Re: WM's Matheology § 030

[Uergil]

In article
<20d2f7a3-d22f-4fcf...@s9g2000vbg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 8 Jun., 22:00, Uergil <Uer...@uer.net> wrote:
>
> > > > So that in WM's "tree" every path eventually has the same tail?
> >
> > > Yes, exactly.
> >
> >  There is an obvious bijection between any set of paths all having the
> > same tail and the set of finite binary sequences, by merely cutting off
> > those identical tails.
>
> Infinite bijections do not exist.
> >
> > Then WM is effectively claiming that there exists a bijection between
> > the set of finite paths, which an be listed lexicographically, and the
> > set of all paths, which cannot be listed at all.
> >
> > Until you can actually display such a bijection, why should anyone
> > listen to you when you claim you can?
>
> Infinite bijections do not exist.

So there is no bijection between the infinite set of even naturals and
the infinite set of odd naturals?

Or from any infinite set to itself?

For a finite set of n elements there are n-factorial bijections to
itself for any set of n elements

I know of one that will work for the 0 origin naturals and another that
will work for the 1-origin naturals, so which noes doe WM claim do not
exist.
>
> Actually infinite paths cannot. Otherwise try it. What cannot be
> distingusihed cannot be put in bijection and cannot exist as a number.

What cannot exist in WM's matheology are often quite common in standard
mathematics.
>
>
> > > Every path will have that tail. And you should guess which one that is
> > > - and what paths are missing.
> >
> > Cut off the tails
>
> No, I don't.

Since the tails are all the same, only a single exemplar is needed,
which can be supplied by that path which is all tail, and then the rest
of them lose nothing by being cut off where their identical tails begin.


So that WM's "tree" is equivalent to one with only finite paths.
>
> >and you can list all your paths lexicographically, but
> > the set of all paths cannot be listed.
> >
> > So until WM can refute the Cantor diagonal proof, his allgedly
> > "complete" infinite binary tree remains incomplete.
>
> Tell me what is missing.

Among other things, a valid refutation of Cantor's diagonal proof.
--
"Ignorance is preferable to error, and he is less
remote from the- truth who believes nothing than
he who believes what is wrong.
Thomas Jefferson

[WM]

On 9 Jun., 02:41, Uergil <Uer...@uer.net> wrote:
> In article
> <20d2f7a3-d22f-4fcf-a8c7-c1d60aeb4...@s9g2000vbg.googlegroups.com>,

So I said. But you would not recognize a difference, when I used all
possible infinite paths!

> only a single exemplar is needed,
> which can be supplied by that path which is all tail, and then the rest
> of them lose nothing by being cut off where their identical tails begin.
>
> So that WM's "tree" is equivalent to one with only finite paths.

Not at all! Every path is infinite.

>
> >
> > Tell me what is missing.
>
> Among other things, a valid refutation of Cantor's diagonal proof.

So you agree that it is impossible (for you) to recognize a real
number by its infinite bit-sequence only? But that's just the argument
of the proponents of uncountably many real numbers: Although it is not
possible to give a finite definition for each number, each number is
"identified" by iths infinite sequence of bits or digits. Now you see
that this assertion is false. Why do you continue to believe that
Cantor's diagonal number could be identified by its actually infinite
sequence of digits???

It's become a habit with you. It's the sluggishness of your brain.

Regards, WM

[WM]

On 9 Jun., 02:41, Uergil <Uer...@uer.net> wrote:

> > Infinite bijections do not exist.
>
> So there is no bijection between the infinite set of even naturals and
> the infinite set of odd naturals?
>
> Or from any infinite set to itself?

Correct. There is no complete infinite set. Hence there is no complete
bijection.

Regards, WM

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:fccd29ff-d2d9-45a5...@l17g2000vbj.googlegroups.com...

> [...] Why do you continue to believe that

> Cantor's diagonal number could be identified by its actually infinite
> sequence of digits???

What is the problem with that? We do have a rule for the digits. Then, of
course, it is "impredicative": the prototypical impredicativity maybe. Then
proofs are easy: assume an impossibility.

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message
news:jqvl9q$m5q$1...@speranza.aioe.org...

P.S. To me (at this point), the diagonal argument tells that there are more
strings than rules. In fact, rules are finite, strings are not.

-LV

[WM]

On 9 Jun., 16:14, "LudovicoVan" <ju...@diegidio.name> wrote:
> "LudovicoVan" <ju...@diegidio.name> wrote in message
>
> news:jqvl9q$m5q$1...@speranza.aioe.org...
>

> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message

> >news:fccd29ff-d2d9-45a5...@l17g2000vbj.googlegroups.com...
>
> >> [...]  Why do you continue to believe that
> >> Cantor's diagonal number could be identified by its actually infinite
> >> sequence of digits???
>
> > What is the problem with that?  We do have a rule for the digits.  Then,
> > of course, it is "impredicative": the prototypical impredicativity maybe.
> > Then proofs are easy: assume an impossibility.
>
> P.S.  To me (at this point), the diagonal argument tells that there are more
> strings than rules.  In fact, rules are finite, strings are not.

If this were so, then you could tell the infinite paths I use for
constructing the complete infinite Binary Tree.

But nobody can tell.

Further: The Cantor diagonal argument does not specify any number
unless all numbers of the complete list are defined (like the digits
of the number 0.123123123...). But then the list as well as its
diagonal belong to a countable set.

Regards, WM

[Virgil]

In article
<6246cc58-d3e4-48fa...@d6g2000vbe.googlegroups.com>,

The identity map on any set bijects it with itself, and any function
from an infinite set onto itself which is given by a finite formula,
such as f(n) = n+1 on the set of all integers, creates a bijection on an
infinite set.

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

[Virgil]

In article
<fccd29ff-d2d9-45a5...@l17g2000vbj.googlegroups.com>,

You have not used all possible infinite paths when every path has the
same tail as every other path.

If all paths have the same tail, they can be arranged lexicographically
into a list by de=tailing them, but the set of all paths cannot be
listed, as Cantor proved and WM has not unproved..

>
> Not at all! Every path is infinite.
> >
> > >
> > > Tell me what is missing.
> >
> > Among other things, a valid refutation of Cantor's diagonal proof.
>
> So you agree that it is impossible (for you) to recognize a real
> number by its infinite bit-sequence only?

Where is that a consequence of anything I have said?

> But that's just the argument
> of the proponents of uncountably many real numbers: Although it is not
> possible to give a finite definition for each number, each number is
> "identified" by iths infinite sequence of bits or digits. Now you see
> that this assertion is false. Why do you continue to believe that
> Cantor's diagonal number could be identified by its actually infinite
> sequence of digits???

Cantor does not have a diagonal "number", he has a rule for finding a
non-member for any given list of infinite binary sequences.

Misrepresenting Cantor, as WM does, does not disprove his proofs.

It's become a habit with WM. It's the sluggishness of his brain.
--

[Virgil]

In article
<7dbb2bcd-5335-45bc...@q2g2000vbv.googlegroups.com>,

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

> On 9 Jun., 16:14, "LudovicoVan" <ju...@diegidio.name> wrote:
> > "LudovicoVan" <ju...@diegidio.name> wrote in message
> >
> > news:jqvl9q$m5q$1...@speranza.aioe.org...
> >
> > > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> > >news:fccd29ff-d2d9-45a5...@l17g2000vbj.googlegroups.com...
> >
> > >> [...]  Why do you continue to believe that
> > >> Cantor's diagonal number could be identified by its actually infinite
> > >> sequence of digits???
> >
> > > What is the problem with that?  We do have a rule for the digits.  Then,
> > > of course, it is "impredicative": the prototypical impredicativity maybe.
> > > Then proofs are easy: assume an impossibility.
> >
> > P.S.  To me (at this point), the diagonal argument tells that there are more
> > strings than rules.  In fact, rules are finite, strings are not.
>
> If this were so, then you could tell the infinite paths I use for
> constructing the complete infinite Binary Tree.

But Wm does not ever construct any complete infinite Binary Trees.
His trees are always as necessarily incomplete as any listing of binary
sequences.

>
> But nobody can tell.
>
> Further: The Cantor diagonal argument does not specify any number
> unless all numbers of the complete list are defined (like the digits
> of the number 0.123123123...).

But it does specify a way to be sure that not all binary sequences are
included in a list, which is all it need to do.

> But then the list as well as its
> diagonal belong to a countable set.

And that very countability of that list guarantees that it is not
complete. I.e., omits at least one sequence. In fact, the argument can
be extended easily to show that any such list omits as many sequences as
it includes.
--

[WM]

On 9 Jun., 21:22, Virgil <vir...@ligriv.com> wrote:

> You have not used all possible infinite paths when every path has the
> same tail as every other path.

But you cannot tell which I used. And that would be possible, if you
could obtain information from infinite sequences.

> Cantor does not have a diagonal "number", he has a rule for finding a
> non-member for any given list of infinite binary sequences.

"given" list! A list is given, if every entry is known, i.e., defined
by a finite definition like this

0.1
0.11
0.111
...

and *countably* many others.

Regards WM

[Virgil]

In article
<3271cdcf-0bbb-4fc8...@w24g2000vby.googlegroups.com>,

The existence of such a list proves the countability of the things
listed, as ,for example, the set of finite binary sequences can be
listed in lexicographical order and thus proven to be only countably
infinite, but the set of infinite binary sequences cannot be so listed.
--

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:7dbb2bcd-5335-45bc...@q2g2000vbv.googlegroups.com...

> On 9 Jun., 16:14, "LudovicoVan" <ju...@diegidio.name> wrote:
>> "LudovicoVan" <ju...@diegidio.name> wrote in message
>> news:jqvl9q$m5q$1...@speranza.aioe.org...
>> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>> >news:fccd29ff-d2d9-45a5...@l17g2000vbj.googlegroups.com...
>>
>> >> [...] Why do you continue to believe that
>> >> Cantor's diagonal number could be identified by its actually infinite
>> >> sequence of digits???
>>
>> > What is the problem with that? We do have a rule for the digits.
>> > Then,
>> > of course, it is "impredicative": the prototypical impredicativity
>> > maybe.
>> > Then proofs are easy: assume an impossibility.
>>
>> P.S. To me (at this point), the diagonal argument tells that there are
>> more
>> strings than rules. In fact, rules are finite, strings are not.
>
> If this were so, then you could tell the infinite paths I use for
> constructing the complete infinite Binary Tree.
>
> But nobody can tell.

I concur with you on that: nobody can tell because the digits are not ever
enough to tell what the rule is, in principle. My point was that the
anti-diagonal is rather not a path of your tree. That is, the objection was
irrelevant.

> Further: The Cantor diagonal argument does not specify any number
> unless all numbers of the complete list are defined (like the digits
> of the number 0.123123123...). But then the list as well as its
> diagonal belong to a countable set.

I'd think that is not a valid objection. "(0)" is a rule, "(123)" is a
rule, 'pi' is a rule, etc. etc., and the anti-diagonal is a rule, too: I
mean, we have *every* digit. OTOH, the anti-diagonal is the one here that
is (possibly) based on an impredicative definition: it is defined in terms
of the whole set of listed strings, to the point that I'd agree that the
anti-diagonal is essentially a non-number. Not that I am any sure of my
positions, but I am conjecturing that the real numbers are rules (a subset
of the surreals), while infinite strings are essentially uncountable (even
the surreals can be diagonalized, right?). And even after the surreals, the
realm of games (at a quick glance) does not seem to add anything
substantially new: infinite strings just cannot be exhausted. -- Or so it
seems from here...

-LV

[WM]

On 10 Jun., 04:41, "LudovicoVan" <ju...@diegidio.name> wrote:

>
> I concur with you on that: nobody can tell because the digits are not ever
> enough to tell what the rule is, in principle.  My point was that the
> anti-diagonal is rather not a path of your tree.  That is, the objection was
> irrelevant.

But the antidiagonal is a real number between 0 and 1. If you express
it in binary, then it is a path in the tree.

>
> > Further: The Cantor diagonal argument does not specify any number
> > unless all numbers of the complete list are defined (like the digits
> > of the number 0.123123123...). But then the list as well as its
> > diagonal belong to a countable set.
>
> I'd think that is not a valid objection.  "(0)"  is a rule, "(123)" is a
> rule, 'pi' is a rule, etc. etc., and the anti-diagonal is a rule, too: I
> mean, we have *every* digit.  OTOH, the anti-diagonal is the one here that
> is (possibly) based on an impredicative definition: it is defined in terms
> of the whole set of listed strings, to the point that I'd agree that the
> anti-diagonal is essentially a non-number.  Not that I am any sure of my
> positions, but I am conjecturing that the real numbers are rules

I agree completely. Every real number, be it 0.25 or 0.333... or
sqrt(17), is a finite rule.

> (a subset
> of the surreals), while infinite strings are essentially uncountable (even
> the surreals can be diagonalized, right?).  And even after the surreals, the
> realm of games (at a quick glance) does not seem to add anything
> substantially new: infinite strings just cannot be exhausted.  --  Or so it
> seems from here...

Here I cannot agree. If infinite strings without any finite definition
should make sense, then it must be possible to use them without a
finite definition and to recognize them as numbers without knowing of
a finite definition. But that's impossible by their very being
infinite. Information transfer under all circumstances must be done by
finite strings of bits. And that holds also for monologizing and
thinking.

