Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: William Hughes on 25 Nov 2006 06:53 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > By induction one can even prove that always a finite number of words > > > has been mapped on the natural numbers. > > > > By induction one can also map an infinite set of words onto the natural > > numbers. > > Then you think the natural numbers are covered by induction? You read > my heart. I knew they were not actually infinite. > > > > > > > Now, after you have "completed" this finite list of words, take it and > > > construct a diagonal number. > > > > I prefer to take the infinite set of words. > > I know, but you don't get it by induction. > > > > > This is always a finite number, as the list is finite. > > > > Maybe yours is, mine isn't. > > > > > Fine. > > > So we have proved that there is a finite word which was missing in the > > > "complete" list. > > > > You haven't proved anything, > > If the set 1,2,3,...,n is finite, then the set 1,2,3,..., n+1 is finite > too. > Therefore, by induction we (that is: those who can think logically) > prove that the generated set is always finite. The two statements: The elements of N satsify property X and The set N satisfies property X are independent.. Induction proves that the elements of N are finite. Induction does not show that N is finite. - William Hughes > > Regards, WM
From: Lester Zick on 25 Nov 2006 13:56 On Sat, 25 Nov 2006 03:12:20 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >In article <lcbem29apkc14rfr1m1jtqaqj2anh1e41l(a)4ax.com> Lester Zick <dontbother(a)nowhere.net> writes: > > On Fri, 24 Nov 2006 01:10:39 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > > wrote: >... > > >Pray, tell me. I would say that a straight line in the Euclidean plane > > >is a line of the form ax + by = c. With the standard formulas for angles > > >it is easy enough to get rectangles. And I would say that a rectangle is > > >a square when the sides have equal length (this is the point where the > > >measure creeps in). So we have a rectangle enclosed by the lines: > > > x + y = 1 > > > x - y = 1 > > > - x + y = 1 > > > - x - y = 1 > > >Now define the Manhattan measure: > > > d((x1,y1), (x2,y2)) = ||x1 - x2| + |y1 - y2|| > > >and we see easily enough that the figure enclosed by the lines above is > > >a square with sides with length 2. > > > > > >A circle is a figure where each point has the same distance to a common > > >centre, and it is also easy to show that the points on the boundary of > > >that square have the same distance to the origin: 0. > > > > Well for one thing points equidistant from any point define a sphere > > not a circle unless one assumes "on a plane" when the Euclidean plane > > isn't defined to begin with. But my question was directed not at the > > definition of a circle or square on a Euclidean plane but at the > > definition of a square with the Manhattan measure. > >You completely misunderstand what I wrote. I *start* with an Euclidean >plane without measure (i.e. distance function). With that we can at >most define a rectangle. Neither a square, nor a circle. Actually I think I understand you very well. How is it exactly you start with a Euclidean plane? The fact that you start without a distance measure doesn't allow you to start with a plane Euclidean or otherwise. If you begin by assuming this you wind up by assuming that and pretty soon you find yourself assuming what you were supposed to demonstrate in the first place, that square circles are conceivable. > > In other words we have two different figures defined in the Euclidean > > metric, one as a curve and one with straight lines. Now I'm not trying > > to quibble over the modern math definition of either figure at the > > moment, just trying to point out that you have certain characteristics > > and properties defined in the Euclidean metric and then apparently > > claim that if you use some other metric and don't use the Euclidean > > metric the two figures are the same. > >I do not use Euclidean metric at all. Where, above, do I use Euclidean >metric? I suspect we're using the phrase "Euclidean metric" in different ways. When I use the term I'm referring not just to measures of distance but to all definitive characteristics which go into definitions of such things as dimensionality and geometric figures in addition to distance measures. For example I see no definition of yours for "plane" which I think would be impossible to define without a Euclidean metric. On the other hand if all you're describing are variable measures of distance then you'd have to explain how you obtain those measures without an underlying Euclidean metric and corresponding assumptions. You can't just assume them as modern mathematikers are wont to do. > > For example is it possible to define plane squares with non Euclidean > > metrics? > >Of course. Well then let's see some definitions for planes, circles, and squares which don't explicitly or implicitly rely on Euclidean assumptions. > > And can we define right angles without the parallel postulate > > people simply ellide when operating with non Euclidean geometries? No. > >I think you can. But I used Euclidean geometry above. But the problem here is what kind of definition for squares doesn't rely on straight lines, right