set of properties are >> the same object. > > But this begs the question: what do we mean, exactly, by a SET of > properties? What exactly are we trying to say when we say "This set of > properties is the same as this other set of properties"? > > Cheers - Chas > Well, what we really mean is that there is a set of universal properties, each of which is a set of values, and that each object is defined by a set of values, one from each set of property values, such that any two distinct objects differ in at least one property value. Was that specific enough? Your other question about 3 needs to be thought about a bit. You could ask Dave Rusin, He confirmed a couple of years ago it was a simple repeating bit pattern. I suppose I need to do my own legowrk in that area now. :) Tony
From: Virgil on 3 Dec 2006 00:18 In article <45725329(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > 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. > > Every line is ordered. Duh. TO is clearly not familiar with projective geometry.
From: Virgil on 3 Dec 2006 00:20
In article <4572549f(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <4570848f(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Bob Kolker wrote: > >>> Tony Orlow wrote: > >>> > >>>> Are x and y ordered? The Cartesian plane is ordered in two dimensions, > >>>> not a linear order, but a 2D ordered plane with origin. > >>> > >>> The plane is not a linearly ordered set of points. > >>> > >>> Bob Kolker > >> That's what I just said. > > > > And then you went on to say that because parts of it are ordered, we can > > treat it as if it were ordered. > > We can treat it as if each dimension were ordered, because each is. Yeah. What order do the orderings separate dimensions place on the ordering of points of a circle centered at origin? If none, then TO is desperately trying to ignore his error. |