Galileo's Paradox
in [Math]
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From: Tony Orlow on 2 Dec 2006 23:24 Virgil wrote: > In article <457059e6(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Lester Zick wrote: >>> On Thu, 30 Nov 2006 09:52:49 +0100, Eckard Blumschein >>> <blumschein(a)et.uni-magdeburg.de> wrote: >>> > > TO verus LZ versus EB. It only wants WM to become the ultimate battle of > the pigmies Isn't that spelled "pygmies"? Would you like to take a walk on my vine bridge, oh Tall One? :0
From: Tony Orlow on 2 Dec 2006 23:26 Virgil wrote: > In article <45705ad3(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Huh! So, what happens if I declare a number, Big'un, and say that that >> is the number of reals in (0,1]? What if I say the real line is >> homogeneous, so every unit interval contains the same number of points? >> And then, what if I say the positive number line is going to include >> Big'un such unit intervals, so it has Big'un^2 reals up to Big'un? Does >> the universe collapse, or all tautologies suddenly become false? > > As long as it is only TO playing his silly games, nobody much cares. > > If TO were ever to produce anything like a coherent system with actual > proofs, we might actually have to pay some attention. Of course, that would be nice, to get some attention around here, especially from Virgil. But, he never pays me no mind. He hardly responds to anything I say. ;) Tony <3
From: Tony Orlow on 2 Dec 2006 23:27 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? > > 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.
From: Tony Orlow on 2 Dec 2006 23:31 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? :) 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.... >> 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....
From: Tony Orlow on 2 Dec 2006 23:31
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. |