Virgil
Posts:
8,833
Registered:
1/6/11


Re: Cantor's first proof in DETAILS
Posted:
Dec 9, 2012 12:56 AM


In article <893b913050b04ffeaff1313d15bfceb4@r10g2000pbd.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson@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! 

