Albrecht wrote:
> On 22 Nov., 04:37, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> > > Those are old hat, the question in which I have particular interest at
> > > this time is if via a method to select at uniform random an element or
> > > infinitely many elements of the real numbers whether via a pretty
> > > simple construction that can be construed as a method to select a
> > > natural integer at uniform random, as I have so described.About this notion of a bijection between the unit interval of reals and
> > N^N, in standard ways, and about some informal consideration of the
> > N-choice sequences or in shorthand N-sequences being uniformly dense
> > within the elements of N^N, I should enumerate an obvious progression
> > of reasoning to that effect.

> > Basically the sketch is this:  the reals in the unit interval biject to
> > N^N, the cartesian product of infinitely many copies of the natural
> > integers.  There exists a uniform probability distribution over the
> > unit interval of reals, and thus the elements of N^N, in that each
> > element of N^N so correlated, informally, has an equal probability of
> > being selected as a sample as does any element in R[0,1].

> > Now, among all those elements of N^N, very very few of them are
> > N-sequences, sequences with each element of the natural integers being
> > an element of the sequence exactly once, where those are ranges of
> > choice functions, well-orderings of the natural integers and have a
> > least, in this case first, element of the sequence.  None of those
> > sequences are identical.  So, sample from N^N at uniform random and
> > discard the sample if it is not an N-sequence, and sample until there
> > is an N-sequence, selected at uniform random from among the
> > N-sequences.

> > Then, each N-sequence in N^N has the same probability of being
> > selected.  As samples of the natural integers, let the N-sequences that
> > begin with n be N(n)-sequences, N sequences that begin with x, y, z
> > being N(x, y, z)-sequences, for and etcetera, for a finite number of
> > variables.  There is an N(n)-sequence for each n, finite natural
> > integer in N.

> > The N(n)-sequences can be considered identical, the elements of the
> > sequence after the first are inconsequential.

> > Then that seems to beg the question of there being a natural uniform
> > probability distribution over the natural integers for there to be one.

> > While that may be so, each of the N(n)-sequences has the same
> > probability of being selected as any other, as each is an element of
> > N^N.  So, it may be disregarded which one it is.

> > I'm not sure about bijecting R[0,1] to N^N, but it seems that if it
> > took R^N to biject to N^N then R would biject to N.  Otherwise there
> > would be cardinals between those of N and P(N).

> > Taking infinitely many samples of R to sample N^N is the same as
> > sampling infinitely many samples of {0,1} to sample R, and that returns
> > infinitely many samples.  That process returning a rational would
> > return infinitely many copies of a variety of rationals, and returning
> > an algebraic irrational would seem to return only algebraics.
> > Returning zero means all zeros.

> > So, yeah, that's why.

> > Ross

> I'm not sure whether I'm understanding this argument completely, so I'm
> interested in comments about it.
> My understanding: Since there are no functions which covers all reals
> we have to use probabilistic methods to find something out. Using this
> ansatz, we find no difference between the frequency of reals and
> naturals.
> Is this, roughly said, correct?
> Comments?

> Best regards
> Albrecht S. Storz