Discussion:
Matheology § 413
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m***@rz.fh-augsburg.de
2013-12-31 11:05:58 UTC
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Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion. It is impossible for an infinite list of decimals to appear in mathematical discourse, dialogue, or monologue other than as the finite rule how to calculate *every* decimal at a finite place but never *all* decimal, since beyond every finite index there are infinitely many further indices. Every decimal that appears in mathematical discourse, dialogue, or monologue belongs to a rational number.

By the way that is also the reason why Cantor's uncountability proofs must fail.

A matheologian (for the definition of matheology see § 1 of
http://www.hs-augsburg.de/~mueckenh/KB/Matheology.PDF
) answered: Both your assertions above are incorrect.

Wouldn't that claim oblige him, in crank-free mathematics, to support his opinion by listing all decimals of a nonterminating decimal representation of a real number of his choice?

Regards, WM
w***@gmail.com
2013-12-31 14:22:09 UTC
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Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion.
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.

William Hughes
m***@rz.fh-augsburg.de
2013-12-31 15:03:35 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion.
It is impossible for an infinite list of decimals to appear in mathematical discourse, dialogue, or monologue other than as the finite rule how to calculate *every* decimal at a finite place
Post by w***@gmail.com
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.
Of course. Why do you say "no"? Just your habit?


Regards, WM
w***@gmail.com
2013-12-31 15:37:31 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion.
It is impossible for an infinite list of decimals to appear in mathematical discourse, dialogue, or monologue other than as the finite rule how to calculate *every* decimal at a finite place
Post by w***@gmail.com
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.
Of course.
Everything that you say cannot be done with
digits can be done with finite definitions of potentially infinite
sequences e.g. diagonalization.

Indeed, it can easily be shown that given a potentially infinite sequence, L,
of finite definitions of potentially infinite 0/1 sequences there is
a finite definition of a potentially infinite 0/1 sequence that is not
an element of L.

William Hughes
m***@rz.fh-augsburg.de
2013-12-31 15:58:50 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion.
It is impossible for an infinite list of decimals to appear in mathematical discourse, dialogue, or monologue other than as the finite rule how to calculate *every* decimal at a finite place
Post by w***@gmail.com
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.
Of course.
Everything that you say cannot be done with
digits can be done with finite definitions of potentially infinite
sequences e.g. diagonalization.
Indeed, it can easily be shown that given a potentially infinite sequence, L,
of finite definitions of potentially infinite 0/1 sequences there is
a finite definition of a potentially infinite 0/1 sequence that is not
an element of L.
You think it is impossible to list all finite definitions?
Then it is impossible to list all terminating rationals.
It is easy to verify that the list of all terminating rationals is in bijection with the list of all finite words. And the list of all finite definitions is a sublist (hence a list) of the latter, and therefore less than uncountable.

Regards, WM
w***@gmail.com
2013-12-31 16:08:54 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever integer-positive-base) expansion.
It is impossible for an infinite list of decimals to appear in mathematical discourse, dialogue, or monologue other than as the finite rule how to calculate *every* decimal at a finite place
Post by w***@gmail.com
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.
Of course.
Everything that you say cannot be done with
digits can be done with finite definitions of potentially infinite
sequences e.g. diagonalization.
Indeed, it can easily be shown that given a potentially infinite sequence, L,
of finite definitions of potentially infinite 0/1 sequences there is
a finite definition of a potentially infinite 0/1 sequence that is not
an element of L.
You think it is impossible to list all finite definitions?
Yes, and so do you, since for a potentially infinite list, L, of finite
definitions of potentially infinite 0/1 sequences there is a finite
definition of a potentially infinite 0/1 sequence not an element of L.
Any putative list you provide is a putative contradiction in
Wolkenmuekenheim.

William Hughes
m***@rz.fh-augsburg.de
2013-12-31 18:03:45 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
You think it is impossible to list all finite definitions?
Yes, and so do you,
No. I have shown a list of all finite words.
Post by w***@gmail.com
since for a potentially infinite list, L, of finite
definitions of potentially infinite 0/1 sequences there is a finite
definition of a potentially infinite 0/1 sequence not an element of L.
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Everybody knows that you are wrong, compare Cantor, Bernays, Weyl, or here Schuette: "Definiert man die reellen Zahlen in einem streng formalen System, in dem nur endliche Herleitungen und festgelegte Grundzeichen zugelassen werden, so lassen sich diese reellen Zahlen gewiß abzählen, weil ja die Formeln und die Herleitungen auf Grund ihrer konstruktiven Erklärungen abzählbar sind."
And in actual infinity of set theory: an infinite list has no finite antidiagonal.

Regards, WM
w***@gmail.com
2013-12-31 19:25:14 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.

William Hughes
Ben Bacarisse
2014-01-01 12:48:03 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
--
Ben.
m***@rz.fh-augsburg.de
2014-01-01 16:36:36 UTC
Permalink
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
WH claims just the same, namely that the list of all finite words has an antidiagonal. Of course that cannot be a finite word.

Regards, WM
w***@gmail.com
2014-01-01 17:57:41 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
WH claims just the same, namely that the list of all finite words has an antidiagonal.
Well I never explicitly claimed this but it is true. Note that this antidiagonal
(unlike others that I did explicitly talk about) is potentially
infinite.

William Hughes
m***@rz.fh-augsburg.de
2014-01-01 18:35:51 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
WH claims just the same, namely that the list of all finite words has an antidiagonal.
Well I never explicitly claimed this but it is true. Note that this antidiagonal
(unlike others that I did explicitly talk about) is potentially
infinite.
The list of finite definitions contains only finite definitions. A potentially infinite antidiagonal is not a finite definition.

Regards, WM
Sam Sung
2014-01-01 18:45:31 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
WH claims just the same, namely that the list of all finite words has an antidiagonal.
Well I never explicitly claimed this but it is true. Note that this antidiagonal
(unlike others that I did explicitly talk about) is potentially
infinite.
he list of finite definitions contains only finite definitions.
LOL, asshole.
Post by m***@rz.fh-augsburg.de
A potentially infinite antidiagonal is not a finite definition.
It is: d((x),(y)) != (x),(y)
Post by m***@rz.fh-augsburg.de
Regards, WM
Go fuck your mother again, stinking asshole
Sam Sung
2014-01-01 18:49:48 UTC
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Post by Sam Sung
Post by m***@rz.fh-augsburg.de
he list of finite definitions contains only finite definitions.
LOL, asshole.
Post by m***@rz.fh-augsburg.de
A potentially infinite antidiagonal is not a finite definition.
It is: d((x),(y)) != (x),(y)
Post by m***@rz.fh-augsburg.de
Regards, WM
Go fuck your mother again, stinking asshole
For mathematicians: not ALL mappings are injective.
w***@gmail.com
2014-01-01 19:48:11 UTC
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WM: the list of all finite words

Look! Over there! A pink elephant!

WM: The list of finite definitions
Ben Bacarisse
2014-01-01 21:23:37 UTC
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Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
Well that made my day (but endangered my keyboard)! I've called it his
"bait and switch" tactic in the past, but this is so much clearer.
--
Ben.
m***@rz.fh-augsburg.de
2014-01-02 17:36:06 UTC
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Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist of the list of all finite word.

No pink required to recognize that.

Further: A potentially infinite antidiagonal is not a finite definition.

