Cantor Confusion
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From: Eckard Blumschein on 7 Dec 2006 06:46 On 12/7/2006 4:38 AM, Dik T. Winter wrote: > In article <virgil-6BADE6.20103906122006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > > In article <4576D4C7.4070207(a)et.uni-magdeburg.de>, > > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > ... > > > Dedekind as well as Cantor started from this idea. Cantor even > > > fabricated the notion cardinality in order to quantify the putative > > > difference in size. Considering the rationals a subset of the reals, > > > dull people conclude that there must be more reals than rationals. > > > > There certainly cannot be fewer, or even just as many, so what other > > option is open/ Logically there is the so called 4th option: neither more nor fewer not even equally many, just incomparable. The three relations belong to finite objects. Are there more integers than odd integers? No. See Galilei. > Why not just as many? Vanishing differences would be a good crutch for those who desparately depend on intuitive suggestions. I myself imagined the rationals so far like discrete in contrast to continuous uncountables. See a brandnew reply of mine below. In a lattice definition of < between sets there > are obviously fewer. In a cardinality definition there are not > obviously fewer, it could be just as many. It all depends on the > definitions of more and fewer and equally many.
From: Bob Kolker on 7 Dec 2006 06:48 mueckenh(a)rz.fh-augsburg.de wrote: > > > Those who pretend to be able of knowing every integer and, therefore, > to imagine the whole actually infinite set. If there were an integer I could not imagine, there would have to be a least such integer. Well, it isn't 0 or 1. So the least integer I could not image has a predecessor. But I can imagine it. I could also imagine adding one to it. Contradiction. Bob Kolker
From: mueckenh on 7 Dec 2006 06:50 cbrown(a)cbrownsystems.com schrieb: > So now you return to claiming that your mapping T /is/ a surjection > from edges onto paths. Perhaps there is some confusion regarding "what > is the domain/range of T?". > > To clarify, let e be any edge; say the first branch to the left in your > original diagram. According to your above statement, e is in the domain > of T. Which path is T(e)? You know that the two edges mapped on a path consist of shares of many edges. Why do you put your question? Why should I name a special edge? I have proved that two edges are collected in form of shares. > > I have proved it by rational relation and by a random mapping. > > You haven't proved it until you provide a surjective function T : > (edges -> paths). (And I am only interested in your "rational relation" > argument). It has been proved. See above. I do not see the necessity to do more than to show that there are enough shares. > > > > > > > > I add an appendix to one of my papers, where this is underlined I (here > > > > the arguing is based on nodes instead of edges, but that doesn't matter > > > 4> much): > > > > > > > > > > Your appendix fails to address the key question: what is the domain of > > > the function T? If e is an edge, what is the set of "shares of the > > > divided edge e"? > > > > What is your problem? The complete set of shares of one edge is the > > full edge. > > So you have a bijection S : (edges -> complete sets of shares of an > edge). But given an edge e, what are the /elements/ of the set of > shares, S(e)? How many elements does S(e) have? 1? 32? A countable > number? An uncountable number? Everything in this tree is countable. That is why I devised it. By the fact that the paths separated up to level n are finitely many, we see that the set of all paths remains countable. lim[n-->oo] 2^n is the cardinal number of a countable set. This continuity of he tree is the clue which cannot be circumvented. All you may argue is that such a tree cannot exist. Regards, WM
From: Bob Kolker on 7 Dec 2006 06:52 mueckenh(a)rz.fh-augsburg.de wrote: > > We know that every line consists of a finite number of elements. That > is the point! That shows that for each line the quantifier can be > reversed. In Euclidean space between any two points on a line is a third point on the line. Which implies that the set of points on a line is infinite. In Euclidean space the line is dense in points. In real life you are just dense. Bob Kolker
From: mueckenh on 7 Dec 2006 06:55
Virgil schrieb: > In article <4576BC30.2000101(a)et.uni-magdeburg.de>, > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > > > It is evident that EB understands "infinity" even less that Cantor. That is a hard attack! To understand the infinite even less than Cantor??? I don't know anybody who understood the infinite better than Cantor, not even approximately as well. Regards, WM |