Date: Dec 4, 2012 4:47 PM
Author: Virgil
Subject: Re: Ross' Delusions re his EF.

In article 
<fca2fc08-f251-436c-9294-de17b9b43872@g7g2000pbi.googlegroups.com>,
"Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:


> Heh, got your goat, there, eh, Hancher. Huff and puff, you. Quit
> with your ad hominem attacks.


What you have called my ad hominem, others recognise as mathematics.
And much of what you have claimed as mathematics is nonsense.

> And Heaviside's step is continuous, now.

While the Heaviside's step function may be represented as the limit of
a sequence of continuous Functions, in order to prove its continuity by
such a limit, it would have to be a UNIFORM limit of those continuous
functions, which it never can be.

Not that for the Heaviside step function H(x),
lim_(x -> 0) H(|x|) = 1
lim_(x -> 0) H(-|x|) = -1
H(0) = 0

But for such a function to be continuous at 0 one needs
lim_(x -> 0) H(|x|) = lim_(x -> 0) H(-|x|) = H(0)

Or is the requirement of continuity at x = 0 different in Rossiana?
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