UNCOUNTABILITY

[Zuhair]

Some people are trying to prove from definability that only countably
many sets can exist. The idea is that a set is an object extension of
a predicate that is definable after a parameter free finite formula,
and since we have countably many such formulas and since under
Extensionality each parameter free formula can define only ONE set,
then there is One-One correspondence between sets and their defining
formulas and accordingly we can only have countably many sets.

In symbols:

X is a parameter free definable set <-> [Exist Phi. for all y. y in X
<-> Phi(y)]

where Phi(y) is a formula having y as the sole free variable.

Now is the above rationale correct?

The answer is NO!

If one insists that EVERY set must be parameter free definable, then
we will end up having UNCOUNTABLY many parameter free definable sets!

Proof:
Lets assume we have countably may parameter free definable sets.
Then the set of all parameter free definable reals must be countable.
Accordingly there exist a bijection between the set N of all naturals
and the set R* of all parameter free definable reals.
Now since EVERY set is parameter free definable, then this bijection
is definable!
By The diagonal argument of Cantor, then the diagonal defined after
that bijection would be PARAMETER FREE definable real that is NOT in
the set of ALL parameter free definable reals. A CONTRADICTION!

Thus there cannot exist any parameter free definable bijection between
the set N and R*.

But since all sets must be parameter free definable (by assumption
above)
Then there cannot exist a bijection between N and R*.

Thus the set of all parameter free definable reals is UNCOUNTABLE!
QED

This is too much of a price to pay for just having our desire for
parameter free definability fulfilled.

There is a natural sense of having countably many parameter free
definable sets, and of course parameter free definable reals. But this
natural sense implies that we must admit the existence of a set that
is not definable in a parameter free manner, and this would be the
case of any set that is a bijection between N and R*.

In other simpler words one must accept that countability of all finite
formulas to be non parameter free definable!

Of course this will open the door wide for accepting the concept of
having some sets that are not parameter free definable, and thus
parameter dependent definability would emerge, and of course with this
later kind of definability of sets one can have an uncountable domain
over which parameters range since for each formula with a parameter, a
set is definable after each assignment given for that parameter, so if
we have uncountably many assignments then this would give rise to
uncountable many sets definable after that SINGLE parametric formula!
So we have Many-One relationship between parameter definable sets and
their defining formulae, so even though we have countably many
parametric formulas (formulas in more than one free variable) still
this doesn't entail that the number of sets definable after them is
countable because how many sets defined after each formula is not
determined by the formula alone it is determined by how many values
the parameters in the formula can range over, so if the parameters
range over uncountably many values then we can get uncountably many
sets each defined after an assignment given to parameters in the
defining formula.

In other words what I want to say is that unlike the case with
parameter free definability were the number of sets definable in that
manner must be equal to the number of parameter free finite formulas
which is of course countable, in the case with parameter dependent
definability there is no limitation on the number of sets definable
after those kinds of formulas, so having uncountably many sets
definable in this manner is not hindered by having countably many such
formulas.

Now the set of all definable reals (whether parameter free or not)
obviously must be Uncountable, this a straightforward corollary of
Cantor's diagonal argument. And as clarified above this is not
hindered by the countability of parametric formulas!

So having parameters in the defining formulas provides the grounds for
POSSIBILITY of having uncountably many sets definable after them. But
what PROVES the existence of uncountabily many parameter definable
reals is of course the diagonal argument of Cantor.

Anyway we turn matters we face uncountability!

The alternative explanation that uncountability only rise from local
defect in the expressive language of a theory and that all sets are
countable in the real world, though possible yet looks far from being
the case. The only cause for believing in such a direction is a bias
towards countability, or to minimalists reductionist approaches more
akin to that of Ockam's razor and the alike, which actually do not
address the strength of the argument of uncountability that almost
pops up so frequently that one can hardly think of it as being just an
anomaly of expressiveness.


Zuhair

[George Greene]

On Dec 19, 4:11 pm, Zuhair <zaljo...@gmail.com> wrote:
> So having parameters in the defining formulas provides the grounds for
> POSSIBILITY of having uncountably many sets definable after them. But
> what PROVES the existence of uncountabily many parameter definable
> reals is of course the diagonal argument of Cantor.

Surely you must realize by now that attacking the diagonal argument
(actually the direct construction of the ANTI-diagonal of a PRESUMED
omegaXomega square) is THE #1 FAVORITE crank pastime around here.
By posting this you are just inviting every crank from WM on down
to do battle with you. You are new enough that this my seem
worthwhile
to you, BUT IT ISN'T. I'm not saying that JUST because I'm burnt out
after 20 years.
I'm saying it because it really is obvious from even casual
consideration
that LOGIC means NOTHING to these people! They DON'T CARE if you/we
have
a proof! THEY have THEIR canards and maxims and precepts, and THAT'S
ALL
THEY care about! Progress is not really possible along a merely
intellectual
front (merely intellectual, our position is simply far too easy to
confirm).
A unified wall of contempt stands a better chance.

[Zuhair]

Yes I agree with you. I just thought the above might meet some of
their concerns like the definability question that they so frequently
raise, and how it is possible to have uncountably many definable sets
while there is only countably many defining formulas, I gave an answer
here to that concern, which is that definability with parameters is
NOT ONE-ONE, and as I explained in the head post there is no problem
with having uncountably many sets if those are definable after
assignments given to parameters, simply because there is nothing to
forbid having uncountably many assignments to parameters. This is a
subtle difference that I felt that I must clarify. Also parameter
definable reals do not in principle suffer from the intellectual
problem of being on distinguishable and the alike raised conundrums,
since they are indeed definable but obligatorily through parameters,
that doesn't make out of them incognitos. I just thought that would be
helpful to remove some of the concerns around uncountability.

Zuhair

[Albrecht]

Your considerations are not stringent without assuming ZF.

The claim, that there are only finitely definable objects in the realm of facts is, at first, a non-axiomatic claim. If someone claims the countability of numbers, your proof doesn't disprove him. Your prove just shows that there is no formula to count the numbers (e.g. teh reals). But your further conclusions are wrong since nobody forces someone to consider only inside ZF or the like.

[Barb Knox]

In article
<6f8b71be-9d79-4d8c...@n8g2000vbb.googlegroups.com>,

Maybe some of them picked up their canards and maxims at a tender age
(like most religious people pick up their beliefs), but I think in most
cases their problem is that they have strong intuitions (such as that
nothing can be bigger than infinity, since that's got to be bigger than
everything else), and their mathematics education is sufficiently
impoverished that they never had to encounter and wrestle with anything
counterintuitive. Their loss, and ours if we expect to "convert" them.

> A unified wall of contempt stands a better chance.

Of what? Abuse by "the establishment" will convince the hard-core even
more of the existence of a vast conspiracy that unjustifiably opposes
their revolutionary insights. A unified wall of *silence* might
discourage some of them from posting.

On the positive side, their is some value for non-cranks in attempting
to understand how it is that the cranks are so stuck. But I agree that
there is very little value in trying to de-crank them.

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------

[Ross A. Finlayson]

Of what?

A unified wall of contempt stands a good chance: of being just that.
Here: of mathematics and mathematical progress.

Defendit numerus, deperdit numerus.

There's progress to be made in mathematics, even in the foundations of
mathematics. Goedel proves this, in modern mathematics.

Obviously nobody expects WM to roll mathematics back, and he won't,
but I'll tell you what: I'm not stopping rolling it forward. And,
that's not just carrying the torch, it's learning where it's from, and
why it is.

Mathematically, all results are due individuals, and those who came
before them, and to pass it on: mathematically.

If you're really so good at finding errors in others, why don't you do
humanity a favor and discover something true in oversights of our
greatest thinkers, and share it. Oh, I guess that's not the same as
following the parade with a bucket. I'll agree that it's a lot easier
to find somebody doing it wrong than to put together the pieces and do
it right: that nobody ever told anybody how to do.

And that's mathematical progress, not calculation, nor the simple (and
of course relevant and valued) industry of instruction. The rubrics
are of knowledge, generally, and historically, not Johnny's chart:
mathematical progress, of mathematics.

And NOTHING means LOGIC.

Regards,

Ross Finlayson

[Graham Cooper]

Pardon my French this is UTTER FNG DRIVEL

The SETS are 1-TO-1 with some PREDICATE.

A *F_I_N_I_T_E* sentence in Predicate Calculus.

|N| = |GODEL NUMBERS|
|GODEL NUMBERS| = |FUNCTIONS|
|FUNCTIONS| = |CHOICE FUNCTIONS|
|CHOICE FUNCTIONS| = |SETS|
|SETS| > |N|

YOU ARE ALL FOXES IN HEADLIGHTS DUMB!

SPOT THE CONTRADICTION!

Herc

[Zuhair]

No we don't need to assume ZF. Actually the diagonal argument of
Cantor's is provable in very weak fragments of second order arithmetic
which a weak fragment of Z which is an extremely weak fragment of ZF.

Of course one can work under different assumptions, for example assume
working within a countable model of ZF, and matters would look somehow
different and supportive of countability, but I already mentioned that
for one to prefer this model over the intended model as to what
reflects what's going on in the real world is really supported by
nothing other than a bias for countability. I already mentioned that
the reason for regarding uncountability true is because it pops up in
various mathematical contexts involved with theories that are capable
of handling reals and higher objects, it pops up so frequently that
disregarding it as some kind of anomaly of expressiveness is really
too biased against it. If one actually regards mathematics as
discourse about POSSIBLE form. Then of course uncountability stands as
a strong possibility, and thus discourse about it is at the essence of
what mathematics is.

Zuhair

[Zuhair]

Just to complete that: I generally don't encourage speech about : THE
real world.
I do encourage speech about: A POSSIBLE world.

Mathematics is about speaking about possible form, and not necessarily
about REAL form.

Within this mainframe I behold speaking about uncountability as fully
justified!

Regards

Zuhair

[Zuhair]

On Dec 20, 7:52 am, George Greene <gree...@email.unc.edu> wrote:

I didn't really read this last line. What do you mean by that?

Zuhair

[WM]

On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:


> Thus the set of all parameter free definable reals is UNCOUNTABLE!
> QED

And it is countable by the simple proof that there is a bijection
between all finite words an the natural numbers. Hence the result of
your argument is not that the other one is wrong but that set theory
is inconsistent, namely the notion of countability that presupposes
finished infinity is contradictory, as everybody with a sober mind
would see immediately, nit cranks as you, Greene or Knox of course.

