Meanwhile I know three mathematicians [1, 2, 3] who deny that the Binary Tree can produce the paths belonging to single real numbers.
There
remain sheaves or bunches of paths, each one containing uncountably many paths which are not further distinguishable in the infinite Binary Tree.
-a/\
/\/\
...
In my opinion this forbids the complete digit sequence of any real
number because a path in the Binary Tree is nothing else than a sequence
of bits. On the other hand Cantor's diagonal argument produces a
complete digit sequence (in the original version [4] a complete bit sequence, using the symbols W M) of a real number, namely the famous diagonal number.
How can this contradiction be resolved?
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce.
There are paths in a binary tree but they don't go
anywhere other
tnan to nodes of the tree. Unless at least some nodes are real numbers
the paths don't go to any real number
There remain sheaves or bunches of paths, each one containing
uncountably many paths which are not further distinguishable in the
infinite Binary Tree.
Every path is distinguished from every other path by any one of the
nodes that one of them contains and the other does not.
-a-a/\
/\/\
...
In my opinion this forbids the complete digit sequence of any real
number because a path in the Binary Tree is nothing else than a
sequence of bits. On the other hand Cantor's diagonal argument
produces a complete digit sequence (in the original version [4] a
complete bit sequence, using the symbols W M) of a real number, namely
the famous diagonal number.
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
The most effective way is to stick to formal proofs that are verified
with a good simple proof checker.
On 07/05/2026 23:48, WM wrote:Note: Cantor considered just the set of all sequences of symbols w m,
[...] On the other hand Cantor's diagonal argument
produces a complete digit sequence (in the original version [4] a
complete bit sequence, using the symbols W M) of a real number, namely
the famous diagonal number.
A bit sequence is useful for proving that the power set of aA good psychiatrist might be helpful.
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
[...] On the other hand Cantor's diagonal argument produces a
complete digit sequence (in the original version [4] a complete bit
sequence, using the symbols W M) of a real number, namely the famous
diagonal number.
Note: Cantor considered just the set of all sequences of symbols w m,
not "real numbers" (in, say, [0, 1]).
How can this contradiction be resolved?
A good psychiatrist might be helpful.
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce. They just are.
There are paths in a binary tree but they don't go anywhere other
tnan to nodes of the tree. Unless at least some nodes are real numbers
the paths don't go to any real number and in any case they don't go
to any other real number.
There remain sheaves or bunches of paths, each one containing
uncountably many paths which are not further distinguishable in the
infinite Binary Tree.
Every path is distinguished from every other path by any one of the
nodes that one of them contains and the other does not.
/\
/\/\
...
In my opinion this forbids the complete digit sequence of any real
number because a path in the Binary Tree is nothing else than a
sequence of bits. On the other hand Cantor's diagonal argument
produces a complete digit sequence (in the original version [4] a
complete bit sequence, using the symbols W M) of a real number, namely
the famous diagonal number.
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
The most effective way is to stick to formal proofs that are verified
with a good simple proof checker.
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
[...] On the other hand Cantor's diagonal argument produces a
complete digit sequence (in the original version [4] a complete bit
sequence, using the symbols W M) of a real number, namely the famous
diagonal number.
Note: Cantor considered just the set of all sequences of symbols w m,
not "real numbers" (in, say, [0, 1]).
Indeed:
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
A good psychiatrist might be helpful.
On 05/08/2026 12:58 AM, Mikko wrote:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce. They just are.
There are paths in a binary tree but they don't go anywhere other
tnan to nodes of the tree. Unless at least some nodes are real numbers
the paths don't go to any real number and in any case they don't go
to any other real number.
There remain sheaves or bunches of paths, each one containing
uncountably many paths which are not further distinguishable in the
infinite Binary Tree.
Every path is distinguished from every other path by any one of the
nodes that one of them contains and the other does not.
/\
/\/\
...
In my opinion this forbids the complete digit sequence of any real
number because a path in the Binary Tree is nothing else than a
sequence of bits. On the other hand Cantor's diagonal argument
produces a complete digit sequence (in the original version [4] a
complete bit sequence, using the symbols W M) of a real number, namely
the famous diagonal number.
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
The most effective way is to stick to formal proofs that are verified
with a good simple proof checker.
About the formalization (basically for theories of types and then
for usual accounts of abstract symbolic notation then for the
course of inference and the equi-interpretability of "proof theory"
and "model theory"), then some examples of theorem-proof-checking
(if not exactly "theorem-proving" systems abount.
Mizar
Metamath
Isabelle/Coq
Lean
...
Usual accounts of solvers like Matlab, Mathematica, Macsyma, Maple,
... "computer algebra systems".
