
Re: Cantor's first proof in DETAILS
Posted:
Dec 7, 2012 12:02 AM


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
* ''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 (0,½).
http://en.wikipedia.org/wiki/Oliver_Heaviside
Looks good to me.
Not so, Hancher: not so.
Regards,
Ross Finlayson