Regards, WM
w***@gmail.com
2014-01-02 18:04:35 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
Stop taking claims I make about the list of
all finite words and applying them to your
putative list of all finite definitions.

(Note the list of finite definitions
does not exist is Wolkenmuekenheim
as there is no finite way of telling whether
a putative definition is actually a definition)

William Hughes
m***@rz.fh-augsburg.de
2014-01-02 21:32:20 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist has never larger cardinality than its list.
Post by w***@gmail.com
Stop taking claims I make
You have no reason to complain since you always try to turn the words in my mouth!
Post by w***@gmail.com
about the list of
all finite words and applying them to your
putative list of all finite definitions.
An antidiagonal of the list of all finite words, if existing, is not a finite word.
And the list of all finite definitions, if existing, has not a finite definition as antidiagonal
Post by w***@gmail.com
(Note the list of finite definitions
does not exist is Wolkenmuekenheim
as there is no finite way of telling whether
a putative definition is actually a definition)
You are wrong again - and you know it, since I told you the finite way to decide for every word whether it is a definition: Read it and decide or have a machine read and decide it.

Regards, WM
Virgil
2014-01-02 23:54:54 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist has never larger cardinality than its list.
Post by w***@gmail.com
Stop taking claims I make
about the list of
all finite words and applying them to your
putative list of all finite definitions.
You have no reason to complain since you always try to turn the words in my mouth!
We only try to turn then from falsehoods into truths, but you keep
turning them back.

But most reals have no individual or finite definition.

Each infinite sequence of decimal digits represents by its finite
initial subsequences a bounded monotone sequence of rational numbers, so
must have a real number limit.

To avoid duplication of limits we can even limit the allowable digits to
0 thru 8 or to 1 thru 9, and we still get uncountably many digit
sequences each producing a real number necessarily different from that
of any other sequence of the same digits.

These are obviously collectively definable as I have just collectively
defined them, but are equally obviously not all of them individually
definable as even WM admits there are just not enough individual
definitions.

And with an obvious bijection between say the 0-8 sequences and the
reals they define, there must be uncountably many reals, a la the Cantor
diagonal argument.
--
w***@gmail.com
2014-01-03 01:29:58 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist
Is not the the list of all finite words.

William Hughes
m***@rz.fh-augsburg.de
2014-01-03 12:17:20 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist
Is not the the list of all finite words.
But has no larger cardinality.
And only that is the topic!

Regards, WM
w***@gmail.com
2014-01-03 12:28:32 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist
Is not the the list of all finite words.
But
Is not the the list of all finite words.

William Hughes
Virgil
2014-01-03 21:03:37 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist
Is not the the list of all finite words.
But has no larger cardinality.
And no smaller. But too small for the set of all real numbers.
Note that there must be real number for every sequence of decimal (or
other base) digits, whether finitely defineable or not, since the finite
initial segments of each such sequence represents a bounded monotone
increasing sequnce whose least upper bound must be a real number.
Post by m***@rz.fh-augsburg.de
And only that is the topic!
Regards, WM
--
w***@gmail.com
2014-01-03 21:15:49 UTC
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Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist
So is not the list of all finite words.
A sublist
Is not the the list of all finite words.
But
Is not the list of all finite words.
Virgil
2014-01-04 00:37:45 UTC
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Post by w***@gmail.com
But
Is not the list of all finite words.
"But" seems far too short to fit into any list of non-finite words, but
my dictionaries all assure me that it is a word.
--
w***@gmail.com
2014-01-03 01:32:06 UTC
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You have solved the halting problem !?!?
m***@rz.fh-augsburg.de
2014-01-03 11:40:54 UTC
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Post by w***@gmail.com
You have solved the halting problem !?!?
There is no halting problem. There is only some misunderstanding. Look at a definition and see hwether you understand it. If not, it defines the empty set.

There is no list of all men, for instance. Does that establish a halting problem or the uncountability of mankind?

Regards, WM
w***@gmail.com
2014-01-03 12:37:30 UTC
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Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
You have solved the halting problem !?!?
There is no halting problem. There is only some misunderstanding. Look at a definition and see hwether you understand it.
A putative definition

The nth digit of the sequences of 0/1 digits is given by the
output of algorithm X when evaluated at n.
I understand this.
There is no finite way to tell (in general)
whether this is an actual definition.
However, we cannot map all such putative definitions to
the empty set as for some algorithms X this is an actual
definition, and if we want every definition we cannot discard
any actual definitions.
William Hughes
m***@rz.fh-augsburg.de
2014-01-03 16:33:45 UTC
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Post by w***@gmail.com
The nth digit of the sequences of 0/1 digits is given by the
output of algorithm X when evaluated at n.
I understand this.
There is no finite way to tell (in general)
whether this is an actual definition.
Then it is not understandable as a definition of a real number. Ununderstandable definitions are not definitions.
Post by w***@gmail.com
However, we cannot map all such putative definitions to
the empty set as for some algorithms X this is an actual
definition, and if we want every definition we cannot discard
any actual definitions.
Everything that you cannot understand within your lifetime is not understandable for you. Of course the notion of definition is relative. But there is no problem to use this definition of definition: Everything that you cannot understand such that you can write the real number after 1 hour is not a definition of a real number with respect to the mathematical discourse that accepts you as a participant.

However, all that evasive arguing does not change the fact that a subset has never larger cardinality than its set. Therefore the present discussion is nothing else good for but to delay the time until you will understand that transfinite set theory is nonsense.

Regards, WM
fom
2014-01-03 16:48:56 UTC
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Post by m***@rz.fh-augsburg.de
However, all that evasive arguing does not change the fact that a subset has never larger cardinality than its set. Therefore the present discussion is nothing else good for but to delay the time until you will understand that transfinite set theory is nonsense.
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?

Exclude, for a moment, transfinite
arithmetic.
m***@rz.fh-augsburg.de
2014-01-03 17:05:10 UTC
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Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it over than I wish to spend. But spontaneously topics like the calculation of cardinalities of sets of angels or devils come to my mind.

Regards, WM
fom
2014-01-03 17:28:59 UTC
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Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it over than I wish to spend. But spontaneously topics like the calculation of cardinalities of sets of angels or devils come to my mind.
Well, what I really had in mind
is language.

Whatever "mathematics proper" may
be, there is a long-standing
relationship with logic. In the
nineteenth century Boole intended
to improve logic by making it more
mathematical while the logicists
who followed intended to repair
mathematics by making it more
logical.

I am ambivalent concerning transfinite
arithmetic because I view it as a
consequence of the intrinsic circularity
of explaining language with language. In
a more acceptable parlance, foundational
mathematics is full of "reflection
principles". So, it is like standing in
a room with mirrors on opposing walls.

This view is neither Platonist nor parochial,
although it is non-standard.

But, it does not exclude mathematicians
from investigating how they can fix the
notions they use in mathematics against
the vagaries that language introduces.

That is why I have pointed out that the
entire effect of transfinite arithmetic
is to merely shift "infinity" to "absolute
infinity".

I am sure that you are familiar with
fixed points. Take a look at the first
paragraph and the line above the "See
Also" section:

http://en.wikipedia.org/wiki/Directed_complete_partial_order#Properties

Omega is just a fixed point and transfinite
arithmetic is just a way to talk about
the mathematics of real numbers in a way that
is based on fixed points.

