On Thursday, June 27, 2013 11:54:40 PM UTC-7, muec...@rz.fh-augsburg.de wrote: > On Thursday, 27 June 2013 20:21:47 UTC+2, Virgil wrote: > > > In article <firstname.lastname@example.org>, muec...@rz.fh-augsburg.de wrote: > On Wednesday, 26 June 2013 17:48:25 UTC+2, dull...@sprynet.com wrote: > > > > > > > >For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3,...,q_kn. >This theorem it is not less important than Cantor's theorem: For all >k: d =/= q_k. Both theorems contradict each other. > > > > > > > Erm, no they don't. If your "for every n there exists k such that" was > > instead "there exists k such that for every n" then they would contradict > > each other. > > > > >> Try to find a d_n that, together with all its predecessors d_j for 0 < j < n, is not in a q_k. Then you may boast "erm". Otherwise learn that, in mathematics, there is no continuation of d beyond every d_n. > > > > > But there is a continuation beyond ANY d_n, which is > > > > which is realized by digits d_n with larger n. You cannot excape. > > > > "There exists k such that for every n: a_kn = d_n" is a contradiction for dummies. > So, you can prove that a collection of finite decimal sequences contains an infinite one.
IF: An infinite collection of finite decimal sequences, when written in one line, makes an actually infinite collection, THEN: an infinite collection of finite decimal sequences written in different lines does the same. (Unioning over subsets cannot not extend a set.)
> > "For every n there exists k such that a_kn = d_n" is a contradiction for mathematicians who take the trouble to try to escape it - without success, of course. > That doesn't imply, "There exist m e N, such that a_m = d."
That is your uneducated or miseducated guess. Prove it by finding any digit of d that is not, together will all its predecessors, in an a_m. You will fail. Therefore your statement is wrong. (It has been transferred without sufficient reason from non-linear sets like knifes and forks. But in this case forking is invalid, as invalid as forcing in general). If there is a set of FISONs that contains every natural number n and all its predecessors, then this set cntains |N. nN? Is there something in |N that is outside of all its FISONs?