"Z", I suppose
I'm curious what would be a, "portable subset" of language of theorem-proof-checkers, or, an "abstraction layer" of the
interface of theorem-proof-checkers, about the "equi-interpretability"
among various theorem-proof-checkers, "TPC's" now, thusly
that "compilers" make for "generating code" that "targets"
various "language runtimes" reliably.
I'm curious a survey of "inference systems" about "inference systems".
Then, the most reliable way is yet rather philosophical account
about what's "true" and "consistent" then as for the "complete"
and "paradox-free", the acounts of reasoning, if though at some
point "good" and "simple" aren't the same thing, then about the
usual idea that it's cheaper to "check" something than "compute"
it, while yet there's a usual idea that in the ephemeral and
temporal that while "measure twice, cut once" that "time's a wasting".
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce.
That is a matter of taste. It is possible to describe what happens when
we go through a mathematical object. Then a sequence can decrease, a sum
or series can grow and a node can produce a sheaf.
-aThey just are.
There are paths in a binary tree but they don't go
In fact, they go? Fast or slow?
anywhere other
tnan to nodes of the tree. Unless at least some nodes are real numbers
No. Nodes are points.
They can be defined by natural numbers.
Real numbers are (represented by) paths. But obviously only countably many can be distinguished by nodes.
the paths don't go to any real number
They are (representing) real numbers.
There remain sheaves or bunches of paths, each one containing
uncountably many paths which are not further distinguishable in the
infinite Binary Tree.
Every path is distinguished from every other path by any one of the
nodes that one of them contains and the other does not.
-a-a/\
/\/\
...
In my opinion this forbids the complete digit sequence of any real
number because a path in the Binary Tree is nothing else than a
sequence of bits. On the other hand Cantor's diagonal argument
produces a complete digit sequence (in the original version [4] a
complete bit sequence, using the symbols W M) of a real number,
namely the famous diagonal number.
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
The most effective way is to stick to formal proofs that are verified
with a good simple proof checker.
But it is obvious that the current formalism is nonsense since it
assumes or even proves that every rational number can be finitely
defined (disproved above) and that uncountably many paths differing by
nodes are existing in the Binary Tree (contradiction accepted by
yourself). So why should anybody depend on that???
On 05/08/2026 05:57 AM, Moebius wrote:
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
[...] On the other hand Cantor's diagonal argument produces a
complete digit sequence (in the original version [4] a complete bit
sequence, using the symbols W M) of a real number, namely the famous
diagonal number.
Note: Cantor considered just the set of all sequences of symbols w m,
not "real numbers" (in, say, [0, 1]).
Indeed:
A bit sequence is useful for proving that the power set of a
countable set is not countable. For uncountablility of reals there is
the problem that bit sequences with only finitely many zeros are
different from bit sequences with only finitely many ones but denote
the same real numbers. This problem is avoided with base 3 or higher.
How can this contradiction be resolved?
A good psychiatrist might be helpful.
"Disambiguating quantifiers".
The idea that in combinatorics that the constant "2" is fundamentally different than the constant "3", yet in asymptotics they run out on
the same orders naively, is for aspects of what's called "Ramsey theory".
Also there's base one and base infinity to consider, when for
an integer is just tally-marks of increment, and a real numbers
is just +- (integer-part) (radix) (non-integer part), that also
the word "radix" fills at least two roles one the idea of the
base of the exponent, the other the divider between integer and
non-integer.
A good psychiatrist is not necessarily a "conscientious logician"
of the competent and thorough sort. The "help" may help, yet,
the conscientious logician has a bit of a bigger brain to satisfy.
So, disambiguating quantifiers is a usual account of de-craze-ifying,
since the crazing leads to the cracking, and the failure.
Having an account of "paradox-free" reason may help.
Anyways that it actually matters for the infinitary reasoning
why the binary representation of numbers and trinary/ternary
representations of numbers have different theorems about them, for
example where the binary anti-diagonal has only and exactly _one_ rule
for making the anti-diagonalization and that to avoid "dual
representation" that the usual account is to make the list in a higher
base and say there are "anti-semi-tri-diagonals", instead of an "anti-diagonal", here is that there are accounts like the "Equivalency Function" that only makes for one rule for an anti-diagonal.
So, that "the problem" isn't solve instead just put off.
Then, "Ramsey theory" is a usual umbrella for independence results
of the non-standard, yet these days it's often reduced to talking
about graph-coloring and arithmetic progressions and Szmeredi's
conjectures of all one kind.
On 08/05/2026 15:46, wm wrote:
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce.
That is a matter of taste. It is possible to describe what happens
when we go through a mathematical object. Then a sequence can
decrease, a sum or series can grow and a node can produce a sheaf.