I am sure you will still see that as objectionable.
But, I thought perhaps it would be a view to
which you are not usually apprised.
m***@rz.fh-augsburg.de
2014-01-03 18:55:58 UTC
Permalink
Post by fom
I am ambivalent concerning transfinite
arithmetic because I view it as a
consequence of the intrinsic circularity
of explaining language with language.
That circularity can only be broken by explaining language with showing objects of dicourse, as it is done in the early childhood.
Post by fom
I am sure you will still see that as objectionable.
Of course. It does not repair the fact that every natural number defines a rational approximation. In order to arrive at omega you must leave the natural numbers, but you cannot claim that omega is nothing but the natural numbers.

Regards, WM
fom
2014-01-03 19:10:11 UTC
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Post by m***@rz.fh-augsburg.de
Post by fom
I am ambivalent concerning transfinite
arithmetic because I view it as a
consequence of the intrinsic circularity
of explaining language with language.
That circularity can only be broken by explaining language with showing objects of dicourse, as it is done in the early childhood.
Yes.

But, that is the difference between, say,
your position in a laser lab where your
personal empirical experience coincides
with scientific activity and the dogmatic
teaching of the facts of science.

I understand your desire to use the very
symbols of language as if that could
substitute for actual objects of empirical
experience. But, however one wishes to
apply that idea, it is still not the same.
It can never get beyond explaining language
with language.

So, even if you could ban all of the books
with disagreeable views today, they would
reappear tomorrow.
Virgil
2014-01-03 21:54:15 UTC
Permalink
Post by m***@rz.fh-augsburg.de
In order to arrive at omega you must leave the natural numbers, but you
cannot claim that omega is nothing but the natural numbers.
WRONG AGAIN, silly boy!

Omega is the standard name for the order type of which the well-ordered
set of natural numbers is the standard exemplar, so in that sense omega
IS nothing but the ordered set of natural numbers as a exemplar of an
order type.
--
m***@rz.fh-augsburg.de
2014-01-03 22:16:39 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
In order to arrive at omega you must leave the natural numbers, but you
cannot claim that omega is nothing but the natural numbers.
Omega is the standard name for the order type of which the well-ordered
set of natural numbers is the standard exemplar, so in that sense omega
IS nothing but the ordered set of natural numbers as a exemplar of an
order type.
Cantor used omega as the ordered set of natural numbers and as a number larger than every natural number. That is a contradiction. Every set of natural numbers counts itself. The number of natural numbers is not omega. But this contradiction is recognized best by the fact that every index of a decimal number indicates a rational approximation whereas all indices shall reduce the error to zero.

Regards, WM
Virgil
2014-01-04 00:27:36 UTC
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Post by m***@rz.fh-augsburg.de
Post by Virgil
Post by m***@rz.fh-augsburg.de
In order to arrive at omega you must leave the natural numbers, but you
cannot claim that omega is nothing but the natural numbers.
Omega is the standard name for the order type of which the well-ordered
set of natural numbers is the standard exemplar, so in that sense omega
IS nothing but the ordered set of natural numbers as a exemplar of an
order type.
Cantor used omega as the ordered set of natural numbers and as a number
larger than every natural number. That is a contradiction.
Not OUTSIDE of Wolkenmuekenheim.
Post by m***@rz.fh-augsburg.de
Every set of
natural numbers counts itself.
WRONG!, In fhe von Numann naturals each one which has a predecessor
counts only that predecessor itiself.
Post by m***@rz.fh-augsburg.de
The number of natural numbers is not omega.
The order type of the set of natural IS omega.
Post by m***@rz.fh-augsburg.de
But this contradiction is recognized best by the fact that every index of a
decimal number indicates a rational approximation whereas all indices shall
reduce the error to zero.
Except that WM repeatedly denies that all indices can ever exist.
Thus in Wolkenmuekenheim only those reals with terminating expansions
are allowed to exist, even though this gives different sets of reals for
different bases.
Post by m***@rz.fh-augsburg.de
Regards, WM
--
m***@rz.fh-augsburg.de
2014-01-04 10:45:00 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
The number of natural numbers is not omega.
The order type of the set of natural IS omega.
In later years Cantor used omega as cardinal too - as is done in modern set theory.
Post by Virgil
Post by m***@rz.fh-augsburg.de
But this contradiction is recognized best by the fact that every index of a
decimal number indicates a rational approximation whereas all indices shall
reduce the error to zero.
Except that WM repeatedly denies that all indices can ever exist.
That is the necessary result. Here we are concerned with the errors of set theory.
Post by Virgil
Thus in Wolkenmuekenheim only those reals with terminating expansions
are allowed to exist,
Only expansions with natural indices are possible. Every natural is finite. All expansions of an irrational number with only natural indices supply values different from the irrational number. It is the great magic trick that an expansion with all indices should supply the correct value. But since matheology is existing in minds that can imagine everything, even if it is a logical contradiction, such things exist in matheology.

Regards, WM
Virgil
2014-01-04 21:09:03 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Only expansions with natural indices are possible.
x = Sum 1/10^(n!) defines by its digits an irratonal number.
And all its indices are naturals.
Post by m***@rz.fh-augsburg.de
Every natural is finite.
But the set of all of them is not!
Post by m***@rz.fh-augsburg.de
All expansions of an irrational number with only natural indices supply
values different from the irrational number.
x = Sum 1/10^(n!) does not!
And all its indices are naturals.
Post by m***@rz.fh-augsburg.de
It is the great magic trick that
an expansion with all indices should supply the correct value.
It works when all but finitely many indices give value 0-, so why not
otherwise?

Those succesive finite approximations formed by truncating an infinite
expansion are a bounded montone sequence which must have a limit among
the real numbers, so every one of them corresponds to a real real number.
Post by m***@rz.fh-augsburg.de
But since matheology
WM's matheoogy, which we call WMytheology , is irrelevant.
--
Apollyon
2014-01-03 17:30:55 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it over than I wish to spend. But spontaneously topics like the calculation of cardinalities of sets of angels or devils come to my mind.
Regards, WM
Well, I can answer how many Angels can dance on the head of a pin (though you did not ask but is inferred by mine self from your comment)
ONE
Because Angels are shy but also fiercely territorial ;-)
fom
2014-01-03 17:41:17 UTC
Permalink
Post by Apollyon
Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it over than I wish to spend. But spontaneously topics like the calculation of cardinalities of sets of angels or devils come to my mind.
Regards, WM
Well, I can answer how many Angels can dance on the head of a pin (though you did not ask but is inferred by mine self from your comment)
ONE
Because Angels are shy but also fiercely territorial ;-)
I suppose, then, that they can be well-ordered...

http://mathworld.wolfram.com/SolitaryNumber.html
m***@rz.fh-augsburg.de
2014-01-03 18:50:11 UTC
Permalink
Post by Apollyon
Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it over than I wish to spend. But spontaneously topics like the calculation of cardinalities of sets of angels or devils come to my mind.
Regards, WM
Well, I can answer how many Angels can dance on the head of a pin (though you did not ask but is inferred by mine self from your comment)
ONE
Because Angels are shy but also fiercely territorial ;-)
That is a valid topic of social studies. But it is not mathematics.

Regards, WM
Virgil
2014-01-03 21:42:57 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it
over than I wish to spend. But spontaneously topics like the calculation of
cardinalities of sets of angels or devils come to my mind.
Regards, WM
WM overlooks the classical example of such a question:
"How many angels can dance on the head of a pin?"