> So having parameters in the defining formulas provides the grounds for
> POSSIBILITY of having uncountably many sets definable after them. But
> what PROVES the existence of uncountabily many parameter definable
> reals is of course the diagonal argument of Cantor.
>
> Anyway we turn matters we face uncountability!
>
> The alternative explanation that uncountability only rise from local
> defect in the expressive language of a theory and that all sets are
> countable in the real world, though possible yet looks far from being
> the case. The only cause for believing in such a direction is a bias
> towards countability,

You are dreaming of parameters that help to define uncountably many
elements. But that is and remains a dream, unless you assume the
existence of uncountably many undefined parameters. That, however, is
matheology, not mathematics. Since parameters do not help anything in
defining uncountably many elements, set theory is contradicted.

Regards, WM

[Ross A. Finlayson]

Of course our esteemed George Greene shouldn't be vilified for
expressing his opinions, either. And, it's fair to note that compared
to mathoverflow, and other available Internet communication platforms,
that quixotic jousting to the infinite is over-represented among
suitable topics on mathematics and the development of mathematics. It
is a fallacy to read too much into his words beyond leaving them as
indeterminate and the common fallacy of overgeneralization, as it
would be to hew to it.

Then while that's so and Greene is respectable, mathematical progress
is by definition different than what there was before, where
mathematics is true: more. And: a stalwart, stoic, static, stoney,
silence on that: yes, there is more to foundations than the modern,
is only that.

Regards,

Ross Finlayson

[Zuhair]

On Dec 20, 4:58 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:
>
> > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > QED
>
> And it is countable by the simple proof that there is a bijection
> between all finite words an the natural numbers. Hence the result of
> your argument is not that the other one is wrong but that set theory
> is inconsistent, namely the notion of countability that presupposes
> finished infinity is contradictory, as everybody with a sober mind
> would see immediately, nit cranks as you, Greene or Knox of course.

The bijection between all finite words and the natural numbers is NOT
parameter free definable.
So if one insist that EVERY set must be parameter free definable, then
this mean that there is no bijection between the set of all finite
words and the naturals. HOWEVER I already said that this result only
stems if one desires that ALL sets must be parameter free definable,
which is as I said a high price to pay, because there is a natural
sense of the existence of a bijection between all naturals and all
finite words. NOW if we follow that natural sense then this mean that
we must give up the concept that all sets are parameter free
definable, because any bijection between the naturals and the set of
all finite words is itself NOT parameter free definable. And this
opens the door wide for accepting sets that are not parameter free
definable. And of course the number of those sets is determined by the
number of assignments given to parameters in the defining formulas,
which is something that range over the whole universe of discourse, so
it is not limited by the countability of those formulas.

Do you think that the bijection between the naturals and all finite
words parameter free definable?

>
> > So having parameters in the defining formulas provides the grounds for
> > POSSIBILITY of having uncountably many sets definable after them. But
> > what PROVES the existence of uncountabily many parameter definable
> > reals is of course the diagonal argument of Cantor.
>
> > Anyway we turn matters we face uncountability!
>
> > The alternative explanation that uncountability only rise from local
> > defect in the expressive language of a theory and that all sets are
> > countable in the real world, though possible yet looks far from being
> > the case. The only cause for believing in such a direction is a bias
> > towards countability,
>
> You are dreaming of parameters that help to define uncountably many
> elements. But that is and remains a dream, unless you assume the
> existence of uncountably many undefined parameters. That, however, is
> matheology, not mathematics. Since parameters do not help anything in
> defining uncountably many elements, set theory is contradicted.
>

Parameters helps in opening the *POSSIBILITY* of having uncountably
many elements. IT doesn't not prove it. What proves uncountability of
the universe of discourse is Cantor's diagonal argument.

Zuhair

[Graham Cooper]

On Dec 21, 2:02 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

Greene, Ullrich, Hughes, Knox

the 4 most vile attackers of free mathematical discussion

ALL GET PAID TO DEFEND THEIR CURRICULUM TEXT BOOKS!

You will GET ZERO* LOGICAL DISCUSSION FROM ANY OF THEM!

*some high level dismissals*

Herc

[Virgil]

In article
<66faa8ed-f190-48e4...@n9g2000vbv.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:
>
>
> > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > QED
>
> And it is countable by the simple proof that there is a bijection
> between all finite words an the natural numbers.

In order to claim that the set of reals is countable reals, WM must show
that the set of reals satisfies the definition of countability.

I.E.,WM must show a surjection from N to R, or an injection from R to N.

Neither of which WM has ever done and CANNOT do.

Hence WM is WRONG ! AGAIN !! AS USUAL!!!
--

[Graham Cooper]

On Dec 21, 7:14 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <66faa8ed-f190-48e4-a600-a65b0e1f0...@n9g2000vbv.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > > QED
>
> > And it is countable by the simple proof that there is a bijection
> > between all finite words an the natural numbers.
>
> In order to claim that the set of reals is countable reals, WM must show
> that the set of reals satisfies the definition of countability.
>
> I.E.,WM  must show a surjection from N to R, or an injection from R to N.
>
> Neither of which WM has ever done and CANNOT do.
>
> Hence  WM is WRONG !  AGAIN !! AS USUAL!!!
>

Your above reasoning is bogus as usual.

Enumeration Vs Enumerable

Herc

[Graham Cooper]

On Dec 21, 7:14 am, Virgil <vir...@ligriv.com> wrote:
>
> > > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > > QED
>
> > And it is countable by the simple proof that there is a bijection
> > between all finite words an the natural numbers.
>
> In order to claim that the set of reals is countable reals, WM must show
> that the set of reals satisfies the definition of countability.
>
> I.E.,WM  must show a surjection from N to R, or an injection from R to N.
>

For the UMPTEENTH TIME!!!!

http://tinyurl.com/blueprints-powerset


Use a Universal Turing Machine to generate POWERSET(N)
Every <algorithm, input> pair that terminates populates a
value into P(N).

INPUT 1 2 3 4 5 6 7 8 9 10 ...
===========================================
TM1 H L H H H L L L L L ...
TM2 H H H H H H H H H H ...
TM3 H L L L L L L L L L ...
TM4 H H H H H H H H H H ...
...

TM1(1) Halts => 1 e POWERSET_1
TM1(2) Loops => 2 !e POWERSET_1
TM1(3) Halts => 3 e POWERSET_1

TM2(1) Halts => 1 e POWERSET_2
TM2(2) Halts => 2 e POWERSET_2
TM2(3) Halts => 3 e POWERSET_2
...

POWERSET(N) = { {1,3,...} {1,2,3...} ... }


ANYBODY READING YET!!!!

INPUT 1 2 3 4 5 6 7 8 9 10 ...
===========================================
TM1 | H L H | H H L L L L L ...
TM2 | H H H | H H H H H H H ...

SEE WHERE THOSE SETS CAME FROM???? ANYONE??? BUELLER???

IF IT HALTS THEN THE INPUT VALUE
IS AN ELEMENT OF THAT SUBSET OF N

FFS! P(N)

You can't fng read Virgil!

Herc

[Albrecht]

You don't have uncountability without implementing uncountability first. Any other view is just believing in wishing wells.

See e.g. http://plato.stanford.edu/entries/paradox-skolem/

"Of course, as Jané notices, there are a number of strategies that we could use to evade this application of Skolem's Paradox: we could use uncountably many axioms to force our models to have uncountable domains, we could appeal to the Upward Löwenheim-Skolem theorem to show that Zermelo's axioms also have uncountable models (see section 1), or we could move to a second-order version of Zermelo's axioms and then prove that these axioms can only be satisfied by models with uncountable domains (see section 2.5). Unfortunately, each of these strategies presupposes that we already have a prior grip on the notion of an uncountable set—e.g., to initially characterize an uncountable set of axioms, to formulate the Upward Löwenheim-Skolem theorem, or to prove that second-order ZFC has only uncountable models. So, none of these strategies can be used to introduce uncountable sets into mathematics in the first place. Or so, at any rate, Jané takes Skolem to be arguing."

[Zuhair]

> See e.g.http://plato.stanford.edu/entries/paradox-skolem/

>
> "Of course, as Jané notices, there are a number of strategies that we could use to evade this application of Skolem's Paradox: we could use uncountably many axioms to force our models to have uncountable domains, we could appeal to the Upward Löwenheim-Skolem theorem to show that Zermelo's axioms also have uncountable models (see section 1), or we could move to a second-order version of Zermelo's axioms and then prove that these axioms can only be satisfied by models with uncountable domains (see section 2.5). Unfortunately, each of these strategies presupposes that we already have a prior grip on the notion of an uncountable set—e.g., to initially characterize an uncountable set of axioms, to formulate the Upward Löwenheim-Skolem theorem, or to prove that second-order ZFC has only uncountable models. So, none of these strategies can be used to introduce uncountable sets into mathematics in the first place. Or so, at any rate, Jané takes Skolem to be arguing."

Correct! and there is no problem with that.

Zuhair

[WM]

On 20 Dez., 22:14, Virgil <vir...@ligriv.com> wrote:

> In order to claim that the set of reals is countable reals, WM must show
> that the set of reals satisfies the definition of countability.

No, I have shown that the notion of countability is nonsense.

> I.E.,WM must show a surjection from N to R, or an injection from R to N.

In order to show that unicorns do not exist, I need not show that all
unicorns have two horns.

Regards, WM

[WM]

On 20 Dez., 19:32, Zuhair <zaljo...@gmail.com> wrote:
> On Dec 20, 4:58 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > > QED
>
> > And it is countable by the simple proof that there is a bijection
> > between all finite words an the natural numbers. Hence the result of
> > your argument is not that the other one is wrong but that set theory
> > is inconsistent, namely the notion of countability that presupposes
> > finished infinity is contradictory, as everybody with a sober mind

> > would see immediately, not cranks as you, Greene or Knox of course.

>
> The bijection between all finite words and the natural numbers is NOT
> parameter free definable.

It need not to be defined. I is proved (iff a complete set of natural
numbers is assumed).

> So if one insist that EVERY set must be parameter free definable, then
> this mean that there is no bijection between the set of all finite
> words and the naturals.

This bijection is provable (iff a complete set of natural numbers is
assumed).

> HOWEVER I already said that this result only
> stems if one desires that ALL sets must be parameter free definable,
> which is as I said a high price to pay, because there is a natural
> sense of the existence of a bijection between all naturals and all
> finite words.