Regards, WM

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:f9a25ad9-2448-44e9...@j9g2000vbk.googlegroups.com...

> On 10 Jun., 04:41, "LudovicoVan" <ju...@diegidio.name> wrote:
>
>> I concur with you on that: nobody can tell because the digits are not
>> ever
>> enough to tell what the rule is, in principle. My point was that the
>> anti-diagonal is rather not a path of your tree. That is, the objection
>> was
>> irrelevant.
>
> But the antidiagonal is a real number between 0 and 1. If you express
> it in binary, then it is a path in the tree.

In depends in which tree, surely not in the OP's: if you diagonalize the
finite strings (up to the dyadic rationals, I am claiming), you will not get
a finite string (you will not get a dyadic rational). But, to the point: to
me it is correct that the anti-diagonal is a digits rule, so a potential
candidate to be a real: then it isn't a real because the rule is
impredicative.

>> > Further: The Cantor diagonal argument does not specify any number
>> > unless all numbers of the complete list are defined (like the digits
>> > of the number 0.123123123...). But then the list as well as its
>> > diagonal belong to a countable set.
>>
>> I'd think that is not a valid objection. "(0)" is a rule, "(123)" is a
>> rule, 'pi' is a rule, etc. etc., and the anti-diagonal is a rule, too: I
>> mean, we have *every* digit. OTOH, the anti-diagonal is the one here
>> that
>> is (possibly) based on an impredicative definition: it is defined in
>> terms
>> of the whole set of listed strings, to the point that I'd agree that the
>> anti-diagonal is essentially a non-number. Not that I am any sure of my
>> positions, but I am conjecturing that the real numbers are rules
>
> I agree completely. Every real number, be it 0.25 or 0.333... or
> sqrt(17), is a finite rule.

OK, but you don't express a position on the idea that the anti-diagonal is
essentially (i.e. not considering the "trivial" lists) a *non-number*. In
fact, that infinite binary strings are in themselves not numbers.

>> (a subset
>> of the surreals), while infinite strings are essentially uncountable
>> (even
>> the surreals can be diagonalized, right?). And even after the surreals,
>> the
>> realm of games (at a quick glance) does not seem to add anything
>> substantially new: infinite strings just cannot be exhausted. -- Or so
>> it
>> seems from here...
>
> Here I cannot agree. If infinite strings without any finite definition
> should make sense, then it must be possible to use them without a
> finite definition and to recognize them as numbers without knowing of
> a finite definition. But that's impossible by their very being
> infinite. Information transfer under all circumstances must be done by
> finite strings of bits. And that holds also for monologizing and
> thinking.

Indeed that is my point: that we cannot use infinite strings unless via
finite rules, so that they are *not* numbers, rather the rule picks some of
them and makes them (representations of) numbers. IOW, infinite strings
belong to another realm altogether, from which we pick (I suppose, in the
end) *numerals*, but of which we can only say that it cannot be
(arithmetically?) exhausted (so it is "uncountable" in the basic sense of
never countable, not even over transfinite ordinals). -- My terminology is
a bit tentative...

-LV

[Virgil]

In article
<f9a25ad9-2448-44e9...@j9g2000vbk.googlegroups.com>,

The Cantor 'antidiagonal' argument is totally conveyed in finitely many
bits.
--

[WM]

On 10 Jun., 20:27, Virgil <vir...@ligriv.com> wrote:

>
> The Cantor 'antidiagonal' argument is totally conveyed in finitely many
> bits.

But the antidiagonal is not, unless the list itself is defined by
finitely many bits.

Regards, WM

[WM]

On 10 Jun., 16:39, "LudovicoVan" <ju...@diegidio.name> wrote:

> Indeed that is my point: that we cannot use infinite strings unless via
> finite rules, so that they are *not* numbers, rather the rule picks some of
> them and makes them (representations of) numbers.  IOW, infinite strings
> belong to another realm altogether, from which we pick

How do you "pick" an infinite string that has not a finite rule?

Regrads, WM

[Virgil]

In article
<4db90d20-d4f5-4978...@n16g2000vbn.googlegroups.com>,

That is the problem only for those who reject Cantor's proof.

But claiming that those strings ARE countable they imply a rule does
exist for counting them, and Cantor merely shows that no such rule can
exist.
--

[Virgil]

In article
<960f4598-12dd-46ae...@p27g2000vbl.googlegroups.com>,

Cantor merely says that no such complete list can exist, as any such
claim of completeness allows proof of its own falsehood.
--

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:4db90d20-d4f5-4978...@n16g2000vbn.googlegroups.com...

As far as I can see, the only way that a rule could be "infinite" is by
impredicativity, such as with the anti-diagonal. But these strings are
never numbers. The diagonal argument tells that there are more infinite
strings than (finite) rules: it does not give a general way to distinguish
these strings, in fact it tells that no such general way can exist.

-LV

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message
news:virgil-92E00A....@bignews.usenetmonster.com...

> In article
> <960f4598-12dd-46ae...@p27g2000vbl.googlegroups.com>,
> WM <muec...@rz.fh-augsburg.de> wrote:
>> On 10 Jun., 20:27, Virgil <vir...@ligriv.com> wrote:
>>
>> > The Cantor 'antidiagonal' argument is totally conveyed in finitely many
>> > bits.
>>
>> But the antidiagonal is not, unless the list itself is defined by
>> finitely many bits.
>

> Cantor merely says that no such complete list can exist, as any such
> claim of completeness allows proof of its own falsehood.

I would not agree: the diagonal argument tells that no complete list of
*infinite strings* can exist, still we can give a (structurally) complete
list of any *numbers*, by construction. The real numbers do not make an
exception: a Cauchy sequence or a Dedekind cut are finite rules, it is the
identity between reals and infinite strings that is untenable. -- Maybe
impredicative definitions correspond to "infinite rules", but these are
essentially non-numbers, namely what we get when "wondering without (out
of/on top of)" the given structures. No?

-LV

[WM]

On 11 Jun., 21:37, "LudovicoVan" <ju...@diegidio.name> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

>
> news:4db90d20-d4f5-4978...@n16g2000vbn.googlegroups.com...
>
> > On 10 Jun., 16:39, "LudovicoVan" <ju...@diegidio.name> wrote:
>
> >> Indeed that is my point: that we cannot use infinite strings unless via
> >> finite rules, so that they are *not* numbers, rather the rule picks some
> >> of
> >> them and makes them (representations of) numbers.  IOW, infinite strings
> >> belong to another realm altogether, from which we pick
>
> > How do you "pick" an infinite string that has not a finite rule?
>
> As far as I can see, the only way that a rule could be "infinite" is by
> impredicativity, such as with the anti-diagonal.  But these strings are
> never numbers.

I fully agree.

>  The diagonal argument tells that there are more infinite
> strings than (finite) rules: it does not give a general way to distinguish
> these strings, in fact it tells that no such general way can exist.

And now it is only a short way to say that strings that cannot be
distinguished do not belong to any language.

Regards, WM

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:5ee4192d-ede0-48a3...@n33g2000vbi.googlegroups.com...

I do not think there is such thing as infinite strings that in principle
cannot be distinguished. But I would agree that infinite strings cannot be
exhausted by any language (of finite expressions over a finite alphabet).

-LV

[Virgil]

In article <jr5hea$qd2$1...@speranza.aioe.org>,

WRONG!

Two infinite binary strings which differ at all will necessarily differ
in a first finite position, and noting that strings which differ at some
postion are dffernt strings is precisely the general method.
--

[Virgil]

In article
<5ee4192d-ede0-48a3...@n33g2000vbi.googlegroups.com>,

Any two binary strings which differ at all , do so at some finite
position, so the problem is not to show that two such strings are
different, that is always possible in finite time if it is true.

The only difficulty would be in proving two strings the same.
--

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message

news:virgil-381C93....@bignews.usenetmonster.com...

It is you who are missing the point: we are talking about distinguishing,
i.e. listing all infinite strings.

> The only difficulty would be in proving two strings the same.

Right: I hope I am not too mistaken in saying that the "set" of infinite
strings is provably not recursively enumerable.

-LV

[Virgil]

In article <jr626n$qu$1...@speranza.aioe.org>,

Why bother trying to do what has been proven cannot be done>

>
> > The only difficulty would be in proving two strings the same.
>
> Right: I hope I am not too mistaken in saying that the "set" of infinite
> strings is provably not recursively enumerable.
>
> -LV
>

--

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message

news:virgil-A43D74....@bignews.usenetmonster.com...
<snip>

> Why bother trying to do what has been proven cannot be done

We were talking about the impossibility of such thing.

You have just entered denial mode: and I'll quit.

-LV

[Virgil]

In article <jr64uk$63j$1...@speranza.aioe.org>,

Giving up!
--

[Alan Smaill]

And, FWIW, the classical presentation of the reals in FOL
does not claim to have names/labels for every real.


> -LV
>
>

--
Alan Smaill

[Alan Smaill]

If it's not countable, then it's not enumerable in any sense.



> -LV
>
>

--
Alan Smaill

[LudovicoVan]

"Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
news:fweboko...@eriboll.inf.ed.ac.uk...
> "LudovicoVan" <ju...@diegidio.name> writes:
<snip>

>> Right: I hope I am not too mistaken in saying that the "set" of
>> infinite strings is provably not recursively enumerable.
>
> If it's not countable, then it's not enumerable in any sense.

Isn't countable equivalent to recursive? If so, not being countable does
not (necessarily) entail not being recursively enumerable.

-LV

[LudovicoVan]

"Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message

news:fwek3zc...@eriboll.inf.ed.ac.uk...
> "LudovicoVan" <ju...@diegidio.name> writes:
<snip>

>> I do not think there is such thing as infinite strings that in
>> principle cannot be distinguished. But I would agree that infinite
>> strings cannot be exhausted by any language (of finite expressions
>> over a finite alphabet).
>
> And, FWIW, the classical presentation of the reals in FOL
> does not claim to have names/labels for every real.

You should say "for every infinite string", i.e. you too are missing the
point: I am claiming (not alone) that real numbers are not infinite strings
tout court, actually that infinite strings tout court are not numbers.
While there are as many numbers as there can be names, as names are finite:
the logic of names does not look essentially different from that of numbers.

I'll say it again in the new style:

INFINITE STRINGS TOUT COURT ARE *NOT* NUMBERS.

-LV

[WM]

On 12 Jun., 12:04, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> And, FWIW, the classical presentation of the reals in FOL
> does not claim to have names/labels for every real.

Numbers belong to a subset of the set of names. "All" real numbers
were considered to exist (although they could not all be identified by
finite names) because they were believed to have their identities in
infinite strings of digits. I have proven that this is wrong. Nobody
can identify missing paths in the Binary Tree. So the last apology of
real numbers, being absent without excuse, has broken down.

Regards, WM

[LudovicoVan]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:81746ffd-2046-4ff2...@3g2000vbx.googlegroups.com...

> Numbers belong to a subset of the set of names.

I am intrigued by this: why do you think there are more names than numbers?
IOW, why do you think numbers and names are not essentially equivalent?

-LV

[WM]

On 12 Jun., 17:46, "LudovicoVan" <ju...@diegidio.name> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

In my opinion "Ludovico" is a name, but not a number. It denotes you
and probably some other men. Also "car" is a name but not a number. It
denotes many objects of a well-known kind. "Many" is a name but not
yet a number. It denotes manifolds of objects. Three is a name and a
number. It denotes threefolds from the holy trinity to sun, moon,
earth, and father, mother, child.

Regards, WM

[Virgil]

In article
<81746ffd-2046-4ff2...@3g2000vbx.googlegroups.com>,

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

> On 12 Jun., 12:04, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > And, FWIW, the classical presentation of the reals in FOL
> > does not claim to have names/labels for every real.
>
> Numbers belong to a subset of the set of names.

Perhaps so in WM's matheology, but mathematics has long been able to
distingish between numerals and numbers.

> "All" real numbers
> were considered to exist (although they could not all be identified by
> finite names) because they were believed to have their identities in
> infinite strings of digits. I have proven that this is wrong.

WM often claims to have proven things, but no one reading his alleged
"proofs" carefully is likely to agree.

> Nobody
> can identify missing paths in the Binary Tree.

In a complete infinite binary tree, there are no paths missing.
In WM's incomplete infinite binary trees most paths are missing, but
since WM will never identify which paths are present, we can't tell them
from the ones missing from his incomplete tree.

> So the last apology of
> real numbers, being absent without excuse, has broken down.

Only absent from WM's mind, as is so much of mathematics.
--

[Virgil]

In article
<2a43cc3b-7559-4a57...@f30g2000vbz.googlegroups.com>,

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

> On 12 Jun., 17:46, "LudovicoVan" <ju...@diegidio.name> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> > news:81746ffd-2046-4ff2...@3g2000vbx.googlegroups.com...
> >
> > > Numbers belong to a subset of the set of names.
> >
> > I am intrigued by this: why do you think there are more names than numbers?
> > IOW, why do you think numbers and names are not essentially equivalent?
>
> In my opinion "Ludovico" is a name, but not a number. It denotes you
> and probably some other men. Also "car" is a name but not a number. It
> denotes many objects of a well-known kind. "Many" is a name but not
> yet a number.

WRONG! In English, names are all required to be nouns or noun phrases.
"MANY" is not one of them, but is an adjective.