angles, planes, and so on? > > When I > > ask a question such as "why are square circles unimaginable" the > > defining metric for "squares" and "circles" is Euclidean and not > > Manhattan and I don't expect answers that if we look at Euclidean > > figures through some other metric we'll find they are imaginable. > >In that case you should use better formulations in your questions. And you should use better formulations in your answers. > And >when I follow-up to point out that in the Manhattan measure all circles >are squares (but not the other way around) you should state that your >formulation was insufficient. Except that your answer relies on non Euclidean assumptions that circles are squares. If I ask why one sided quadrangles are unimaginable and you reply that they aren't if you start counting from four would you consider your answer responsive to the question asked? All you're doing is answering a question that wasn't asked in terms employed by the original question. I can make up private definitions just like everyone else does but that doesn't make definitions true. In fact my personal favorite private definition for distance metrics is one I made up for the real number line which runs 1, 2, e, 3, pi, 4, 5, . . but I don't try to pretend that when I'm trying to analyze real numbers that that is a true definition. Besides as far as I can tell you still haven't answered my question as to whether there are any curves at all with the Manhattan measure. In fact I can't even find it reprinted above. ~v~~
From: Virgil on 25 Nov 2006 13:59 In article <1164446020.298851.293790(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > By induction one can even prove that always a finite number of words > > > has been mapped on the natural numbers. > > > > By induction one can also map an infinite set of words onto the natural > > numbers. > > Then you think the natural numbers are covered by induction? You read > my heart. I knew they were not actually infinite. If your heart says that induction, in the sense of Peano, is limited to the members of a finite set, your heart is stupid. > > > > > > > Now, after you have "completed" this finite list of words, take it and > > > construct a diagonal number. > > > > I prefer to take the infinite set of words. > > I know, but you don't get it by induction. Then maybe you don't know how to use induction, as those who do know how to use it can get get an infinite set of results from it. The form of induction I use is based on having, as one has within ZF and NBG, an infinite set of finite naturals. If one has the first natural and one has the successor to every natural, then one has all of the members of that infinite set of finite naturals. If WM's version of induction is less effective, it is not our fault. > > If the set 1,2,3,...,n is finite, then the set 1,2,3,..., n+1 is finite > too. > Therefore, by induction we (that is: those who can think logically) > prove that the generated set is always finite. Provided one assumes that that process has a last step, one can conclude that that process has a last step, but absent that assumption, one cannot reach that conclusion. And I, for one, do not make that assumption.
From: Virgil on 25 Nov 2006 14:03 In article <1164446180.867559.37730(a)l39g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Philosophy of mathematics belongs to mathematics. If that is so then those like WM and EB who try to impose the philosophy of physics on math are out of bounds.
From: Virgil on 25 Nov 2006 14:04
In article <1164446332.276791.309770(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1164126713.968092.237570(a)k70g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > There are infinite paths > > > > > > in your tree, but they do not contain a node that represents (for > > > > > > instance) 1/3. So, if the nodes represent numbers (as you have > > > > > > said), > > > > > > > > > > Do you have a reference? > > > > > > > > Not needed. Just above you state that the nodes represent the bits 0 > > > > or > > > > 1. I have shown how you could concatenate the representation of a > > > > node > > > > with the representations of its parent nodes to get a number. > > > > > > A number consists of bits. Some numbers consist even of one bit. But > > > you must not mix up these terms. > > > A number like 1/2 consist of the bit sequence 0.1000.... That is a path > > > in my tree: > > > > As an infinite bit sequence, yes, but there is *no* node in your tree > > that represents that bit sequence. > > > > > > > > 1/3 is not in your tree. > > > > > > Of course it is, like 0.333... is in Cantor's list of decimals. > > > > No, there is *no* node in your tree that represents 1/3. Because there > > is *no* node in your tree that represents an infinite sequence. On the > > other hand, Cantor's diagonal proof is about infinite sequences. > > You do not require that one digit represents the number 1/3 in Cantor's > list. > You do not require that any digit there represents an infinite > sequence. > > But you require that one digit (node) represents 1/3 in the tree? > But you require require that one digit (node) in my tree represents an > infinite sequence? > > Are you trolling, Dik? It is WM who seems to be trolling by demanding things of his nodes that they are obviously incapable of giving. |