In article <b6b07da2-13e2-4172-91ab-d2cf297c1522@y11g2000yqm.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 15 Jun., 21:03, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > >On 15 Jun., 18:46, stevendaryl3...@yahoo.com (Daryl McCullough) wrote: > > >> I'm not sure what you are saying. The fact is, we can prove > > >> that for every real r_n on the list, d is not equal to r_n. > > > > >Of course. Every real r_n belongs to a finite initial segment of the > > >list. > > >That does not yield any result about the whole list > > > > On the contrary, the definition of "d is on the list" > > is that "there exists a natural number n such that r_n = d". > > We proved "forall n, r_n is not equal to d". So that > > means "there does not exist a natural number n such that r_n = d", > > so that means "d is not on the list". > > > > We have thus proved something about the whole list. > > > > >> That means that d is not on the list. There is no extrapolation > > >> involved. > > > > >Look here: We can prove for any finite segment > > >{2, 4, 6, ..., 2n} > > >of the ordered set of all positive even numbers that its cardinal > > >number is surpassed by some elements of the set. > > > > >Nevertheless this appears not be a proof that the cardinal number of > > >the whole set is less than some elements of the set. > > > > So there we have an example of an illegitimate extrapolation. > > If you prove Phi(n) for an arbitrary natural number n, then you are allowed > > to conclude: > > > > forall natural numbers n, Phi(n). > > > > So you can conclude: > > > > forall natural numbers n > 0, the set of all even numbers less than or > > equal to 2n has a cardinality less than 2n. > > > > That's true. That's a legitimate proof. On the other hand, it is not > > legitimate to conclude: > > > > The set of all even numbers has a cardinality that is less than > > some even number. > > > > That's an unwarranted extrapolation. > > > > So there are legitimate proofs, and there are bogus proofs. > > And Cantor's proof shows something for every natural number: Every > line numerated with a natural number from 1 to n does not contain the > diagonal.
Not just from 1 TO n, but from 1 PAST every n.
> But to conclude that the whole set, i.e., the infinite list, does not > conclude the diagonal is a bogus conclusion.
Is that "conclude" supposed to b "contain"? I will presume so.
So that WM regards as sometimes valid a statement that what is true for each member of a set need not be true for every member of that set?
