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

On Oct 19, 10:54 am, David R Tribble <da...@xxxxxxxxxxx> wrote:
Ross A. Finlayson wrote:
Then, this thread ends in a similar fashion as to how the rational/
irrational thread of late 2005 ended: illustrating a perceived
inconsistency in ZFC from the uncountability of irrationals and
denseness of rationals and irrationals in the reals.

MoeBlee wrote:
Yes, excellent work, Ross. You have successfully proven that you have
a perception that ZFC is inconsistent.

Ross A. Finlayson wrote:
MoeBlee wrote:
Yes, it's rather clear since being proven mathematically.

So what so you plan to do with the money you get when
you accept the Fields Medal?

I figured I'd endow a much smaller prize.

Consider this, for some irrational number p, the interval (0,p).
Then, for each irrational p_alpha greater than zero and less than p,
take the union of those intervals (0, p_alpha). Is that union equal
to (0,p)?

If so, then an intersection of non-empty nested intervals (p_alpha, p)
is empty.

Yet, each interval is pairwise non-disjoint with each other interval,
and non-empty. (Each p_alpha is strictly less than p.)

Then, for an interval (0,r), r e R, it is equal to: the union of (0,
r_alpha)'s for real 0 < r_alpha < r, the union of (0, p_alpha)'s for 0
< p_alpha < r, and the union of (0, q_alpha)'s for 0 < q_alpha < r,
for rational q_alpha's. For any set dense in the reals X, the union
of (0, x_alpha) for x e X, 0 < x < r, equals (0,r).

There is trichotomy in the standard reals, each x_alpha is strictly
less than r.

Via induction, for each finite alpha, (0,p) \ (0,p_alpha) is not
empty. Now, consider two cases, one with the rationals and one with
the irrationals. If the rationals are countable, then assign to each
q_alpha a natural integer.

Then, with this notion of defining a real number r via unions of
nested intervals less than r, any real number can be represented by a
countable collection of rational numbers, or uncountable (in ZFC,
assuming ZFC is consistent) collection of irrationals. So, the
rational numbers in a sense have much more _weight_ than the
irrational numbers, because it takes so many less to encapsulate the
meaning of a real number, any real number.

That's a casual notion, here to relate to the consideration of above
about sampling real numbers, and about how sampling a particular
rational number is at once a sample an infinite family of rational
numbers, while sampling an irrational number samples infinitely many
distinct, and not necessarily related in a neat algebraic number-
theoretic manner, irrational numbers. In marking the points on the
line, the rational points are darker.

That's some suet for the mill.

With regards to the proof (largely in sketch, but assembled piece by
piece) of the existence via choice of a transfinite "sequence" of
descending irrationals, I would like to hear more arguments as to why
that is not possible, so they can be addressed.

Ross

--
Finlayson Consulting

.

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