Or when in Wolkenmuekenheim, "Which is the last/largest natural number?"
--
Apollyon
2014-01-04 15:33:22 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
Post by fom
Do you think that there are topics
toward which mathematicians ought
not direct their efforts?
Exclude, for a moment, transfinite
arithmetic.
That is a very philosophical question. It would require more time to think it
over than I wish to spend. But spontaneously topics like the calculation of
cardinalities of sets of angels or devils come to my mind.
Regards, WM
"How many angels can dance on the head of a pin?"
One.
They use man's measuring devices, therefore they would not be the size of a diatom in order to cram many small ones. And they are too territorial to allow others with them, unless so ordered. Assuming they can dance on a pin at all, of course ;-)
http://biblehub.com/revelation/21-17.htm
They don't dance much, so they don't dance very well, so they don't want people watching.
"They Dance Alone"
Sting


Why are these women here dancing on their own?
Why is there this sadness in their eyes?
Why are the soldiers here
Their faces fixed like stone?
I can't see what it is that they despise
They're dancing with the missing
They're dancing with the dead
They dance with the invisible ones
Their anguish is unsaid
They're dancing with their fathers
They're dancing with their sons
They're dancing with their husbands
They dance alone, they dance alone
http://sting.com/discography/lyrics/lyric/song/253
Post by Virgil
Or when in Wolkenmuekenheim, "Which is the last/largest natural number?"
--
Move along, nothing to see here.
w***@gmail.com
2014-01-03 18:52:32 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Of course the notion of definition is relative.
And hence your definition of definition is useless.

William Hughes

P.S . You are still stuck believing that given a list of finite
definitions there is a finite definition not on the list, yet
despite this that there is a list of all finite definitions.
m***@rz.fh-augsburg.de
2014-01-03 19:01:42 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Of course the notion of definition is relative.
And hence your definition of definition is useless.
Everything in mathematical discourse is in a not negligible way relative, namely relative to the used model (for instance uncountability - according to Skolem) relative to the used language, and relative to the participants of the discourse.
Post by w***@gmail.com
P.S . You are still stuck believing that given a list of finite
definitions there is a finite definition not on the list, yet
despite this that there is a list of all finite definitions.
In set theory there is a list of all finite definitions, namely a sublist of the list of all finite words. This list contains all finite definitions and has cardinality aleph_0.

Regards, WM
w***@gmail.com
2014-01-03 19:42:48 UTC
Permalink
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of the list of all finite words. This list contains all finite definitions
And despite this you believe there is a finite definition not on this list.
Keep taking those Red Queen lessons.

William Hughes
m***@rz.fh-augsburg.de
2014-01-03 22:14:40 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of the list of all finite words. This list contains all finite definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption that there could be an actually infinite set, list, sequence or whatever.
Post by w***@gmail.com
Keep taking those Red Queen lessons.
We can avoid them when leaving matheology and returning to sober mathematics.

Regards, WM
Virgil
2014-01-04 00:15:35 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist
of the list of all finite words. This list contains all finite
definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption that there
could be an actually infinite set, list, sequence or whatever.
That may be the only remedy inside of Wolkenmuekenheim but is a remedy
at all anywhere else.
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Keep taking those Red Queen lessons.
We can avoid them when leaving matheology
Unfortuntately for WM , he has proved unable to leave his WMatheology.

The rest of us never enter it but stick with standard mathematics.
--
w***@gmail.com
2014-01-04 00:23:33 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of the list of all finite words. This list contains all finite definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption that there could be an actually infinite set
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
The contradiction is in Wolkenmuekenheim.

William Hughes
Sam Sung
2014-01-04 00:30:52 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely
a sublist of the list of all finite words. This list contains
all finite definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption
that there could be an actually infinite set
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
The contradiction is in Wolkenmuekenheim.
William Hughes
Yes - so omega is the cardiality of omega.
Sam Sung
2014-01-04 01:11:01 UTC
Permalink
Post by Sam Sung
Yes - so omega is the cardiality of omega.
Yes - so omega is the card(iNality) of omega.

sorry


Virgil
2014-01-04 02:33:41 UTC
Permalink
Post by Sam Sung
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely
a sublist of the list of all finite words. This list contains
all finite definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption
that there could be an actually infinite set
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
The contradiction is in Wolkenmuekenheim.
William Hughes
Yes - so omega is the cardiality of omega.
More properly, aleph_0 is the cardinality of omega,
while omega is the ordinality of omega.
--
Sam Sung
2014-01-04 02:38:07 UTC
Permalink
Post by Virgil
Post by Sam Sung
Yes - so omega is the cardiality of omega.
More properly,
aleph_0 is the cardinality of omega,
so aleph_0 equals = cardinality(omega)
Post by Virgil
while omega is the ordinality of omega
which equals = cardinality(omega)?

+
http://youtu.be/WXJrySutxLY
m***@rz.fh-augsburg.de
2014-01-04 10:44:51 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of the list of all finite words. This list contains all finite definitions
And despite this you believe there is a finite definition not on this list.
If so, then the only remedy consists of dropping the assumption that there could be an actually infinite set
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
Except that potential infinity has no completed infinite sets and therefore has no cardinal numbers of such sets.

Regards, WM
Sam Sung
2014-01-04 11:28:11 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
Except that potential infinity
It's you who "has" got each screw lose.
Post by m***@rz.fh-augsburg.de
has no completed infinite sets
The mankind uses to define the class of isomorphic sets
N := {0 which owns exactly 1 successor /AND any element
which owns exactly 1 predecessor and 1 successor}
which is very comlete because EACH ELEMENT next to 0
has got THE SAME properties. There is NO BETTER completion.

And you motherfucker don't belong to the mankind, which uses
to utilize abstractions without the permission of you artard
and without to ask you any questions. Furthermore, NOBODY,
especially no mathematician wants NONE of your crock excretion.
Post by m***@rz.fh-augsburg.de
and therefore has no cardinal numbers of such sets.
Piss off you asshole and go perish already.
Post by m***@rz.fh-augsburg.de
Regards, WM
We spit into your ugly motherfucker face.
w***@gmail.com
2014-01-04 13:27:18 UTC
Permalink
<snip> ... potential infinity has no completed infinite sets and therefore has no cardinal numbers of such sets.
We can and do ask whether a set can be listable by a potentailly
infinite list. (Cardinal numbers are not part of the argument).


William Hughes
m***@rz.fh-augsburg.de
2014-01-04 17:08:29 UTC
Permalink
Post by w***@gmail.com
<snip> ... potential infinity has no completed infinite sets and therefore has no cardinal numbers of such sets.
We can and do ask whether a set can be listable by a potentailly
infinite list. (Cardinal numbers are not part of the argument).
The set of every man, for instance, is not listable. The set of every finite word including every finite definition including every definition of every real number is listable.

The set of every finite definition only is not listable unless you have a way to decide what a definition is.

The set of the numbers of terrestrial hamsters living by the last day of the years 3000 to 2000 BC, 12 am, is not listable for the same reason.

Regards, WM
w***@gmail.com
2014-01-04 17:33:27 UTC
Permalink
Post by w***@gmail.com
<snip> ... potential infinity has no completed infinite sets and therefore has no cardinal numbers of such sets.
We can and do ask whether a set can be listable by a potentailly
infinite list. (Cardinal numbers are not part of the argument).
You insist there is a list L, of all finite definitions
and at the same time there is a finite definition not in L.
This contradiction is part of Wolkenmuekenheim and has nothing to do with cardinality of infinite (either actual or potential) sets.