> It denotes manifolds of objects. Three is a name and a
> number.

Wrong again! At least in English, "three" is the name but three is a
number, the thing named.
--

[Alan Smaill]

No, it isn't --
the set of Turing machines that terminate on empty input
is countable, but not recursive. It's the effectiveness
of the enumeration that is at issue (in the usual story).

The claim that S is recursively enumerable is equivalent to the claim
that the there is an effective way of enumerating S, so that S has
to be the range of a function defined over the naturals; that restricts
the possible cardinality (in the usual story).

> -LV
>
>

--
Alan Smaill

[Alan Smaill]

"LudovicoVan" <ju...@diegidio.name> writes:

> "Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
> news:fwek3zc...@eriboll.inf.ed.ac.uk...
>> "LudovicoVan" <ju...@diegidio.name> writes:
> <snip>
>
>>> I do not think there is such thing as infinite strings that in
>>> principle cannot be distinguished. But I would agree that infinite
>>> strings cannot be exhausted by any language (of finite expressions
>>> over a finite alphabet).
>>
>> And, FWIW, the classical presentation of the reals in FOL
>> does not claim to have names/labels for every real.
>
> You should say "for every infinite string", i.e. you too are missing
> the point: I am claiming (not alone) that real numbers are not
> infinite strings tout court, actually that infinite strings tout court
> are not numbers.

It is the case, though, that the classical presentation of the reals in FOL
does not claim to have names/labels for every real. It also does
not have a term in the language for every infinite string;
but the primary objects of interest are the reals, rather
than the strings.

> While there are as many numbers as there can be
> names, as names are finite: the logic of names does not look
> essentially different from that of numbers.

But if there are reals without names, there will be more numbers than
names.

> I'll say it again in the new style:
>
> INFINITE STRINGS TOUT COURT ARE *NOT* NUMBERS.

As a slogan, that is already part of the standard story;
there is a bit of work to do to show that a real has an expansion
to any base.

> -LV
>
>

--
Alan Smaill

[WM]

On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> "LudovicoVan" <ju...@diegidio.name> writes:
>
> It is the case, though, that the classical presentation of the reals in FOL
> does not claim to have names/labels for every real.

It does not claim to have *finite* names/labels for every real. It
claims that reals are different from one another. Otherwise they could
not be elements of a set.

> > While there are as many numbers as there can be
> > names, as names are finite: the logic of names does not look
> > essentially different from that of numbers.
>
> But if there are reals without names, there will be more numbers than
> names.

Could they not be distinguished by digits, then Cantor's argument was
wrong. It works with digits. Real numbers must be distinguishable by
their infinite strings. But as I have proven, they are not. Cp. the
Binary Tree. Nobody can say what paths I use. Therefore most real
numbers are not distinct. They cannot belong as different elements to
a set.

Regards, WM

[Alan Smaill]

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

> On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> "LudovicoVan" <ju...@diegidio.name> writes:
>>
>> It is the case, though, that the classical presentation of the reals in FOL
>> does not claim to have names/labels for every real.
>
> It does not claim to have *finite* names/labels for every real.

Right.

> It claims that reals are different from one another.
>
> Otherwise they could not be elements of a set.

It claims 0.99999...... = 1.000000 ,
and they can be members of a set.

>> > While there are as many numbers as there can be
>> > names, as names are finite: the logic of names does not look
>> > essentially different from that of numbers.
>>
>> But if there are reals without names, there will be more numbers than
>> names.
>
> Could they not be distinguished by digits, then Cantor's argument was
> wrong.

Yes, every real has at least one decimal expansion.

> It works with digits. Real numbers must be distinguishable by
> their infinite strings.

The claim is that each real is distinguishable from each other in
differing in at least one digit (being careful to take into account
equivalent representations).

> But as I have proven, they are not.

What are you claiming?

That reals are not effectively distinguishable one from another,
in classical set theory? But that is well known, and does
not constitute a contradiction.



> Regards, WM

--
Alan Smaill

[WM]

On 13 Jun., 16:00, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >> "LudovicoVan" <ju...@diegidio.name> writes:
>
> >> It is the case, though, that the classical presentation of the reals in FOL
> >> does not claim to have names/labels for every real.
>
> > It does not claim to have *finite* names/labels for every real.
>
> Right.
>
> > It claims that reals are different from one another.
>
> > Otherwise they could not be elements of a set.
>
> It claims 0.99999......  = 1.000000 ,
> and they can be members of a set.

You should distinguish: Taken as real numbers "they" are one and the
same number with only two names like 1 and one.

Taken as digit sequences they are two elements thar easily can be
distinguished. Both have a finite definition.

>
> >> > While there are as many numbers as there can be
> >> > names, as names are finite: the logic of names does not look
> >> > essentially different from that of numbers.
>
> >> But if there are reals without names, there will be more numbers than
> >> names.
>
> > Could they not be distinguished by digits, then Cantor's argument was
> > wrong.
>
> Yes, every real has at least one decimal expansion.

Then tell me which reals I have used to construct the Binary Tree that
I recently pubslkished. I used only countably many. If there are
uncountably many different decimal expansions, you should be able to
tell that.

>
> > It works with digits. Real numbers must be distinguishable by
> > their infinite strings.
>
> The claim is that each real is distinguishable from each other in
> differing in at least one digit (being careful to take into account
> equivalent representations).
>
> > But as I have proven, they are not.
>
> What are you claiming?
>
> That reals are not effectively distinguishable one from another,
> in classical set theory?    But that is well known, and does
> not constitute a contradiction.

It has not been known. It has been claimed, as you did above, that
every one can be distinguished from all others. Why would you believe
in uncountably many different numbers, if you could not prove that
they are different? And even if it had been known, that fact would not
remove the contradiction with extensionality but would only show that
addicts do everything to maintain their drugs.

Regards, WM

[LudovicoVan]

"Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message

news:fwevciv...@eriboll.inf.ed.ac.uk...

> "LudovicoVan" <ju...@diegidio.name> writes:
>> "Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
>> news:fwek3zc...@eriboll.inf.ed.ac.uk...
>>> "LudovicoVan" <ju...@diegidio.name> writes:
>> <snip>
>>
>>>> I do not think there is such thing as infinite strings that in
>>>> principle cannot be distinguished. But I would agree that infinite
>>>> strings cannot be exhausted by any language (of finite expressions
>>>> over a finite alphabet).
>>>
>>> And, FWIW, the classical presentation of the reals in FOL
>>> does not claim to have names/labels for every real.
>>
>> You should say "for every infinite string", i.e. you too are missing
>> the point: I am claiming (not alone) that real numbers are not
>> infinite strings tout court, actually that infinite strings tout court
>> are not numbers.
>
> It is the case, though, that the classical presentation of the reals in
> FOL
> does not claim to have names/labels for every real.

If, as I am claiming, the real numbers are the set of (finite) rules that
output infinite strings in some fixed finite alphabet (so that, if I am not
mistaken, they are basically equivalent to the standard computable reals),
then there is a name for every real. For instance, just take the statement
of the rule as its name. Of course, I am reading the diagonal argument
differently: to me, it applies to infinite strings to show that *these* are
uncountable. In fact, I am taking some kind of constructive stance as to
what can legitimately be called a *number*. The standard reals are, in this
sense, an invalid construct: self-contradictorily both the set of rules and
the collection of infinite strings.

-LV

[LudovicoVan]

"Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message

news:fwe395z...@eriboll.inf.ed.ac.uk...

> "LudovicoVan" <ju...@diegidio.name> writes:
>> "Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
>> news:fweboko...@eriboll.inf.ed.ac.uk...
>>> "LudovicoVan" <ju...@diegidio.name> writes:
>> <snip>
>>
>>>> Right: I hope I am not too mistaken in saying that the "set" of
>>>> infinite strings is provably not recursively enumerable.
>>>
>>> If it's not countable, then it's not enumerable in any sense.
>>
>> Isn't countable equivalent to recursive? If so, not being countable
>> does not (necessarily) entail not being recursively enumerable.
>
> No, it isn't --
> the set of Turing machines that terminate on empty input
> is countable, but not recursive.

Isn't it not even recursively enumerable? Namely, how can it be countable
when you cannot tell which elements it has?

> It's the effectiveness
> of the enumeration that is at issue (in the usual story).

I am still rookie with these definitions, but I am (roughly) guessing that
the notion of non-computable function may be as valid as that of uncountable
number...

-LV

[Virgil]

In article
<117abba7-c448-4d24...@m10g2000vbn.googlegroups.com>,

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

> On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > "LudovicoVan" <ju...@diegidio.name> writes:
> >
> > It is the case, though, that the classical presentation of the reals in FOL
> > does not claim to have names/labels for every real.
>
> It does not claim to have *finite* names/labels for every real. It
> claims that reals are different from one another. Otherwise they could
> not be elements of a set.
>
> > > While there are as many numbers as there can be
> > > names, as names are finite: the logic of names does not look
> > > essentially different from that of numbers.
> >
> > But if there are reals without names, there will be more numbers than
> > names.
>
> Could they not be distinguished by digits, then Cantor's argument was
> wrong.

Cantor's diagonal argument is not about numbers at all, but about
mappings from N to some fixed two element set such as {m,w}.




> But as I have proven

Often claimed by WM, but rarely achieved to the satisfaction of anyone
but WM himself.
--

[Virgil]

In article <fwed353...@eriboll.inf.ed.ac.uk>,

Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> WM <muec...@rz.fh-augsburg.de> writes:
>
> > On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >> "LudovicoVan" <ju...@diegidio.name> writes:
> >>
> >> It is the case, though, that the classical presentation of the reals in FOL
> >> does not claim to have names/labels for every real.
> >
> > It does not claim to have *finite* names/labels for every real.
>
> Right.
>
> > It claims that reals are different from one another.
> >
> > Otherwise they could not be elements of a set.
>
> It claims 0.99999...... = 1.000000 ,
> and they can be members of a set.

Numerals like "0.99999" and "1.000000" are not numbers but merely names
for them.

And "0.99999...... = 1.000000" merely means that the two names identify
the same number.
>
--

[Virgil]

In article
<41f197c4-d1be-478b...@x21g2000vbc.googlegroups.com>,

And YOUR binary trees are provably incomplete, as there are paths
(infinite binary sequences) missing from any countable set of paths.
--

[YBM]

Virgil a �crit :

Yep, just like "WM" and "a german wanna-be mathematician stupid
dishonest asshole" identify the same person.

[Alan Smaill]

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

> On 13 Jun., 16:00, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 13 Jun., 11:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

...

>> >> It is the case, though, that the classical presentation of the
>> >> reals in FOL does not claim to have names/labels for every real.
>>
>> > It does not claim to have *finite* names/labels for every real.
>>
>> Right.
>>
>> > It claims that reals are different from one another.
>>
>> > Otherwise they could not be elements of a set.
>>
>> It claims 0.99999......  = 1.000000 ,
>> and they can be members of a set.
>
> You should distinguish: Taken as real numbers "they" are one and the
> same number with only two names like 1 and one.
>
> Taken as digit sequences they are two elements thar easily can be
> distinguished. Both have a finite definition.

Right -- and in either case we have a set of the elements.
So your inability to tell which of the views I was thinking
about does not stop these reals from being members of a set.

Your objection about extensionality is not coherent.
If all the reals were one and the same, that would not break
extensionality; there would still be a set of all reals
(with one element).

>> >> > While there are as many numbers as there can be
>> >> > names, as names are finite: the logic of names does not look
>> >> > essentially different from that of numbers.
>>
>> >> But if there are reals without names, there will be more numbers than
>> >> names.
>>
>> > Could they not be distinguished by digits, then Cantor's argument was
>> > wrong.
>>
>> Yes, every real has at least one decimal expansion.
>
> Then tell me which reals I have used to construct the Binary Tree that
> I recently pubslkished. I used only countably many. If there are
> uncountably many different decimal expansions, you should be able to
> tell that.

I have no idea what reals you used.

>> > It works with digits. Real numbers must be distinguishable by
>> > their infinite strings.
>>
>> The claim is that each real is distinguishable from each other in
>> differing in at least one digit (being careful to take into account
>> equivalent representations).
>>
>> > But as I have proven, they are not.
>>
>> What are you claiming?
>>
>> That reals are not effectively distinguishable one from another,
>> in classical set theory? But that is well known, and does
>> not constitute a contradiction.
>
> It has not been known.

You may be unaware of it, of course. The computable reals are reals,
even according to the drug-crazed cantorists. But there is no
algorithmic way to tell if two such reals are the same or not -- even
wikipedia has this information. A fortiori, there is no algorithm in
general. (I think this goes back to Turing 1936.)

"The order relation on the computable numbers is not computable"
"The same holds for the equality relation :
the equality test is not computable."

http://en.wikipedia.org/wiki/Computable_number

> It has been claimed, as you did above, that
> every one can be distinguished from all others.

As often, you misrepresent the view of others.
I said that the usual view (not my view) claims that the reals are
different from each other (not distinguishable).

You are the one who introduced the terminology of "distinguishable" --
it's up to you to explain the term. In English, it is not synonymous
with "different", nor does it play a role in the usual exposition
of real arithmetic.

> Why would you believe
> in uncountably many different numbers, if you could not prove that
> they are different?

Now you say "different"!