There is no "natural sense", but a proof. And there is no
mathematician refuting that. Take for instance:

If we pursue the thought that each real number is defined by an
arithmetical law, the idea of the totality of real numbers is no
longer indispensable. (Bernays)

Definiert man die reellen Zahlen in einem streng formalen System, in
dem nur endliche Herleitungen und festgelegte Grundzeichen zugelassen
werden, so lassen sich diese reellen Zahlen gewiß abzählen, weil ja
die Formeln und die Herleitungen auf Grund ihrer konstruktiven
Erklärungen abzählbar sind. (Schütte)

> NOW if we follow that natural sense then this mean that
> we must give up the concept that all sets are parameter free
> definable, because any bijection between the naturals and the set of
> all finite words is itself NOT parameter free definable.

And they are not definable with parameters either.

> And this
> opens the door wide for accepting sets that are not parameter free
> definable. And of course the number of those sets is determined by the
> number of assignments given to parameters in the defining formulas,
> which is something that range over the whole universe of discourse, so
> it is not limited by the countability of those formulas.

You cannot define something by undefined notions. The universe of
discourse is for you something like the heaven is for religious
people. But mathematics is not religion.

>
> Do you think that the bijection between the naturals and all finite
> words parameter free definable?

I am not interested in the question whether it is definable, but I
know that this bijection can be proved, for instance by the following
list:
0
1
00
01
10
11
000
...

> Parameters helps in opening the *POSSIBILITY* of having uncountably
> many elements. IT doesn't not prove it. What proves uncountability of
> the universe of discourse is Cantor's diagonal argument.

That does only prove that matheologians have a blind spot.

Proof: Cantor assumed a list that is enumerated by all natural
numbers, i.e., which is complete with respect to its enumeration.
Then
he "proved" that there is always a real number that is not in the
list. Then he proved that this real number can be inserted as another
entry into the list without removing another entry. The new list is
again complete and does not contain more entries than before. In
effect he proved that the first list had not been complete with
respect to the enumeration, i.e., he proved that his crucial
assumption had been wrong. But as every logician knows, from a wrong
assumption everything can be concluded.

Regards, WM

[Jesse F. Hughes]

Graham Cooper <graham...@gmail.com> writes:

> Pardon my French this is UTTER FNG DRIVEL
>
> The SETS are 1-TO-1 with some PREDICATE.
>
> A *F_I_N_I_T_E* sentence in Predicate Calculus.
>
> |N| = |GODEL NUMBERS|
> |GODEL NUMBERS| = |FUNCTIONS|
> |FUNCTIONS| = |CHOICE FUNCTIONS|
> |CHOICE FUNCTIONS| = |SETS|
> |SETS| > |N|
>
> YOU ARE ALL FOXES IN HEADLIGHTS DUMB!
>
> SPOT THE CONTRADICTION!

I know that you're literally delusional, so perhaps your stubborn
ignorance isn't really your fault, but you've been told repeatedly that
the second equation up there is false. Goedel numbers are in
correspondence with *computable* functions N^<w -> N, not with the set
of *all* functions.

--
Jesse F. Hughes
"There's a thrill that's gone that I'll probably not have in quite the
same way again. After all, FLT was a unique animal, and we had a
great dance." -J.S. Harris on "proving" Fermat's last theorem

[Albrecht]

Am Freitag, 21. Dezember 2012 11:55:22 UTC+1 schrieb Zuhair:
>
> > See e.g.http://plato.stanford.edu/entries/paradox-skolem/
>
> >
>
> > "Of course, as Jané notices, there are a number of strategies that we could use to evade this application of Skolem's Paradox: we could use uncountably many axioms to force our models to have uncountable domains, we could appeal to the Upward Löwenheim-Skolem theorem to show that Zermelo's axioms also have uncountable models (see section 1), or we could move to a second-order version of Zermelo's axioms and then prove that these axioms can only be satisfied by models with uncountable domains (see section 2.5). Unfortunately, each of these strategies presupposes that we already have a prior grip on the notion of an uncountable set—e.g., to initially characterize an uncountable set of axioms, to formulate the Upward Löwenheim-Skolem theorem, or to prove that second-order ZFC has only uncountable models. So, none of these strategies can be used to introduce uncountable sets into mathematics in the first place. Or so, at any rate, Jané takes Skolem to be arguing."
>
>
>
> Correct! and there is no problem with that.
>
>

Nobody is forced by logic to think that there are uncountable many objects anywhere. I thought you had claimed such a thing as the contrary of that. But if not, we agree.

[Zuhair]

Yes we agree, what I claimed is about uncountability being a
POSSIBILITY, and not necessarily a reality.

Zuhair

[Zuhair]

> know that this bijection can be proved.

It is not a matter of what you are interested in. I asked a simple
question, and the answer to it is either YES or NO, or if you don't
know then just say that you don't.

The question is:

Is the bijection between the set of all finite words and the set of
all naturals definable in a parameter free manner or not?

A simple question that demands a simple answer: YES or NO?

Zuhair

[Zuhair]

On Dec 21, 2:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> And they are not definable with parameters either.

Do you mean that the bijection between the set of all naturals and the
set of all finite words is NOT definable at all, whether with or
without parameters?

Zuhair

[WM]

On 21 Dez., 17:34, Zuhair <zaljo...@gmail.com> wrote:

> > I am not interested in the question whether it is definable, but I
> > know that this bijection can be proved.
>
> It is not a matter of what you are interested in. I asked a simple
> question, and the answer to it is either YES or NO, or if you don't
> know then just say that you don't.
>
> The question is:
>
> Is the bijection between the set of all finite words and the set of
> all naturals definable in a parameter free manner or not?

Of course it is. You just defined it. Otherwise we could not talk
about it.

>
> A simple question that demands a simple answer: YES or NO?

Yes it is definable. It has been defined. Nevertheless it does not
exist, because the sets do not exist.

Regards, WM

[WM]

No, this bijection has obviously been defined. What is no definable is
a set with uncountably many elements, or, better: What is not
definable are uncountably many elements. (The set is definable, yet
not exististing free of contradictions.)

Note finally: Every Cantor diagonal r differs from any other real
number by a finite initial segment n(r) of its string of digits. That
is not possible with the Binary Tree. A diagonal does not differ from
all finite paths, i.e., for every initial segment n(r) of every real
number r there exists a finite path of the Binary Tree that is n(r).
You may consider actual infinity as well as uncountable languages, but
that does not change the fact that Cantor's argument does not apply.

Regards, WM

[WM]

On 20 Dez., 05:52, George Greene <gree...@email.unc.edu> wrote:

> Progress is not really possible along a merely
> intellectual
> front (merely intellectual, our position is simply far too easy to
> confirm).
> A unified wall of contempt stands a better chance.

All Caesars, all Napoleons, and all Einsteins take that position.
There are cases where it is more promising than arguments. Matheology
must be one of them.

Regards, WM

[Virgil]

In article
<3e4d24ce-00db-4865...@17g2000vba.googlegroups.com>,

WM is clearly misrepresenting Einstein, by lumping him in with the
Caesars and Napoleon.

And WM is the sole WMytheologist here.
--

[Alan Smaill]

WM allies himself with Cantor:
the prophet in the wilderness.

> Regards, WM
>
>

--
Alan Smaill

[Virgil]

In article
<07a35fee-e2e2-4d53...@v7g2000yqv.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 20 Dez., 19:32, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 20, 4:58 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 19 Dez., 22:11, Zuhair <zaljo...@gmail.com> wrote:
> >
> > > > Thus the set of all parameter free definable reals is UNCOUNTABLE!
> > > > QED
> >
> > > And it is countable by the simple proof that there is a bijection
> > > between all finite words an the natural numbers. Hence the result of
> > > your argument is not that the other one is wrong but that set theory
> > > is inconsistent, namely the notion of countability that presupposes
> > > finished infinity is contradictory, as everybody with a sober mind
> > > would see immediately, not cranks as you, Greene or Knox of course.
> >
> > The bijection between all finite words and the natural numbers is NOT
> > parameter free definable.
>
> It need not to be defined. I is proved (iff a complete set of natural
> numbers is assumed).
>
> > So if one insist that EVERY set must be parameter free definable, then
> > this mean that there is no bijection between the set of all finite
> > words and the naturals.
>
> This bijection is provable (iff a complete set of natural numbers is
> assumed).

Not without using what Zuhair is calling parameter dependent definitons.

.......

> But mathematics is not religion.

Your WMYtheology is!

> >
> > Do you think that the bijection between the naturals and all finite
> > words parameter free definable?
>
> I am not interested in the question whether it is definable, but I
> know that this bijection can be proved, for instance by the following
> list:
> 0
> 1
> 00
> 01
> 10
> 11
> 000
> ...
>

WMatheological nonsense as usual.

> Proof: Cantor assumed a list that is enumerated by all natural
> numbers, i.e., which is complete with respect to its enumeration.

ON the contrary, that is exactly what Cantor did not assume.

What Cantor stated, and proved, was that if such a list did exist it
could not be a complete listing of all binary sequences.

Since countability of a set requires, by definition, such a list to
exist, those like WM who claim such lists cannot exist are claiming
uncountability of those unlistable sets.

WM, and many others, wrongly describe Cantor's argument as an indirect
one, as a proof by contradiction, but Cantor did not like indirect
proofs and avoided them whenever possible, and di so in his dialgonal
argument.

> Then
> he "proved" that there is always a real number that is not in the
> list. Then he proved that this real number can be inserted as another
> entry into the list without removing another entry. The new list is
> again complete and does not contain more entries than before. In
> effect he proved that the first list had not been complete with
> respect to the enumeration, i.e., he proved that his crucial
> assumption had been wrong. But as every logician knows, from a wrong
> assumption everything can be concluded.

The assumption of countability, according to the very definition of
countability, which WM carefully ignores, would justify an assumption of
listability, but Cantor did not even need to make that assumption.


Cantor proved:
If such a list (an injection from N to to a complete set of
binary sequences) COULD exist, it still could not be a surjection.

His proof does not actually require that any such list does exist.

And once again WM's WMytheological monomania had tripped him up.
--

[Virgil]

In article
<6cfff32f-e65f-468a...@hf3g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 21 Dez., 17:36, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 21, 2:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> > > And they are not definable with parameters either.
> >
> > Do you mean that the bijection between the set of all naturals and the
> > set of all finite words is NOT definable at all, whether with or
> > without parameters?
>
> No, this bijection has obviously been defined. What is no definable is
> a set with uncountably many elements, or, better: What is not
> definable are uncountably many elements. (The set is definable, yet
> not exististing free of contradictions.)

I ask again: what is WM's definition of countability of a set?

The usual definition, as used by Canter an everyone since, except
possibly WM, is that there must be either a surjection from N to the set
or an injection from the set to N in order to establish countability,
and the provable inability to create either mapping makes a set
uncountable.

WM seems to be using some different definition, but will not say what it
is.