-a-aThey just are.
There are paths in a binary tree but they don't go
In fact, they go? Fast or slow?
In Common Language the word "go" has a different meaning when the
subject is "path"
They can be defined by natural numbers.
Nodes can even be natural numbers. That is the simplest way to relate
nodes to natural numbers.
Real numbers are (represented by) paths. But obviously only countably
many can be distinguished by nodes.
That is not obvious and not true.
the paths don't go to any real number
They are (representing) real numbers.
Without a mapping from paths to real numbers a path cannot represent
a real number.
But it is obvious that the current formalism is nonsense since it
assumes or even proves that every rational number can be finitely
defined (disproved above) and that uncountably many paths differing by
nodes are existing in the Binary Tree (contradiction accepted by
yourself). So why should anybody depend on that???
Every rational number can be represented by a pair of integer numbers
where the send one is greater than zero. Every integer can be
represented as a finite string of characters of a finite alphabeth.
Therefore every rational number can be represented as a finite string.
You may wtite whatever humbug you want but that does not change the
facts.
Am 09.05.2026 um 09:59 schrieb Mikko:
On 08/05/2026 15:46, wm wrote:
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the
Binary Tree can produce the paths belonging to single real numbers.
Mathematical objecs like a binary tree don't produce.
That is a matter of taste. It is possible to describe what happens
when we go through a mathematical object. Then a sequence can
decrease, a sum or series can grow and a node can produce a sheaf.
-a-aThey just are.
There are paths in a binary tree but they don't go
In fact, they go? Fast or slow?
In Common Language the word "go" has a different meaning when the
subject is "path"
Same with produce.
They can be defined by natural numbers.
Nodes can even be natural numbers. That is the simplest way to relate
nodes to natural numbers.
Yes:
-a-a 0
-a 1 2
3 4 5 6
Real numbers are (represented by) paths. But obviously only countably
many can be distinguished by nodes.
That is not obvious and not true.
Even you have understood it:"Nodes further down separate further but
only to infinite subsets." Countably many nodes can produce only
countably many sheaves.
the paths don't go to any real number
They are (representing) real numbers.
Without a mapping from paths to real numbers a path cannot represent
a real number.
They do. The nodes are the bits.
You can't prove about any rational number that it is "undefined" so yourBut it is obvious that the current formalism is nonsense since it
assumes or even proves that every rational number can be finitely
defined (disproved above) and that uncountably many paths differing
by nodes are existing in the Binary Tree (contradiction accepted by
yourself). So why should anybody depend on that???
Every rational number can be represented by a pair of integer numbers
where the send one is greater than zero. Every integer can be
represented as a finite string of characters of a finite alphabeth.
Therefore every rational number can be represented as a finite string.
You may wtite whatever humbug you want but that does not change the
facts.
The facts are that between two defined rational numbers there are
infinitely many undefined rational numbers. That cannot be remedied.
On 10/05/2026 00:20, WM wrote:Then you should learn that the Binary Tree has nodes which produce
Am 09.05.2026 um 09:59 schrieb Mikko:
On 08/05/2026 15:46, wm wrote:
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the >>>>>> Binary Tree can produce the paths belonging to single real numbers. >>>>>Mathematical objecs like a binary tree don't produce.
That is a matter of taste. It is possible to describe what happens
when we go through a mathematical object. Then a sequence can
decrease, a sum or series can grow and a node can produce a sheaf.
-a-aThey just are.
There are paths in a binary tree but they don't go
In fact, they go? Fast or slow?
In Common Language the word "go" has a different meaning when the
subject is "path"
Same with produce.
The word "produce" has various meanings in Common Language but as far
as I know none of them is analogous to "go" as in "the way goes to the town".the
There is no way to map paths to nodes so that every path is mapped to
a different node. Therefore the set of paths is uncountable.
The important point is that every real
number is represented by at least one path.
You cannot map nodes to real numbers so that no real number remains
unmapped.
The facts are that between two defined rational numbers there areYou can't prove about any rational number that it is "undefined"
infinitely many undefined rational numbers. That cannot be remedied.
Am 10.05.2026 um 09:25 schrieb Mikko:
On 10/05/2026 00:20, WM wrote:
Am 09.05.2026 um 09:59 schrieb Mikko:
On 08/05/2026 15:46, wm wrote:
Am 08.05.2026 um 09:58 schrieb Mikko:
On 07/05/2026 23:48, WM wrote:
Meanwhile I know three mathematicians [1, 2, 3] who deny that the >>>>>>> Binary Tree can produce the paths belonging to single real numbers. >>>>>>Mathematical objecs like a binary tree don't produce.
