Date: Dec 8, 2012 11:31 PM
Subject: Re: Cantor's first proof in DETAILS
On Dec 6, 9:24 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> "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.
> > > --
> 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
> One wonders whether Ross knows what continuity reall is all about.
> > 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 I wonder is this: where did Heavside himself draw it as a
step. Because, it seems he'd be happy enough to to draw it, without
the wasted effort of lifting the pencil. Defining Heaviside's H(0) as
zero or one has it left- or right-continuous, defining it as the
vertical step has it left-, ad right-, ntinuous.
We might consider a survey of what defines "function" and what defines
"continuous" over time, then as well to the modern definitions of
The set of points that define the line is continuous, horizontal or
vertical, each topologically continuous, as sets of points. (Under
rotation they're the same curve.) Basically the horizontal is
continuous with limits in x and vertical continuous with limits in y.
Then, continuous functions need not be smooth. For example, y = |x|
is not smooth at zero, but it's continuous.
Here, with H(0) being the vertical line, rotating H any but multiples
of pi/2 radian (multiples of 90 degrees) has the rotated function
continuous. This is just like y =|x| is except rotated odd multiples
of pi/4, and the line is except rotated odd multiples of pi/2:
standardly continuous. So, topologically, there's an argument for its
There's no gap in this Heaviside step with connecting H(0+) and H(0-)
with a simple line segment. There is no point in it such that, not in
the function, it is the only point in all neighborhoods of any two
points in the function, not in the function (not even a point
discontinuity). Here "in the function" is each (x,y) in the combined
coordinate image or co-range, with the function defined by the points
in it. The two points are from: te left and right limit sequences,and
the points on the asymptote. (The contrapositive is a strong
rationale not all find.)
http://www.cogsci.ucsd.edu/~nunez/web/edMC.pdf - "What Did Weierstrass
Really Define? The Cognitive Structure of Natural and epsilon-delta
Continuity", Nunez, Lakoff, 1998:
'Euler referred to a continuous curve as "a curve described by freely
leading the hand."' ... 'We call this everyday idea of continuity,
Here then at least those authors may have H, connected, being
"naturally continuous", as would pretty much everybody before
Weierstrass. As well with a definition of gaplessness for example as
above or would fit in alternative definitions of continuity, functions
that are strongly continuous, under any rotation, are naturally
Then, in as to whether a function is left- and right- continuous at
each point and to the same limit point, that being a strong definition
of continuity as is suitable for use in many of methods of real
analysis, there as well weaker definitions of continuity suitable for
use in related methods of real analysis, rigorously.
Here, it is seen that a function being left- and right-continuous at
each point, with the limits from either side being the same, is
continuous, as well, there are reasonable definitions for continuity,
suitable for application, that are not that definition.
Then, fine, Heaviside's step, connected, is "naturally" continuous, or
under reasonable alternative topological definitions with correct
relevance to application, "continuous".
So, I'll totally agree that, the definition of continuous (vis-a-vis,
say, piecewise continuous) has definite import to our methods of
integration and other useful methods of mathematics, and that's
Heaviside's step, being at least piece-wise continuous, has its given
analytical properties. And that, connected, and perhaps a scholar of
Heaviside might know how he would have it, it's naturally continuous.
Here it is in the vagaries of definition that stronger and weaker
conditions of continuity are both applied as adjectives, leaving it to
the experts. The weaker conditions may well be more general, the
stronger more specific, in other cases the opposite. We have our
natural plain language and mathematical terms within it: those
establishing strong and for that matter concrete abstract symbolic
So, again, I'll agree that it is not so that although H-connected is
left- and right-continuous at each point besides those in H(0), that
the limits are the same from each limit.
Then, relevant to continuity and the discrete: the elements of the
range of EF are dense in [0,1]. And I'll have that it's not-a-real-
function, though all its properties hold as modeled by real functions,
and each of its elements is in [0,1], for the standard real numbers.
This non-real function only has values in R[0,1].
Nyquist, Shannon, and Huffman called, they ask: where do continuous
and discrete meet. Where do they part.