The usual, and also the constructive view, is that we cannot list
the real numbers; in both cases there is a proof that any listing
(effective in the latter case) is missing a real that is provably
different from the given reals. (And this is consistent with
the fact that the equality relation is not computable *in general*.)

> And even if it had been known, that fact would not
> remove the contradiction with extensionality but would only show that
> addicts do everything to maintain their drugs.

You have shown no contradiction with extensionality.

[Graham Cooper]

On Jun 20, 8:47 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> Your objection about extensionality is not coherent.
>

> You may be unaware of it, of course.  The computable reals are reals,
> even according to the drug-crazed cantorists.  But there is no
> algorithmic way to tell if two such reals are the same or not -- even
> wikipedia has this information.  A fortiori, there is no algorithm in
> general. (I think this goes back to Turing 1936.)
>
>   "The order relation on the computable numbers is not computable"
>   "The same holds for the equality relation :
>           the equality test is not computable."
>
>  http://en.wikipedia.org/wiki/Computable_number
>

THE SAME WAY YOU DETERMINE EQUALITY IN ZFC

ALGORITHMIC EXTENSIONALITY
ALL(x) ALL(y) ( ALL(a) TM-x(a)=TM-y(a) ) -> x == y
(equivalent programs have the same output for every input)


You are in ERROR ALL OF YOU insisting that one MUST accept LEVELS OF
INFINITY to complete a list of reals.

SIMPLE AS THAT!

You can match ALL COMPUTABLE REALS TO ALL N.

BUT! THERE'S A CAVEAT!

ARE YOU EVEN LISTENING?

WHY BOTHER EXPLAINING TO THE DIM WITTED SO DIM THEY EVEN CLOSE THE
DOOR IN ON THEMSELVES.

Herc

[Alan Smaill]

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

> On Jun 20, 8:47 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> Your objection about extensionality is not coherent.
>>
>> You may be unaware of it, of course.  The computable reals are reals,
>> even according to the drug-crazed cantorists.  But there is no
>> algorithmic way to tell if two such reals are the same or not -- even
>> wikipedia has this information.  A fortiori, there is no algorithm in
>> general. (I think this goes back to Turing 1936.)
>>
>>   "The order relation on the computable numbers is not computable"
>>   "The same holds for the equality relation :
>>           the equality test is not computable."
>>
>>  http://en.wikipedia.org/wiki/Computable_number
>>
>
> THE SAME WAY YOU DETERMINE EQUALITY IN ZFC
>
> ALGORITHMIC EXTENSIONALITY
> ALL(x) ALL(y) ( ALL(a) TM-x(a)=TM-y(a) ) -> x == y
> (equivalent programs have the same output for every input)

And what program can determine if this is the case or not?

> WHY BOTHER EXPLAINING TO THE DIM WITTED SO DIM THEY EVEN CLOSE THE
> DOOR IN ON THEMSELVES.

No need to shout ...

> Herc

--
Alan Smaill

[Graham Cooper]

On Jun 20, 10:08 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> Graham Cooper <grahamcoop...@gmail.com> writes:
> > On Jun 20, 8:47 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >> Your objection about extensionality is not coherent.
>
> >> You may be unaware of it, of course.  The computable reals are reals,
> >> even according to the drug-crazed cantorists.  But there is no
> >> algorithmic way to tell if two such reals are the same or not -- even
> >> wikipedia has this information.  A fortiori, there is no algorithm in
> >> general. (I think this goes back to Turing 1936.)
>
> >>   "The order relation on the computable numbers is not computable"
> >>   "The same holds for the equality relation :
> >>           the equality test is not computable."
>
> >>  http://en.wikipedia.org/wiki/Computable_number
>
> > THE SAME WAY YOU DETERMINE EQUALITY IN ZFC
>
> > ALGORITHMIC EXTENSIONALITY
> > ALL(x) ALL(y) ( ALL(a) TM-x(a)=TM-y(a) ) -> x == y
> > (equivalent programs have the same output for every input)
>
> And what program can determine if this is the case or not?
>

Ah right! Well you'd have to compare 2 infinite strings

so probably a mix of BLACK BOX COMPARISON and WHITE BOX INSPECTION.

e.g.

TMx = 0.333333..MOREDIGITS
TMy = 0.333333..MOREDIGITS

6 digits match, possible x==y

TMx = WRITE 3 REPEAT
TMy = DIVIDE(1/3)

some sort of analysis of the DIVIDE function should result in a proof
of equivalence to TMx.

------

But if you're going to tell me:

START: IF NOT(INFINITE-LOOP(START)) THEN GOTO START

forces an uncomputability on the function INFINITE-LOOP

I'm simply not going to accept your definition of uncomputable.

I'd jab Turing with Oestrogen myself if he wrote that proof!

Herc

[WM]

On 20 Jun., 12:47, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:

> As often, you misrepresent the view of others.
> I said that the usual view (not my view) claims that the reals are
> different from each other (not distinguishable).

What means to be different, if you cannot eplain the difference. Twi
sets are the same if they have the same elements. How do you judge
about two sets if you cannot find out whether elements are different?

>
> You are the one who introduced the terminology of "distinguishable" --
> it's up to you to explain the term.  In English, it is not synonymous
> with "different",

What kind of difference can exist that cannot be recognized?

>
> >  Why would you believe
> > in uncountably many different numbers, if you could not prove that
> > they are different?
>
> Now you say "different"!
>
> The usual, and also the constructive view, is that we cannot list
> the real numbers;

That's not been asked.

>
> > And even if it had been known, that fact would not
> > remove the contradiction with extensionality but would only show that
> > addicts do everything to maintain their drugs.
>
> You have shown no contradiction with extensionality.

But you cannot find ouit in most cases whether two sets are the same
or not.

Look at the example of the Binary Tree. You cannot find out whether
two Binary Trees are identical (by paths, by nodes all are identical).
So your numbers are useless.

And look here:
http://math.stackexchange.com/questions/132022/formalizing-an-idea
My answer proves that uncountability is in contradiction with
continuity of the continuum.

Regards, WM

[Virgil]

In article
<da6b738c-13f4-4ac8...@m10g2000vbn.googlegroups.com>,

If the nodes, including parentage connections, are the same in two trees
the trees are, at least up to isomorphism, identical.

But in the matter of paths, in WM's versions of an infinite binary tree,
there must exist many sets of nodes satisfying every requirement for
being a path to which WM denies pathhood.

>
> And look here:
> http://math.stackexchange.com/questions/132022/formalizing-an-idea
> My answer proves that uncountability is in contradiction with
> continuity of the continuum.

Whereas, at least in standard mathematics, the uncountability of the
reals is an unavoidable consequence of the reals being a complete
Archimedean ordered field, and having the LUB and GLB properties.

So that WM's "reals" are unreal.
--

[WM]

On 20 Jun., 22:43, Virgil <vir...@ligriv.com> wrote:

> > Look at the example of the Binary Tree. You cannot find out whether
> > two Binary Trees are identical (by paths, by nodes all are identical).
> > So your numbers are useless.
>
> If the nodes, including parentage connections, are the same in two trees
> the trees are, at least up to isomorphism, identical.

Right!

>
> But in the matter of paths, in WM's versions of an infinite binary tree,
> there must exist many sets of nodes satisfying every requirement for
> being a path to which WM denies pathhood.

At least when constructing the complete Binary Tree these paths are
not used.

>
>
>
> > And look here:
> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> > My answer proves that uncountability is in contradiction with
> > continuity of the continuum.
>
> Whereas, at least  in standard mathematics, the uncountability of the
> reals is an unavoidable consequence of the reals being a complete
> Archimedean ordered field, and having the LUB and GLB properties.

This shows that your matheology is self-contradictory.

Regards, WM

[Virgil]

In article
<680cf537-2356-4c5f...@fr28g2000vbb.googlegroups.com>,

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

> On 20 Jun., 22:43, Virgil <vir...@ligriv.com> wrote:
>
> > > Look at the example of the Binary Tree. You cannot find out whether
> > > two Binary Trees are identical (by paths, by nodes all are identical).
> > > So your numbers are useless.
> >
> > If the nodes, including parentage connections, are the same in two trees
> > the trees are, at least up to isomorphism, identical.
>
> Right!
> >
> > But in the matter of paths, in WM's versions of an infinite binary tree,
> > there must exist many sets of nodes satisfying every requirement for
> > being a path to which WM denies pathhood.
>
> At least when constructing the complete Binary Tree these paths are
> not used.

But they are there, and they satisfy every requirement of being paths,
even if WM doesn't believe they are there or are paths.

The definition of a (non-empty) path in any tree is a set of nodes such
that (1) it contains the root node, and (2) contains exactly one child
node each of its parent nodes, and nothing else.

And EVERY set of nodes satisfying that definition in any tree is a path
of that tree.

So what WM is doing is claiming that there are sets of nodes satisfying
all those requirements for being paths which are, for some unexplained
and unexaplainable reason, not paths.

> >
> >
> >
> > > And look here:
> > >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> > > My answer proves that uncountability is in contradiction with
> > > continuity of the continuum.
> >
> > Whereas, at least  in standard mathematics, the uncountability of the
> > reals is an unavoidable consequence of the reals being a complete
> > Archimedean ordered field, and having the LUB and GLB properties.
>
> This shows that your matheology is self-contradictory.

Not nearly so much so as WM's matheology, which is self-contradictory
and nonsensical and also anti-mathematical.
--

[Alan Smaill]

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

> On 20 Jun., 12:47, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>
>> As often, you misrepresent the view of others.
>> I said that the usual view (not my view) claims that the reals are
>> different from each other (not distinguishable).
>
> What means to be different, if you cannot eplain the difference. Twi
> sets are the same if they have the same elements. How do you judge
> about two sets if you cannot find out whether elements are different?

You introduced the terminology of "distinguishable";
I'm asking you to explain what difference, if any, *you* intend
when you use that terminology.

>> You are the one who introduced the terminology of "distinguishable" --
>> it's up to you to explain the term.  In English, it is not synonymous
>> with "different",
>
> What kind of difference can exist that cannot be recognized?

Same answer as above.

>> >  Why would you believe
>> > in uncountably many different numbers, if you could not prove that
>> > they are different?
>>
>> Now you say "different"!
>>
>> The usual, and also the constructive view, is that we cannot list
>> the real numbers;
>
> That's not been asked.

It remains the case.

>> > And even if it had been known, that fact would not
>> > remove the contradiction with extensionality but would only show that
>> > addicts do everything to maintain their drugs.
>>
>> You have shown no contradiction with extensionality.
>
> But you cannot find ouit in most cases whether two sets are the same
> or not.

But there is simply no such claim in the classical view;
therefore there is no contradiction.

You may propose a different view of what sets should be,
of course. Your own view remains mostly reactionary, and obscure, IMHO.
But you also claim that the classical view leads to contradiction
internally, and obviously. This you fail yet again to show.

> Look at the example of the Binary Tree. You cannot find out whether
> two Binary Trees are identical (by paths, by nodes all are identical).
> So your numbers are useless.

These are not "my" numbers.

> And look here:
> http://math.stackexchange.com/questions/132022/formalizing-an-idea
> My answer proves that uncountability is in contradiction with
> continuity of the continuum.

Claimed but not proved.

Consider that Leibniz, Newton, Euler before the fall of mathematics into
the hands of the crazy folk made free use of infinitesimals; and Euler
of higher infinities, using them to work out things like power series
expansions, which allow numerical approximation.

[Alan Smaill]

Yes, this is a formulation of computable reals.

> For instance,
> just take the statement of the rule as its name. Of course, I am
> reading the diagonal argument differently: to me, it applies to
> infinite strings to show that *these* are uncountable. In fact, I am
> taking some kind of constructive stance as to what can legitimately be
> called a *number*. The standard reals are, in this sense, an invalid
> construct: self-contradictorily both the set of rules and the
> collection of infinite strings.

Clearly this is in contradiction with the usual story --
but when you say it is self-contradictory, it looks like you are making
a claim that the usual story is in contradiction with itself,
without saying why.

After all, the rules you mention can be used to produce successive
approximations as decimal expansions, so in both cases there is
an associated expansion. And how far does your constructive stance go?
What is it to give a list of reals in the first place --
is that given by a rule also, as would be natural here?



> -LV
>
>

--
Alan Smaill

[Alan Smaill]

"LudovicoVan" <ju...@diegidio.name> writes:

> "Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
> news:fwe395z...@eriboll.inf.ed.ac.uk...
>> "LudovicoVan" <ju...@diegidio.name> writes:
>>> "Alan Smaill" <sma...@SPAMinf.ed.ac.uk> wrote in message
>>> news:fweboko...@eriboll.inf.ed.ac.uk...
>>>> "LudovicoVan" <ju...@diegidio.name> writes:
>>> <snip>
>>>
>>>>> Right: I hope I am not too mistaken in saying that the "set" of
>>>>> infinite strings is provably not recursively enumerable.
>>>>
>>>> If it's not countable, then it's not enumerable in any sense.
>>>
>>> Isn't countable equivalent to recursive? If so, not being countable
>>> does not (necessarily) entail not being recursively enumerable.
>>
>> No, it isn't --
>> the set of Turing machines that terminate on empty input
>> is countable, but not recursive.
>
> Isn't it not even recursively enumerable? Namely, how can it be
> countable when you cannot tell which elements it has?