William Hughes
m***@rz.fh-augsburg.de
2014-01-04 22:17:20 UTC
Permalink
Post by w***@gmail.com
You insist there is a list L, of all finite definitions
if infinity is completed. Otherwise we have only a list of every finite definition.
Post by w***@gmail.com
and at the same time there is a finite definition not in L.
No. There is no finite definition outside the list, i.e., every finite definition has its place in the list of all finite words.
Post by w***@gmail.com
This contradiction is part of
your misunderstanding, intentionally or not.

Regards, WM
w***@gmail.com
2014-01-04 22:32:49 UTC
Permalink
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
And at the same time you note that one
can use the list to construct a finite
definition not in the list.

William Hughes
Sam Sung
2014-01-05 08:36:13 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
And at the same time you note that one
can use the list to construct a finite
definition not in the list.
Of course - otherwise you love birdies could not continue
to nurse each others so tenderly forever - same as they
DID 7+ years long in the german math news group.
Virgil
2014-01-05 01:36:17 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
You insist there is a list L, of all finite definitions
if infinity is completed. Otherwise we have only a list of every finite definition.
But a list of every but only finite definitions, since it must contain
every terminating decimal, and every terminatng expansion in every other
of infinitely many bases, must be actually infinite.
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
and at the same time there is a finite definition not in L.
No. There is no finite definition outside the list, i.e., every finite
definition has its place in the list of all finite words.
There are definitions that require more than one word each, at least in
every language but German, so how is WM going to guarantee that in an
arbitrary list of words that every sequence of words which is a
definition will occur and will occur in the proper order in WM's list?
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
This contradiction is part of
WM'smisunderstanding, intentionally or not.
--
w***@gmail.com
2014-01-05 17:13:00 UTC
Permalink
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
Let the list of finite words be W
Let the list of every finite definition be L'
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').
However, L' contains every finite definition in W.
Contradiction.

William Hughes
WM
2014-01-06 12:31:20 UTC
Permalink
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
Let the list of finite words be W
For example:
0
1
00
10
01
11
000
...
Post by w***@gmail.com
Let the list of every finite definition be L'
Assume every word starting with 0 is a finite definition.

0
00
01
000
...
Post by w***@gmail.com
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Why?
Where does it end?
Post by w***@gmail.com
Use induction to show that for every n in |N
d_L is not the nth element of L'
Here is L'

0
00
01
000
...

It has no diagonal. Already the third line is too short.
Please construct the antidiagonal.
Post by w***@gmail.com
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
If it is a definition, it is also in L' - by definition.
Post by w***@gmail.com
but not an element of L').
However, L' contains every finite definition in W.
Contradiction.
Perhaps you can answer my above questions?

Regards, WM
w***@gmail.com
2014-01-06 13:46:33 UTC
Permalink
Post by WM
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
Let the list of finite words be W
0
1
00
10
01
11
000
...
Post by w***@gmail.com
Let the list of every finite definition be L'
Assume every word starting with 0 is a finite definition.
Nope. L' is a list of finite definitions of potentially infinite
0/1 sequences. The L' you decribe is e.g.

0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0's
...

L, the list of potentially infinite 0/1 sequences
is

00...
010...
010...
...

(Do not confuse L' and L)

Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Post by WM
Why?
Where does it end?
The definition ends at " "od2)". The potentially infinite
sequence defined by the definition does not end.
Do not confuse definitions with the potentially infinite
0/1 sequences they define.

Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').

However, by defintion, d_L is in L'.
Contradiction.

William Hughes
WM
2014-01-06 14:52:28 UTC
Permalink
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
Let the list of finite words be W
0
1
00
10
01
11
000
...
Post by w***@gmail.com
Let the list of every finite definition be L'
Assume every word starting with 0 is a finite definition.
Nope. L' is a list of finite definitions of potentially infinite
0/1 sequences. The L' you decribe is e.g.
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0's
...
That is all in my list, encoded in bits.
Post by w***@gmail.com
L, the list of potentially infinite 0/1 sequences
is
L is the list of all finite words! Please don't switch meanings.
Post by w***@gmail.com
00...
010...
010...
...
(Do not confuse L' and L)
In fact, do not confuse it.
Post by w***@gmail.com
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Post by WM
Why?
Where does it end?
The definition ends at " "od2)". The potentially infinite
sequence defined by the definition does not end.
Do not confuse definitions with the potentially infinite
0/1 sequences they define.
In fact.
Post by w***@gmail.com
Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').
However, by defintion, d_L is in L'.
Contradiction.
You are confused.

The list of all finite words is, in binary notation:

0
1
00
...

The list of all finite definitions is a subset. Of course it will contain the sequence of letters:
"Let d_L(n) = (f(n,n) +1) (mod 2)"
like every other finite word.

Alas the antidiagonal d_L defined by it does not exist, because there are many too short lines, like:
"0 followed by the binary expression for 0 followed by 0's."

Concluding: Neither is the definition of the antidiagonal missing nor is the construction of the antidiagonal possible. Your argument fails.

Regards, WM
w***@gmail.com
2014-01-06 14:58:31 UTC
Permalink
Post by WM
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
No. There is no finite definition outside the list,
Let the list of finite words be W
0
1
00
10
01
11
000
...
Post by w***@gmail.com
Let the list of every finite definition be L'
Assume every word starting with 0 is a finite definition.
Nope. L' is a list of finite definitions of potentially infinite
0/1 sequences. The L' you decribe is e.g.
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0'
...
That is all in my list, encoded in bits.
Nope. You forgot the trailing 0's
Post by WM
Post by w***@gmail.com
L, the list of potentially infinite 0/1 sequences
is
L is the list of all finite words! Please don't switch meanings.
Nope. W is the list of all finite words.
Post by WM
Post by w***@gmail.com
00...
010...
010...
...
(Do not confuse L' and L)
In fact, do not confuse it.
Post by w***@gmail.com
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Post by WM
Why?
Where does it end?
The definition ends at " "od2)". The potentially infinite
sequence defined by the definition does not end.
Do not confuse definitions with the potentially infinite
0/1 sequences they define.
In fact.
Post by w***@gmail.com
Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').
However, by defintion, d_L is in L'.
Contradiction.
William Hughes
WM
2014-01-06 16:59:54 UTC
Permalink
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0'
...
That is all in my list, encoded in bits.
Nope. You forgot the trailing 0's
You seem to think that my list contains numbers. It contains finite words (like "0 followed by the binary expression for 0 followed by 0's" in an alphabet of only two letters. In order to remove that misunderstanding, consider this list of all finite words:

a
b
aa
ab
ba
bb
aaa
...

All numerals and signs including the point are described by letters, for convenience I use only two instead of 30.
Post by w***@gmail.com
Post by WM
L is the list of all finite words! Please don't switch meanings.
Nope. W is the list of all finite words.
Agreed. W is the list of finite words like that one I gave above.
And the list of finite definitions is a sublist, for instance it could be the list of all finite words that have an a on third position.

