Re: An uncountable countable set
- From: Virgil <virgil@xxxxxxxxxxx>
- Date: Sat, 11 Nov 2006 12:40:10 -0700
In article <4556122c@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
David R Tribble wrote:
I've given a simple example many times. Starting with 0, your
rules will never generate 3, 1/3, or any integer multiple or
power of 3. You still have not responded to that little flaw,
which would go a long way towards convincing others that
you might be on to something.
Why is that a flaw in the H-riffics, and not in the digital reals? You
cannot express 1/3 as a decimal fraction either.
Not as a terminating decimal fraction, but every rational can be
represented by either a terminating or an eventually repeating decimal
fraction.
Do the decimal
fractions constitute a countable set? That's counter to Cantor's
Diagonal Argument for the uncountability of the reals.
TO, as usual, puts his mind into neutral and his typing into high gear
before posting.
The terminating decimals, and even the terminating together with
repeating decimals (which is no more than the rationals), form countable
sets. If TO want a set of decimals which is uncountable, he must include
non-terminating and nonrepeating ones. And the set of all such
represents the reals but no countable subset of them does.
The set of all
digital fractions is uncountable. The set of all H-riffics is.
uncountable. Every digital fraction has successor and predecessor as
well, when the bits are mirrored to the other side of the digital point,
and ordered as naturals. Ah, you say, but 0.333... does not have a
natural mirror, since ...333 is not a natural. And yet, I say to you,
that is every digit in that string is in a finite position with respect
to the digital point, then there is no point in that string where it
achieves an infinite value, and so it represents some sort of finite but
unbounded quantity. Since any such unending strings also has a successor
and a predecessor (except perhaps for ...000 and ...999), it would seem
that in two significant ways such numbers are very like the terminated
finite strings you call naturals. You might be reassured by keeping in
mind that the set of such finite string representations of infinite
strings is still countable, since they depend on a repetition of a
finite string of digits.
David - If every element in the reals has at least one successor, what
does that say to you?
Not much, apart from the fact that some countable subset of the reals
is covered by your generating rules.
Your rules produce a countable set by their very nature. For any
H-riffic H_n, I can produce H_2n and H_2n+1 from it. Which means
that I can map every natural k to some H-riffic, and vice versa.
Hence the H-riffics are a countable set.
But there is no r in R for which you can say there is no successor. How
do you partition R into countable subsets? Which r in R does not have
two successors?
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