Date: Jan 4, 2013 9:23 PM
Subject: Re: Uncountably Nested Intervals

On Jan 4, 11:04 am, fom <> wrote:
> On 1/4/2013 10:52 AM, Ross A. Finlayson wrote:

> > On Jan 3, 9:32 pm, Virgil <> wrote:
> >> In article
> >> <>,
> >>   "Ross A. Finlayson" <> wrote:

> >>> On Jan 3, 7:02 pm, Virgil <> wrote:
> >>>> In article
> >>>> <>,
> >>>>   "Ross A. Finlayson" <> wrote:

> >>>>> On Jan 3, 9:07 am, "Ross A. Finlayson" <>
> >>>>> wrote:

> >>>>>> On Jan 2, 12:48 am, Virgil <> wrote:
> >>>>>>> In article
> >>>>>>> <>,
> >>>>>>>   "Ross A. Finlayson" <> wrote:

> >>>>>>>> On Jan 1, 11:22 pm, Virgil <> wrote:
> >>>>>>>>> In article
> >>>>>>>>> <

> >>>>>>>>>> ,
> >>>>>>>>>    "Ross A. Finlayson" <> wrote:
> >>>>>>>>>> On Jan 1, 8:59 pm, Virgil <> wrote:
> >>>>>>>>>>> In article
> >>>>>>>>>>> <5e016173-aa1b-4834-9d70-0c6b08f19...@jl13g2000pbb.googlegroup
> >>>>>>>>>>> s.
> >>>>>>>>>>> com>, "Ross A. Finlayson" <> wrote:

> >>>>>>>>>>>> On Jan 1, 7:29 pm, Virgil <> wrote:
> >>>>>>>>>>>>> In article But in that proof Cantor does not require a
> >>>>>>>>>>>>> well
> >>>>>>>>>>>>> ordering of the reals, only an arbitrary sequence of
> >>>>>>>>>>>>> reals
> >>>>>>>>>>>>> which he shown cannot to be all of them, thus no such
> >>>>>>>>>>>>> "counting" or sequence of some reals can be a count or
> >>>>>>>>>>>>> sequnce of all of them. --

> >>>>>>>>>>>> Basically
> >>>>>>>>>>> Nonsense deleted! --
> >>>>>>>>>> Nonsense deleted, yours?
> >>>>>>>>> Nope! --
> >>>>>>>> Great:  from demurral to denial.
> >>>>>> Seems clear enough:  in ZFC, there are uncountably many irrationals,
> >>>>>> each of which is an endpoint of a closed interval with zero.  And,
> >>>>>> they nest.  Yet, there aren't uncountably many nested intervals, as
> >>>>>> each would contain a rational.
> >>>>>> To whit:  in ZFC there are and there aren't uncountably many
> >>>>>> intervals.
> >>>>>> Then, with regards to Cantor's first for the well-ordering of the
> >>>>>> reals instead of mapping to a countable ordinal, there are only
> >>>>>> countably many nestings in as to where then, the gap is plugged (or
> >>>>>> there'd be uncountably many nestings).  Then, due properties of a well-
> >>>>>> ordering and of sets defined by their elements and not at all by their
> >>>>>> order in ZFC, the plug can be thrown to the end of the ordering, the
> >>>>>> resulting ordering is a well-ordering.  Ah, then the nesting would
> >>>>>> still only be countable, until the plug was eventually reached, but,
> >>>>>> then that gets into why the plug couldn't be arrived at at a countable
> >>>>>> ordinal.  Where it could be, then the countable intersection would be
> >>>>>> empty, but, that doesn't uphold Cantor's first proper, only as to the
> >>>>>> finite, not the countable.  So, the plug is always at an uncountable
> >>>>>> ordinal, in a well-ordering of the reals.  (Because otherwise it would
> >>>>>> plug the gap in the countable and Cantor's first wouldn't hold.)