>
> Note finally: Every Cantor diagonal r differs from any other real
> number by a finite initial segment n(r) of its string of digits. That
> is not possible with the Binary Tree. A diagonal does not differ from
> all finite paths

Every infinite path differs from every finite path at all but finitely
many nodes. But even in WM's incomplete infinite binary tree there are
absolutely no finite paths. every path has infinitely many nodes in it.

So any reference to "finite paths" is at best misleading, and more
likely deliberately false.

> i.e., for every initial segment n(r) of every real
> number r there exists a finite path of the Binary Tree that is n(r).

And uncountably many infinite sequnces that begin with n(r), as many as
sequences as in the entire tree, as an easy bijection proves.

> You may consider actual infinity as well as uncountable languages, but
> that does not change the fact that Cantor's argument does not apply.

It still applies quite nicely everywhere except possibly in WMythology.
--

[Virgil]

In article
<d0fea249-ff50-40ce...@n9g2000vbv.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 21 Dez., 17:34, Zuhair <zaljo...@gmail.com> wrote:
>
> > > I am not interested in the question whether it is definable, but I
> > > know that this bijection can be proved.
> >
> > It is not a matter of what you are interested in. I asked a simple
> > question, and the answer to it is either YES or NO, or if you don't
> > know then just say that you don't.
> >
> > The question is:
> >
> > Is the bijection between the set of all finite words and the set of
> > all naturals definable in a parameter free manner or not?
>
> Of course it is. You just defined it.

Referring to such an alleged bijection does not define it unless only
one such bijection could ever be possible, which is definitely not the
case here, so that WM must be confused again, this time about what the
word :definition" means.

> Otherwise we could not talk about it.

WM manages to talk about all sorts of things that do not exist, like his
alleged complete infinite binary tree whose set of paths he claims are
listable. Though he has proved that set listable for any truly complete
such tree.

> >
> > A simple question that demands a simple answer: YES or NO?
>
> Yes it is definable. It has been defined.

I see nothing that in any sane sense can be called a definition

> Nevertheless it does not
> exist, because the sets do not exist.

A set exists, at least as far as standard mathematics is concerned as
soon as it is defined to the point of being able to tell its members
from non-members.

Apparently in WMytheology things arre different
>
> Regards, WM
--

[Virgil]

In article
<d1c92b8a-9177-41a6...@h2g2000yqa.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 20 Dez., 22:14, Virgil <vir...@ligriv.com> wrote:
>
> > In order to claim that the set of reals is countable reals, WM must show
> > that the set of reals satisfies the definition of countability.
>
> No, I have shown that the notion of countability is nonsense.

NOt anywhere nearly as clearly as others have shown many of the notions
of your WMYtheology are nonsense.

Countability has a standard definition: A set S is countable if and only
if there is an injective mapping from N to S or a surjective mapping
from S to N, and S is otherwise uncountable.

That such a definition gets WM's bowels in a knot is irrelevant.
It is, and remains, a standard definition in standard mathematics.
And if WM cannot stand it, he should restrict himself to such
mathematics, if any, that does not get his bowels all knotted up, rather
than continuously inflicting such pain on himself.

>
> > I.E.,WM must show a surjection from N to R, or an injection from R to N.
>
> In order to show that unicorns do not exist, I need not show that all
> unicorns have two horns.

To show an injection from N to R or a surjection from R to N is simple.

The proven impossibility of any surjection from N to R, or injection
from R to N means that R totally satisfies the definition of being an
uncountable set according to the standard mathematical definition of
uncountability.

Learn to live with it, WM, or learn to avoid it entirely.
--

[William Hughes]

> Yes it is definable. It has been defined. Nevertheless it does not
> exist, because the sets do not exist.
>

It is definable which means that WM can use it to prove that the
bijection exists, but it does not exist.

Outside Wolkenmuekenheim, the fact that it does not exist means the
bijection does not exist.

WM wants to have his cake and eat it.

[Graham Cooper]

On Dec 22, 7:17 am, Virgil <vir...@ligriv.com> wrote:
> In article

> <d1c92b8a-9177-41a6-a9f8-f1ab678d9...@h2g2000yqa.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 20 Dez., 22:14, Virgil <vir...@ligriv.com> wrote:
>
> > > In order to claim that the set of reals is countable reals, WM must show
> > > that the set of reals satisfies the definition of countability.
>
> > No, I have shown that the notion of countability is nonsense.
>
> NOt anywhere nearly as clearly as others have shown many of the notions
> of your WMYtheology are nonsense.
>
> Countability has a standard definition: A set S is countable if and only
> if there is an injective mapping from N to S or a surjective mapping
> from S to N, and S is otherwise uncountable.
>

This is all countable subsets of N.

TM1(1) Halts => 1 e POWERSET_1
TM1(2) Loops => 2 !e POWERSET_1
TM1(3) Halts => 3 e POWERSET_1
TM2(1) Halts => 1 e POWERSET_2
TM2(2) Halts => 2 e POWERSET_2
TM2(3) Halts => 3 e POWERSET_2
...

POWERSET(N) = { {1,3,...} {1,2,3...} ... }

1 <=> {1,3,..}
2 <=> {1,2,3,...}
...


What are you guys missing here, you want the 1st million elements of
the 1st million subsets of N printed out??

What method for counting functions do you prefer?

Turing machines?

TM-1
S-00L->S
S-10L->S
TM-2
S-00L->S
S-10R->S
...
TM-8
S-11R->S
S-01R->S
--------
Next comes TM-9, the first 2 State TM.
TM-9
S-00L->1
S-10L->1
1-00L->S
1-10L->S
..


LISP strings? lambda expressions, LR grammar reduction,

my cellular Half-TM computer model?
www.tinyurl.com/blueprints-turing

Unify Fetch cycle, incremental 3GL .EXE files, Supercombinators,
neural nets, ...

What's your preferred computational method for listing all subsets of
N?



Herc

--
S: if stops(S) gosub S
This example proves stops() must be un-computable!
~ George Greene

[Graham Cooper]

On Dec 22, 7:17 am, Virgil <vir...@ligriv.com> wrote:

> The proven impossibility of any surjection from N to R, or injection
> from R to N means that R totally satisfies the definition of being an
> uncountable set according to the standard mathematical definition of
> uncountability.
>
> Learn to live with it, WM, or learn to avoid it entirely.
>

Every time your Bluff is challenged you cannot prove it.

You all claim for the last 10 years you have formal proofs

instead you post drivel nonsense


AD(X) = FLIP(D(X))

AD(1) = FLIP(D(1))
AD(2) = FLIP(D(2))
AD(3) = FLIP(D(3))
...

AND SO ON...

THERFORE

AD(1) =/= D(1)
AD(2) =/= D(2)
AD(3) =/= D(3)
...

AND SO ON

===========

That is not a formal proof you keep regurgitating!

FAR FAR FAR ..... FAR FROM IT!

Herc

[Zuhair]

When I asked my question I thought you were already familiar with what
I meant by definable in a parameter free manner, I have already given
the definition of that in a prior post, I'll re-iterate it again.

Definition:
x is said to be parameter free definable iff there exist a parameter
free formula phi(y) such that:
For all y. y in x iff phi(y)

And a parameter free formula means a formula in ONE free variable, so
when we say phi(y) is a parameter free formula then this means that y
is the only variable occurring free in phi(y).

When I asked my above question I meant specifically that concept
definability.,

So to re-iterate my question again:

Is the bijection between the set N of all natural numbers and the set
of all finite words parameter free definable or not?

If you say yes, then please show me the proof of that, you must show
that there must exist a parameter free formula phi(y) such that a
bijection f from N to the set of all finite words have it true that
for all y. y in f <-> phi(y).

Zuhair

[Zuhair]

NO, Cantor's diagonal argument construct a diagonal r that differs
from every element of a COUNTABLE set of reals by a finite initial
segment n(r). You are just not getting Cantor's argument. Cantor's
argument is not about diagonalizing the set of ALL reals since that is
clearly not possible, Cantor's argument is about diagonalizing any
COUNTABLE set of reals. So again for any *countable* set S of reals
there is a diagonal r that differs from each real in S by a finite
initial segment n(r).
This has been PROVED by Cantor. This logically entails that the set R
of ALL reals cannot be countable, since otherwise this leads to the
obvious contradiction that R is missing a member of it which cannot
be.

The complete infinite binary tree have paths that can represent any
real! So no diagonal path can be defined over the whole set, this is
clear. But again also using Cantor's argument we can prove that for
Any subtree T of the complete infinite binary tree if T has countably
many paths then we can define a diagonal path that is missing from T,
i.e. a diagonal path that belongs to the complete infinite binary tree
but yet missing from T. Thus the infinite binary tree itself cannot be
countable!

I reviewed your writings about the infinite binary tree. You want to
prove that if we assume completed infinity (which you don't believe it
to be a consistent assumption, so you don't believe even in the
possible existence of such objects that are completed infinite sets)
then we will arrive at a contradiction, this contradiction is the
Cantor-WM contradiction, that is:

If we assume that the sets N and R of all naturals and reals
respectively are completed infinite sets, then it follows that

(1) R is strictly bigger than N by Cantor's proof
(2) R is as big as N by the WM infinite binary tree proof

Thus R is both as big and not as big as N. A flagrant contradiction!

Thus our assumption that N and R are completed infinite sets is FALSE.

That was the line of your thought.

But unfortunately you couldn't prove (2), your alleged infinite binary
tree proof is simply flawed. I have examined it and reviewed your
writing and replies to posters here and there, all of them do not show
any proof. You have many many errors in your logical steps, one of
them is the one I just showed above. Of course If you manage to prove
that R is as big as N by any argument, then of course you'll be wright
in saying that no completed infinite sets of naturals and reals could
ever exist. But so far, you couldn't.

The main problem with your binary tree proof is that you keep
insisting on the obvious error of identifying the complete infinite
binary tree by its nodes, while a tree is generally identified by the
ordered pair of the set of all its nodes and the set of all its EDGES.
You think that once you've discerned the nodes of your tree then
that's it any tree that have those nodes would be it. Which is wrong,
I can simply construct another tree that completely differ from the
infinite complete binary tree and yet using the nodes of the complete
infinite binary tree themselves. So what identifies the complete
infinite binary tree, and any tree actually, and even any graph is the
set of all nodes and the set of all Edges of it, you keep forgetting
the edges.