Regards, WM
w***@gmail.com
2014-01-06 17:11:24 UTC
Permalink
Post by WM
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0'
...
That is all in my list, encoded in bits.
Nope. You forgot the trailing 0's
You seem to think that my list contains numbers.
Nope, Your list is L'. The elements of L' are not numbers.
Every element of L' is a definition of
a potentially infinite 0/1 sequence.
Post by WM
Post by w***@gmail.com
Post by WM
L is the list of all finite words! Please don't switch meanings.
Nope. W is the list of all finite words.
Agreed. W is the list of finite words like that one I gave above.
And the list of finite definitions is
L' (ell prime). The list L is not the list L'.


Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').
However, L' contains every finite definition in W.
Contradiction.

William Hughes
WM
2014-01-06 17:28:52 UTC
Permalink
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0'
...
That is all in my list, encoded in bits.
Nope. You forgot the trailing 0's
You seem to think that my list contains numbers.
Nope, Your list is L'. The elements of L' are not numbers.
Every element of L' is a definition of
a potentially infinite 0/1 sequence.
How did you get to ´know that I forgot trailing 0's?
Post by w***@gmail.com
L' (ell prime). The list L is not the list L'.
What is L?
Post by w***@gmail.com
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
What if line number n does not have an nth element?

Regards, WM
Sam Sung
2014-01-06 17:32:32 UTC
Permalink
WM drivels
Post by WM
Post by w***@gmail.com
L' (ell prime). The list L is not the list L'.
What is L?
What if line number n does not have an nth element?
After 10...20 years hard work you could know if you were
not such a abhoring motherfucker.
Post by WM
Regards, WM
Piss off and go croak, verrecke!
w***@gmail.com
2014-01-06 22:27:37 UTC
Permalink
Post by WM
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
Post by WM
Post by w***@gmail.com
0 followed by the binary expression for 0 followed by 0's
0 followed by the binary expression for 1 followed by 0's
0 followed by the binary expression for 2 followed by 0'
...
That is all in my list, encoded in bits.
Nope. You forgot the trailing 0's
You seem to think that my list contains numbers.
Nope, Your list is L'. The elements of L' are not numbers.
Every element of L' is a definition of
a potentially infinite 0/1 sequence.
How did you get to ´know that I forgot trailing 0's?
I may have been wrong. It does not matter. What matters
is that every element of your L' is a finite definition
of a potentially infinite 0/1 sequence.
Post by WM
Post by w***@gmail.com
L' (ell prime). The list L is not the list L'.
What is L?
The list of potentially infinite 0/1 sequences. The nth
row of L is the potentially infinite 0/1 sequence defined
by the nth element of L'. Note, it is L not L' which
has a diagonal.
Post by WM
Post by w***@gmail.com
Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'.
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
What if line number n does not have an nth element?
d_L is defined in terms of the elements of L'
(or equivalently the rows of L).
Each element of L' defines a potentially infinite
0/1 sequence, which does have an nth element.

[Note I am assuming that a finite 0/1 sequence is not
a potentially infinite 0/1 sequence. If this is not your
view take the elements of L' to be finite definitions
of non-finite potentially infinite sequences.]

Define a function of two integers f by
f(n.m) = the mth digit defined by the nth element of L'. .
Let d_L(n) = (f(n,n) +1) (mod 2)
Then d_L is a finite definition.
Use induction to show that for every n in |N
d_L is not the nth element of L'
Use an indirect proof to show there is no n in |N
such that d_L is the nth element of L'.
(Note that d_L is an element of W,
but not an element of L').
However, L' contains every finite definition in W.
Contradiction.

William Hughes
Sam Sung
2014-01-06 17:21:49 UTC
Permalink
Post by w***@gmail.com
Nope. You forgot the trailing 0's
You seem to think that my list contains numbers...
a
b
aa
ab
ba
bb
aaa
...
All numerals and signs including the point are described by letters,
for convenience I use only two instead of 30.
So there would just be a letter for "no letter", eg an "x":
No problem to make d.
ax...
bx...
aax...
abx...
bax...
bbx...
aaax...
...
You are so fucking stupid - the most primitive motherfucker
in the newsnet ever. Piss off and go PERISH! (Verrecke!)
Regards, WM
We spit into your stinking ugly face
Virgil
2014-01-06 23:03:46 UTC
Permalink
Post by WM
0
1
00
...
The list of all finite definitions is a subset.
Definitions are rarely single words, at least in English, they may even
require sentences, or several paragraphs.
Post by WM
Of course it will contain the
"Let d_L(n) = (f(n,n) +1) (mod 2)"
As far as I can see, your list above can contain only the "1", from
"Let d_L(n) = (f(n,n) +1) (mod 2)", but nothing else.
Post by WM
like every other finite word.
In English, "Let d_L(n) = (f(n,n) +1) (mod 2)" is not a word.
Post by WM
Alas the antidiagonal d_L defined by it does not exist
Some of WM's many lies snipped.
--

Virgil
2014-01-04 21:27:16 UTC
Permalink
Post by m***@rz.fh-augsburg.de
We can and do ask whether a set can be listable by a potentially
infinite list. (Cardinal numbers are not part of the argument).
The set of every man, for instance, is not listable.
The set of every man living at a particular time is at least
theoreticaly listable, thoughthere are serious practical difficulties
involved in actually doing it..
Post by m***@rz.fh-augsburg.de
The set of every finite
word including every finite definition including every definition of every
real number is listable.
In English, and I suspect most languages other than German, most
definitions can not be given by single words so would not appear in any
list of all words except by extremely rare accident, and certainly not
all of them in any one word list unless lots of words were allowed to
appear repeatedly in that list.
Post by m***@rz.fh-augsburg.de
Regards, WM
--
Virgil
2014-01-04 21:13:30 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Nope, all arguments used potential infinity only.
Assuming there is no actually infinite set changes nothing.
Except that potential infinity has no completed infinite sets and therefore
has no cardinal numbers of such sets.
Potential infinity also does not support inductive proof, nor a real
real number system in which bounded monotone sequences all have limits.
--
Virgil
2014-01-03 21:38:52 UTC
Permalink
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of
the list of all finite words. This list contains all finite definitions and
has cardinality aleph_0.
"The list of all finite words" need not itself be finite, thus need not
exist at all in WM's wild weird world of WMytheology.

Similarly "The list of all finite definitions " need not itself be
finite, thus need not exist at all in WM's wild weird world of
WMytheology.

WM often outdoes the Red Queen by believing six or more mutually
incompatible things before breakfast.
--
m***@rz.fh-augsburg.de
2014-01-03 22:16:17 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of
the list of all finite words. This list contains all finite definitions and
has cardinality aleph_0.
"The list of all finite words" need not itself be finite,
It cannot be finite.
Post by Virgil
Similarly "The list of all finite definitions " need not itself be
finite, thus
WH's claim that the diagonal of one of these lists is another example of an entry is false.

Regards, WM
Virgil
2014-01-04 00:32:36 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by Virgil
Post by m***@rz.fh-augsburg.de
In set theory there is a list of all finite definitions, namely a sublist of
the list of all finite words. This list contains all finite definitions and
has cardinality aleph_0.
"The list of all finite words" need not itself be finite,
It cannot be finite.
But in Wolkenmuekenheim, there cannot be anything actually infinite.

