Date: Dec 9, 2012 1:07 AM
Author: ross.finlayson@gmail.com
Subject: Re: Cantor's first proof in DETAILS

On Dec 8, 9:56 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <893b9130-50b0-4ffe-aff1-313d15bfc...@r10g2000pbd.googlegroups.com>,
>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
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> > On Dec 6, 9:24 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <97829085-9b08-479c-b693-fde704b4f...@nl3g2000pbc.googlegroups.com>,
> > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:

>
> > > > On Dec 5, 9:05 pm, Virgil <vir...@ligriv.com> wrote:
> > > > > In article
> > > > > <5312c40d-7490-4838-b49c-573a9f2e1...@i2g2000pbi.googlegroups.com>,
> > > > > "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.
>
> > > > >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 non-crossing 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!
> --


Euler: and he was blind.

http://en.wikipedia.org/wiki/Leonhard_Euler

Regards,

Ross Finlayson