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Re: The Distinguishability argument of the Reals.
Posted:
Jan 6, 2013 9:43 PM
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On Jan 6, 5:19 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <90d0b714-c19e-406c-ba9b-28ba2650c...@r10g2000pbd.googlegroups.com>,
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
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> > On Jan 6, 2:59 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <7a163160-c36a-46d0-ab7a-97cf0fa11...@q16g2000pbt.googlegroups.com>,
> > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
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> > > > On Jan 6, 11:55 am, Virgil <vir...@ligriv.com> wrote:
> > > > > In article
> > > > > <1038fe29-f169-4511-bd13-c7ade7fd1...@pd8g2000pbc.googlegroups.com>,
> > > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
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> > > > > > > lim_(d -> oo) f(n,d) = 0 for every n in N
> > > > > > > --
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> > > > > > No
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> > > > > If, as Ross defined it, f(n,d) = n/d for all d in N and all n in
> > > > > {0,1,2,...,d}, then for any n, lim_(d -> oo) f(n,d) = 0
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> > > > > And no amount of denial by Ross will alter that fact.
> > > > > --
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> > > > Wait, aren't you going to misquote Einstein? Because, you have quite
> > > > the practice of misquoting me. Now, I'm no Einstein, but, I generally
> > > > heartily agree with him, of the rather conscientious sort.
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> > > > d/d = 1
> > > > lim_(n->d) n/d = 1
> > > > lim_(n->d, d->oo) n/d = 1
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> > > That last one is false because
> > > lim_(n -> d) [ lim_(d -> oo) n/d ] =lim_(n -> oo) 0 = 0
> > > lim_(d -> oo) [ lim_(n -> d) n/d ] =lim_(n -> oo) 1 = 1
> > > SO
> > > lim_(d->oo)(lim_(n->oo) n/d =/= lim_(n->oo)(lim_(d->oo) n/d
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> > > SO Ross's lim_(n->d, d->oo) n/d does not exist!
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> > > At least in stndard mathematics.
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> > > But he might try in in WMytheology or RAFeology.
> > > --
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> >http://en.wikipedia.org/wiki/Order_of_integration_(calculus)
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> > It's not quantifier dyslexia so much as that there's only one free
> > parameter modeling the function: d.
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> > There d, for denominator, for n, numerator: is free and unbounded, as
> > is n, as it ranges through elements simply enough in d. For all
> > values of d, the range is [0,1].
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> For each value of d, the range is a finite subset of d+1 values in [0,1].
> You may properly may say that the codomain for each d is [0,1], but the
> range is the set of values values actual taken by the function, and for
> each value of d the range is a set of d+1 values, not an interval.
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> Wrong again, Ross!
> --
Ah, that's (partly) fair: the co-domain (variously image) is [0,1]
for each value of d, and as d is unbounded. The range is in [0,1].
(For "all" d, the union of for each finite d is the rationals, as d is
unbounded the function takes its valid form, in a similar vein as the
tree and the paths, from each finite, the infinite.)
Then, to be the interval [0,1], the values of the range would satisfy
gaplessness, in the standard reals. Here, there are ready definitions
of continuity, that see the elements of the range continuous. This is
where, there is the symmetry of the elements of the range, from zero
and one, about one-half. As they go from zero to one, the reflection
goes from one to zero, exactly as it does. And, the constant
difference between values, non-zero as they sum to one, is smaller
than any standard value, i.e., it's infinitesimal, and non-zero.
This is a thread of distinguishability of reals, and I thank the
readers for tolerating this course, re the distinguishability of
reals, with this construction the unit interval's reals are
distinguished as the naturals are distinguished. Then, as forming the
same continuum as the complete ordered field or its segment in the
unit interval, they are distinguished this way.
Of course, another familiar notion is that the elements of the range
wouldn't maintain the character of taking all values of the reals,
because in ZF (or via ZF) the reals are uncountable. However, again
to be noted: this function sees a different result than others re
nested intervals and the antidiagonal argument: they don't apply to
this function, i.e., the contradictions of those developments don't
follow.
So, the range of EF(n) is the unit interval: [0,1].
Then, for what value of n is EF(n) 1/2? d/2. For what value is it a/
b? d a/b. These results aren't directly retrievable from the
standard natural integers, without a simple scalar infinity, instead,
it is the property of the function that guarantees their existence.
So, the function:
a) sees contradictions via hypothesis to uncountability not follow
b) has co-domain [0,1]
c) is symmetrical about y = 1/2
d) sees the range meet definitions of continuity throughout [0,1]
e) has range R[0,1]
Then, the points of the line segment are so valued, distinguished: as
are the natural integers.
Regards,
Ross Finlayson
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