> >>>>>> Then, that's to strike this:
> >>>>>> "So, there couldn't be uncountably many nestings of the interval, it
> >>>>>> must be countable as there would be rationals between each of those.
> >>>>>> Yet, then the gap is plugged in the countable: for any possible value
> >>>>>> that it could be.  This is where, there aren't uncountably many limits
> >>>>>> that could be reached, that each could be tossed to the end of the
> >>>>>> well-ordering that the nestings would be uncountable.  Then there are
> >>>>>> only countably many limit points as converging nested intervals, but,
> >>>>>> that doesn't correspond that there would be uncountably many limit
> >>>>>> points in the reals. "
> >>>>>> Basically that the the gap _isn't_ plugged in the countable.

> >>>>>> Then, there are uncountably many nested intervals bounded by
> >>>>>> irrationals, and there aren't.

> >>>> Yes there are, as I pointed out in a posting that Ross has carefully
> >>>> snipped entirely.

> >>>> The set of intervals {  [-x,x] : x is a positive irrational}  is one
> >>>> such set of uncountably many nested intervals bounded by irrationals.

> >>>> A simple, and obvious, example of what Ross claims does not exist.
> >>>>> Point being there are uncountably many disjoint intervals defined by
> >>>>> the irrationals of [0,1]:  each non-empty disjoint interval contains a
> >>>>> distinct rational.  Thus, a function injects the irrationals into a
> >>>>> subset of the rationals.

> >>>> This too is false.
> >>>> {  [x,1-x] : x is an irrational between 0 and 1/2} being an explicit
> >>>> counterexample. And as there are way more such intervals than rationals
> >>>> in their union, no such injection from intervals as Ross claims to
> >>>> rationals can exist.

> >>>> And Ross is totally wrong again!!!
> >>>> And Ross will, no doubt, snip all of this proof of his errors too, just
> >>>> as he did the last one, if he repies at all.
> >>>> --

> >>> That example contains zero, a rational, no?
> >> No! For z between 0 and 1/2, no interval from x to 1-x will contain 0.
> >>> What, that is news?  Once again your plain arguments against the man
> >>> instead of for the argument show your lack of argumentative ability,
> >>> and responsibility, and poor form.

> >> Since my ARGUMENT was entirely a mathematical example refuting your own
> >> mathematical claim, it is ad mathematics not ad hominem.

> >> Though I did enjoy being able to show your mathematics to be totally
> >> wrong!

> >>>   But, for me to note that, is it ad
> >>> hominem, to note ad hominem?

> >> It is certainly an ad hominem to claim it when it did not exist, as you
> >> did.

> >>> See, for that I would refrain:  because
> >>> it's less than perfectly ethical to argue ad hominem.

> >> Particularly when you are in the wrong and trying to cover your ass.
> >>> Also quit
> >>> bullying me, I'm bigger than you.  A suitable change of topic for the
> >>> thread, to respect the time of readers, is more along the lines of
> >>> "Uncountably Nested Intervals".

> >>> Uncountably many nested intervals, each pairwise disjoint contains two
> >>> rationals, or rather as nested their disjoint contains a rational.

> >> "Uncountably many nested intervals, each pairwise disjoint"?
> >> Nested intervals are not pairwise disjoint, at least in any real world.
> >>> The rationals are dense in the reals.  Deal with it.
> >> What misleads Ross onto thinking I don't already?
> >> Re the original issue:
> >> There are countably infinite sequences of nested intervals with rational
> >> endpoints. For example { (-1/n,1/n) : n in |N }, but obviously no
> >> uncountable nested set of such sequences.

> >> Ross then claimed that there could not be any uncountable set of nested
> >> intervals with irrational endpoints, which is trivially false:
> >> { (-x, x) : x is a positive irrational} is just such an uncountable but
> >> nested set of intervals with irrational endpoints as Ross had claimed
> >> did not exist.

> >> So Ross was wrong, and too chicken to own up.
> >> --

> > No, what I said was there are and aren't.  I simply constructed
> > examples where there are and examples where there could not be.
> > That's a poor and ungenerous representation.  Your argument is simply
> > fallacious, where generally your mathematical content can be replaced
> > with a text reader.  You should speak well of yourself, I don't care
> > if you do or don't.  Obviously enough I was thinking of the
> > irrationals in [-1/2, 1/2] containing zero.

> > Here, the conundrum is that in ZFC there are uncountably many
> > irrationals, that there be uncountably many disjoint intervals, that
> > each contains a rational, which are countable in ZFC, which would be a
> > contradiction.