Now the reals correspond to PATHS of the complete infinite binary
tree, that's correct! And there is no proof whatsoever up till today
that proves the number of those paths to be countable. We know that
the number of NODES of the complete binary tree is countable, we know
that you can identify every finite path of that tree (that is uni-
directional and that begins with the root node) with the end node of
it, and thus we know that we do have countably many finite paths of
the complete infinite binary tree. But that is not enough! for your
purposes you need to prove that the set of ALL PATHS whether finite or
infinite of the complete infinite binary tree is countable. Which you
haven't done. At last you had to resort to countability of infinite
paths by linking them to finitely many words that can describe them,
which is an argument that has nothing to do with the structure of the
infinite binary tree.And to address it Cantor was not speaking about
the subset of reals that is definable or actually describable by
finite words, he was speaking of the reals whether they are so
definable or not. And I showed you that via parameters there is no
upper limit on the number of reals other than the size of the universe
of discourse itself, and thus this precludes another argument of
impossibility of having uncountably many reals, since there is no
prior limit set on the size of the universe of discourse!

A final word about your project. You tried to prove an "inconsistency"
with the assumption of uncountability of reals, which you couldn't
manage to carry out. And nobody managed and nobody could ever manage
because uncountability of reals is provable in very weak fragments of
Z actually of second order arithmetic, and those are PROVED to be
consistent. However what should that mean! This mean that
uncountability of the reals is a consistent assumption! and thus it is
POSSIBLE that the reals are uncountable, and thus mathematics is to
cover this possibility since it is interesting really. It doesn't mean
at all that we are forced to acknowledge that the reals are
uncountable in the REAL world, no this is not what is entailed. All
what we are presuming here is that we can speak of a model in which
the reals are uncountable, and that is a POSSIBLE model. That's all.

I keep saying that: Mathematics is Discourse about POSSIBLE form. And
I'll speak about that in details in a separate post. Note the emphasis
on "possible". And what determines "possibility" is simply absence of
a contradiction, i.e. consistency. So if you provide a consistent
discourse then you are speaking about a model that might possibly
exist! And thus it would be a legitimate piece of mathematical query.

One would wonder saying is it the job of mathematics to investigate
all possibilities about form even if it flies high up in fantasy. The
answer is yes and no. It is yes in the sense that those possibilities
are mathematical, and it is no in the sense that it is not practical
to do so, so mathematical research will search only those that are
valued as "interesting", and those are matters that rise in
mathematical discourse in almost recursive manner, starting from
obviously interesting issues that seems highly possible to be real,
and then investigating problems of them which would inherit that
interest, and possible solutions to those would be interesting and
those lead to investigating other problems with those solutions which
would become interesting raising possible solutions...problems... and
so on.

Does uncountability of reals raise within this context of mathematical
interest. The answer is YES.

Zuhair

[Virgil]

In article
<c30be873-3190-4c3d...@v7g2000yqv.googlegroups.com>,

I think you meant to say

(2) R is no bigger than N by the WM infinite binary tree proof'
What you actually said would, at least in standard English, allow R and
N still to satisfy (1).

--

[Zuhair]

On Dec 22, 11:09 am, Virgil <vir...@ligriv.com> wrote:
> In article

> <c30be873-3190-4c3d-a11a-67660228c...@v7g2000yqv.googlegroups.com>,

Note 'strictly' in (1).

Zuhair

[WM]

On 21 Dez., 21:05, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

Have you meanwhile understood that Cantor multiplies transfinite
numbers with numbers that are not cardinals or ordinals?

Regards, WM

[WM]

On 22 Dez., 00:34, William Hughes <wpihug...@gmail.com> wrote:
> > Yes it is definable. It has been defined. Nevertheless it does not
> > exist, because the sets do not exist.
>
> It is definable which means that WM can use it to prove that the
> bijection exists, but it does not exist.

The bijection of all finite words with all natural numbers has been
defined in binary:

0
1
00
01
10
11
000

and so on.

From this definition the natural number belonging to any desired
finite word can be obtained. It can easily be translated into any
other language. Or do you need some help?

Nevertheless there is no set of all natural numbers and no set of all
infinite words.

It is the same with pi. The (potentially) infinite string of digits of
pi can be defined. In fact there are (potentially) infinitely many
definitions. Nevertheless there is no actually infinite string
expressing pi.

Yes, I know that is not easy to understand. That's why so many
mathematicians have gone astray.

Regards, WM

[WM]

On 22 Dez., 05:02, Zuhair <zaljo...@gmail.com> wrote:

> > > A simple question that demands a simple answer: YES or NO?
>
> > Yes it is definable. It has been defined. Nevertheless it does not
> > exist, because the sets do not exist.
>
> When I asked my question I thought you were already familiar with what
> I meant by definable in a parameter free manner,

A definition in a parameter free manner is a sentence from which your
correspondent can recognize what you mean. Of course a definition can
only explain something by other words that are already known. Example:
When I define x = SUM 1/n! then you should know what I mean by x.

>
> Definition:
...

> you must show
> that there must exist a parameter free formula phi(y) such that a
> bijection f from N to the set of all finite words have it true that
> for all y. y in f <-> phi(y).

I need not do anything of that kind. In particular because it does not
change anything with respect to the fact that there are not more
mathematical objects than finite words.

Regards, WM

[WM]

On 22 Dez., 06:05, Zuhair <zaljo...@gmail.com> wrote:

> You are just not getting Cantor's argument.

You have not got my argument. The list is complete with respect to its
*enumeration*! There is no more place to insert another real number!

> Cantor's
> argument is not about diagonalizing the set of ALL reals since that is
> clearly not possible, Cantor's argument is about diagonalizing any
> COUNTABLE set of reals. So again for any *countable* set S of reals
> there is a diagonal r that differs from each real in S by a finite
> initial segment n(r).

And that shows that the Binary Tree should contain for every real a
*finite* path that differs from a finite initial segment of the real.
Alas there are only countably many finite paths. Or do you object this
simple truth? On the other hand we know that the Binary Tree contains
all possible finite initial segments. So there cannot be any finite
initial segment of any real that differs from all finite paths of the
Binary Tree. Hence there is a contradiction.

> Thus the infinite binary tree itself cannot be
> countable!

Then you should be able to find out *by means of nodes* what is
missing in the Binary Tree that I construct from countably many paths.

>
> Of course If you manage to prove
> that R is as big as N by any argument, then of course you'll be wright
> in saying that no completed infinite sets of naturals and reals could
> ever exist. But so far, you couldn't.

You are in error. Otherwise you could point out *by means of nodes at
finite positions* (others do not exist, which paths are missing in the
Binary Tree constructed from countably many paths). But you cannot.
Therefore all your attempts to "prove" anything else are condemned to
fail.

> The main problem with your binary tree proof is that you keep
> insisting on the obvious error of identifying the complete infinite
> binary tree by its nodes, while a tree is generally identified by the
> ordered pair of the set of all its nodes and the set of all its EDGES.

Every edge is in diect correspondence with the node at its end. You
should be able to recognize that.

> You think that once you've discerned the nodes of your tree then
> that's it any tree that have those nodes would be it. Which is wrong,
> I can simply construct another tree that completely differ from the
> infinite complete binary tree and yet using the nodes of the complete
> infinite binary tree themselves.

But how would you prove that it is completely different? A Binary Tree
is defined by its nodes only (or, what is tantamount) by its edges.
Every node except the root node is the end of an edge).

> So what identifies the complete
> infinite binary tree, and any tree actually, and even any graph is the
> set of all nodes and the set of all Edges of it, you keep forgetting
> the edges.

Are you really insisting that the edges are not isomorphic to the
nodes where they end?

>
> Now the reals correspond to PATHS of the complete infinite binary
> tree, that's correct! And there is no proof whatsoever up till today
> that proves the number of those paths to be countable.

Again: The proof is that you and nobody else can discern which paths
are missing when I construct or cover the complete Binary Tree by
means of countably many paths.

> I keep saying that: Mathematics is Discourse about POSSIBLE form.

Not quite correct. Mathematics is *possible* discourse about possible
froms.

> Does uncountability of reals raise within this context of mathematical
> interest. The answer is YES.

As you are unable to discern, from the (kept secret) bunch of paths
that I use to construct all possible finite paths of the Binary Tree,
which further paths are missing, your belief in uncountably many paths
is purely unmathematical nonsense.

Regards, WM

[WM]

On 22 Dez., 06:05, Zuhair <zaljo...@gmail.com> wrote:

> On Dec 21, 8:44 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> Cantor's argument is about diagonalizing any
> COUNTABLE set of reals. So again for any *countable* set S of reals
> there is a diagonal r that differs from each real in S by a finite
> initial segment n(r).

But each of these different initial segments n(r) is contained in
another place of *that* Binary Tree which contains only all finite
paths - hence is countable even in your opinion.

Therefore Cantor's argument here fails in case of an obviously
countable set.

> This has been PROVED by Cantor

not considering the Binary Tree!

> This logically entails that the set R
> of ALL reals cannot be countable,

and that it must be countable.

!
>
> A final word about your project. You tried to prove an "inconsistency"
> with the assumption of uncountability of reals, which you couldn't
> manage to carry out. And nobody managed and nobody could ever manage
> because uncountability of reals is provable in very weak fragments of
> Z actually of second order arithmetic, and those are PROVED to be
> consistent.

From that you can see what a mess your "proofs" are. It's matheology.
No serious person will give a dime for that nonsese.

Regards, WM

[Virgil]

In article
<27b9237c-12bb-4467...@10g2000yqo.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 22 Dez., 06:05, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 21, 8:44�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Cantor's argument is about diagonalizing any
> > COUNTABLE set of reals. So again for any *countable* set S of reals
> > there is a diagonal r that differs from each real in S by a finite
> > initial segment n(r).
>
> But each of these different initial segments n(r) is contained in
> another place of *that* Binary Tree which contains only all finite
> paths - hence is countable even in your opinion.

Actually each finite initial segment will be contained in uncountably
many complete paths. But you tree is provably incomplete, since, if
there are only countably many, one must by the very definition of
countability be able to list its members and thus by Cantor's diagonal
argument show that a least one is missing.,

>
> Therefore Cantor's argument here fails in case of an obviously
> countable set.

You countable set of paths or binary sequences is provably incomplete by
means of the very proof of its countability:
To becoutnatble requires being listable, and for listable sets of binary
sequences, Cantor proved inocmpleteness.

>
> > This has been PROVED by Cantor
>
> not considering the Binary Tree!

Actually Cantor considered the set of all binary sequences which is the
the same as a set of all paths in a Complete Infinite Binary Tree.

>
> > This logically entails that the set R
> > of ALL reals cannot be countable,
>
> and that it must be countable.

