
Re: Cantor's first proof in DETAILS
Posted:
Dec 9, 2012 1:07 AM


On Dec 8, 9:56 pm, Virgil <vir...@ligriv.com> wrote: > In article > <893b913050b04ffeaff1313d15bfc...@r10g2000pbd.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > > > On Dec 6, 9:24 pm, Virgil <vir...@ligriv.com> wrote: > > > In article > > > <978290859b08479cb693fde704b4f...@nl3g2000pbc.googlegroups.com>, > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > On Dec 5, 9:05 pm, Virgil <vir...@ligriv.com> wrote: > > > > > In article > > > > > <5312c40d74904838b49c573a9f2e1...@i2g2000pbi.googlegroups.com>, > > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > On Dec 4, 1:15 pm, Virgil <vir...@ligriv.com> wrote: > > > > > > > In article > > > > > > > <42cabcca089d456f837ac1d789bda...@jj5g2000pbc.googlegroups.com>, > > > > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > > > > > And Heaviside's step is continuous, > > > > > > > > now. For that matter it's a real function. > > > > > > > > I already said that the step function is a real function, I only > > > > > > > objected to your claim that it was a continuous function. > > > > > > >  > > > > > > > Heh, then you said it wasn't, quite vociferously > > > > > > I objected to it being called continuous. possibly vociferously, but > > > > > your claim that it was continuous deserved vociferous objection. > > > > > > : you were wrong > > > > > Don't you wish! > > > > > > , and > > > > > > > within the course of a few posts wrote totally opposite things. Your > > > > > > memory fails and that's generous, not to mention you appear unable to > > > > > > read three posts back. > > > > > > > And everybody sees that. > > > > > > > Then as noted Heaviside's step, a real function, can be simply drawn > > > > > > classically: without lifting the pencil. > > > > > > Not outside of Rossiana. > > > > > >http://en.wikipedia.org/wiki/Heaviside_step_function > > > > > The Heaviside step function, or the unit step function, usually denoted > > > > > by H (but sometimes u or ?), is a discontinuous function whose value is > > > > > zero for negative argument and one for positive argument. It seldom > > > > > matters what value is used for H(0), since H is mostly used as a > > > > > distribution. > > > > > > It's continuous that way. > > > > > > Not according to Wiki, whom EVERONE here, except possibly WM, trusts far > > > > > more than they trust Ross. > > > > > > See that phase "discontinuous function"? > > > > > > Or maybe your as blind as you are thick. > > > > >  > > > > >http://en.wikipedia.org/wiki/Heaviside_step_function > > > > From wiki: > > > The Heaviside step function, or the unit step function, usually denoted > > > by H (but sometimes u or ?), is a DISCONTINUOUS function whose value is > > > zero for negative argument and one for positive argument. It seldom > > > matters what value is used for H(0), since H is mostly used as a > > > distribution. Some common choices can be seen below. > > > > > * ''H''(0) can take the values zero through one as a removal of the > > > > point discontinuity, preserving and connecting the neighborhoods of > > > > the limits from the right and left, and preserving rotational symmetry > > > > about (0, ). > > > > Except that the value of the Heaviside step function AT zero cannot be > > > chosen so as to make its limit as x increases towards zero though > > > negative values become equal to the limit as x decreases through > > > positive values towards zero, which would be necessary to make the > > > function continuous at zero according to every standard definition of > > > continuity. > > > > One wonders whether Ross knows what continuity reall is all about. > > > > >http://en.wikipedia.org/wiki/Oliver_Heaviside > > > > > Looks good to me. > > > > Try getting your eyes tested, and if that doesn't clear things up, get > > > your brain tested. > > >  > > > You describe a particularly strong condition of continuity, there are > > weaker ones, that leave the classical notion that if you can draw it > > in one noncrossing stroke it's continuous. Heaviside's step, > > connected, is in a sense continuous. > > Then what is your example of a discontinuity? > > My "strict" one is the only level of "condition of continuity" that I > have found in any elementary calculus text that I have, or in any > advanced text, for that matter. For example, none of the several texts > on calculus written by Tom Apostol, accept anything less than the sort > of definition I would require: > for a real function, f, defined on an open set containing 0 > to be continuous at zero, it is necessary that > (1) the function must have a value, f(0), at zero. > (2) the function must have a finite limit as x increases to 0. > (2) the function must have a finite limit as x decreases to 0. > (4) both limits must equal f(0). > > I Googled for "continuity at a point" and found over 20 million sites. > Those I sampled all agreed with me And Apostol, and I dare sat that none > of them disagree to the point of calling any function which has value > zero at every negative argument and value one at every positive argument > can be anything but definitely DIScontinuous at zero. > > It you ever find one of those 20 million sites, not written by > yourself,which says otherwise, please post its URL in refutation. > Otherwise stop lying! > 
Euler: and he was blind.
http://en.wikipedia.org/wiki/Leonhard_Euler
Regards,
Ross Finlayson