> ZFC does not say anything about the topology of the
> real line.
> There are better topologists out here than I.  But, while
> the reals are not compact, they are sigma compact.  Or, if
> I have that wrong, they have whatever compactness condition
> is required so that the second-countability of the space
> applies to your example.
> The identity criterion in set theory is not the identity
> criterion of the real numbers.  The latter come from the
> topological completeness by which is meant completeness
> relative to the rationals.  Thus, trichotomy of the
> rationals yields trichotomy in the reals.  Hence, the
> reals have an identity criterion based on their order
> relation
> x=y <-> (x<=y /\ x>=y)
> If you want to think about how the identity criterion
> of a logical system like ZFC is related to topology
> look at the difference between the axioms of a metric
> and a pseudometric.  For one you have,
> x=y<->d(x,y)=0
> and for the other you have
> x=y->d(x,y)=0
> The latter statement is what is involved with attaching
> metric structure (and, therefore, a metric based
> topology) to a logical system.
> The next step is to look at uniform spaces defined
> relative to uniformities.   That will give the first
> conditions for a system of relations to have
> a topology that may be metrizable.
> Finally, find a proof for the metrization of relations
> so that you can see what is involved with attaching a
> pseudometric to a logical system.  You can find one
> in Kelley's "General Topology".
> Curiously, it is circular when speaking to the
> metrization of the reals.  It invokes the least
> upper bound property.  Still, it is instructive.

These topological properties of density in the reals of the
irrationals and rationals build up from the density of the rationals,
the non-continuity of the rationals, and the irrationals being the
complement of the rationals, in the reals.

The cardinal properties of the rationals and irrationals (here often
specifically for the unit interval instead of the entire set) follow
from ZFC and cardinality of the reals, and the general definition of
the rationals as integer ratios and reals as sufficiently represented
as infinite expansions or as Cauchy or as equivalence classes of
Cauchy sequences.

I'm not familiar yet with "second-countability". It might seem from
its statement an order type property, of the rationals as having order
type 2^w. (As would (0,1)^w have were it countable.)

Second-countability is a property of a space that it is completely
separable, that the topology of the space has a basis that is
countable. This basically has that there aren't any open subsets that
aren't countable unions of some countable collection of open

Then, while there's a countable cover of open subsets of R, here of
intervals, the point is that there are uncountably many points in R,
in ZFC, each defines an interval, closed with endpoints open without,
of the set between zero and it. Where these are the irrationals then
R though completely separable, _does_ have uncountable unions of open
sets the union of which may be an open set that is not the entire
set. It still may only take a countable union of open subsets to form
any open subset, but, open subsets of irrationals, can also be
considered uncountable unions of open sets, bounded by an irrational
but not rationals, simply as they would be. Similarly closed sets
bounded on at least one side by an irrational may be the union of
uncountably many closed sets, bounded by an irrational but not
rationals (as there aren't uncountably many rationals).

The reals as space aren't first-countable because there are
uncountably many neighborhoods for each point (in ZFC). Here, each of
those (uncountably-many) delta-neighborhoods, open or closed, contains
rationals that each epsilon<delta-neighborhood does not. Then we
would be arguing about first-countability. There are uncountably many
neighborhoods of each point in topology in ZFC each contains rationals
that narrower neighborhoods do not, thus it is a contradiction that
there aren't uncountably many of those, at least a distinct one for

There are and aren't: contradiction.

Is there a neighborhood with standardly-valued radius r of a point in
R without uncountably many neighborhoods with radius < r? No. Is it
so that for each ordinal gamma > beta that for diminishing values of r
through uncountably many ordinals that r_gamma contains irrationals
not in r_beta? Directly, yes, circuitously, no. (This is where for
each r_beta there exists uncountably many values r_gamma that have via
Choice one of them, quantification over them.)

So, are there:
a) uncountably many neighborhoods, some without distinct rationals?
b) countably many neighborhoods?
c) some other option consistent with the real space in topology and
sets of numbers in ZFC?
d) other?

Because, to be consistent: _all_ the properties hold _all_ the time.

Countable additivity, of non-finite differentiable regions, is enough
for the integral calculus and its results that agree with geometry.


Ross Finlayson