Cantor Confusion
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From: Virgil on 31 Oct 2006 17:03 In article <1162299756.756951.78990(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > For the diagonal number of Cantor's list it is nnot sufficient to come > > > arbitrarily close to a number which is different from any list number. > > > > Sorry, but numbers are fixed and not variable. The sentence "number ... > > to come arbitrarily close" is nonsense. Numbers do not come arbitrarily > > close to each other, it is sequences that can come arbitrarily close to > > each other. > > Irrational numbers have no last digit. Irrational numbers are not the same as decimal representations of these numbers. An irrational can exist without having a known decimal representation in all senses of "exist" except that of having a decimal representation. > Therefore, with a sequence of > digits like the diagonal number is, one can never have a completed > number but only come as close as possible to any number --- or avoid to > do so. For the Cantor diagonal proof, that is quite sufficient. > > > > > > > I do not understand why you want to tie that list in with the > > > > > > definition of the reals? > > > > > > > > > > > Because the real numbers are used in that list. > > > > > > > > Representatives of the equivalence classes that actually are the real > > > > numbers. > > > > > > How many different representatives can be chosen in Cantor's list for > > > one equivalence class? > > > > As many as you want. > > Not so. Of course we talk about a fixed base like 10. > > > But if you want to make the diagonal argument to > > work it makes sense to select particular representatives for each > > equivalence class. So selecting for some a base pi notation and for > > others a base (1 + i) notation is not likely to work. But again (and > > we have discussed this before), in my opinion, Cantor's original > > diagonal proof was not about the reals at all. It was about the > > theorem he states in the first part of the paragraph for which you > > wish only to quote the second part: "there are sets that have larger > > cardinality than the naturals" and uses the set of infinite sequences > > of two objects, called 'm' and 'w'. > > Let us drop that. > > How many different representatives can be chosen in Cantor's list > (which needs a fixed base) for > one equivalence class? Only one need be chosen for each listed number, in any base 4 or larger, and the nth listed need only be known at the nth place.
From: Virgil on 31 Oct 2006 17:10 In article <1162300306.820436.173330(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Again, this statement is completely in contradiction with the axiom of > > infinity. If you want to talk about mathematics without that axiom, > > please do so, but do not suggest that you are talking about mathematics > > with that axio. > > I use the simple fact that aleph 0 or omega is a number which is larger > than any natural number. > A list of all finite words has no word which has omega letters. > Therefore it is impossible to exchange omega letters in a diagonal. A list of finite words may have the nth word contain at least n characters, in which case WM is, as usual< wrong again. The > diaogonal cannot be roader than the list. But a list of finite words need not have bounded "broadness". >The length of the diagonal is > the minimum of width and length. This knowledge is prior to your > axioms. Neither "width" not "length" need be bounded. > A matrix with width A and length B has a diagonal which has min(A,B) > elements. If your axiom contradicts this, then the axiom contradicts > mathematics and should be abolished. A list with endlessly many strings of endlessly increasing lengths trivially and obviously does not contradict anything except WMspeak.
From: Virgil on 31 Oct 2006 17:15 In article <1162300936.776151.45540(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > Attention: 0.111... has only finite initial segments - and nothing > > > more. Only those can be indexed. > > > > You keep stating that, without proof. That it has only finite initial > > segments, I agree. Not the remainder. > > It is you who denies proof. Give an example of a digit which does not > belong to a finite sequence. If you cannot do so, then every digit > belongs to a finite sequence. Give an example of a digit that does not belong to an infinite sequence, WM. If you cannot do so, then every digit belongs to am infinite sequence. > > > You need no transfinity to show that lim [n-->oo] 1/2^n = 0 and that > > > lim [n-->oo] (1 - 1/2^n) / (1 - 1/2) = 2. Because that was known before > > > transfinity was introduced. > > > > But I never argued that. > > But it was argued in connection with my infinite binary tree. It's relevance to binary trees was argued, not its validity as a limit. And we still say that it is irrelevant. > Anyhow, in the binary tree there is no transfinity required. In the infinite binary tree, uncountably many paths will be made out of countably many edges.
From: Virgil on 31 Oct 2006 17:17 In article <1162301133.288356.234700(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1162139168.281239.183260(a)b28g2000cwb.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > You need not tell that to me. You should tell that to Wolfgang > > > > Mueckenheim who insists that sqrt(2) does not exist because it is > > > > impossible to know all the decimals in its decimal expansion. > > > > > > It does not exist as a number. It has no b-adic representation. It > > > exists as a geometric entity. It is an idea. > > > > It exists as the continued fraction [1,2,2,2,2,...]. It also exists > > as 1 in the base sqrt(2) notation. > > And it exists in many other disguising. But it does not exist as a > number which can be put in trichotomy with all other numbers. No number can be "put in trichotomy" with all other numbers.
From: Virgil on 31 Oct 2006 17:23
In article <1162302941.268560.153350(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Every mathematician whose mind has not yet suffered the drill of set > > > theory you probably were exposed to knows that my argument is correct. > > > > How could they know it is correct regarding set theory if they don't > > know set theory? > > It is enough to know that it is correct. > > Every path is related to two edges which are not related to any other > path. And that is easy enough to see. In an infinite binary tree, there are infinitely many paths through each edge and through each node of that tree. So how does WM isolate one of those paths from all the others to be related to only that one edge and no other edges? > > If set theory denies that, then set theory is wrong. If WM claims that then WM is wrong! > > Regards, WM |