Uniform Convergence for 0.999...

Finite(smal) equality does not imply Infinite(smal) equality

[Expanded from my sci.math post, 2/13/96 - Nota Bene: I am its original and sole author]

We learned in elementary school that 0.999... = 1 , as 9.999...-0.999... = 9 . Well and good, but that equality is merely valumetric: we say 0.999=0.999999, as well, in our valid mathematical notations of significant figures, because any difference is beyond our ability to distinguish it, and so we determine it is equal. [In some cases a dot is put over the '=' equals sign: ≐]

But this is not always enough for the student: for in the theory of irrational numbers the student learns a 'trick' for deciding that the irrationals are uncountably more than the rationals which are countable (as every rational can be aligned with a corresponding integer, and so counted, all),-- and this 'trick' relies on the ability to infinitesmally distinguish numbers in their decimal notation ... The 'trick' aligns each irrational number into a list, and invents one that cannot be in the list because it is constructed of digits differing in some corresponding [digit] place with each number considered in the list .... It assumes that ∞+1 > ∞ in certain cases: and oddlier still, that all digits can be specified for any single fraction, but not for that fraction's radix-reversed 'infinite' number (*).

* (The 'trick' is likenable to 'proving' the counting numbers uncountable because a number can be invented -picked- larger-than each number in an ascending list and pretending it is not any in the list: the invention is just as rigorous, and just as specious though lacking finitesmal convergence giving it an air of realism, using addition rather than digit assignment, yet requiring just as countably-infinite-many digits specified, to be, equal, or not .... The fact is that 1 is a discontinuity for 0.999...→, comparable to 1/0 being neither ±∞.)

The first trouble is that the chosen 'different' number may be indirectly in the list by its equivalent, like 0.999... vs. 1.000... and the choice of 'different' was improperly constructed; There may be no way of knowing whether the 'different' was properly constructed before it was discovered wrong ... and the student must ponder whether there is an answer, or a paradox. (It might be easy to force in single-choice binary, easier than in multiple-choice decimal.)

The correlative trouble is that the trick asserts that decimal fractions are unique by their digits,- yet this contradicts the claim that 0.4999... equals 0.5000.... It is counter-intuitive to suppose that ending-in-nines collapses onto the adjacent ending-in-zeros. (It may be notable that this situation was discovered in rational division, where in fact the borrow-bit was never fully resolved but neglected tossed onward, as-if-under-the-never-finished-infinite-carpet: there was always a leftover infinitesimal undiscussed; 0.999...=1.000... only on any 'valumetric' finite, scale, not on any infinitesmal scale such as tangential-adjacency-of-points.)

The second correlative trouble is that if numbers are deletable by such equationing, then in different radixes, it's different numbers; and different numbers of numbers....

The second trouble is that a counter argument can be constructed to force inclusion of those very 'different' numbers, by herding; and there are only countably many [rational] constructions of such 'different' numbers: And therefore it is a paradox, not an answer.

And its correlative trouble is that the All declaration of digits is specious: Infinite All is not the same-as nor extension-of, finite all.

The third trouble is that the suggestion that a number can be specified for all its digits to be sufficiently different, is intrinsically a Zenoic paradox: At every stage in the development its 'discovered' number may be further in the list ... Merely to assume the infinite all can compass itself to exclusion to break-through or-not in the infinite-all case, is not a proof but a self-paradoxic assumption: For example of the paradox itself consider the infinite sequence, .0, .10, .110, .1110, .11110, ... and 'different' number .11111..., and the test-question is whether the sequence includes it:- Certainly no finite position in the sequence is it, but the sequence is infinite and has just as many numbers as there are digits in .11111..., whence if .11111... can be specified 'for-all-digits', then that sequence can be specified 'for-all-numbers', which means that there are numbers in the infinite sequence that are as long as the sequence infinitely long: and the sequence must include .11111... in its 'infinite all' numbers as surely as .11111... exists only in all its digits ... It is simply 'in there'.

Regarding the convergence of 0.999...→1, I find the pointwise convergence insufficient, and propose the lack of uniform convergence necessarily un-equates the two:

Case #1. 'uniform convergence' by digit-specification [*]

Let 0.999... = lim 0.ddd... as d→9. Notice that there is no digitwise convergence toward 1.000... but only toward 0.999..., as for d<10 there is no carry (nor carry-specification) approaching 'yes' in the limit, and whence the limit never involves 1.000... for =0.999....

Case #2. 'uniform convergence' by radix-specification [*]

Let 0.999... = lim 0ρNNN... as ρ→radix ten and N=ρ-1 ('nine' in radix ρ). In particular let ρ=10*(1-ε) as ε→0. Translate this to radix ten:0.N1N2N3.... Then N1 = ρ-1 = (9-10ε)*(1-ε)^-1 = 9-ε-εε-εεε-..., while N2 = 9+8ε+7εε+... which is dominated by 9+8ε for small ε>0, whence N2>9. But notice that far N's blow-up: some #>10 delivers a guaranteed carry to propagate all the way up to N1, and as ε→0, N1+1→10, thus involving 0.999...→1.000.... But alas, further N's blow up with finite excess: some #>11 no matter what ε>0 you choose, whence there is no uniform convergence for 0.999... (it is an open number, not closed).

NB. This technique also points to another discontinuity: that the calculation of 0ρNNN... = 1 , holds in all radixes until ρ=1, where N=0, where it does-not hold even in the limit, but would hold for ρ=1+ð, ð infinitesimal, if ð and the count of digits were inversely relatable.

In summarial remedy, It is probably best were mathematicians to distinguish Any and All, and not identicably, in the Infinite case.
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See also a discussion of transfinitesimal qualification.

* [Uniform Convergence is usually presented in Real Analysis as a zero-convergent sequence of integrable functions squeezing standup area along the functions' pointwise infinite mesh into the interval boundary: never reaching 'all-zero'; here squeezing to infinitesimal]
* [Recognize that these are as proofs-by-contradiction, that, Uniformly handling all, digit places, does not yield uniform convergence]
* [I argued this out before I ever posted 2/13/96 on the Internet where it has been argued repeatedly: even after I gave my proof]

Topic note: The meaning of "real" number has principally the mathematical sense here, and whence can deal with infinitesimals,- as the cosmological sense of "real" is lower-bounded by aether-noodle granulation at its 'next-level' of 'under-standing' and 'oper-ation'.

A premise discovery under the title,

Grand-Admiral Petry
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