Not outside of WMytheology

> !
> >
> > A final word about your project. You tried to prove an "inconsistency"
> > with the assumption of uncountability of reals, which you couldn't
> > manage to carry out. And nobody managed and nobody could ever manage
> > because uncountability of reals is provable in very weak fragments of
> > Z actually of second order arithmetic, and those are PROVED to be
> > consistent.
>
> From that you can see what a mess your "proofs" are. It's matheology.

Which is far, far better than WMythology.

> No serious person will give a dime for that nonsese.

No serious person would give two cents for anything that needs your
WMytheology to support it.

But many would give as much or more for such things as ZFC in which your
WMytheology is naturally forbidden.
--

[William Hughes]

On Dec 22, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 22 Dez., 00:34, William Hughes <wpihug...@gmail.com> wrote:
>
> > > Yes it is definable. It has been defined. Nevertheless it does not
> > > exist, because the sets do not exist.
>
> > It is definable which means that WM can use it to prove that the
> > bijection exists, but it does not exist.
>
> The bijection of all finite words with all natural numbers has been
> defined in binary:

However, the bijection you need is all definitons with a subset
of the natural numbers. And there is no way to define this subset.
If you put restrictions on the 0/1 sequences you allow to exist you
put restrictions on the subsets you allow to exist. Note, that
subcountable
does not mean countable.

[William Hughes]

On Dec 22, 10:03 am, WM <mueck...@rz.fh-augsburg.de> wrote:

<snip>

> In particular because it does not
> change anything with respect to the fact that there are not more
> mathematical objects than finite words.

This only shows that the mathematical objects are subcountable.
You insist that they are also countable. For this you need to
demonstrate the existence of a bijection.

[Virgil]

In article
<3dfdf6df-a13d-4567...@z8g2000yqo.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 22 Dez., 06:05, Zuhair <zaljo...@gmail.com> wrote:
>
> > You are just not getting Cantor's argument.
>
> You have not got my argument. The list is complete with respect to its
> *enumeration*! There is no more place to insert another real number!

But there still provably exist at least as many unplaced reals as have
been placed in any given list.

>
>
> > Cantor's
> > argument is not about diagonalizing the set of ALL reals since that is
> > clearly not possible, Cantor's argument is about diagonalizing any
> > COUNTABLE set of reals. So again for any *countable* set S of reals
> > there is a diagonal r that differs from each real in S by a finite
> > initial segment n(r).
>
> And that shows that the Binary Tree should contain for every real a
> *finite* path that differs from a finite initial segment of the real.

Actually, one does not need Cantor's results to deduce that.

Given any real in, say, decimal format, EVERY real that differes from it
does so at some finite digit postion.

So similarly, any binary tree with more than one path does what WM says.

> Alas there are only countably many finite paths.

Irrelevant, as usual!

> Or do you object this
> simple truth?

Only to is being in any way relevant.

> On the other hand we know that the Binary Tree contains
> all possible finite initial segments. So there cannot be any finite
> initial segment of any real that differs from all finite paths of the
> Binary Tree. Hence there is a contradiction.

Only in WMytheology. The Cantor argument re binary sequences is totally
independent of WM's delusions about what it implies.

>
> > Thus the infinite binary tree itself cannot be
> > countable!
>
> Then you should be able to find out *by means of nodes* what is
> missing in the Binary Tree that I construct from countably many paths.

Once you tell me WHICH countably many paths, with proof of the set being
countable, I can easily provide you with just as many paths which you
missed, but until you tell me which ones you have, I cannot determine
which ones you do not have.

>
> >
> > Of course If you manage to prove
> > that R is as big as N by any argument, then of course you'll be wright
> > in saying that no completed infinite sets of naturals and reals could
> > ever exist. But so far, you couldn't.
>
> You are in error. Otherwise you could point out *by means of nodes at
> finite positions* (others do not exist, which paths are missing in the
> Binary Tree constructed from countably many paths). But you cannot.

Since you remain so careful not to tell anyone which paths you have, we
have no way of telling which ones you do not have, as the ones you
missed are determined entirely by the ones you did not miss,

> Therefore all your attempts to "prove" anything else are condemned to
> fail.

Only in WMytheology. In standard mathematics, standard proofs still
prove things.

>
> > The main problem with your binary tree proof is that you keep
> > insisting on the obvious error of identifying the complete infinite
> > binary tree by its nodes, while a tree is generally identified by the
> > ordered pair of the set of all its nodes and the set of all its EDGES.
>
> Every edge is in diect correspondence with the node at its end. You
> should be able to recognize that.
>
> > You think that once you've discerned the nodes of your tree then
> > that's it any tree that have those nodes would be it. Which is wrong,
> > I can simply construct another tree that completely differ from the
> > infinite complete binary tree and yet using the nodes of the complete
> > infinite binary tree themselves.
>
> But how would you prove that it is completely different? A Binary Tree
> is defined by its nodes only (or, what is tantamount) by its edges.
> Every node except the root node is the end of an edge).

It is neither the nodes in isolation nor thee edges in isolation, but
whether a node, other than the root node is a laft cild or a right child
of its parent node or whether an edge is a left branch or a right branch
from its parent node to its child node.

>
> > So what identifies the complete
> > infinite binary tree, and any tree actually, and even any graph is the
> > set of all nodes and the set of all Edges of it, you keep forgetting
> > the edges.
>
> Are you really insisting that the edges are not isomorphic to the
> nodes where they end?

Depends very much on what form of "isomorphism" you are claiming.

Certainly there is no ismorphism of any sort which bijects edges with
the nodes where they end.

> >
> > Now the reals correspond to PATHS of the complete infinite binary
> > tree, that's correct! And there is no proof whatsoever up till today
> > that proves the number of those paths to be countable.
>
> Again: The proof is that you and nobody else can discern which paths
> are missing when I construct or cover the complete Binary Tree by
> means of countably many paths.

Only because you refuse to tell us which paths you have.
The only way to tell which paths are missing is by knowing which paths
are not missing, but you are careful to keep that secret.

If you flip an honest coin and look at the result but keep it secret,
while I can know it to be either a head or a tail I cannot be sure which.

If WM creates a binary tree with only countably many infinite paths but
does not tell anyone else which paths he has included, then no one else
can tell WHICH paths are missing even though most of them will be.

>
>
> > I keep saying that: Mathematics is Discourse about POSSIBLE form.
>
> Not quite correct. Mathematics is *possible* discourse about possible
> froms.
>
> > Does uncountability of reals raise within this context of mathematical
> > interest. The answer is YES.
>
> As you are unable to discern, from the (kept secret) bunch of paths
> that I use to construct all possible finite paths of the Binary Tree,
> which further paths are missing, your belief in uncountably many paths
> is purely unmathematical nonsense.

It is your argument above which is, like so many of your other
arguments, unmathematical nonsense.

It is the fact of your keeping your set of paths SECRET that prevents
us from telling whether any particular path is or is not in it.

Any listing of them, and such a listing must be possible if there are
PROVABLY only countably many of them, itself provides proof that your
listing is incomplete.

That is why you must hide your list of paths!

Revealing it would prove you wrong!
--

[Virgil]

In article
<c53a1784-2d1f-4e8e...@f4g2000yqh.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 22 Dez., 05:02, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > A simple question that demands a simple answer: YES or NO?
> >
> > > Yes it is definable. It has been defined. Nevertheless it does not
> > > exist, because the sets do not exist.
> >
> > When I asked my question I thought you were already familiar with what
> > I meant by definable in a parameter free manner,
>
> A definition in a parameter free manner is a sentence from which your
> correspondent can recognize what you mean. Of course a definition can
> only explain something by other words that are already known. Example:
> When I define x = SUM 1/n! then you should know what I mean by x.

When countability and uncountability of a set were defined, it was in
terms or whether one could prove a surjection from N to the set or prove
that so such surjection can exist. Absence either proof for a set,
neither countability nor uncountability can be claimed.

Those definitions are far too clear and concise for use in WMytheology,
so that WM avoids them at all cost.

WM has claimed that he has a Complete Infinite Binary Tree with only
countably many paths but has carefully never tested whether his set of
paths satisfied the definition above of countability by proving the
necessary surjection.

WM has claimed that our definition of a Complete Infinite Binary Tree
cannot have a set of uncountably many paths even though we, and others,
have both shown repeatedly that such a set of paths does satisfy the
above definition of uncountability.

Thus we (not just me, but all those in opposition to WM's
anti-mathematical WMytheology) have been working with standard
mathematical techniques while WM has been fervently avoiding them.
--

[Virgil]

In article
<9687d371-bbee-46ee...@u19g2000yqj.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 22 Dez., 00:34, William Hughes <wpihug...@gmail.com> wrote:
> > > Yes it is definable. It has been defined. Nevertheless it does not
> > > exist, because the sets do not exist.
> >
> > It is definable which means that WM can use it to prove that the
> > bijection exists, but it does not exist.
>
> The bijection of all finite words with all natural numbers has been
> defined in binary:
>
> 0
> 1
> 00
> 01
> 10
> 11
> 000
> and so on.

While I can see some binaries, I do not see any words being paired off
with them. Thus WM's claim fails.

>
> From this definition the natural number belonging to any desired
> finite word can be obtained. It can easily be translated into any
> other language. Or do you need some help?

What is the natural number of the word "help"?

>
> Nevertheless there is no set of all natural numbers and no set of all
> infinite words.

In standard mathematics, like in ZFC, for example, there is a set of all
natural numbers, or at least a set having all the properrties one wants
or needs for a set of natural numbers.

>
> It is the same with pi. The (potentially) infinite string of digits of
> pi can be defined. In fact there are (potentially) infinitely many
> definitions. Nevertheless there is no actually infinite string
> expressing pi.

On the other hand, the ratio of the circumference of a circle to its
diameter is perfectly well defined.

>
> Yes, I know that is not easy to understand.

WM certainly illustrates his own difficulty in understanding quite
clearly!
--

[Virgil]

In article
<b8e8e12a-d8dc-4a5c...@u19g2000yqj.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Have you meanwhile understood that Cantor multiplies transfinite
> numbers with numbers that are not cardinals or ordinals?

I have certainly seen no evidence that he has ever multiplied any number
that is not also considered either a cardinal or an ordinal by any
transfinite number.
--

[camg...@hush.com]

On Dec 22, 7:17 am, Virgil <vir...@ligriv.com> wrote:

> The proven impossibility of any surjection from N to R, or injection

drivel nonsense S-PROOF!

AD(X) = FLIP(D(X))
AD(1) = FLIP(D(1))
AD(2) = FLIP(D(2))
AD(3) = FLIP(D(3))
...
AND SO ON...