No, it's recursively enumerable.
If a Turing machine terminates, you can show this by waiting
long enough as it runs. Find a fair way to interleave
the execution of all these Turing machines, and if a particular
one terminates, you will find that out eventually.

>> It's the effectiveness
>> of the enumeration that is at issue (in the usual story).
>
> I am still rookie with these definitions, but I am (roughly) guessing
> that the notion of non-computable function may be as valid as that of
> uncountable number...

Is that a typo?
It's the set of reals that is claimed to be uncountable, not individual
real numbers.

> -LV
>
>

--
Alan Smaill

[Graham Cooper]

On Jun 22, 3:21 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > Isn't it not even recursively enumerable?  Namely, how can it be
> > countable when you cannot tell which elements it has?
>
> No, it's recursively enumerable.
> If a Turing machine terminates, you can show this by waiting
> long enough as it runs.  Find a fair way to interleave
> the execution of all these Turing machines, and if a particular
> one terminates, you will find that out eventually.

No this is not a work-around of the halt problem.

You won't be able to SPECIFY the pattern of which digits are included,
hence they won't all be enumerated.

Say the complexity of a REAL is EXPONENTIAL for every consecutive
digit you calculate.

0.1 takes 10 processing cycles
0.12 takes 100 processing cycles
0.126 takes 1000 processing cycles
..

Then say you WEAVE around the TMS using UTM(real#,digit#) using a
POLYNOMIAL scheduling sequence - the decision to include a row, in a
preliminary listing.

All the reals with complexity > polynomial will be perpetually pushed
down the queue while you wait for them to halt.

Herc

[Alan Smaill]

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

> On Jun 22, 3:21 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>>
>> > Isn't it not even recursively enumerable?  Namely, how can it be
>> > countable when you cannot tell which elements it has?
>>
>> No, it's recursively enumerable.
>> If a Turing machine terminates, you can show this by waiting
>> long enough as it runs.  Find a fair way to interleave
>> the execution of all these Turing machines, and if a particular
>> one terminates, you will find that out eventually.
>
>
> No this is not a work-around of the halt problem.

Clearly not.
But it does show that the set is recursively enumerable,
which was the claim in question.

> All the reals with complexity > polynomial will be perpetually pushed
> down the queue while you wait for them to halt.

Nothing is perpetually pushed down the queue;
finding a "fair way" to interleave the execution I leave up to you.

>
> Herc

--
Alan Smaill

[WM]

On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

>
> > And look here:
> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> > My answer proves that uncountability is in contradiction with
> > continuity of the continuum.
>
> Claimed but not proved.

Proved by the fact, that an uncovered irrational x is not a singleton,
unless there is a rational that makes it being a singleton.
Uncountably many singletons cannot exist as singletons without
uncountably many other points between them. But every "other point"
q_n is covered by an interval I_n. This proof is so simple that
everybody with modest intelligence and not perverted by matheological
dogmas understands it easily. I have made this experience quite often
and never failed (except with matheologians).

Regards, WM

[Alan Smaill]

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

> On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
>>
>> > And look here:
>> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
>> > My answer proves that uncountability is in contradiction with
>> > continuity of the continuum.
>>
>> Claimed but not proved.
>
> Proved by the fact, that an uncovered irrational x is not a singleton,
> unless there is a rational that makes it being a singleton.

It is your insistence in stating what you call "facts", but which
are in fact in English statements vague to the point of meaninglessness,
that stops you from producing anything like a convincing proof.

Simply repeating yourself does not make what you say more convincing;
it just gives the impression that you prefer propaganda to logic.

> I have made this experience quite often
> and never failed (except with matheologians).

No doubt your students bow down to the Scripture According To Mueckenheim.

[WM]

On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> >> > And look here:
> >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> >> > My answer proves that uncountability is in contradiction with
> >> > continuity of the continuum.
>
> >> Claimed but not proved.
>
> > Proved by the fact, that an uncovered irrational x is not a singleton,
> > unless there is a rational that makes it being a singleton.
>
> It is your insistence in stating what you call "facts", but which
> are in fact in English statements vague to the point of meaninglessness,
> that stops you from producing anything like a convincing proof.

What makes a singleton a singleton?

>
> Simply repeating yourself does not make what you say more convincing;
> it just gives the impression that you prefer propaganda to logic.

What makes a singleton a singleton?

>
> > I have made this experience quite often
> > and never failed (except with matheologians).
>
> No doubt your students  bow down to the Scripture According To Mueckenheim.

What makes a singleton a singleton?

Can you answer this question?

Regards, WM

[Jesse F. Hughes]

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

> On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>>
>> >> > And look here:
>> >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
>> >> > My answer proves that uncountability is in contradiction with
>> >> > continuity of the continuum.
>>
>> >> Claimed but not proved.
>>
>> > Proved by the fact, that an uncovered irrational x is not a singleton,
>> > unless there is a rational that makes it being a singleton.
>>
>> It is your insistence in stating what you call "facts", but which
>> are in fact in English statements vague to the point of meaninglessness,
>> that stops you from producing anything like a convincing proof.
>
> What makes a singleton a singleton?

A rational! Of course! It's all clear now.

Before, I was rather muddle-headed, and thought that what makes a
singleton a singleton was the fact that it contains only a single
element.

But now I see. Thanks, much!

--
Meaningless movies
on the screen behind the band that's blowing Waterboys,
throwing shapes "My Love is My Rock
Half of the music is on tape in the Weary Land"

[Virgil]

In article
<ab18866c-033a-4724...@d6g2000vbe.googlegroups.com>,

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

> On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> >
> > > And look here:
> > >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> > > My answer proves that uncountability is in contradiction with
> > > continuity of the continuum.
> >
> > Claimed but not proved.
>
> Proved by the fact, that an uncovered irrational x is not a singleton,
> unless there is a rational that makes it being a singleton.

That is not a fact until proved, and has not been proved.

It is much more likely that the set of uncovered irrationals looks like
an irregular Cantor set with countably many open intervals leaving
uncovered the uncountably many members of the Cantor set.

> Uncountably many singletons cannot exist as singletons without
> uncountably many other points between them.

"Singleton" is a bit ambiguous in WM's hands, so we would like a formal
definition of what HE means by it.

And it may well be that, like for members of the Cantor set, between
some pairs of uncovered irrationals there will be uncountably many
uncovered irrationals.

> But every "other point"
> q_n is covered by an interval I_n.

Since the example of the Cantor set refutes your claim in general, you
must prove that it holds in your particular cse or concede defat.

> This proof is so simple that
> everybody with modest intelligence and not perverted by matheological
> dogmas understands it easily.

I understand that it is not a valid proof when a valid counterexample,
in the form of the Cantor set, exists.

I have made this experience quite often
> and never failed (except with matheologians).

Then the vast majority of mathematicians are what WM falsely calls
matheologians.
--

[Virgil]

In article
<6d7248b6-da75-4144...@l5g2000vbo.googlegroups.com>,

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

> On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >
> > >> > And look here:
> > >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> > >> > My answer proves that uncountability is in contradiction with
> > >> > continuity of the continuum.
> >
> > >> Claimed but not proved.
> >
> > > Proved by the fact, that an uncovered irrational x is not a singleton,
> > > unless there is a rational that makes it being a singleton.
> >
> > It is your insistence in stating what you call "facts", but which
> > are in fact in English statements vague to the point of meaninglessness,
> > that stops you from producing anything like a convincing proof.
>
> What makes a singleton a singleton?

What is YOUR definition of that term, WM?

Mine would be that every uncovered irrational is separated by every ote
uncovered irrational by countably many rationals, though not necessarily
in less than countably many clusters.

> >
> > Simply repeating yourself does not make what you say more convincing;
> > it just gives the impression that you prefer propaganda to logic.
>
> What makes a singleton a singleton?

What is YOUR definition of that term, WM?

Mine would be that every uncovered irrational is separated by every ote
uncovered irrational by countably many rationals, though not necessarily
in less than countably many clusters.

>
> >
> > > I have made this experience quite often
> > > and never failed (except with matheologians).
> >
> > No doubt your students  bow down to the Scripture According To Mueckenheim.
>
> What makes a singleton a singleton?
>
> Can you answer this question?

What is YOUR definition of that term, WM?

Mine would be that every uncovered irrational is separated by every
other uncovered irrational by countably many rationals, including at
least one cluster, though not necessarily by less than countably many
clusters.

http://en.wikipedia.org/wiki/Cantor_set

Consider the Cantor set on [0,1]! It contains no rationals and
uncountably many irrationals, Its compliment in [0,1] is a family of
open intervals like clusters. Between every two members of the set lie
at least one such open interval, and possibly many more.

How does the Cantor set differ from WN's model?

It does't!
--

[WM]

On 23 Jun., 02:03, Virgil <vir...@ligriv.com> wrote:

> "Singleton" is a bit ambiguous in WM's hands, so we would like a formal
> definition of what HE means by it.

Singleton is used in topology. I think there is no further definition
required.

>
> And it may well be that, like for members of  the Cantor set, between
> some pairs of uncovered irrationals there will be uncountably many
> uncovered irrationals.

That is true for intervals of irrationals too. My question is: What
makes the singleton a singleton?

>
> > But every "other point"
> > q_n is covered by an interval I_n.
>

> I understand that it is not a valid proof when a valid counterexample,
> in the form of the Cantor set, exists.

My example refutes the Cantor set. So the ´Cantor set is not valid.

Regards, WM

[WM]

On 23 Jun., 05:09, Virgil <vir...@ligriv.com> wrote:

> > What makes a singleton a singleton?
>
> What is YOUR definition of that term, WM?
>

> Mine would be that every uncovered irrational is separated by every other

> uncovered irrational by countably many rationals, though not necessarily
> in less than countably many clusters.

That means, there are never two uncovered irrationals without an
interval between them. That means, for every uncovered irration al
there is a neighborhood that does not contain another uncovered
irrational. The radius of the neighborhood being half the measure of
the smaller of the two neighboring intervals.

> > > What makes a singleton a singleton?
>
> What is YOUR definition of that term, WM?
>
> Mine would be that every uncovered irrational is separated by every ote
> uncovered irrational by countably many rationals, though not necessarily
> in less than countably  many clusters.

I agree with yours. See above.

>
> Consider the Cantor set on [0,1]! It contains no rationals

it contains a lot of rationals.

> and
> uncountably many irrationals, Its compliment in [0,1] is a family of
> open intervals like clusters. Between every two members of the set lie
> at least one such open interval, and possibly many more.
>
> How does the Cantor set differ from WN's model?

it contains a lot of rationals.

>
> It does't!

it contains a lot of rationals.

Regards, WM

[WM]

On 23 Jun., 01:47, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >> WM <mueck...@rz.fh-augsburg.de> writes:
> >> > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> >> >> > And look here:
> >> >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
> >> >> > My answer proves that uncountability is in contradiction with
> >> >> > continuity of the continuum.
>
> >> >> Claimed but not proved.
>
> >> > Proved by the fact, that an uncovered irrational x is not a singleton,
> >> > unless there is a rational that makes it being a singleton.
>
> >> It is your insistence in stating what you call "facts", but which
> >> are in fact in English statements vague to the point of meaninglessness,
> >> that stops you from producing anything like a convincing proof.
>
> > What makes a singleton a singleton?
>
> A rational!  Of course!  It's all clear now.

In my example, there is no rational. There are only intervals of
finite measure containing infinitely many rationals and uncountably
many irrationals.

>
> Before, I was rather muddle-headed, and thought that what makes a
> singleton a singleton was the fact that it contains only a single
> element.

Why should an uncovered irrational number be single? Why should there
not be two or more irrationals numbers as twingletons or mingletons?
Yes, there must be a preventing principle. And that is an interval.

Every uncovered irrational number x is the neighbour of two intervals
in the sense that there is no uncovered irrational number y between x
and the neighbouring intervals.

Regards, WM

[Jesse F. Hughes]

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

> On 23 Jun., 01:47, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> >> WM <mueck...@rz.fh-augsburg.de> writes:
>> >> > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>>
>> >> >> > And look here:
>> >> >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
>> >> >> > My answer proves that uncountability is in contradiction with
>> >> >> > continuity of the continuum.
>>
>> >> >> Claimed but not proved.
>>
>> >> > Proved by the fact, that an uncovered irrational x is not a singleton,
>> >> > unless there is a rational that makes it being a singleton.
>>
>> >> It is your insistence in stating what you call "facts", but which
>> >> are in fact in English statements vague to the point of meaninglessness,
>> >> that stops you from producing anything like a convincing proof.
>>
>> > What makes a singleton a singleton?
>>
>> A rational!  Of course!  It's all clear now.
>
> In my example, there is no rational. There are only intervals of
> finite measure containing infinitely many rationals and uncountably
> many irrationals.

Mighty confusing, this is.

>> Before, I was rather muddle-headed, and thought that what makes a
>> singleton a singleton was the fact that it contains only a single
>> element.
>
> Why should an uncovered irrational number be single? Why should there
> not be two or more irrationals numbers as twingletons or mingletons?
> Yes, there must be a preventing principle. And that is an interval.
>
> Every uncovered irrational number x is the neighbour of two intervals
> in the sense that there is no uncovered irrational number y between x
> and the neighbouring intervals.

I've no idea what you're going on about now, but what makes a singleton
a singleton is the fact that it contains only one element.


