Cantor Confusion
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From: Virgil on 29 Oct 2006 12:49 In article <1162135345.044588.103450(a)k70g2000cwa.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. It is. however, quite sufficient that the constructed number be different from the nth listed number by at least 1/10^n > > How many different representatives can be chosen in Cantor's list for > one equivalence class? One is enough. > > In Cantor's list there are omega irrational numbers. It isn't Cantor's list, it is any list of reals presented to Cantor. > Cantor uses it. Otherwise everything was finite. Only in WM's world is omega itself needed to make things not finite. Everywhere else, one can get by with using only members of omega. > > > > > > > You will need it in order to construct a real number and its > > > > > decimal > > > > > representation for a Cantor list. > > > > > > > > No. By the theory, each decimal number is a representative of an > > > > equivalence class of sequences of rational numbers. By the > > > > construction > > > > we get another decimal number that is also a representative of an > > > > equivalence class of sequences of rational numbers. No omega is > > > > needed. > > > > > > Why then do you think omega is needed at all in mathematics? > > > > Because it comes in handy on many occasions. What *is* needed is the > > axiom of infinity, because that guarantees that you can even talk about > > infinite sequences. > > Omega is but a convenient name of the set which exists according to the > axiom of infinity. The axiom of infinity merely says at least one set, and in no way requires only one set satisfying its properties. In fact, if any one such set exists, there must be many others as well. > > > (And ultimately about limits.) Without that axiom > > there is no way to prove that infinite sets do or do not exist. > > Correct. And with this axiom the first infinite entity which is proven > is omega. And then the digits of any irrational are enumerated. Their > number is omega. Their number is the cardinality of omega. "WMueckenheim" really should learn the difference between ordinal numbers and their cardinality. > > > > I would not know how to do that. Perhaps that is possible, but in that > > case you would need to reformulate the limit concept. > > Why then do you try to dispute that omega does play a role in > constructing he reals? The axiom of infinity essentially says: There is > the set omega. It says both more and less than that. > > > > > Omega is introduced only by the axiom of infinity. > > > > No. It is *defined* using properties obtained through the use of that > > axiom. In some fields of mathematics you do not need omega at all, but > > only the axiom of infinity (analysis, and I think also algebra, number > > theory, and a host of other fields). > > > The axiom of infinity states that there is a set with such and such > properties... And this set is omega. NO! Those properties are held by every limit ordinal, of which there are many besides omega.
From: Virgil on 29 Oct 2006 12:58 In article <1162135573.031943.175480(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1161861368.657796.161130(a)k70g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > > As you allow constructions by lists, and as there are uncountably many > > > > lists, there are also uncountably many WM-constructible numbers. > > > > > > Where are those list? How are uncountably many different lists are > > > constructed by countably many words? > > > > Any infinite sequence of decimal digits is a list. > > Correct. And this sequence has to be defined somehow. But there are > only countably many definitions. But a listed real does not have to have every decimal digit knowable, it only have to have the as many digits as its position in the list knowable. > Forget the Turing machines. The diagonal of the list of all finite > sequences (words) of a finite alphabet is a finite sequence because the > diagonal cannot have more places than the words in the list. But if the nth word has, say, n+1 places, then one has a list of finite sequences which are unbounded. There is a vital difference between a list of merely finite sequences and a list of uniformly bounded sequences ( all being bounded by the sasme bound) which allow the former to have sequences of arbitrarily large, but still finite, lengths.
From: Virgil on 29 Oct 2006 13:03 In article <1162137834.999423.231310(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > All 1's can be indexed in 0.1 and in 0.11, and so on. This does *not* > show that all ones in 0.111... can be indexed. Each 1 in 0.111... can be indexed. Whether "all" can be depends on whether one means simultaneously or sequentially.
From: Virgil on 29 Oct 2006 13:10 In article <1162138418.513488.262830(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > There is indeed a list of all possible finite constructions or > > definitions. Call this list A. However, this list must contain things > > that look > > like definitions but are not because the method given to produce > > a number does not halt. We cannot get a diagonal number from A, > > because some of the members of A do not give numbers. > > So there is no contradiction. > > A contains all finite words (construction fromulas, theorems). The > diagonal (cannot be longer than the lines and hence) is a finite word > too. That is enough to obtain a contradiction. WM is making an essential error here by assuming that a list of finite strings is necessarily bounded. If one considers the nth string to be of length n, then every string is finite but the diagonal formed of last letters need not be. > > Funny, but not sufficient to confuse a clear thinking mind > sufficiently. WM does not have a c lear thinking mind or he would not have made the fundamental error which I have pointed out above.
From: Virgil on 29 Oct 2006 13:15
In article <1162138639.029269.192410(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Sebastian Holzmann schrieb: > > > Virgil <virgil(a)comcast.net> wrote: > > > In article <1161883732.413718.244570(a)k70g2000cwa.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > >> What you propose, namely the infinity of ZF without the axiom INF would > > >> not be an advance. But meanwhile you may have recognized that your > > >> assertion (ZF even without INF is not finite) is false. > > > > > > It is, however, quite true that ZF without INF need not be finite. > > > > It is, more than that, quite true that ZF without INF _is_ infinite > > Are you really sure? > > > (the axiom schema of separation alone provides infinitely many axioms). > > Are you really sure? > > > The point is: ZF without INF does not prohibit the existence of infinite > > sets, nor does it force them to exist. > > It prohibits to speak of infinite sets and to recognize such sets. So > one cannot be sure, but you are? WM may want to prohibit speaking of infinite sets, but nothing in ZF absent INF itself requires such a prohibition in either speaking of or recognizing such sets. What WM may want he does not always get. |