In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 10 Apr., 14:50, Dan <dan.ms.ch...@gmail.com> wrote: > > > That situation is better described as: > > > "Therefore we have a contradiction". > > > > We have a contradiction only if we believe with religious fervor that > > "it should" be on the list. > > No, we can prove that for all n: (d_1, ..., d_n) is infinitely often > repeated in the sequence of all rational numbers - with mathematical > precision. What do you not understand with this result?
Two things: Where is your proof of it. Why you think it implies anything else. > > > > Your conclusion is wrong. Something is not in the list for some k and > > > for all n in the list: > > > > It always fascinating how one can manage to string up words together > > in sentences that are grammatically correct but don't hold a shred of > > meaning, semantically speaking . > > Yes I know.
Then why do you keep ding it?
> Many set theorists do so when claiming that an irrational > number in decimal representation "is more" than all its finite initial > segments
It is certainly different from all its finite initial segments, and, at least in the sense of being longer is certainly "more".
And if all those finite initial segments are truncations, rather than roundoffs, then for each positive irrational it is more than each and every one of those finite initial segments.
WM is again ignoring the fact that, for a strictly increasing infinite sequence of any sort, the limit, of one exists, cannot be a member of the sequence. One would think that something so obvious would hold even in Wolkenmuekenheim.
> since this "more" is required to argue that d "is more" than > all its digits d_n.
It is more than any FIS, by the property that it is the limit of a sequence of FISs.
> Or do you think
We do. WM doesn't!
> If you tend to think so, remember that *for all* Cantor-arguments the > digit substitution a_nn --> d_n takes place at a finite index n. All > these substitutions belong to a finite initial segment of lines 1 to n > of the list, namely to the decimal representation of a terminating > decimal d_1, d_2, d_3, ..., d_n. > > The "beyond" is not included in Cantor's argument.
What IS included is that, by construction, the anti-diagonal differs from the nth listed entry at its nth position, and thus cannot be equal to any entry.
And nothing that WM has said revokes that simple truth.
> This easily > explained by the observation that *for all n* we have 1/n =/= 2/n, but > *n the limit* the inequality fails: lim 1/n = lim 2/n for n --> oo. So > even if the "limit beyond all finite n" is considered, Cantor is not > saved.
That different sequences may have the same limit does not counter Cantor. --