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

On Oct 25, 4:03 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Oct 25, 3:19 pm, lwal...@xxxxxxxxx wrote:

On Oct 25, 9:25 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
I must thank you and MoeBlee for the proof.

I'm a bit frustrated that I didn't quite finish my proof. It just
needs the right lemma (something quite ordinary I bet).

Come to think of it, I should have realized that RAF's
statement was false. And the first proof of it was given
not by you or Mr. Hughes -- but, albeit unwittingly,
by RAF himself!

After all, RAF assumed there existed an uncountable
set wellordered by < or >, then derived a contradiction --
because no such set exists! RAF's mistake, of course,
was then to conclude that therefore ZFC is inconsistent.

Here is RAF's proof in his own words:

RAF:
That is where the desired property for a given ordinal (that there are
more elements in the interval (0, p_alpha) for ordinal alpha less than
the cardinality of the irrationals) holds for ordinals less than or
equal to the cardinality of the irrationals, where for higher ordinals
the property would not consistently hold, but it is not necessary that
it does. So, in the course of values over all ordinals, for ordinals
less than or equal to the cardinality of the irrationals there are at
least that many remaining in the interval (p_alpha, 0). (Otherwise,
there wouldn't be that many in the interval.) For ordinals greater
than the cardinality of the irrationals, they as well satisfy the
property in being greater than the cardinality of the irrationals.
Then, there are as many elements p_alpha as there are ordinals alpha
that are less than or equal to the cardinality of the irrationals.
Then, that holds for sufficiently many irrationals, for each of which
can be displayed a distinct rational, that theorem contradicts another
in the theory.

Of course, RAF wrote this proof under the assumption
that > is a wellorder on the set of irrationals. We now
rewrite RAF's proof, but instead of the irrationals, we
let X be a subset of R on which > is a wellorder. Our
task is to prove that X is countable.

Without loss of generality, assume that the cardinality
of X is aleph_1. (I'm not trying to assume CH here. I'm
actually trying to avoid the problems that occur if the
continuum is a singular cardinal, a cardinal whose
cofinality is not itself. If CH is false and X has cardinality
greater than aleph_1, then we only consider the first
aleph_1 elements of X under the wellorder >). Just as
RAF does, we enumerate the elements of X as
p_0, p_1, p_2, p_3, ..., p_omega, p_(omega+1), ...,
p_alpha, for every countable ordinal alpha.

Now RAF appears to assume that zero is the greatest
lower bound of X. But this version of the proof will work
even if X has no lower bound.

If M is the greatest lower bound of X, then we let X_n be
the set of all elements of X greater than M+1/n. Clearly
X_n is a subset of X -- indeed, a proper subset, since
M is the greatest upper bound. So the restriction of >
to X_n is still a wellorder -- and it is isomorphic to a
proper initial segment of aleph_1 (regarded as the
ordinal omega_1). Since every _proper_ initial segment
of aleph_1 is countable (this is the most important step
RAF wanted to emphasize), X_n must be countable also.

If X has no lower bound, instead we define X_n as
the set of all elements of X greater than -n.

Then X itself would be the union of all the X_n's. So X is
the countable union of countable sets, hence X itself
must be countable. QED

(Here the Axiom of Choice is invoked, but of course AC
was already assumed, since the original context of the
proof mentioned the existence of a wellorder of R.)

So I was foolish to attempt to RAF's claim, since RAF
himself had already proved it to be false.

.

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