Date: Dec 6, 2012 12:05 AM
Author: Virgil
Subject: Re: Cantor's first proof in DETAILS

In article 
<5312c40d-7490-4838-b49c-573a9f2e136e@i2g2000pbi.googlegroups.com>,
"Ross A. Finlayson" <ross.finlayson@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.

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.
--