THERFORE
AD(1) =/= D(1)
AD(2) =/= D(2)
AD(3) =/= D(3)
...
AND SO ON

===========

> from R to N means that R totally satisfies the definition of being an
> uncountable set according to the standard mathematical definition of
> uncountability.

Like Totally man! Whoo Wow MORE THAN INFINITY yeh dude!

>
> Learn to live with it, WM, or learn to avoid it entirely.
>

Uncountable Infinity is easy to live with as it has about the same utility as a mobius strip!


Herc
--
www.microPROLOG.com

[camg...@hush.com]

All countable subsets of N.

TM1(1) Halts => 1 e POWERSET_1
TM1(2) Loops => 2 !e POWERSET_1
TM1(3) Halts => 3 e POWERSET_1
TM2(1) Halts => 1 e POWERSET_2
TM2(2) Halts => 2 e POWERSET_2
TM2(3) Halts => 3 e POWERSET_2
...
POWERSET(N) = { {1,3,...} {1,2,3...} ... }

1 <=> {1,3,..}
2 <=> {1,2,3,...}
...

Herc
--
www.microPROLOG.com

[Virgil]

In article
<2263ace1-dfa7-466d...@v7g2000yqv.googlegroups.com>,

William Hughes <wpih...@gmail.com> wrote:

> Note, that
> subcountable
> does not mean countable.

I am not at all sure of what you mean by subcountable.

Or even by countable if your subcountable does not imply countable.

The definition of a set being countable that I prefer is that a set is
countable provided that one can show a surjection from the set of
naturals, N, to the set in question.

Thus any subset of N is trivially countable but the set of all subsets
of N is not, and the set of all functions from N to a set of more than
one element also is not.
--

[|-|erc]

On Dec 23, 9:04 am, Virgil <vir...@ligriv.com> wrote:
> The definition of a set being countable  that I prefer is that a set is
> countable provided that one can show a surjection from the set of
> naturals, N, to the set in question.
>
> Thus any subset of N is trivially countable but the set of all subsets
> of N is not,

[William Hughes]

On Dec 22, 7:04 pm, Virgil <vir...@ligriv.com> wrote:
> In article

> <2263ace1-dfa7-466d-8341-c50692402...@v7g2000yqv.googlegroups.com>,

>  William Hughes <wpihug...@gmail.com> wrote:
>
> > Note, that
> > subcountable
> > does not mean countable.
>
> I am not at all sure of what you mean by subcountable.

A set X is subcountable if we can associate a different natural number
with every element x of X, call it f(x) In classical mathematics
subcountable
implies countable because f(X) must be a subset of the natural
numbers.
However, if we take a contructivist viewpoint, then we do not know
that f(X) is a subset (it may not be contructable). So in
constructive
mathematics the fact that X is subcountable, does not mean we can
find a bijection between X and some subset of the naturals, so X might
not be countable. E.g. in constructive mathematics the (constructive)
reals
are subcountable but not countable.

So the fact that a set is uncountable need not mean it is "bigger"
than
the natural numbers.

[Graham Cooper]

Herc turns his mobius strip over... even more to see there!

Herc

[Virgil]

In article
<ee1d96e2-fbb9-4c48...@s14g2000yqg.googlegroups.com>,

But the constraints of your "constructive mathematics' are not required
in classical mathematics when not doing your constructive mathematics,
so are not relevant in classical mathematics.
--

[WM]

On 22 Dez., 21:52, Virgil <vir...@ligriv.com> wrote:

> Actually each finite initial segment will be contained in uncountably
> many complete paths.

And none of them will contain more than finite initial segments. I.e.,
every path consists only of finite initial segments!

> But you tree is provably incomplete, since, if
> there are only countably many,

There is no doubt among informed scholars that the set of finite
initial segments is countable and that no infinite path contains more
than countably many finite initial segments.

> one must by the very definition of
> countability be able to  list its members and thus by Cantor's diagonal
> argument show that a least one is missing.,

Not by Cantor's diagonal argument, but by pointing to a path that is
not in the Binary Tree. You must be able to identify *by nodes* (or
edges, what is the same) a path that is not in the CIBT that I have
constructed from countably many paths.

>
> > Therefore Cantor's argument here fails in case of an obviously
> > countable set.
>
> You countable set of paths or binary sequences is provably incomplete by
> means of the very proof of its countability:
> To becoutnatble requires being listable, and for listable sets of binary
> sequences, Cantor proved inocmpleteness.

And I proved by your very inability to identify a missing path that
Cantor's proof (precisely his notion of countability i.e. finished
infinity) is wrong.

Regards, WM

[WM]

On 22 Dez., 22:06, William Hughes <wpihug...@gmail.com> wrote:
> On Dec 22, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:

> > The bijection of all finite words with all natural numbers has been
> > defined in binary:
>
> However, the bijection you need is all definitons with a subset
> of the natural numbers.

Even such a bijection would not support your case because the diagonal
of the infinite list of finite definitions is not a finite word. It
does not even exist!

>  And there is no way to define this subset.
> If you put restrictions on the 0/1 sequences you allow to exist you
> put restrictions on the subsets you allow to exist.   Note, that
> subcountable
> does not mean countable.

Then you give up the essence of set theory. Then you give up the
theorem of Schröder-Bernstein. Then set theory and the hierarchy of
infinities need not be contradicted. They simply do not exist. Set
theory demands: Every subset of a countable set is countable. To prove
that does not require a bijection.

Regards, WM

[WM]

When did you convert?
Note: I have refuted classical transfinite set theory. New messing
around will always appear. It is impossible to extinguish human
nonsense without extinguishing humans.

Regards, WM

[Albrecht]

On Saturday, December 22, 2012 10:10:25 PM UTC+1, William Hughes wrote:
> On Dec 22, 10:03 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
> <snip>
>
>
>
>
>
> > In particular because it does not
>
> > change anything with respect to the fact that there are not more
>
> > mathematical objects than finite words.
>
>
>
> This only shows that the mathematical objects are subcountable.

That's the only truth: The mathematical objects which ever could be considered are members of a subset of the class the of possible words.

>
> You insist that they are also countable. For this you need to
>
> demonstrate the existence of a bijection.

If you use multiple semantics you may biject any word with a mathematical object, and vic versa. But the formulation of that bijection is as complex as the formulation of the well-ordering of the reals, maybe.

[Albrecht]

On Saturday, December 22, 2012 6:05:34 AM UTC+1, Zuhair wrote:
> On Dec 21, 8:44 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>

> > On 21 Dez., 17:36, Zuhair <zaljo...@gmail.com> wrote:
>
> >
>
> >
>
> > Note finally: Every Cantor diagonal r differs from any other real
>
> > number by a finite initial segment n(r) of its string of digits. That
>
> > is not possible with the Binary Tree. A diagonal does not differ from
>
> > all finite paths, i.e., for every initial segment n(r) of every real
>
> > number r there exists a finite path of the Binary Tree that is n(r).
>
> > You may consider actual infinity as well as uncountable languages, but
>
> > that does not change the fact that Cantor's argument does not apply.
>
> >
>
>
>
> NO, Cantor's diagonal argument construct a diagonal r that differs
>

> from every element of a COUNTABLE set of reals by a finite initial
>
> segment n(r). You are just not getting Cantor's argument. Cantor's

>
> argument is not about diagonalizing the set of ALL reals since that is
>
> clearly not possible, Cantor's argument is about diagonalizing any
>
> COUNTABLE set of reals. So again for any *countable* set S of reals
>
> there is a diagonal r that differs from each real in S by a finite
>
> initial segment n(r).
>

> This has been PROVED by Cantor. This logically entails that the set R
>
> of ALL reals cannot be countable, since otherwise this leads to the
>
> obvious contradiction that R is missing a member of it which cannot
>
> be.
>

But how is ensured that any list of reals is actually provided? There is always the "emergency exit" that the list of all reals is not giveable, not because they are more than the naturals but because it just lacks the bijection (the algorithm, the construction, the realization of that bijection, whatever) to the naturals.

That's the interpretation with relativity in axiomatic. As we agreed, uncountability is only there if uncountability is claimed. As infinite sets are only there if infinite sets are claimed.

[William Hughes]

On Dec 23, 6:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
<snip>

> Set
> theory demands: Every subset of a countable set is countable.

However, given a collection X and a rule f such that
we can associate a different natural, f(x) to every
x in X, we cannot conclude that f(X) *is* a subset
(it may not be computatable, and we may only
allow computable objects). If you disallow arbitrary
0/1 sequence you must disallow arbitrary subsets.

[Virgil]

In article
<4a94bfd4-b26b-49f2...@b8g2000yqh.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Note: I have refuted classical transfinite set theory.

Not to the satisfaction of anyone but yourself.

Thus not outside Wolkenmuekenheim!
--

[Virgil]

In article
<8587c29c-635b-4eab...@a2g2000yqh.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:


> Set theory demands: Every subset of a countable set is countable. To
> prove that does not require a bijection.

Then WM concedes infinite sets and all that they imply, which includes
the falsehood of his own WMytheology and the uncountability of sets like
the set of all paths of a complete infinite binary tree.
--

[Virgil]

In article
<b48ee3dd-66ed-4c00...@f4g2000yqh.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 22 Dez., 21:52, Virgil <vir...@ligriv.com> wrote:
>
> > Actually each finite initial segment will be contained in uncountably
> > many complete paths.
>
> And none of them will contain more than finite initial segments. I.e.,
> every path consists only of finite initial segments!

Each path in an actual Complete Infinite Binary Tree is, effectively,
the same as a mapping from |N to {Left-child, right-child}.

That WM's tree does not contain any such paths proves that it is not a
Complete Infinite Binary Tree at all.

>
> > But you tree is provably incomplete, since, if
> > there are only countably many,
>
> There is no doubt among informed scholars that the set of finite
> initial segments is countable and that no infinite path contains more
> than countably many finite initial segments.

Which, while true, is totally irrelevant to the issue that WM is
disputing.

Each complete path is the union of infinitely many "nested" finite
initial segments, none of which is a path or even determinesany
particular path on its own.

In fact each finite initial segment of any path is the finite initial
segment of uncountably many paths, as many as in the whole tree, as is
proved by a simple bijection between the sets of its tails and those
paths.

>
> > one must by the very definition of
> > countability be able to �list its members and thus by Cantor's diagonal
> > argument show that a least one is missing.,
>
> Not by Cantor's diagonal argument, but by pointing to a path that is
> not in the Binary Tree. You must be able to identify *by nodes* (or
> edges, what is the same) a path that is not in the CIBT that I have
> constructed from countably many paths.