Presumably including lists of words, and lists of numbers.
WM again blows hot and cold in the same breath!
--
Virgil
2014-01-03 21:34:15 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Everything that you cannot understand such that you can write the real
number after 1 hour is not a definition of a real number
There are lots of valid definitions of real numbers that WM cannot
understand even after a full day.
--
Virgil
2014-01-03 21:30:05 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
You have solved the halting problem !?!?
There is no halting problem.
There is outside of WM's wild weird world of WMytheology.
Post by m***@rz.fh-augsburg.de
There is only some misunderstanding. Look at a
definition and see hwether you understand it. If not, it defines the empty
set.
The empty set does describe Wolkenmuekenheim.
Post by m***@rz.fh-augsburg.de
There is no list of all men, for instance.
There could be, if it were timestamped, but not otherwise, because at
any given time, the set of men is fixed and finite, but over even short
periods of time changes.
Post by m***@rz.fh-augsburg.de
Does that establish a halting
problem or the uncountability of mankind?
It establishes the unaccountability of WM's math.
--
Virgil
2014-01-02 19:03:22 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist of the list of all finite word.
But most reals have no individual definition.

Each infinite sequence of decimal digits represents by its finite
subsequences a bounded monotone sequence of rational numbers, so must
have a real number limit.

To avoid duplication of limits we can even limit the allowable digits to
0 thru 8 or 1 thru 9, and we still get uncountably many digit sequences
each producing a real number necessarily different from that of any
other sequence.

These are obviously collectively definable as I have just collectively
defined them, but are equally obviously not all individually definable
as there are just not enough individual definitions.

And with an obvious bijection between say the 0-8 sequences and the
reals they define, uncountably many reals, a la the Cantor diagonal
argument.
--
m***@rz.fh-augsburg.de
2014-01-02 22:03:16 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist of the list of all finite word.
But most reals have no individual definition.
How do you put such a real number into a list or get it as an antidiagonal?

Regards, WM
Virgil
2014-01-02 23:47:57 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by Virgil
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
WM: the list of all finite words
Look! Over there! A pink elephant!
WM: The list of finite definitions
is a sublist of the list of all finite word.
But most reals have no individual definition.
How do you put such a real number into a list or get it as an antidiagonal?
Regards, WM
What makes you think that you have to be able to get EVERY real into a
list?

But most reals have no individual definition.

Each infinite sequence of decimal digits represents by its finite
subsequences a bounded monotone sequence of rational numbers, so must
have a real number limit.

To avoid duplication of limits we can even limit the allowable digits to
0 thru 8 or 1 thru 9, and we still get uncountably many digit sequences
each producing a real number necessarily different from that of any
other sequence.

These are obviously collectively-definable as I have just collectively
defined them, but are equally obviously are not all individually
definable as there are just not enough individual definitions.

And with an obvious bijection between say the 0-8 sequences and the
reals they define, we see that there must be uncountably many reals,
a la the Cantor diagonal argument.
--
m***@rz.fh-augsburg.de
2014-01-03 12:17:14 UTC
Permalink
Post by Virgil
Post by m***@rz.fh-augsburg.de
Post by Virgil
But most reals have no individual definition.
How do you put such a real number into a list or get it as an antidiagonal?
What makes you think that you have to be able to get EVERY real into a
list?
All definable reals belong to one and the same countable set of definable real numbers. How do you introduce undefinable reals in a list or get them as antidiagonals?
Post by Virgil
But most reals have no individual definition.
Reals which can appear in mathematics, I mean in any discourse, have to be defined. How else should they be addressed?
Post by Virgil
Each infinite sequence of decimal digits represents by its finite
subsequences a bounded monotone sequence of rational numbers, so must
have a real number limit.
Of course. Each infinite sequence is defined. Otherwise one would not know that there is a sequence.
Post by Virgil
To avoid duplication of limits we can even limit the allowable digits to
0 thru 8 or 1 thru 9,
Only if all are defined. For undefinable reals it should not make a difference if they are not uniquely defined,should it?
Post by Virgil
and we still get uncountably many digit sequences
each producing a real number necessarily different from that of any
other sequence.
Please produce such a real number that does not belong to the countable set of all real numbers.
Post by Virgil
These are obviously collectively-definable as I have just collectively
defined them, but are equally obviously are not all individually
definable as there are just not enough individual definitions.
So you cannot use them or obtain them from the diagonal argument.

That means, all antidiagonals that can be part of mathematical discourse belong to a countable set.
Post by Virgil
And with an obvious bijection between say the 0-8 sequences and the
reals they define, we see that there must be uncountably many reals,
a la the Cantor diagonal argument.
The latter only shows that all antidiagonals that can result from defined lists belong to a countable set. But can undefined lists be considered as lists? And can they have antidiagonals?

Regards, WM
Virgil
2014-01-03 21:23:00 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by Virgil
Post by m***@rz.fh-augsburg.de
Post by Virgil
But most reals have no individual definition.
How do you put such a real number into a list or get it as an antidiagonal?
What makes you think that you have to be able to get EVERY real into a
list?
All definable reals belong to one and the same countable set of definable
real numbers. How do you introduce undefinable reals in a list or get them as
antidiagonals?
Post by Virgil
But most reals have no individual definition.
Reals which can appear in mathematics, I mean in any discourse, have to be
defined. How else should they be addressed?
Collectively, as that is the only way most of them can be addressed.

Note that there must be real number for every sequence of decimal (or
other base) digits, whether a finitely defineable sequence or not,
since the finite initial segments of each such sequence represent a
bounded monotone increasing sequnce whose least upper bound must be a
real number.
Post by m***@rz.fh-augsburg.de
Of course. Each infinite sequence is defined. Otherwise one would not know
that there is a sequence.
Not individually, but only collectively.
Post by m***@rz.fh-augsburg.de
Post by Virgil
To avoid duplication of limits we can even limit the allowable digits to
0 thru 8 or 1 thru 9,
Only if all are defined. For undefinable reals it should not make a
difference if they are not uniquely defined,should it?
All reals are defineable, but most do not have individual definitions,
only collective ones defining sets of them.
Post by m***@rz.fh-augsburg.de
Post by Virgil
and we still get uncountably many digit sequences
each producing a real number necessarily different from that of any
other sequence.
Please produce such a real number that does not belong to the countable set
of all real numbers.
Only when WM can show a last natural number, as he keeps claiiming.
Post by m***@rz.fh-augsburg.de
Post by Virgil
These are obviously collectively-definable as I have just collectively
defined them, but are equally obviously are not all individually
definable as there are just not enough individual definitions.
So you cannot use them or obtain them from the diagonal argument.
Not individually.
Post by m***@rz.fh-augsburg.de
That means, all antidiagonals that can be part of mathematical discourse
belong to a countable set.
That would imply that the set of all lists is countable, which I do not
claim, and WM certainly cannot prove.
Post by m***@rz.fh-augsburg.de
Post by Virgil
And with an obvious bijection between say the 0-8 sequences and the
reals they define, we see that there must be uncountably many reals,
a la the Cantor diagonal argument.
The latter only shows that all antidiagonals that can result from defined
lists belong to a countable set
Outside of Wolkenmuekenheim, it shows no such thing.
Outside of Wolkenmuekenheim there is no stricture on sets to be
countable, and
outside of Wolkenmuekenheim the power set of any set has a larger
cardinality than the original set has.



I
--
Virgil
2014-01-01 21:29:37 UTC
Permalink
Post by m***@rz.fh-augsburg.de
The list of finite definitions contains only finite definitions
And a list of real numbers can never contain all real numbers.
--
m***@rz.fh-augsburg.de
2014-01-02 17:36:00 UTC
Permalink
Post by Virgil
And a list of real numbers can never contain all real numbers.
Do you agree that a list can contain all real numbers that can appear as individuals in mathematical discourse?
Do you agree that a list can contain all real numbers that can appear as individuals in lists?

