Virgil
Posts:
4,483
Registered:
1/6/11
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Re: Cantor's first proof in DETAILS
Posted:
Dec 9, 2012 1:35 AM
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In article
<2e2a7c1e-973c-4b07-b8f1-cce24261a550@jl13g2000pbb.googlegroups.com>,
"Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> 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:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > 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.c
> > > > > > > > om>,
> > > > > > > > "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
I have looked at your reference to Euler and found nothing in it that in
any way contradicts the definition of continuity I expressed above.
So that, since I am not blind, Ross must be, to be able to se what is
not there.
>
> Regards,
>
> Ross Finlayson
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