
Re: Joel David Hamkins on definable real numbers in analysis
Posted:
Jun 28, 2013 4:34 AM


On Thursday, June 27, 2013 11:54:40 PM UTC7, muec...@rz.fhaugsburg.de wrote: > On Thursday, 27 June 2013 20:21:47 UTC+2, Virgil wrote: > > > In article <28b6c17f095046dea12570baec6a079f@googlegroups.com>, muec...@rz.fhaugsburg.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 nonlinear 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?
Regards, WM