Regards, WM
Virgil
2014-01-01 21:33:08 UTC
Permalink
Post by m***@rz.fh-augsburg.de
WH claims just the same, namely that the list of all finite words has an
antidiagonal.
But every list of real numbers does define a real number not in the list.
--
m***@rz.fh-augsburg.de
2014-01-02 17:35:52 UTC
Permalink
Post by Virgil
But every list of real numbers does define a real number not in the list.
Only if the real numbers are given as infinite strings of digits. Not if they are given by finite definitions.

Regards, WM
Virgil
2014-01-02 19:17:36 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by Virgil
But every list of real numbers does define a real number not in the list.
Only if the real numbers are given as infinite strings of digits. Not if they
are given by finite definitions.
Regards, WM
Since there clearly is a different real number for every infinite string
of decimal digits not including 9 ( or not including 0, and even for
most of those allowing all ten digits), since the finite subsequences of
any such sequence represent a bounded monotone increasing sequence of
rationals which perforce must have a real number limit, there are
clearly more real numbers required to exist than there can be finite
names to name them with.

At least as long as every bounded monotone increasing sequence of
rationals defines its least upper bound and limit as a real number there
will be more such numbers than finite names for them.

Thus "uncountability"!
--
Albrecht
2014-01-02 09:14:53 UTC
Permalink
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
If you are unable to grasp arguments, why pose with this fact?

Note: if there is a list of _all_ natural numbers and you are able to
process _all_ of them, what hinders you to gather a n+1 or something
else, which is different from _all_ of them? Since n stands for _all_ of
them. And why should the result not be natural since
natural+natural=natural?

That is clearly nonsense but only because the word and the concept
"_all_" is harmed by the invention of infinite sets. And Cantors
prestidigitation uses this confusion.

Regards
Albrecht
Ben Bacarisse
2014-01-02 14:21:16 UTC
Permalink
Post by Albrecht
Post by Ben Bacarisse
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Nope.
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
If you are unable to grasp arguments, why pose with this fact?
Note: if there is a list of _all_ natural numbers and you are able to
process _all_ of them, what hinders you to gather a n+1 or something
else, which is different from _all_ of them? Since n stands for _all_
of them. And why should the result not be natural since
natural+natural=natural?
Does your weird mathematics permit unbounded sums? I.e. can one write

e = sum(n=0,1,...)[ 1/n! ]

in your mathematics?

<snip>
--
Ben.
Ben Bacarisse
2014-01-03 21:54:46 UTC
Permalink
<snip>
Post by Ben Bacarisse
Post by Albrecht
Post by Ben Bacarisse
Quite. Someone tried that on here recently: listing N and then claiming
that the fact that the anti-diagonal is not in N (which it isn't) meant
something.
If you are unable to grasp arguments, why pose with this fact?
Note: if there is a list of _all_ natural numbers and you are able to
process _all_ of them, what hinders you to gather a n+1 or something
else, which is different from _all_ of them? Since n stands for _all_
of them. And why should the result not be natural since
natural+natural=natural?
Does your weird mathematics permit unbounded sums? I.e. can one write
e = sum(n=0,1,...)[ 1/n! ]
in your mathematics?
Nothing?
--
Ben.
Virgil
2013-12-31 21:49:01 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
You think it is impossible to list all finite definitions?
Yes, and so do you,
No. I have shown a list of all finite words.
No you have not.

WM has repeatedly denied the existence of any actually infinite lists,
of which a list of all finite words would certainly be one, and now WM
claims they actually do exist and that he has provided one of them.

WM blows hot and cold with the same breath!
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
since for a potentially infinite list, L, of finite
definitions of potentially infinite 0/1 sequences there is a finite
definition of a potentially infinite 0/1 sequence not an element of L.
Any putative list you provide is a putative contradiction
There is no difference with the list of all natural numbers.
Not every word names a natural number and not every natural number can
be named by one world. While there may be a bijection between the SET of
all natural numbers and the SET of all finite words, only WM would ever
be so arrogant as to claim that every finite word IS a natural number
and that every natural number IS a finite word.
Everybody knows that WM is wrong.
Post by m***@rz.fh-augsburg.de
And in actual infinity of set theory: an infinite list has no finite antidiagonal.
WM has no idea what goes on or does not go on with actual infinities.
His mind is much too small.

Besides, no one other than WM himself ever claims finiteness for the
number of digits of an antidiagonal, but its definition requires it to
have a value between 0 and 1, which is eminently finite.
--
Virgil
2013-12-31 21:29:31 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Post by w***@gmail.com
Post by m***@rz.fh-augsburg.de
On Tuesday, December 31, 2013 7:05:58 AM UTC-4,
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever
integer-positive-base) expansion.
It is impossible for an infinite list of decimals to appear in
mathematical discourse, dialogue, or monologue other than as the finite
rule how to calculate *every* decimal at a finite place
No you need a finite definition of a potentially infinite decimal sequence.
This can define an irrational number.
Of course.
Everything that you say cannot be done with
digits can be done with finite definitions of potentially infinite
sequences e.g. diagonalization.
Indeed, it can easily be shown that given a potentially infinite sequence, L,
of finite definitions of potentially infinite 0/1 sequences there is
a finite definition of a potentially infinite 0/1 sequence that is not
an element of L.
You think it is impossible to list all finite definitions?
It is certainly impossible for WM to do it so long as he rejects the
possiblity of actual infinteness.
Post by m***@rz.fh-augsburg.de
Then it is impossible to list all terminating rationals.
Every rational terminates in some base.


Except that a listing of ALL rationals has been done by any of a huge
number of well-orderings of them
Post by m***@rz.fh-augsburg.de
It is easy to verify that the list of all terminating rationals is in
bijection with the list of all finite words.
Note how WM often claims things are easy, but never manages ever to do
any of them, just as WM often claims things are impossible to do when
others have actually done them.
--
Virgil
2013-12-31 21:17:00 UTC
Permalink
Post by m***@rz.fh-augsburg.de
It is impossible for an infinite list of decimals
Only in Wolkenmuekenheim

x = Sum 1/10^(n!) defines by its digits one irrational number for each base in |N\{1}
--
Virgil
2013-12-31 21:12:58 UTC
Permalink
Post by m***@rz.fh-augsburg.de
Irrational numbers have no decimal (or binary or whatever
integer-positive-base) expansion.
x = Sum 1/10^(n!) IS the expansion of a different irrational number in
every different base.

So in one line there are infinitely many counter-examples to WM's claim.
Post by m***@rz.fh-augsburg.de
It is impossible for an infinite list of
decimals to appear in mathematical discourse, dialogue, or monologue other
than as the finite rule how to calculate *every* decimal at a finite place
but never *all* decimal, since beyond every finite index there are infinitely
many further indices.
So in one line there are infinitely many counter-examples to WM's claim.
nswered: Both your assertions above are incorrect.
Post by m***@rz.fh-augsburg.de
Wouldn't that claim oblige him, in crank-free mathematics, to support his
opinion by listing all decimals of a nonterminating decimal representation of
a real number of his choice?
A finite rule for finding all infinitely many is quite sufficient
everywhere in the crank-free mathematics that exists only outside the
wild weird world of WMytheology.
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