Date: Dec 7, 2012 12:02 AM
Subject: Re: Cantor's first proof in DETAILS
On Dec 5, 9:05 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> > On Dec 4, 1:15 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <42cabcca-089d-456f-837a-c1d789bda...@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.
> 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
> 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.
* ''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
Looks good to me.
Not so, Hancher: not so.