
Re: Irrefutable proofs that both Dedekind and Cauchy did not produce any valid construction of the mythical "real" number
Posted:
Oct 4, 2017 7:47 AM


Den onsdag 4 oktober 2017 kl. 12:48:01 UTC+2 skrev genm...@gmail.com: > On Tuesday, 3 October 2017 23:10:37 UTC4, Zelos Malum wrote: > > Den tisdag 3 oktober 2017 kl. 19:16:15 UTC+2 skrev John Gabriel: > > > On Tuesday, 3 October 2017 12:32:26 UTC4, Markus Klyver wrote: > > > > Den fredag 29 september 2017 kl. 14:06:42 UTC+2 skrev John Gabriel: > > > > > https://drive.google.com/open?id=0BmOEooW03iLSTROakNyVXlQUEU > > > > > > > > > > Comments are unwelcome and will be ignored. > > > > > > > > > > Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics. > > > > > > > > > > gilstrang@gmail.com (MIT) > > > > > huizenga@psu.edu (HARVARD) > > > > > andersk@mit.edu (MIT) > > > > > david.ullrich@math.okstate.edu (David Ullrich) > > > > > djoyce@clarku.edu > > > > > markcc@gmail.com > > > > > > > > Those are not Dedekind cuts. > > > > > > Of course they are monkey! > > > > Of course they aren't, because as said, we can show, trivially, it is not using even the most general definition!. > > > Hey Stupid. Even "Me" has finally understood that my definition is a D. Cut. Ask him to explain to you moron! > > L={1 < x < pi} and R={pi < x < 4} where x \in Q > > is a valid D Cut. > > You can choose any other elements m and n such that m < pi < n and it will conform as follows: > > L={m < x < pi} and R={pi < x < n} where x \in Q > > END OF DISCUSSION.
And none of those will be a cut because
(L^u)^l=(pi,oo)^l)=(oo,pi) \neq L
So L is not a lower cut as for it to be, we must have (L^u)^l=L, which it clearly doesn't.