--
"And that's what's wrong with Usenet. You people [...] can reply to a
post [...] and then convince yourselves that you're great. Because
you can open your mouths you think what you have is worth saying that
you have proven your value." -- James S. Harris masters self-diagnosis

[WM]

On 23 Jun., 15:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> I've no idea what you're going on about now, but what makes a singleton
> a singleton is the fact that it contains only one element.

The real axis contains many irrational numbers. How can an irrational
number become a singleton? Why is there a number that is a singleton?

Regards, WM

[Jesse F. Hughes]

This question makes no particular sense at all. Numbers do not become
singletons[1].

Look, I know that thinking about sets is challenging for you, but {pi},
{0} and {sqrt(2)} are all singletons. Duh.

Footnotes:
[1] Ignoring for a moment odd facts about various representations of
real numbers as sets.

--
"Another factor one has got to look at is the amount of liquidity in
the system. In other words, is there enough liquidity to enable
markets to be able to correct? And I am told there is enough liquidity
in the system to enable markets to correct." -- Guess who.

[WM]

On 23 Jun., 18:24, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 23 Jun., 15:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> >> I've no idea what you're going on about now, but what makes a singleton
> >> a singleton is the fact that it contains only one element.
>
> > The real axis contains many irrational numbers. How can an irrational
> > number become a singleton? Why is there a number that is a singleton?
>
> This question makes no particular sense at all.

That depends on your intelligence. Try harder.

>  Numbers do not become
> singletons[1].

That is blatantly wrong. The real numbers are not a completely
disconnected before all rationals are covered by intervals. After that
the uncovered real numbers a compkletely disconnected space,
consisting of singletons. So they have become singletons, no?

Regards, WM

[Virgil]

In article
<5401879d-e5a6-4b5e...@d17g2000vbv.googlegroups.com>,

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

> On 23 Jun., 05:09, Virgil <vir...@ligriv.com> wrote:
>
> > > What makes a singleton a singleton?
> >
> > What is YOUR definition of that term, WM?
> >
> > Mine would be that every uncovered irrational is separated by every other
> > uncovered irrational by countably many rationals, though not necessarily
> > in less than countably many clusters.
>
> That means, there are never two uncovered irrationals without an
> interval between them.

Unproven and requires proof, since while it is true for the Cantor set,
every point of the Cantor set is a limit point of the set, so NO point
of it is an isolated point, and depending on what WM's definition of
"singleton" is, it is likely that no point of the Cantor set is a
singleton.


Thus there is no reason to suppose that any uncovered irrational of WM's
set is a singleton.

> That means, for every uncovered irration al
> there is a neighborhood that does not contain another uncovered
> irrational.

Not necessarily, since for the Cantor set it is totally false:
Every neighborhood of every point in the C-set contains other points of
it.

>
> > > > What makes a singleton a singleton?
> >
> > What is YOUR definition of that term, WM?
> >
> > Mine would be that every uncovered irrational is separated by every ote
> > uncovered irrational by countably many rationals, though not necessarily
> > in less than countably  many clusters.
>
> I agree with yours. See above.
> >
> > Consider the Cantor set on [0,1]! It contains no rationals
>
>
> it contains a lot of rationals.

My error, but that is not critical to my claim that removing countably
many open intervals from [0, 1], even if they cover a set of points
dense in [0, 1], does not necessarily leave a residue of ONLY countably
many one-point intervals, as you claim.

The Cantor set construction does the same thing but leaves the Cantor
set which is not countable.

So your claim still requires a proof which you have not provided.
--

[Virgil]

In article
<c3f74d1a-c9da-45d1...@q2g2000vbv.googlegroups.com>,

HOW does your 'example' refute the Cantor set?

In my view it is the other way round, the Cantor set shows that your
sort of construction does not NECESSARILY leave a residue of only
countably many points or singletons (one-member sets).

Strong assertions, like WM's, require strong proofs, not merely
assertions, such as he makes, however strongly asserted.

But WM is, as usual, incapable of any proofs, much less strong ones.
--

[Virgil]

In article
<5e756cbe-707e-42ba...@5g2000vbf.googlegroups.com>,

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

> On 23 Jun., 01:47, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > Before, I was rather muddle-headed, and thought that what makes a
> > singleton a singleton was the fact that it contains only a single
> > element.
>
> Why should an uncovered irrational number be single?

Because between it and any other irrational, uncovered or not, lies at
least one rational which must have already been deleted.

> Why should there
> not be two or more irrationals numbers as twingletons or mingletons?
> Yes, there must be a preventing principle. And that is an interval.
>
> Every uncovered irrational number x is the neighbour of two intervals
> in the sense that there is no uncovered irrational number y between x
> and the neighbouring intervals.

Actually, that need not be the case.

Consider the Cantor set in which each point is an accumulation point of
other points of the set.

Thus between any point of the set and any interval of its compliment
there are other points of the set, in fact infinitely many of them.

So if WM's claims are to prevail, he must sow why the same thing cannot
happen in his construction.

So far WM has signally failed to do so.
--

[Virgil]

In article
<f85fbea7-429b-4adf...@x21g2000vbc.googlegroups.com>,

There are all sorts of numbers that are members of singletons.
In WM's construction, every non-empty interval of uncovered irrationals
is necessarily a singleton, because if it contained more than one
element it would have to also contain a rational between them.

But WM's construction is alleged to have covered all rationals.
--

[WM]

On 23 Jun., 23:21, Virgil <vir...@ligriv.com> wrote:
> In article
> <5e756cbe-707e-42ba-a6ea-e81812f23...@5g2000vbf.googlegroups.com>,

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

> > Why should an uncovered irrational number be single?
>
> Because between it and any other irrational, uncovered or not, lies at
> least one rational which must have already been deleted.

This rational now is member of an interval. So between it and any
other uncovered irrational lies at least one interval-

> > Every uncovered irrational number x is the neighbour of two intervals
> > in the sense that there is no uncovered irrational number y between x
> > and the neighbouring intervals.
>
> Actually, that need not be the case.

You contradict yourself.

Regards, WM

[Virgil]

In article
<cf2c619e-49a0-499b...@cu1g2000vbb.googlegroups.com>,

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

> On 23 Jun., 18:24, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 23 Jun., 15:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> > >> I've no idea what you're going on about now, but what makes a singleton
> > >> a singleton is the fact that it contains only one element.
> >
> > > The real axis contains many irrational numbers. How can an irrational
> > > number become a singleton? Why is there a number that is a singleton?
> >
> > This question makes no particular sense at all.
>
> That depends on your intelligence. Try harder.
>
> >  Numbers do not become
> > singletons[1].
>
> That is blatantly wrong.

If one uses the standard meaning of "singleton" as being a set with
exactly one member:
http://en.wikipedia.org/wiki/Singleton_(mathematics)
then numbers are NOT singletons (with the possible exception of the
natural number 1, when it is defined as {{}} ).

But real numbers can be members of singletons, e.g., closed intervals of
zero length.

The real numbers are not a completely
> disconnected before all rationals are covered by intervals. After that
> the uncovered real numbers a compkletely disconnected space,
> consisting of singletons. So they have become singletons, no?

No! To be precise, the maximal connected subsets of a completely
disconnected space are singleton sets. By a common abuse of language,
which should be avoided in the vicinity of WM, as he will always get
muddled, one might casually refer to the member of the singleton as
being the singleton.
--

[Jesse F. Hughes]

Again, I can't seem to make any particular sense of this blather.

I think there's no need to continue this.

--
Jesse F. Hughes

"There's absolutely no information here."
-- Hank Hill, on blogging and information theory.

[Mike Terry]

"Jesse F. Hughes" <je...@phiwumbda.org> wrote in message
news:874nq1i...@phiwumbda.org...

> WM <muec...@rz.fh-augsburg.de> writes:
>
> > On 23 Jun., 18:24, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> WM <mueck...@rz.fh-augsburg.de> writes:
> >> > On 23 Jun., 15:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >>
> >> >> I've no idea what you're going on about now, but what makes a
singleton
> >> >> a singleton is the fact that it contains only one element.
> >>
> >> > The real axis contains many irrational numbers. How can an irrational
> >> > number become a singleton? Why is there a number that is a singleton?
> >>
> >> This question makes no particular sense at all.
> >
> > That depends on your intelligence. Try harder.
> >
> >> Numbers do not become
> >> singletons[1].
> >
> > That is blatantly wrong. The real numbers are not a completely
> > disconnected before all rationals are covered by intervals. After that
> > the uncovered real numbers a compkletely disconnected space,
> > consisting of singletons. So they have become singletons, no?
>
> Again, I can't seem to make any particular sense of this blather.

Not that I think you should continue, because in fact I think there you
would be wasting your time (like trying to help Herc). Still, here is what
I think he means, based on previous matheology threads...

He considers the unit interval [0,1], and removes a countable sequence of
intervals of length say 1/10^n, each centred on a rational q_n, for some
enumeration of the rationals in [0,1].

The remaining points are all clearly irrational, and if we look at the
connected components of this set they are "singleton intervals" of the form
[x,x], i.e. SINGLETONs! :-)

OTOH if we look at the connected components of the union of all intervals
removed, they are (of course) intervals, which have been referred to as
"clusters" in other threads. The clusters are all open intervals with
irrational endpoints, which belong to the complement and so are examples of
the "irrational singletons". (Note not all the irrationals in the
complement are endpoints of clusters, although WM does not believe this.)

I think WM believes there are only countably many singletons, and his
argument for this was that each singleton is adjacent to a cluster, so there
is some kind of correspondence between them. He's been told this is wrong
of course, and has not provided any proof of this correspondence, so there's
not much to be said, other than he's wrong! Of course each cluster is
"adjacent" to two singletons, but not vice-versa, and it's been explained to
him how for certain irrationals there are *sequences* of clusters which
converge to the irrational, with no single cluster being adjacent to it...

Anyway, when he asks "what makes a singleton a singleton?" I think he wants
you to say "the clusters that have been removed", and then he will say
"aha - that means that there must be clusters adjacent to it!" which of
course is just muddled thinking...


Mike.

[Virgil]

In article
<08ea3c13-2a27-47b8...@d6g2000vbe.googlegroups.com>,

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

> On 23 Jun., 23:21, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <5e756cbe-707e-42ba-a6ea-e81812f23...@5g2000vbf.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > Why should an uncovered irrational number be single?
> >
> > Because between it and any other irrational, uncovered or not, lies at
> > least one rational which must have already been deleted.
>
> This rational now is member of an interval. So between it and any
> other uncovered irrational lies at least one interval-

What does your "it" refer to? Your "it" seems to refer to the rational
of the previous sentence, which would be nonsense. If you meant some
other uncovered irrational, it is decidedly unclear from what you have
written.

>
> > > Every uncovered irrational number x is the neighbour of two intervals
> > > in the sense that there is no uncovered irrational number y between x
> > > and the neighbouring intervals.
> >
> > Actually, that need not be the case.
>
> You contradict yourself.

Not at all. It must be the case that some uncovered irrationals are
limit points of the set of all such uncovered irrationals, in which case
between such a limit point uncovered irrational and other uncovered
irrationals there will be infinitely many other uncovered irrationals.

Just as between points of the Cantor set.
--

[WM]

On 24 Jun., 02:12, "Mike Terry"

> I think WM believes there are only countably many singletons, and his
> argument for this was that each singleton is adjacent to a cluster, so there
> is some kind of correspondence between them.  He's been told this is wrong
> of course, and has not provided any proof of this correspondence,

The proof is that an uncovered irrational is a singleton, i.e. it is
adjacent to intervals I_n. Otherwise it was not a singleton.

> so there's
> not much to be said, other than he's wrong!  Of course each cluster is
> "adjacent" to two singletons, but not vice-versa,

Do you believe in singletons that are adjacent to other singletons
without a rational between them? That would simply be the consequence.

> and it's been explained to
> him how for certain irrationals there are *sequences* of clusters which
> converge to the irrational, with no single cluster being adjacent to it...

Cantor's axiom says that every irrational x is a point on the real
axis. Whether this point is a singleton, is not at all determined by
distant intervals or clusters, but only by intervals that are adjacent
to x. And if there are no such intervals, then x is not a singleton.

>
> Anyway, when he asks "what makes a singleton a singleton?" I think he wants
> you to say "the clusters that have been removed", and then he will say
> "aha - that means that there must be clusters adjacent to it!" which of
> course is just muddled thinking...

Otherwise you must believe in singletons that are not adjacent to
intervals. Or do you see another alternative?

But try to comprehend: Cauchy sequences and clustering and the like do
not help you, if Cantor's axiom is valid, which says that every real
number corresponds to a point of the real axis. Then *only* the direct
surrounding decides about the character of a singleton. In fact there
is a neighbourhood for every singleton, such that the neighbourhood
does not contain any further uncovered irrationals, i.e. any further
singleton.

Regards, WM

[WM]

On 24 Jun., 03:56, Virgil <vir...@ligriv.com> wrote:
> In article

> <08ea3c13-2a27-47b8-baf1-35bda5b2b...@d6g2000vbe.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 23 Jun., 23:21, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <5e756cbe-707e-42ba-a6ea-e81812f23...@5g2000vbf.googlegroups.com>,
>
> > >  WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > Why should an uncovered irrational number be single?
>

> > > Because between ***it*** and any other irrational, uncovered or not, lies at

> > > least one rational which must have already been deleted.
>

> > This rational now is member of an interval. So between ***it*** and any
> > other uncovered irrational lies at least one interval.

