Re: Proof 0.999... is not equal to one.

"T.H. Ray" <thray123@xxxxxxx> writes:

Yes, we know (by the Continuum Hypothesis, Cantor)
that
betweeness is infinite--not just "huge." Suppose
we reject the CH? Every sequence is then finite
and differs from another sequence that infinitely
approaches, but does not reach it, by an
infinitesimal
margin. (This is the insight, of course, that led
to the development of analysis--the study of
continuous> > functions.)

Sorry, but I have no idea what you are trying to
express here. How is
the Continuum Hypothesis relevant here?

--
"But remember, as long as one human being follows the
rules of
mathematics, then mathematics as a human discipline
survives.
Right now I'm that one human being, so mathematics
survives."
-- James S. Harris

Look at the context of the OP's original claim, along
with the rest of my explanation.

We can do without the CH and have finite betweeness
(Weyl's term)or we can employ the CH and have infinite
betweeness. But we cannot have terms that are
simultaneously infinitely between, and finite,
between the same pair of integer terms (in the OP's
case, between 0 and 1). Our sequences
have to be either analytical with limits
(in which case 1.000... = 0.999...) or rational
numbers with terminating point. These are differentiable.
The OP allows no means to differentiate a finite sequence
from an infinite series. Both the analysis of continuous
functions, and Cantor's theory, do incorporate such means.

The point is, geometrical constraint can be either
analytically measured in limit points of a continuous
function, or transcendentally counted and measured, but
not both (as the OP would have it).

Tom
.

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