Cantor Confusion
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From: Virgil on 24 Nov 2006 17:23 In article <45676677.1000403(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 11/24/2006 9:17 PM, Virgil wrote: > > >> As ambiguous as "hat" in the sense of as many red as you like and "hat" > >> in the sense the quality green includes something irrational: all of > >> indefinitely much. > > > > It would appear that EB cannot distinguish between red hats and green > > ones. Not only innumerate but colorblind. > > Not enough phantasy? EB had not enough carrots as a child.
From: Dik T. Winter on 24 Nov 2006 21:58 In article <456692F9.7070103(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > On 11/24/2006 2:25 AM, Dik T. Winter wrote: > > > > Mathematicians like you are hopefully aware of the trifle that the > > > relation >= cannot be applied to the really real numbers. > > > > Oh. What are "really real numbers"? > > Those, like the imagined numerical solution to the task pi, which are > just fictions. Those which are assumed as basis for DA2, In that case what are the "not-really real numbers"? > > What you are missing is that mathematics provides an idealisation of > > arithmetic (amongst others), and from that background provides > > processes to do "real" calculations to get results in (amongst others) > > the physical world. The whole field of numerical mathematics could > > barely have been build without the idealised mathematical background. > > No no. I do not deny that numbers can come as close as you like to the > ideal concept of infinity and continuum. Rational numbers already do > that job. You completely misunderstood what I wrote. Cholesky decomposition of positive definite matrices will not work when you are using rationals only. But it is commonly used. The reason is that in ideal mathematics the decomposition will work, and numerical mathematics provides the additional information you need to get it to work in finite precision. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 24 Nov 2006 22:12 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. > 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? > For example is it possible to define plane squares with non Euclidean > metrics? Of course. > 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. > 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 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Ross A. Finlayson on 24 Nov 2006 22:56 MoeBlee wrote: > > No, you're ignorant. It is a theorem of Z set theory that for all x the > identity function on x exists. > For all x of what? There is no universe in ZF(C). Ross
From: mueckenh on 25 Nov 2006 04:08
Dik T. Winter schrieb: > In article <1164281187.964067.115190(a)m7g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > Cantor wrote: "It is remarkable that removing a countable set leaves > > the plaine *being connected*." But it is not remarkable that removing > > an uncountable set leaves the plaine being connected? What a foolish > > assertion. > > But removing an uncountable set can leave the plane either connected or > disconnected. What is remarkable about that? Nothing. In order to discuss Cantors paper one should know it. Of course Cantor knew your trivial example and those proposed by Mr. Bader, who is seems to be used to take trivialities for the main issue. > Remove the line x=0 from > the plane. You remove an uncountable set and the result is clearly > disconnected. Yes, but Cantor considered only point sets which are dense. "Betrachten wir irgendeine Punktmenge (M), welche innerhalb eines n-dimensionalen stetig zusammenhängenden Gebietes A überalldicht verbreitet ist und die Eigenschaft der Abzählbarkeit besitzt,..." I did not quote this sentence in my paper because I considered only algebraic and transcendental point sets which automatically are dense."The manifold R^n (with n ï½ï 2) remains continuous if the set of points with purely algebraic coordinates is taken off." Cantor wrote that the countability is an important condition for his proof: "Es genügt, diesen Satz für den Fall n = 2 als richtig zu erkennen; sein Beweis beruht wesentlich auf dem in Art. 1 bewiesenen Satze, daÃ, wenn irgendeine gesetzmäÃige Reihe reeller GröÃen ... vorliegt, in jedem noch so kleinen willkürlich gegebenen Intervalle reelle GröÃen gefunden werden können, die in jener Reihe nicht vorkommen. And this condition (of countability) is obviously not necessary. Regards, WM |