Since your hidden construction does not reveal which paths occur in your
tree, it also does not reveal the ones not in your tree, nor prove that
your set of paths is countable.

To prove that your set of paths is countable you must present us with a
complete listing of them, as that is the only legitimate way to justify
your claim of countability.

And when, if ever, you do that, your will simultaneously have proved
your listing is incomplete.

Thus WM loses again/

> >
> > > Therefore Cantor's argument here fails in case of an obviously
> > > countable set.
> >
> > You countable set of paths or binary sequences is provably incomplete by
> > means of the very proof of its countability:

> > To be coutnatble requires being listable, and for listable sets of binary

> > sequences, Cantor proved inocmpleteness.
>
> And I proved by your very inability to identify a missing path that
> Cantor's proof (precisely his notion of countability i.e. finished
> infinity) is wrong.

WM has a set of paths whose individual identities he will not reveal,
which he claims he can list and challenges others to find a nonmember
path.

Well, if that set is listable, for any such listing Cantor has proved
the existence of non members.

So apparently WM is claiming the existence of a set which he can prove
to be countable but which he can also prove not to be listable.

But in the real world of real mathematics, rather than WM's WMytheology,
a set being countable is DEFINED to mean that there is a surjection from
N to that set.

So that whatever definition of "countable" WM is using, it is certainly
not valid in the real world of real mathematics.
--

[William Hughes]

On Dec 23, 1:12 am, Virgil <vir...@ligriv.com> wrote:
> In article

> <ee1d96e2-fbb9-4c48-a02e-cef0a3204...@s14g2000yqg.googlegroups.com>,

Indeed. However the original post in this thread was concerned
with the affect of definability on Cantor's argument.
I note that Cantor's theorem is perfectly valid with the
assumption that no unconstructable object exists,
there is no (contructable) list of all (contructable) reals,
so the reals remain uncountable. My remarks are aimed
at the obvious question, "If every constructable number is given by
a string, is there not an injection from the constructable numbers
to the naturals, and hence are the constructable numbers
not countable?" The problem is the collection of all naturals
which represent constructable numbers is not a constructable
subset of the naturals.

I am not a constructivist, though I see the appeal.

[Graham Cooper]

On Dec 24, 10:50 am, William Hughes <wpihug...@gmail.com> wrote:
>
> there is no (contructable) list of all (contructable) reals

yes there is!

All countable subsets of N.
TM1(1) Halts => 1 e POWERSET_1
TM1(2) Loops => 2 !e POWERSET_1
TM1(3) Halts => 3 e POWERSET_1
TM2(1) Halts => 1 e POWERSET_2
TM2(2) Halts => 2 e POWERSET_2
TM2(3) Halts => 3 e POWERSET_2
...
POWERSET(N) = { {1,3,...} {1,2,3...} ... }

1 <=> {1,3,..}
2 <=> {1,2,3,...}
...

A semi-decidable set of all subsets of N.

What subset of N is missing?

Herc
--
www.CAMGIRLS.com
TOTAL: $2834 2012-12-21 Fri
Not the End of the World!

[WM]

I do not disallow anaything. But you are unable to identify the
corresponding paths in the Binary Tree. And as nobody can accomplish
that task, there are no such paths that could be used in mathematics.

By the way, if you adhere to constructivism, then there is no
uncountable set at all. So, however you may turn it, uncountability is
disproved.

Regards, WM

[WM]

There is no list asked for. What should that strawman be good for?
There is only the question whether the cardinality of the reals can be
larger than that of the natural numbers. The answer is no.

And in constructivism, there is nothing uncountable because everything
that can be constructed belongs to a countable set.

> My remarks are aimed
> at the obvious question, "If every constructable number is given by
> a string, is there not an injection from the constructable numbers
> to the naturals, and hence are the constructable numbers
> not countable?"  The problem is the collection of all naturals
> which represent constructable numbers is not a constructable
> subset of the naturals.

And nobody asks this silly question - except some matheologians who
want to cheat the constructivists.

>
> I am not a constructivist,

Then you should see that a subset of a countable set is a countable
set.

Regards, WM

[Virgil]

In article
<0a42746a-e21c-4e7b...@x3g2000yqo.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 24 Dez., 01:50, William Hughes <wpihug...@gmail.com> wrote:

> > On Dec 23, 1:12嚙窮m, Virgil <vir...@ligriv.com> wrote:
> >
> >
> >
> >
> >
> > > In article
> > > <ee1d96e2-fbb9-4c48-a02e-cef0a3204...@s14g2000yqg.googlegroups.com>,

> > > 嚙磕illiam Hughes <wpihug...@gmail.com> wrote:

> >
> > > > On Dec 22, 7:04嚙緘m, Virgil <vir...@ligriv.com> wrote:
> > > > > In article
> > > > > <2263ace1-dfa7-466d-8341-c50692402...@v7g2000yqv.googlegroups.com>,

> > > > > 嚙磕illiam Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > > Note, that
> > > > > > subcountable
> > > > > > does not mean countable.
> >
> > > > > I am not at all sure of what you mean by subcountable.
> >
> > > > A set X is subcountable if we can associate a different natural number

> > > > with every element x of X, call it f(x) 嚙瘢n classical mathematics

> > > > subcountable
> > > > implies countable because f(X) must be a subset of the natural
> > > > numbers.
> > > > However, if we take a contructivist viewpoint, then we do not know

> > > > that f(X) is a subset 嚙�it may not be contructable). 嚙磅o in

> > > > constructive
> > > > mathematics the fact that X is subcountable, does not mean we can
> > > > find a bijection between X and some subset of the naturals, so X might

> > > > not be countable. 嚙瘟.g. in constructive mathematics the (constructive)

> > > > reals
> > > > are subcountable but not countable.
> >
> > > > So the fact that a set is uncountable need not mean it is "bigger"
> > > > than
> > > > the natural numbers.
> >
> > > But the constraints of your "constructive mathematics' are not required
> > > in classical mathematics when not doing your constructive mathematics,
> > > so are not relevant in classical mathematics.
> >

> > Indeed. 嚙瘡owever the original post in this thread was concerned

> > with the affect of definability on Cantor's argument.
> > I note that Cantor's theorem is perfectly valid with the
> > assumption that no unconstructable object exists,
> > there is no (contructable) list of all (contructable) reals,
> > so the reals remain uncountable.
>
>
> There is no list asked for.

The claim of countability for any set requires a surjection from N to
that set claimed to validate the claim of countability, and such a
surjectiin IS a list

> What should that strawman be good for?

Your strawmen are , as usual, good for nothing.

> There is only the question whether the cardinality of the reals can be
> larger than that of the natural numbers. The answer is no.

The only legitimate proof of such a claim would be presentation of a
surjection from N to R, as cardinalities are defined in terms of such
surjections.

Card(A) >= Card (B) if and only if A surjects to B

But WM seems not to understand that rejecting such definition is both
WMytheological and illogical.

>
> And in constructivism, there is nothing uncountable because everything
> that can be constructed belongs to a countable set.
>

> >嚙瞎y remarks are aimed

> > at the obvious question, "If every constructable number is given by
> > a string, is there not an injection from the constructable numbers
> > to the naturals, and hence are the constructable numbers

> > not countable?" 嚙確he problem is the collection of all naturals

> > which represent constructable numbers is not a constructable
> > subset of the naturals.
>
> And nobody asks this silly question - except some matheologians who
> want to cheat the constructivists.

"Constructionists" like WM cheat themselves quite effectively enough,
and need no external aid to mess things up for them even more.

But outside of Wolkenmuekenheim his own type is "constrictionist"!
--

[Virgil]

In article
<249ebc03-0d10-4866...@f8g2000yqa.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 23 Dez., 21:32, William Hughes <wpihug...@gmail.com> wrote:
> > On Dec 23, 6:06�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > <snip>
> >
> > > Set
> > > theory demands: Every subset of a countable set is countable.
> >
> > However, given a collection X and a rule f such that
> > we can associate a different natural, f(x) to every
> > x in X, we cannot conclude that f(X) *is* a subset
> > (it may not be computatable, and we may only
> > allow computable objects). �If you disallow arbitrary
> > 0/1 sequence you must disallow arbitrary subsets.
>
> I do not disallow anaything.

Except logic and truth.

> But you are unable to identify the
> corresponding paths in the Binary Tree. And as nobody can accomplish
> that task, there are no such paths that could be used in mathematics.

Until WM can list the whole set of paths in his allegedly Complete
Infinite Binary Tree, or at least show that it can be listed. he has no
evidence that it is countable.

But as soon as he does, that will prove it's incompleteness.
--

[Alan Smaill]

It remains that for any enumeration of reals,
there is a real not in the range of the enumeration.




>
> Regards, WM

--
Alan Smaill

[Virgil]

WM <muec...@rz.fh-augsburg.de> writes:

What WM is careful not ever to write is what the definition of, and test
for, a set being countable is

At least outside of WMytheology it is:

"A set A is countable if and only if one can prove a surjection from the
set of naturals, |N to the set A."

Or words to the same effect.

One cannot claim countability without being able to present such an
injection (or an equivalent injection from A to |N) and the
establishment of such a mapping also establishes countability.

And also if one can show that no map from |N to A CAN be a surjection,
one has established that A is not countable, or more briefly, is
uncountable.

What constitutes countability or uncountability in WMytheology no one
but WM knows because WM will not say, but WM does go so far as to deny
that it is the same as in standard mathematics.
--

[WM]

On 24 Dez., 23:54, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

>
> It remains that for any enumeration of reals,
> there is a real not in the range of the enumeration.

It remains that in every enumeration of whatever there is a natural
number missing.

Regards, WM

[Virgil]

In article
<c95026cf-78c6-444e...@x10g2000yqx.googlegroups.com>,

Then WM must have a badly screwed up definition of ennumeration.

Ennumeration of a set only requires a rule by which each of the members
of the set to be ennumerated is assigned a natural number, it even
allows such a member to be assigned more than one natural number as long
as each such member is assigned at least one natural.

And it does not even require that every natural number be used in the
process.

The trouble with WM's thinking, if one can call it that, is that he has
only the vaguest grasp of what the formal definitions of standard
mathematics actually say and an ever vaguer grasp of what they mean.

For example "countability" or "non-countability" of a set.

The formal definitions say that a set, S, is countable if and only if
there exists a surjective mapping from |N to S, and is otherwise
uncountable.

But WM's usage of countable and uncountable never mention whether such a
surjection exists or not, but always goes off on tangents unrelated to
the actual definitions of countable and uncountable.

Thus none of his claims of countability or uncountability are at all
related to the defined meanings of the words.
--