Re: Rational numbers, irrational numbers: each dense in real numbers

From: "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx>
Among the notions of why there are "more" irrationals than rationals
is the (not fundamentally) heuristic notion that if one were to try to
sample at uniform random from the real numbers in [0,1] by flipping
fair coins (independent Bernoulli trials) to form the binary expansion
of a real number, that it is extremely unlikely to have the sequence
terminate in ending with all zeros or ones, or some repeating
sequence.

What you said there is empirically meaningless. There is no way
even in principle to determine whether an infinite sequence of coin
flips eventually enters a repeating loop. Even if you use a
quantum-mechanical process to emulate coin flips, repeating the
starting conditions for a quantum flip *exactly*, such as inside a
Shroedinger-cat box totally isolated from outside influences,
cooled to Bose condensate to eliminate noise, still the theory of
quantum mechanics AFAIK is consistent with that experiment
eventually going into a loop. For example, the whole Universe could
degenerate into a thermal equilibrium (outside that somehow
magically protected Shroedinger-cat box) and then some earlier
state of the Universe could repeat, and QM could be a function of
the whole-Universe state, whereupon the whole Universe, including
the innerds of the Shroedinger-cat box, begin repeating exactly the
same activities that they previously did from the same initial
state before.

From an empirical viewpoint, the best you can learn is an initial
segment of the sequence, with no knowledge of what will follow
later.

I personally consider it unlikely that a sequence of identical QM
flips will produce a rational number, but I'm not a god who can
*know* such a thing. I personally consider it reasonable that a
sequence of identical QM flips would have some *computable* number
pre-programmed into it, hence the totality of such possible
sequences would be countable
- in the weak sense that it's a Turing-undecideable subset of a
countable set (the set of all syntactically-correct coin-flip
algorithms can be enumerated, but only some of those algorithms
actually generate an infinite sequence of flips, the rest either
halt or run forever after generating their very last coin-flip
event, and it's undecideable whether a given algorithm will indeed
generate an infinite sequence of flips or not),
- or in the other weak sense that for any machine that will at some
point produce its very last coin-flip we simply define it to
produce an infinite sequence of all-heads flips after that point,
in which case we have no decision procedure for knowing where we
need to do this trick and where we should just be more patient.
And in either case, there is no decision procedure for whether two
different algorithms produce forever-identical or
eventually-different output, hence no way to eliminate duplicate
sequences in order to achieve a true enumeration of all computable
sequences. So in fact there is no Cantor-rigor method to enumerate
the computable reals. (There *is* a Cantor-rigor method to
enumerate the rationals, a zigzag path through Z x Z+ where any
fraction with nontrivial GCD of numerator and denominator is
eliminated to get rid of duplicates.)

each particular sequence has the same probability of selection as
any other

Such a probability distribution, with every possible event having
nonzero (positive) probability, is impossible for any except finite
or countably-infinite sets. In the case of uncountably-infinite
sets, probabilities of all but a finite or countably-infinite
subset must be exactly zero, effectively reducing the distribution
to a finite or countably-infinite probability space.

is said that there exists a uniform probability distribution of
the reals of the unit interval

Anyone who says that is wrong. The best that can be defined is a
set of nested finite distributions, with increasing number of
options at each nesting level. For example, it's possible to set up
such a nested set of finite distrubitions such that the probability
of nesting to within epsilon of any number is exactly 2/epsilon
except at either endpoint where the probability is 1/epsilon.
Still the probability of hitting any point exactly is zero.
This is similar to the idea of *measure* (Lebesgue for example),
where the measure of any single point or any countable subset is
exactly zero, and the total measure of a set is *not* the sum of
all the single-point measures (it can be nonzero). Measure is
necessarily additive (whole equals sum of parts) only for cases
where the whole and each of the parts is a "measurable" set IIRC.
For most theoretical subsets of the unit interval, the subset isn't
"measurable" and there's no measure for it defined in the first
place so you can't even assign meaning to the "whole equals sum of
parts" if the whole or one of the parts is not measurable, IIRC.

... a random sample of the second binary digit (bit) of a real
number's expansion is as well independently a sample of another
real number's first digit.

Yes. Any uniform nested-finite-distribution system which is
"uniform" has that property. This is a measure-theoretic, or
equivalently metric-theoretic, space, not a probability space on
the individual real numbers themselves. All you can say is that for
any interval of length w, your "uniform" sequence generator will
have probability w of ending up within that interval, probability
1-w of ending up outside that interval, and probability 0 of
hanging forever on one endpoint of that interval so that we can
never know by experiment that it goes in or out and in fact that it
converges to exactly that endpoint. Open or closed intervals makes
no difference, since the probability is zero that we have a
problem, an undecideable inside point (endpoint of closed interval)
or an undecideable outside point (endpoint of open interval).

there is a bag (multiset) of i random real numbers

There's no such thing as random real numbers, except in the trivial
case where all but a finite countably-infinite set are excluded
from consideration, or where it's impossible to determine whether
that's the case or not.

In sampling one real number

It's not possible to sample one real number, except in the trivial
case where all but a finite countably-infinite set are excluded
from consideration (because you're using a determinstic algorithm
to generate the nested intervals), or where it's impossible to
determine whether that's the case or not (because you're using
quantum mechanics and have no way to run an experiment forever and
then come back to observe the entire result).

Suppose you had a magic genie that gave you random real numbers,
uniformly distributed? The probability of any single number is
zero, so applying a test of likelihood we have a bad result at the
very first number and it only gets worse as one after another
zero-probability event occurs. I can accept that zero-probability
events might sometimes happen, after all they are *possible* even
if very very unlikely. But if they happen over and over and over, I
dismiss the genie as a fraud.

But how is the genie supposed to establish that even one of the
numbers presented isn't computable, hence selected from a known
countable subset of the reals? If the genie admits she/he is
producing only computable numbers, which have been manipulated to
be uniformly distributed, I would have to go back to my original
proof to refute that: There is no infinite sequence of identical
positive terms that converges. So if you enumerate them (with the
Turing oracle to decide which algorithms halt and which don't, and
among those which don't halt which produce infinite digits and
which hang after their last output), still you aren't consistent.

The rest of what you say is nonsense, based on your false
assumption that it's possible to sample actual real numbers, as
opposed to merely intervals of arbitrarily small size.

Summary:
- If your sample space is finite, it can be uniform, p=1/n.
- If your sample space is countably infinite, it can be
all-nonzero, but it can't be uniform, it has to be some
(absolutely-)convergent series, which implies terms arbitrarily
close to zero as you go deeper down the sequence that are summed.
- If your sample space is uncountably infinite, at best it can be
countably-infinite nonzero, reducing to one of the previous cases.
- If you have more than one sample space, such as nested intervals,
or non-nested intervals with Cauchy bounds on diameters, you can
get what's called a "distribution" or a "measure", but binning
per each level of the nesting follows one of the first two cases above.
.

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