>
> What does your "it" refer to?

It is marked above: uncovered irrational number.

>
>
> > > > Every uncovered irrational number x is the neighbour of two intervals
> > > > in the sense that there is no uncovered irrational number y between x
> > > > and the neighbouring intervals.
>
> > > Actually, that need not be the case.
>
> > You contradict yourself.
>
> Not at all. It must be the case that some uncovered irrationals are
> limit points of the set of all such uncovered irrationals, in which case
> between such a limit point uncovered irrational and other uncovered
> irrationals there will be infinitely many other uncovered irrationals.

Forget limit points. Cantor's axiom says that the number x corresponds
to a point of the real axis. Whether this point is a singleton is only
decided by its immediate neighbourhood.

Regards, WM

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:c7bfbccd-fdb3-4fbe...@e20g2000vbm.googlegroups.com...

> On 24 Jun., 02:12, "Mike Terry"
>
> > I think WM believes there are only countably many singletons, and his
> > argument for this was that each singleton is adjacent to a cluster, so
there
> > is some kind of correspondence between them. He's been told this is
wrong
> > of course, and has not provided any proof of this correspondence,
>
> The proof is that an uncovered irrational is a singleton, i.e. it is
> adjacent to intervals I_n. Otherwise it was not a singleton.

No - that is not the definition of a singleton. I don't think you
understand what is meant by describing the uncovered irrationals as
singletons, even though you introduced the wording - if you did you would
see the need for a proof of a claim such as this, rather than repeating the
claim over and over.

It is disappointing that when challenged to prove a claim, your proof turns
out to be:

Claim: Every uncovered x is the endpoint of some cluster.
Proof: 1. x is a singleton
2. all singletons are adjacent to intervals I_n.
3. therefore x is adjacent to an interval I_n.
4. (i.e. x is the endpoint of the interval I_n.)

Do you really believe that constitutes a mathematical proof? (In case you
genuinely do, the problem is with (2), so please come back and prove (2).)

>
> > so there's
> > not much to be said, other than he's wrong! Of course each cluster is
> > "adjacent" to two singletons, but not vice-versa,
>
> Do you believe in singletons that are adjacent to other singletons
> without a rational between them? That would simply be the consequence.

No I don't, and no it wouldn't. (Not that it matters what I think - we're
waiting for you to provide a proof of your claims, which clearly you will
never do.)

>
> > and it's been explained to
> > him how for certain irrationals there are *sequences* of clusters which
> > converge to the irrational, with no single cluster being adjacent to
it...
>
> Cantor's axiom says that every irrational x is a point on the real
> axis. Whether this point is a singleton, is not at all determined by
> distant intervals or clusters, but only by intervals that are adjacent
> to x. And if there are no such intervals, then x is not a singleton.

No that's plain wrong - x "is a singleton" if x belongs to the complement,
and there is no larger connected component containing x within the
complement. The situation of having a larger connected component within the
complement can be prevented by there being a *succession* of clusters
approaching x arbitrarily closely, but no cluster being adjacent to x.

> >
> > Anyway, when he asks "what makes a singleton a singleton?" I think he
wants
> > you to say "the clusters that have been removed", and then he will say
> > "aha - that means that there must be clusters adjacent to it!" which of
> > course is just muddled thinking...
>
> Otherwise you must believe in singletons that are not adjacent to
> intervals. Or do you see another alternative?
>
> But try to comprehend: Cauchy sequences and clustering and the like do
> not help you, if Cantor's axiom is valid, which says that every real
> number corresponds to a point of the real axis. Then *only* the direct
> surrounding decides about the character of a singleton.

There is no "direct surrounding" for a single point x on the real line, in
the sense of there being points directly adjacent to x. What we have at x
is a "neighbourhood system" which characterises closeness to x in a sense.
What is true, is that every neighbourhood of an uncovered x contains a point
belonging to one of the clusters. So we can say this is what makes x
isolated.

Note the order of quantification here: given a neighbourhood of x there is
a cluster intersecting that neighbourhood. This cluster depends on the
choice of neighbourhood. It is NOT saying that there is a single cluster
that intersects with every neighbourhood of x. (The latter is true for
some, but not all the uncovered x.) I only mention this because I know from
other posts that you are confused on this point...

> In fact there
> is a neighbourhood for every singleton, such that the neighbourhood
> does not contain any further uncovered irrationals, i.e. any further
> singleton.

No, this is wrong. It is essentially what you've been asked to prove, but
have so far only claimed. Of course, feel free to submit a proof of this
claim if you believe it to be true, or withdraw the claim if you realise you
were mistaken. Otherwise just repeating it over and over will result in
people concluding you're a crank...


Mike.

[WM]

On 24 Jun., 17:42, "Mike Terry"
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

>
> news:c7bfbccd-fdb3-4fbe...@e20g2000vbm.googlegroups.com...
>
> > On 24 Jun., 02:12, "Mike Terry"
>
> > > I think WM believes there are only countably many singletons, and his
> > > argument for this was that each singleton is adjacent to a cluster, so
> there
> > > is some kind of correspondence between them. He's been told this is
> wrong
> > > of course, and has not provided any proof of this correspondence,
>
> > The proof is that an uncovered irrational is a singleton, i.e. it is
> > adjacent to intervals I_n. Otherwise it was not a singleton.
>
> No - that is not the definition of a singleton.

The definition of a singleton is a set that contains only one element
because it is separated from other elements of the same kind.

> It is disappointing that when challenged to prove a claim, your proof turns
> out to be:
>
>    Claim:     Every uncovered x is the endpoint of some cluster.
>    Proof:     1.  x is a singleton
>               2.  all singletons are adjacent to intervals I_n.
>               3.  therefore x is adjacent to an interval I_n.
>               4.  (i.e. x is the endpoint of the interval I_n.)
>
> Do you really believe that constitutes a mathematical proof?

Of course. Here is a proof of (2) by contradiction: If x is not
adjacent to a covered interval I_n, then there is another uncovered
irrational y separating x and I_n. (There is no further alternative
separating x from a rational and hence from I_n.) This contradicts the
assumption that x is a singleton.

>  (In case you
> genuinely do, the problem is with (2), so please come back and prove (2).)

Done.

>
> > Do you believe in singletons that are adjacent to other singletons
> > without a rational between them? That would simply be the consequence.
>
> No I don't, and no it wouldn't.

There is no thirs alternative. A singleton either is adjacent to a
rational or to an irrational. There are no other numbers on the real
axis. But all rationals are in intervals.

>  (Not that it matters what I think - we're
> waiting for you to provide a proof of your claims, which clearly you will
> never do.)

Better: You clearly will never accept any such proof because your
brain has been deformed accordingly.

>
>
> > > and it's been explained to
> > > him how for certain irrationals there are *sequences* of clusters which
> > > converge to the irrational, with no single cluster being adjacent to
> it...
>
> > Cantor's axiom says that every irrational x is a point on the real
> > axis. Whether this point is a singleton, is not at all determined by
> > distant intervals or clusters, but only by intervals that are adjacent
> > to x. And if there are no such intervals, then x is not a singleton.
>
> No that's plain wrong - x "is a singleton" if x belongs to the complement,
> and there is no larger connected component containing x within the
> complement.  The situation of having a larger connected component within the
> complement can be prevented by there being a *succession* of clusters
> approaching x arbitrarily closely, but no cluster being adjacent to x.

If we look at points on the real lines then remote intervals and
clusters have the same influence as Venus and Mars have on your fate.

>
>
>
>
>
> > > Anyway, when he asks "what makes a singleton a singleton?" I think he
> wants
> > > you to say "the clusters that have been removed", and then he will say
> > > "aha - that means that there must be clusters adjacent to it!" which of
> > > course is just muddled thinking...
>
> > Otherwise you must believe in singletons that are not adjacent to
> > intervals. Or do you see another alternative?
>
> > But try to comprehend: Cauchy sequences and clustering and the like do
> > not help you, if Cantor's axiom is valid, which says that every real
> > number corresponds to a point of the real axis. Then *only* the direct
> > surrounding decides about the character of a singleton.
>
> There is no "direct surrounding" for a single point x on the real line, in
> the sense of there being points directly adjacent to x.

There is a surrounding, namely the minimum measure of the finite
intervals of measure I_n > 0 adjacent to this point. In this
surrounding of the uncovered irrational x there is no other uncovered
irrational. Otherwise x was not a singleton.

> Note the order of quantification here:

Points on a line do not obey any quantification rules. They are there
or they are not there. There is no "if" and "given that" for
stationary points.

>  given a neighbourhood of x there is
> a cluster intersecting that neighbourhood.  This cluster depends on the
> choice of neighbourhood.  It is NOT saying that there is a single cluster
> that intersects with every neighbourhood of x.  (The latter is true for
> some, but not all the uncovered x.)

Se my proof of (2) above in case you really are able to sober
thinking.
It is really ridiculous how you try to argue that far remote points
and seqeunces should act on point x!

>
> > In fact there
> > is a neighbourhood for every singleton, such that the neighbourhood
> > does not contain any further uncovered irrationals, i.e. any further
> > singleton.
>
> No, this is wrong.

There is a completely disconnected space consisting of separated
intervals, so called singletons containing only one point, but not all
of these points are separated from each other by the only possible
separators, namely rational numbers and hence intervals. They are
separated by distant clusters! Can't you really imagine how silly this
arguing must sound to anybody not deformed by matheology?

Regards, WM

[WM]

On 24 Jun., 17:42, "Mike Terry"

> Note the order of quantification here:  given a neighbourhood of x there is

> a cluster intersecting that neighbourhood.  This cluster depends on the
> choice of neighbourhood.  It is NOT saying that there is a single cluster
> that intersects with every neighbourhood of x.

Addition: Have you ever understood that there is no infinite sequence
(a_n)that defines its limit a? You can see this by leaving out the
first n terms a_n - and that holds for every n. (You cannot specify
any n such that the a_n would be required. Therefore note: The remote
clusters do not have any influence.) The limit a is defined by the
finite definition that defines the sequence - not at all by any of its
terms. Therefore your imagined clustering is not an "in fact" but
simply nonsense.

Regards, WM

[Virgil]

In article
<22ae51ac-1727-4047...@e20g2000vbm.googlegroups.com>,

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

> > > > > Every uncovered irrational number x is the neighbour of two intervals
> > > > > in the sense that there is no uncovered irrational number y between x
> > > > > and the neighbouring intervals.
> >
> > > > Actually, that need not be the case.
> >
> > > You contradict yourself.
> >
> > Not at all. It must be the case that some uncovered irrationals are
> > limit points of the set of all such uncovered irrationals, in which case
> > between such a limit point uncovered irrational and other uncovered
> > irrationals there will be infinitely many other uncovered irrationals.
>
> Forget limit points.

When there are infinitely many uncovered irrationals in the finite [0,1]
interval, some of them must be limit points so that limit points MUST be
considered in any complete analysis of the situation.

More pertinently, one might ask whether in the set of all of them there
must be any uncovered irrationals which are NOT limit points.

Note that for a Cantor set the answer is NO.

> Cantor's axiom says that the number x corresponds
> to a point of the real axis. Whether this point is a singleton is only
> decided by its immediate neighbourhood.

What Cantor "axiom" is that?

Analytic geometry, which preceded Cantor by a considerable period, is
the usual basis for having numbers correspond with points on a "real
axis".

A "singleton" is , by standard definition, a set with exactly one
member, but you are saying that a point may be a set.


What WM appears to be asking is whether every uncovered irrational is an
ISOLATED point in the set of all such uncovered irrational points.

And the clear answer is NO, because there must be infinitely many of
them in a finite interval.
--

[Virgil]

In article
<87faf737-3c2f-4a3f...@v9g2000vbc.googlegroups.com>,

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

> On 24 Jun., 17:42, "Mike Terry"
>
> > Note the order of quantification here:  given a neighbourhood of x there is
> > a cluster intersecting that neighbourhood.  This cluster depends on the
> > choice of neighbourhood.  It is NOT saying that there is a single cluster
> > that intersects with every neighbourhood of x.
>
> Addition: Have you ever understood that there is no infinite sequence
> (a_n)that defines its limit a? You can see this by leaving out the
> first n terms a_n - and that holds for every n. (You cannot specify
> any n such that the a_n would be required. Therefore note: The remote
> clusters do not have any influence.)

No one said they did.

But there may be a sequence of uncovered irrationals separated by
clusters but still converging to an uncovered irrational.

In which case, every neighborhood of that limit point necessarily
intersects a cluster, but for every cluster there can be a neigborhood
of the limit point that does NOT intersect that cluster.

> The limit a is defined by the
> finite definition that defines the sequence - not at all by any of its
> terms. Therefore your imagined clustering is not an "in fact" but
> simply nonsense.

Only nonsense in WM's matheological miasma.

In standard mathematics it is merely way over WM's head.

