Re: Is continuum completely filled up?

If all infinite decimals have distinct value, why is it that 0.999... = 1,
because all numbers have the same right? We can not know whether irrationals
have distinct value or not.
It seems like that rationals don't fill a line. Can we say that irrationals
are the limit of rationals? In traditional set theory, infinite set of
naturals exist, but infinite size of natural number doesn't exist.
Infinite number of rationals exist, but rational number which have
infinitely long unit of random digits doesn't exist. The unit of repetition
is finite, and its size can be regarded as natural number.
The circumstences is the same as that naturals have infinite member, but
don't have infinite size of member.
Naturals and rationals are pertialy treated, but reals are not.
Can we compare their amount in this condition? Line cannot be filled only by
rationals, therefore we must assume that it is filled with reals already.
Digits of decimals don't have the least. Then what decide their value?
Geometrically square root 2 has difinit position on coordinate.
Square root 2 is proven not to be rational number. In finite condition,
reals > rationals > naturals are clearly distinguished. But if the unit of
repetition become large unlimitedly, where is boundery between reals and
rationals?
If a sequence of rationals and its limit is infinitely apart (as the
explanation of zeno ), there is a gap between them. If a sequence is
smoothly
changed and connected to the limit, probably we can choose out them
individually. But is it possible that a sequence smoothly translate to the
limit?
The unit of repetition become large limitlessly like naturals.
Is it possible to cut off omega from infinite sequence of naturals, like to
remove a point from a line? Omega -1 is not natural number, and there are no
natural number which becomes omega by adding 1.
Then why naturals is infinite? If one accept N being infinite, then should
omega be included in N too? And the boundery between rationals and reals
would be lost.
We don't need to take in account of uncomputable numbers, because we don't
need to consider reals to be uncountable. It doesn't matter whether we
assume them or not.
The existence of diagonal number in the proof of Cantor showes that reals
are uncomputable except those which can be listed.
..Moreover when line is constructed only with computable number, I assume
that they cannot be distinguishable at the limit of decreasing distence.
And when we think reals in this way, the explanation about that 0.999... = 1
is easy.

Ozaki Toshiaki




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Relevant Pages

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