Courage?
in [Logic]
|
Prev: arithmetic in ZF
Next: Derivations
From: Eckard Blumschein on 13 Apr 2005 13:12 On 4/13/2005 3:11 PM, Matt Gutting wrote: > Eckard Blumschein wrote: >> On 4/12/2005 11:46 PM, Will Twentyman wrote: >> >> >>>>Cantor was mislead by his intuition. >>>>I do not attribute the difference between countable and non-countable to >>>>the size of the both infinite sets. >>> >>>What do you view the difference between them to be? >> >> >> The difference resides in the property of each single real number >> itself. Cantor assumed his list represents all real numbers. Actually, >> nobody can provide any list of real numbers, not even two subsequent of >> them can be named. >> > > Cantor assumed such a list could be made, in order to prove that "nobody > can provide any list of real numbers". I could not find this sentence in his original papers. > You appear to agree, then, with the conclusion of his proof. I disagree with the conclusion that there are real numbers in excess of his allegedly complete list. Actually, his list was not complete because nobody can really write such list. >> This was your statement. Where is your evidence for it or at least some >> justification? Simply tell me the successor of pi. > > What do countability and successorship have to do with each other? Any countable set is bijective to the natural numbers. Tell me any natural number, and I will tell you its successor. Please tell me the successor of sqrt(2). > The rationals > can be ordered in such a way that each one has a defined successor. That's it. > They can > also be ordered in such a way that none has a defined successor. This ordering > has nothing to do with whether the set is countable or not. This is a typical fallacy. > > Further, you are asking Will to tell you the successor of pi. This request seems > to indicate that you think he believes the reals to be countable. No. I just argue that reals deserve to be treated differently from ordinary numbers. Eckard
From: Chris Menzel on 13 Apr 2005 12:36 On Wed, 13 Apr 2005 15:40:21 +0200, David Kastrup <dak(a)gnu.org> said: > Matt Gutting <tchrmatt(a)yahoo.com> writes: > >> Eckard Blumschein wrote: >> >>> This was your statement. Where is your evidence for it or at least >>> some justification? Simply tell me the successor of pi. >> >> What do countability and successorship have to do with each other? > > Uh, everything? It seems to me that Gutting's point is only that the uncountability of the reals is independent of whether or not they can be well-ordered. Blumschein seems to be confused on this point. >> The rationals can be ordered in such a way that each one has a >> defined successor. They can also be ordered in such a way that none >> has a defined successor. This ordering has nothing to do with >> whether the set is countable or not. > > It pretty much has everything to do with it. If you can make a subset > of the rationals obey the Peano axioms, then this subset is in > one-to-one correspondence with the natural numbers. If you can make > the entire rationals obey the Peano axioms by choosing a different > successor relation), then the rationals are in one-to-one > correspondence with the naturals. I'm not sure I understand the argument. You can, for example, make the set of countable ordinals obey the Peano axioms by choosing a different successor relation, but of course we don't conclude from that that the set of countable ordinals is in one-to-one correspondence with the naturals. So it is not simply in virtue of their serving as a model of PA that we conclude the rationals are countable. But I don't see what other facts you are appealing to here than that. Sorry if I've missed the point. Chris Menzel
From: Will Twentyman on 13 Apr 2005 13:04 David Kastrup wrote: > Will Twentyman <wtwentyman(a)read.my.sig> writes: > > >>Definition: a set A is said to be countable if it is finite or if >>there exists a bijection between A and N. >> >>Theorem: R is not countable. >> >>Proof: assume R is countable, then there exists a bijection f mapping >>N to R. Consider a real number r specified as follows: if the nth >>decimal digit of f(n) is 5, the nth decimal digit of r is 1, otherwise >>the nth decimal digit of r is 5. r cannot be in the image of f, >>because for all n, f(n) differs from r at the nth decimal digit. This >>is a contradiction, therefor R is not countable. >> >>The successor of pi could be pi+1, but that is not relevant to the >>reals being uncountable. > > > Of course it is. > > If I have a set and a mapping with the following properties: > > The set contains a dedicated member. > > For each of its members, the set contains the mapping of the member. > > Different members have different mappings. > > The dedicated member is not the mapping of any other member. > > A set containing the dedicated member, and containing for each of its > members also its mapping, contains all members of the given set. > > So if I can find _any_ successor relation with the given properties, > this is completely relevant to the reals being countable, since it > puts them into a one-to-one correspondence with the naturals. Ok, that I can agree with, since it would provide the mapping between R and N. It was not clear to me that Eckard intended for a successor function to have all those properties, however. -- Will Twentyman email: wtwentyman at copper dot net
From: Will Twentyman on 13 Apr 2005 13:06 worldsofsolution(a)yahoo.com wrote: > There is something I've been wondering about cantor's proof: the > decimal number generated to prove the contradiction, was taken to be a > real number. There is a tacit assumption that all decimal numbers > represent a real number. Does that not require a proof? It depends somewhat on your definition of a real number. Often I use a "all decimals" as my definition of the real numbers. -- Will Twentyman email: wtwentyman at copper dot net
From: David Kastrup on 13 Apr 2005 13:08
Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > On 4/13/2005 3:11 PM, Matt Gutting wrote: > >> You appear to agree, then, with the conclusion of his proof. > > I disagree with the conclusion that there are real numbers in excess > of his allegedly complete list. Actually, his list was not complete > because nobody can really write such list. And exactly that, and nothing else, is what he has set out and accomplished to prove. Really, you have no clue what you are babbling about. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |