Cantor Confusion
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From: Lester Zick on 24 Nov 2006 13:23 On Fri, 24 Nov 2006 01:10:39 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >In article <34i6m255sfv5upv6nenqm6menle2gp8l8b(a)4ax.com> Lester Zick <dontbother(a)nowhere.net> writes: > > On Tue, 21 Nov 2006 03:04:42 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > > wrote: > > >In article <8m8pl2pj1icbeven2hq7rp4hq1rufqh1u2(a)4ax.com> Lester Zick <dontbother(a)nowhere.net> writes: > > >... > > > > Why are square circles unimaginable? > > > > > >They are not. With the Manhattan measure of the plane, each circle is > > >a square. > > > > Then what is a square? > >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. It looks to me that you've just defined a square with the Manhattan metric with the properties of a circle in Euclidean plane metric. What's the point of that if you don't define a square with different properties in the Manhattan metric? 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 suppose I should ask instead are there any Euclidean curves in the Manhattan metric? If not the issue is moot. But my primary concern is that I see people all the time defining lines, figures, etc. with the Euclidean metric then going on to do non Euclidean mathematics with them. I mean Euclidean definitions require the Euclidean metric. And if you want to use some other metric you need some other definition using that metric. In which case my original question should be rephrased "why are square circles unimaginable in the plane Euclidean metric"? For example is it possible to define plane squares with non Euclidean metrics? And can we define right angles without the parallel postulate people simply ellide when operating with non Euclidean geometries? No. Then people complain that what they do gives the same answers. But that's only because they're using the same words yet asking different questions, with concepts defined in the Euclidean metric but operated on in some non Euclidean metric but not defined in that metric. 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. ~v~~
From: Eckard Blumschein on 24 Nov 2006 14:39 On 11/23/2006 6:55 PM, William Hughes wrote: > Eckard Blumschein wrote: >> >> Before I may decide I would like to clarify the meaning of the term set. >> Cantor's definition of a set has been confessed invalid at least since >> 1923 with no possiblity of correction in case of infinite sets. Why? >> >> Fraenkel confessed paradoxa due to this definition. The reason behind >> them is ambiguity. Cantor's definition claims to provide each singe >> element of the set as well as simultaneously the infinite set as an >> entity. So it is a chimera. >> > > Why? What is there about providing the infinite set as an entitiy > which precludes providing each single element of the set? We may suspect but not dirctly verfy the existence of each element. It can never be reached.
From: Virgil on 24 Nov 2006 15:07 In article <1164373730.838645.12940(a)l39g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > > > > > > The two statments > > > > > > > > > > > > i: P(n) is true for every n an element of N. > > > > > > ii: P(N) is true > > > > > > > > > > > > are not the same (trivial example P(x) is true iff x is an element > > > > > > of > > > > > > N). > > > > > > Induction can be used to prove statements of the form i. > > > > > > (eg all elements of N are finite). Induction cannot be used > > > > > > to prove statements of the form ii (e.g. N is finite). > > > > > > > > > > Here again your mathelogy comes to the surface. N is nothing but the > > > > > collection of all natural numbers. They count themselves. If all are > > > > > finite, then all are finite, > > > > > > > > Yes > > > > > > > > > i.e., then N is finite. > > > > > > > > No see above. The fact that all elements of N are finite > > > > does not mean that N is finite. > > > > > > It does. (The EIT proves it.) > > > > No it doesn't. Your claim is that assuming > > > > There exists an infinite set all of whose elements > > are finite > > > > leads to a contradiction. But you have yet > > to show a contradiction that doesn't require assuming > > > > There does not exist an infinite set all of whose elements > > are finite > > Please learn: I use the finiteness of *every* line to change the > quantifiers, because for linear and finite sets, this is unrefutably > true. And unrefuted every line has a finite number of elements. The > usual escape that every number is finite but that there are infinitely > many numbers (this paradigm of nonsensical thinking) does not help > here. In a line there are not infinitely many elements! Since in the diagonal of your EIT there ARE infinitely many elements, one must include consideration of infiniteness. Unless WM can refute "For every line there is a term in the diagonal not in that line", WM must concede that the diagonal is longer than each line. If WM then continues to assert that there is a line as long as the diagonal, he is asserting the truth of "P and not P".
From: Virgil on 24 Nov 2006 15:11 In article <1164373800.790031.60620(a)j44g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > > If there is no bijection with one line, then there must be an element > > > > > of the diagonal outside of every line. > > > > > > > > No. Not unless the one line > > > > contains every element from every line. > > > > > > This holds for every finite line. Every finite line contains every > > > element from every preceding line. There are only finite lines. > > > > > > I claim > > > > There does not exists a line L_1 such that L_1 contains every > > element from every line. > > > > You counter with > > > > For every line L_1, L_1 contains every element from every line > > preceding L_1. > > > > However, these two statments are not contradictory. > > > > "every element from every line" > > > > is not the same thing as > > > > "every element from every line preceding L_1" > > > > > > Yes it is true that > > > > For any element of the diagonal, d_nn, there exists a line > > L_2, such that L_2 contains d_nn. > > > > However, the statement > > > > There exists a line L_1, such that > > L_1 contains every element from every line > > preceding L_1. > > > > is not the same as the statement > > > > There exists a line L_1, such that > > L_1 contains every element from every line > > preceding L_2. > > > > So we cannot use > > > > There exists a line L_1, such that > > L_1 contains every element from every line > > preceding L_1. > > > > to show that > > > > There exists a line L_1, such that L_1 > > contains every element from every line > > For the lines, each of with has a finite number of elements, this *is > the same*. But in EIT, for every line L_1 there is a successor line containing an element not in L_1. > > If you have a linearly ordered set with a finite number of elements, > then there is a maximum. But this does not happen in EIT. > Therefore you cannot counter with the usual argument that the elements > of any line are all finite but that there are infinitely many of them. In EIT one can. It is your model which h disproves your own claims.
From: Virgil on 24 Nov 2006 15:15
In article <45672DF1.6080105(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 11/23/2006 10:47 PM, Virgil wrote: > > In article <4565D569.9080200(a)et.uni-magdeburg.de>, > > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > > > >> On 11/23/2006 9:16 AM, Virgil wrote: > >> > In article <456556CD.4030107(a)et.uni-magdeburg.de>, > >> > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > >> > > >> > > >> >> Virgil> While all sequences are sets ( as functions from N to some set > >> >> of > >> >> > values) not all sets are sequences. > >> >> > >> >> So you could be correct if referring to sets that are no sequences? > >> >> Look above: I referred to something going on forever. > > > > > > Which "forever"? In time, in space, or in something else. > > Neither nor. Counting is just a repetitious operation. > > > > > As neither time not space, in any physical sense, exists in mathematics, > > it must be in something, as yet unexplained, else. > > Not unexplained but best imagined via the application on time or space. Certainly unexplained by EB. And one can easily imagine it as, say, as the succession of points (n-1)/n in the rational open interval (0,1) which is "completed" in the rational interval [0,1]. |