> On 25 Jul 2006 15:01:29 -0700, Gc <Gcut667@hotmail.com> said: > > > > Chris Menzel kirjoitti: > > > >> On 25 Jul 2006 14:13:23 -0700, Gc <Gcut667@hotmail.com> said: > >> > Gc kirjoitti: > >> > > >> >> Aatu Koskensilta kirjoitti: > >> >> > Well, there isn't. No countable model contains all the reals. > >> >> > >> >> How can the countable model then satisfy all the theorems about the > >> >> reals? > >> > > >> > And especially the theorem "There is uncountably many reals."! > >> > >> What that theorem really says is that there is no function from the set > >> of natural numbers onto the set of real numbers. In a countable model > >> in which that statement is true, all functions from the set playing the > >> role of N onto the set playing the role of R have simply been removed. > >> So, in the model, there is no function from the former onto the latter, > >> i.e., in the model, the sentence "there are uncountably many reals" is > >> true. > > > > Let`s call elements playing reals in countable model kvasi-reals. > > Ok, so you are reasoning in ZF about models of ZF now, right? I.e., you > are saying, in ZF: let M be a countable model of ZF (axioms being coded > as sets in some standard fashion, and truth of a (coded) sentence in a > model being understood in the usual way), let R_M be the set in M that > plays the role of the real numbers in M, and call the elements of R_M > "kvasi-reals_M". So far so good.
Yes, that was how I thought it.
> > From the axioms of set theory follows that there is no certain > > bijection, > > Well, in ZF we'd be able to prove that there is no bijection from the > set N_M playing the role of the natural numbers in M onto R_M, but we'd > also be able to prove that there *is* such a bijection outside M > (trivially, given the assumption that M is countable (I guess we'd need > that it's hereditarily countable)).
> > and we get "There are uncountably many kvasi-reals" by Dedekind`s > > definition > > Nope, we don't. In M, being a model of ZF, the sentence "there are > uncountably many reals" will be true because there will be no function > in M from N_M onto R_M. But we will be able to prove "there are > countably many kvasi-reals_M" in ZF. > > > and the law of excluded middle. BUT there really are only countably > > many kvasi-reals. > > Yeah, which we knew at the outset.