On Tue, 25 Jun 2013 13:43:36 -0700 (PDT), email@example.com wrote:
>On Tuesday, 25 June 2013 17:07:36 UTC+2, dull...@sprynet.com wrote: >>> There are not uncountably many reals, neither in the natural order nor in the well-order. > > >> That's very funny. > >That's easily provable: > >Consider a Cantor-list that contains a complete sequence (q_k) of all >rational numbers q_k. The first n digits of the anti-diagonal d are >d_1, d_2, d_3, ..., d_n. It can be shown *for every n* that the Cantor- >list beyond line n contains infinitely many rational numbers q_k that >have the same sequence of first n digits as the anti-diagonal d. > >Proof: There are infinitely many rationals q_k with this property. All >are in the list by definition. At most n of them are in the first n >lines of the list. Infinitely many must exist in the remaining part of >the list. So we have obtained: > >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.
Don't feel bad about this. Confusing the order of quantifiers that way is very common among people who haven't learned to reason carefully.
>with the result that finished >infinity as presumed for transfinite set theory is not a valid >mathematical notion. > >Regards, WM