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

On Nov 4, 3:16 am, hagman <goo...@xxxxxxxxxxxxx> wrote:
On 15 Okt., 05:44, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:
[snipalot]



Browsing Metamath, consider theorem elisseti: http://us.metamath.org/mpegif/elisseti.html
(element is set), that a member of a class is a set, with the class of
ordinals member of class V.

Sorry, I could not find any theorem in metamath that claimed
that the class of ordinals were a member of V (that would
be "On e V", wouldn't it?).
Can you give a reference to an existing metamath proof or at least
a hint what such a proof might look like?
Be sure to readhttp://us.metamath.org/mpegif/df-v.htmlcarefully.

That seems quite controversial, to say
that if a class is a member of a class that it's then a set. Maybe
instead it's just a faulty description. I track back through the
developments, eghttp://us.metamath.org/mpegif/elisseti.html, and,
it seems to say that, uh, if a class A is an element of a class B then
it's a set, using class equality instead of set equality. So, in
Metamath, the universal class is the collection of all elements
satisfying identity, and, in Metamath, all classes satisfy identity,
and, in Metamath, an element of a class is a set, and, each class is
an element of the universal class. Thus, in Metamath, each class, for
example the class of Ordinals, is a set.

As you notice yourself, onprc claims exaclty the opposite.
If you WERE able to produce a proof of "On e V" in metamath,
that WOULD be quite interesting.



Then, in onprc, it is shown that no set contains all the ordinal
numbers. So, then On =/= On,
WRONG
else it would be an element of V
WRONG
and a
set, where V is simply defined as the class of all classes,
WRONG
sets, that
are equal to themselves. (Identity, of an object being itself, is
generally assumed to hold.) As well, it is stated that universal
quantification is unrestricted.

http://us.metamath.org/mpegif/con0.html

http://us.metamath.org/mpegif/df-cleq.htmlhttp://us.metamath.org/mpegif/df-v.html





Second, then there is to be described an infinite set with the well-
ordering thus that via transfinite induction/recursion it exists and
is uncountable. Then, the property to show that holds for transfinite
induction is that for a given ordinal, either it is less or equal than
the cardinality of the irrationals and thus there exists uncountably
many irrationals left from which to select, or it is greater than the
cardinality of the irrationals.

http://us.metamath.org/mpegif/tfi.html

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.

You have consistently been unable to define p_alpha.
You never managed to give a valid recursive definition of p_alpha,
i.e. using only the previously defined p_beta for beta<alpha.
You only did that for alpha=0 and for successor ordinals.



Basically for each partition of the irrationals intersecting the
interval (p_alpha, 0) into (p_alpha, p_alpha+) and (p_alpha+, 0), each
partition has the same cardinality.

As long as p_alpha < p_alpha+ < 0, this at last sounds correct.



The rationals and irrationals are each dense in the reals.

That's correct as well.



So, ZFC is inconsistent.

Non sequitur.



Ross

--
Finlayson Consulting

hagman

The issue I noted with the definition of the universal class is that
it defines its elements by any class, not set but class, that
satisfies class identity, not just set identity. Each class satisfies
class identity, where the universe is defined as something along the
lines of: U (V, L) being {x: x = x}, here casually, and in Metamath
explicitly using class identity. Ah, there is a set variable in the
class builder, not a class variable. So, by preservation of type,
there are only sets, not classes, satisfying the predicate of
equalling themselves that comprise V, in Metamath's definition. Maybe
it's just that using eventually class equality from upcasted sets
gives a ready definition of the collection of classes.

So, I see that the definition of V the class of all sets in Metamath
is of sets. Yet, using class equality to build, with the class
variables and class equality, etcetera, in the class builder, instead
of set terms, the same primitives are available. Obviously there is
no collection of classes in ZF with classes, yet the classes in
Metamath are built as collections of classes in the class builder,
class element-of is defined with class variables, from set variables.
There are class primitives defined.

ZFC's universe is the Russell set, from Russell's well-known paradox,
in a theory with theorems decided by less axioms than ZFC's, eg ZFC -
R, with unrestricted comprehension. Thus it contains itself. So ZFC
is inconsistent.

Hagman, I defined p_alpha+ in a variety of different ways, including a
recursive definition in a "course-of-values" transfinite recursion
scheme, a general type preserving monotonicity.

Here there is the consideration of the mutual denseness of each of the
rationals and irrationals in the reals. Between any two irrationals,
there lie, lay, infinitely many rationals, and between each two of
those, infinitely many irrationals, between each two of those,
infinitely many rationals, etcetera ad infinitum, in the standard real
numbers. (A particular meaningful difference between the verbs lie
and lay may be reasonable to consider as technical.)

Then in consideration of strictly monotone sequences vis-a-vis
critical points, well there are uncountably many strictly monotone
sequences, and correspondingly that many critical points. (The
critical points are unfortunately not monotone.) Here the
consideration of strictness in monotonicity can be ignored, because
the values are already unequal. Between each pair of monotone
subchains there is a critical point, but between critical points there
aren't necessarily non-zero ascending chains. That gets into whether
to divide, to have the boundary, to partition, at the point or between
the points. In the normal ordering of the reals the boundary is
always at a point.

So anyways a well-ordering of the real numbers has the property that
it is a bijection between the set of real numbers and some ordinal
equivalent to the initial ordinal of c the cardinality of the
continuum. If there is an uncountable segment that has preservation
of monotonicity in the normal total ordering of the real numbers, then
as Hughes showed then via a compactness-type argument that would be
contradictory. Yet, the well-ordering is the union of monotone
subchains, where these critical points as subchains of length one can
be considered, variously removing the initial element of the following
monotone subchain/critical point. Defining the critical points as
simply initial elements of a monotone subchain just leads to many
smaller monotone sequences, if they were adjacent, or larger subchains
if they were between subchains of opposite monotonicity. (Again there
is a variation depending on whether each limit ordinal is defined as a
critical point.) In that sense the well-ordering is a union of
monotone subchains, where as a collection of ordered pairs, the well-
ordering's subsets for given ordinal ranges are all defined.

Currently I'm looking at all these but the necessary countability of a
strictly descending uncountable chain ((r_alpha, ..., r_beta), alpha <
beta) of irrationals seems to belie what is otherwise the
uncountability of the irrationals, given a transfinite course-of-
values schema defining them. Otherwise there don't exist choice
functions on R, which do exist in ZFC.

Then there was further consideration and multiple heuristic examples
of why the rationals carry as much weight in the reals as the
irrationals. For example, unless the (0,p) \ U (0, p_i), 0 < p_i < p
is non-empty, then for any real number it is defined as the set of
either the rationals, or the set of irrationals, less than it. If
that set-difference was non-empty then it would contain a rational,
and a distinct rational, for p. If it's empty, then it is so for the
construction using rationals as well, with there not being non-
rational irrationals between a rational and all different rationals
(completeness of rationals), which is not considered to hold in the
standard real numbers. It takes many less rationals than irrationals
to fully specifiy a real number. Another example considered
convergent sequence representation of reals, and of how each of
infinitely many finite subsequences generated a distinct real, and the
infinitely many generated only another.

The point of this is to find more analytical reckonings to extend the
integral calculus, obivously, in redifferintegrosystems, basically for
finding classes of non-real functions with reasonable analytical
properties.

It's said sometimes that well-orderings of the reals are "totally
random", but any enumeration of a subset of the reals forms a subchain
of a well-ordering of the reals. In the standard reals, there are no
ordering-sensitive properties of the elements, thus that is so.

Ross

--
Finlayson Consulting

.

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