Wm again produces nonsense as an objection to sense.
--

[Virgil]

In article
<a429c669-1bfd-4a4b...@x21g2000vbc.googlegroups.com>,

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

> On 24 Jun., 17:42, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> > news:c7bfbccd-fdb3-4fbe...@e20g2000vbm.googlegroups.com...
> >
> > > On 24 Jun., 02:12, "Mike Terry"
> >
> > > > I think WM believes there are only countably many singletons, and his
> > > > argument for this was that each singleton is adjacent to a cluster, so
> > there
> > > > is some kind of correspondence between them. He's been told this is
> > wrong
> > > > of course, and has not provided any proof of this correspondence,
> >
> > > The proof is that an uncovered irrational is a singleton, i.e. it is
> > > adjacent to intervals I_n. Otherwise it was not a singleton.
> >
> > No - that is not the definition of a singleton.
>
> The definition of a singleton is a set that contains only one element
> because it is separated from other elements of the same kind.

Nothing in the STANDARD definition of "singleton" requires anything but
the set itself to separate its member from other objects.

>
> > It is disappointing that when challenged to prove a claim, your proof turns
> > out to be:
> >
> >    Claim:     Every uncovered x is the endpoint of some cluster.
> >    Proof:     1.  x is a singleton
> >               2.  all singletons are adjacent to intervals I_n.
> >               3.  therefore x is adjacent to an interval I_n.
> >               4.  (i.e. x is the endpoint of the interval I_n.)
> >
> > Do you really believe that constitutes a mathematical proof?
>
> Of course. Here is a proof of (2) by contradiction: If x is not
> adjacent to a covered interval I_n, then there is another uncovered
> irrational y separating x and I_n. (There is no further alternative
> separating x from a rational and hence from I_n.) This contradicts the
> assumption that x is a singleton.
>
> >  (In case you
> > genuinely do, the problem is with (2), so please come back and prove (2).)
>
> Done.

Wrong!

Example: let x = 1/sqrt(5) and a_n = (-1/5^n)/
then the sequence x + a_n of irrationals converges to the
irratinal 1/sqr(5)
For a construction similar to WM's, the open intervals between
successive points on the line could be clusters, with the
points, AND THEIR LIMT, x, being uncovered irrationals.

This is precisely the sort of thing that WM claims cannot happen.
But unless and until WM can come p which a valid argument why it cannot
happen even once, much less infinitely often in his example, his claims
bsed on his "example" all fail.

--

[Virgil]

In article
<c7bfbccd-fdb3-4fbe...@e20g2000vbm.googlegroups.com>,

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

> On 24 Jun., 02:12, "Mike Terry"
>
> > I think WM believes there are only countably many singletons, and his
> > argument for this was that each singleton is adjacent to a cluster, so there
> > is some kind of correspondence between them.  He's been told this is wrong
> > of course, and has not provided any proof of this correspondence,
>
> The proof is that an uncovered irrational is a singleton, i.e. it is
> adjacent to intervals I_n. Otherwise it was not a singleton.

Actually no irrational can be "adjacent" to any closed interval like one
of the I_n's.
Perhaps you meant adjacent to a cluster.

Such carelessness does not support your arguments.

And on the contrary, if WM's set of "singletons" has an irrational
accumulation point, and it must have at least one accumulation point,
then that point proves WM wrong.

>
> > so there's
> > not much to be said, other than he's wrong!  Of course each cluster is
> > "adjacent" to two singletons, but not vice-versa,
>
> Do you believe in singletons that are adjacent to other singletons
> without a rational between them? That would simply be the consequence.

No it would not. An irrational can be both the limit of an increasing
sequence of uncovered irrationals and the limit of a decreasing sequence
of uncovered irrationals. Such a point cannot be on the boundary of any
cluster.

>
> > and it's been explained to
> > him how for certain irrationals there are *sequences* of clusters which
> > converge to the irrational, with no single cluster being adjacent to it...
>
> Cantor's axiom says that every irrational x is a point on the real
> axis.

What "Cantor axiom" says anything like that?

You are not allowed to misquote Cantor in your vain attempts to refute
him.

> Whether this point is a singleton, is not at all determined by
> distant intervals or clusters, but only by intervals that are adjacent
> to x. And if there are no such intervals, then x is not a singleton.

Whether or not it is what WM calls a singleton, an uncovered irrational
can be a limit of uncovered irrationals and not an endpoint of any open
cluster.

> >
> > Anyway, when he asks "what makes a singleton a singleton?" I think he wants
> > you to say "the clusters that have been removed", and then he will say
> > "aha - that means that there must be clusters adjacent to it!" which of
> > course is just muddled thinking...
>
> Otherwise you must believe in singletons that are not adjacent to
> intervals. Or do you see another alternative?

An irrational which is both the limit of an increasing sequence of
uncovered irrationals and the limit of an decreasing sequence of
uncovered irrationals is a very realistic alternative.
>
> But try to comprehend

Do not, as some ungracious pastors do,
Show me the steep and thorny way to heaven,
Whiles, like a puffed and reckless libertine,
Himself the primrose path of dalliance treads,
And recks not his own rede.
--

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:a429c669-1bfd-4a4b...@x21g2000vbc.googlegroups.com...

Well yes, of course. (Right so far!)

> This contradicts the
> assumption that x is a singleton.

Um, no it doesn't. Remember that I_n and y are both a finite distance away
from x, and so their existence (taken by themselves) does not say anything
about whether or not x belongs to a singleton connected component. That is
a property that ties in to the whole neighbourhood system at x - i.e. it
depends on the properties of intervals "arbitrarily small" around x.

So still no proof yet...

Hmmm, you should have realised that your approach couldn't conceivably work,
because if x is an uncovered irrational, it will always be possible to
identify a *specific* interval I_n which is not adjacent to x, no? (E.g.
one separated from x by more than 1/2). Then even if x WERE the endpoint of
some other interval as you would like, your argument above would still
apply! (Except of course the argument is wrong wrong wrong and so doesn't
apply to anything :)

> Se my proof of (2) above in case you really are able to sober
> thinking.
> It is really ridiculous how you try to argue that far remote points
> and seqeunces should act on point x!

What I said was that x was isolated due to the presence of a SEQUENCE of
clusters which CONVERGE on x. Obviously ONE remote cluster in isolation
cannot cause this situation, and I never claimed it did - but the clusters
in the SEQUENCE become arbitrarily close to x, and so prevent x being
extended by any finite amount without intersecting one of the clusters.

In this situation, we can apply the correct bit of your proof logic to each
I_n in the sequence. The conclusion is that for each I_n there will be an
uncovered irrational y_n between I_n and x. These y_n will of course
converge to x. (And between the y_n there will be further covered
intervals.) However, none of this contradicts x being a singleton connected
component.


Regards,
Mike.

[Ross A. Finlayson]

On Jun 24, 2:34 pm, Virgil <vir...@ligriv.com> wrote:

>
>
> > But try to comprehend
>
> Do not, as some ungracious pastors do,
> Show me the steep and thorny way to heaven,
> Whiles, like a puffed and reckless libertine,
> Himself the primrose path of dalliance treads,
> And recks not his own rede.
> --

Yeah, we know you're fond of that, although it's unclear in as to
whether you do.

"It follows from the density in [0, 1] of the rationals." -- you

What you're calling "singletons" already have a name as intervals:
degenerate. They're not really intervals, those points. Singletons
are sets with one element.

Mueckenheim, that's not just pre-modern but pre-classical, there's
thousands of years of productive mathematics with infinity in it
already. While I appreciate some notions of that the the paths are
the tree and they go through nodes, the limit is the sum and no finite
approximation of a non-constant sequence's value as real number is the
sum. I don't care if you eschew transfinite cardinals, calculus
doesn't need it, and there's no greatest finite integer, or add one.

So, where are the avenues that Godel guarantees exists of new true
statements about the numbers beyond modern set theory? Get some.

Regards,

Ross Finlayson

[Alan Smaill]

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

> On 22 Jun., 19:46, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:

>> > On 21 Jun., 13:56, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>>
>> >> > And look here:
>> >> >http://math.stackexchange.com/questions/132022/formalizing-an-idea
>> >> > My answer proves that uncountability is in contradiction with
>> >> > continuity of the continuum.
>>
>> >> Claimed but not proved.
>>
>> > Proved by the fact, that an uncovered irrational x is not a singleton,
>> > unless there is a rational that makes it being a singleton.
>>
>> It is your insistence in stating what you call "facts", but which
>> are in fact in English statements vague to the point of meaninglessness,
>> that stops you from producing anything like a convincing proof.
>

> What makes a singleton a singleton?

The problem is your belief that your terminology of
a "rational that makes it a singleton".

What do *you* mean when you say "a rational that makes it a
singleton"? If you can do this, in such a way that the rational
is unique for each irrational, then yes you have your contradiction.
(Others have pointed out what they think is a flaw --
but all you have to do is fill in the details.)

So, spell out clearly what the property is question is
(personally I find this in English vague to the point of
meaninglessess), and show that you can set up an injective map this
way.

>> Simply repeating yourself does not make what you say more convincing;
>> it just gives the impression that you prefer propaganda to logic.
>
> What makes a singleton a singleton?

Simply repeating yourself does not make what you say more convincing;
it just gives the impression that you prefer propaganda to logic.

>> > I have made this experience quite often
>> > and never failed (except with matheologians).
>>
>> No doubt your students  bow down to the Scripture According To Mueckenheim.
>
> What makes a singleton a singleton?
>

Simply repeating yourself does not make what you say more convincing;
it just gives the impression that you prefer propaganda to logic.


> Regards, WM

--
Alan Smaill

[Alan Smaill]

"Jesse F. Hughes" <je...@phiwumbda.org> writes:

> WM <muec...@rz.fh-augsburg.de> writes:
>
>> On 23 Jun., 15:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>>
>>> I've no idea what you're going on about now, but what makes a singleton
>>> a singleton is the fact that it contains only one element.
>>
>> The real axis contains many irrational numbers. How can an irrational
>> number become a singleton? Why is there a number that is a singleton?
>
> This question makes no particular sense at all. Numbers do not become
> singletons[1].
>
> Look, I know that thinking about sets is challenging for you, but {pi},
> {0} and {sqrt(2)} are all singletons. Duh.
>
> Footnotes:
> [1] Ignoring for a moment odd facts about various representations of
> real numbers as sets.

Once upon a time, WM (wrongly) claimed that the German language made no
distinction between set membership and set inclusion ...

--
Alan Smaill

[WM]

On 25 Jun., 00:18, "Mike Terry"

<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

> > Of course. Here is a proof of (2) by contradiction: If x is not
> > adjacent to a covered interval I_n, then there is another uncovered
> > irrational y separating x and I_n. (There is no further alternative
> > separating x from a rational and hence from I_n.)
>
> Well yes, of course. (Right so far!)
>
> > This contradicts the
> > assumption that x is a singleton.
>
> Um, no it doesn't. Remember that I_n and y are both a finite distance away
> from x, and so their existence (taken by themselves) does not say anything
> about whether or not x belongs to a singleton connected component.

No. If y would be a finite distance away from x, then there must a
rational number between x and y. But that is excluded because there is
no rational outside an interval I_n.

That is
> a property that ties in to the whole neighbourhood system at x - i.e. it
> depends on the properties of intervals "arbitrarily small" around x.
>
> So still no proof yet...

Note that your assertion has been falsified.

>
> Hmmm, you should have realised that your approach couldn't conceivably work,
> because if x is an uncovered irrational, it will always be possible to
> identify a *specific* interval I_n which is not adjacent to x, no?

That is as obvious as irrelevant. Every interval I_n can be specified.
There are only countably many.

(E.g.
> one separated from x by more than 1/2). Then even if x WERE the endpoint of
> some other interval as you would like, your argument above would still
> apply!

No. You should say in what way it would apply.

> (Except of course the argument is wrong wrong wrong and so doesn't
> apply to anything :)

Trying to gain confidence?

>
> > Se my proof of (2) above in case you really are able to sober
> > thinking.
> > It is really ridiculous how you try to argue that far remote points
> > and seqeunces should act on point x!
>
> What I said was that x was isolated due to the presence of a SEQUENCE of
> clusters which CONVERGE on x.

x is a point that is not interested in remote intervals.

> Obviously ONE remote cluster in isolation
> cannot cause this situation, and I never claimed it did - but the clusters
> in the SEQUENCE become arbitrarily close to x, and so prevent x being
> extended by any finite amount without intersecting one of the clusters.

Here we deal with points and not with sequences.

>
> In this situation, we can apply the correct bit of your proof logic to each
> I_n in the sequence. The conclusion is that for each I_n there will be an
> uncovered irrational y_n between I_n and x.

But that is obvious. Otherwise all intervals would fall together.

> These y_n will of course
> converge to x. (And between the y_n there will be further covered
> intervals.) However, none of this contradicts x being a singleton connected
> component.

So no counter arguments as yet.

Regards, WM

[WM]

On 25 Jun., 12:02, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:

> > What makes a singleton a singleton?
>
> The problem is your belief that your terminology of
> a "rational that makes it a singleton".
>
> What do *you* mean when you say "a rational that makes it a
> singleton"?  If you can do this, in such a way that the rational
> is unique for each irrational,

Oh no, why should I try?

It is very simple. If the uncovered irrational x is not adjacent to an
interval I_n, then it must be adjacent to another uncovered irrational
y, because there is nothing else on the real line. But two irrationals
without rational between them is impossible.

Regards, WM

[WM]

On 25 Jun., 12:06, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> Once upon a time, WM (wrongly) claimed that the German language made no
> distinction between set membership and set inclusion ...
>

German and English language make the following difference:
A gehört zu B, A belongs to B.

Regards, WM