That is a matter of taste. It is possible to describe what happens
when we go through a mathematical object. Then a sequence can
decrease, a sum or series can grow and a node can produce a sheaf.
-a-aThey just are.
There are paths in a binary tree but they don't go
In fact, they go? Fast or slow?
In Common Language the word "go" has a different meaning when the
subject is "path"
Same with produce.
The word "produce" has various meanings in Common Language but as far
as I know none of them is analogous to "go" as in "the way goes to the
town".
Then you should learn that the Binary Tree has nodes which produce
sheaves although the Binary Tree is an invariable system.
There is no way to map paths to nodes so that every path is mapped to
a different node. Therefore the set of paths is uncountable.
You have understood that every node produces only one sheave and that
there are no more paths that differ by nodes.
Therefore ther is a contradiction in set theory.
The important point is that every real
number is represented by at least one path.
Yes. And there are only countably many paths that differ by nodes.
You cannot map nodes to real numbers so that no real number remains
unmapped.
I can. But that is irrelevant here.
If you cannot prove about at least one of them that it is thereThe facts are that between two defined rational numbers there areYou can't prove about any rational number that it is "undefined"
infinitely many undefined rational numbers. That cannot be remedied.
Of course every rational number that can be a subject of proof is
defined. But there are infinitely many rational numbers undefinable.
On 10/05/2026 16:56, WM wrote:
Then you should learn that the Binary Tree has nodes which produce
sheaves although the Binary Tree is an invariable system.
It rarely is useful to learn nonsense.
There is no way to map paths to nodes so that every path is mapped to
a different node. Therefore the set of paths is uncountable.
You have understood that every node produces only one sheave and that
there are no more paths that differ by nodes.
Whether there are more paths than sheaves is a question that we have
not even tried to answer.
Therefore ther is a contradiction in set theory.
Perhaps in your set theory but not in Cantor's or ZF.
The important point is that every real
number is represented by at least one path.
Yes. And there are only countably many paths that differ by nodes.
So you say but you can't prove that there is a mapping from paths
to nodes that does not map any path to a node that another path
is mapped to. Without such mapping paths are not countable.
You cannot map nodes to real numbers so that no real number remains
unmapped.
I can. But that is irrelevant here.
All your claims in this discussions are relevant. Otherwise you would
not have presented them.
The facts are that between two defined rational numbers there areYou can't prove about any rational number that it is "undefined"
infinitely many undefined rational numbers. That cannot be remedied.
Of course every rational number that can be a subject of proof isIf you cannot prove about at least one of them that it is there
defined. But there are infinitely many rational numbers undefinable.
then your claim has no connection to any truth.
Am 11.05.2026 um 09:51 schrieb Mikko:
On 10/05/2026 16:56, WM wrote:
Then you should learn that the Binary Tree has nodes which produce
sheaves although the Binary Tree is an invariable system.
It rarely is useful to learn nonsense.
Your head is filled with nonsense, for instance the idea that every
rational could be named.
There is no way to map paths to nodes so that every path is mapped to
a different node. Therefore the set of paths is uncountable.
You have understood that every node produces only one sheave and that
there are no more paths that differ by nodes.
Whether there are more paths than sheaves is a question that we have
not even tried to answer.
It is sufficient to know that the set of sheaves is countable. It is an
idle question whether there are more paths because they could not be distinguished by nodes/bits.
Therefore ther is a contradiction in set theory.
Perhaps in your set theory but not in Cantor's or ZF.
Cantor "proved" uncountably many reals/paths that can be distinguished
by bits/nodes.
The important point is that every real
number is represented by at least one path.
Yes. And there are only countably many paths that differ by nodes.
So you say but you can't prove that there is a mapping from paths
to nodes that does not map any path to a node that another path
is mapped to. Without such mapping paths are not countable.
Without such mapping nodes are countable, and distinguishable paths are
as many as nodes.
You cannot map nodes to real numbers so that no real number remains
unmapped.
I can. But that is irrelevant here.
All your claims in this discussions are relevant. Otherwise you would
not have presented them.
I map every node on a paths containing it.
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any proof
of the existece or non-existence unnameable numbers. However, from the
usual axioms follow that unnameable rationals exist if and only if
unnameable natural numbers exist.
There is no way to map paths to nodes so that every path is mapped to >>>>> a different node. Therefore the set of paths is uncountable.
You have understood that every node produces only one sheave and
that there are no more paths that differ by nodes.
Whether there are more paths than sheaves is a question that we have
not even tried to answer.
It is sufficient to know that the set of sheaves is countable. It is
an idle question whether there are more paths because they could not
be distinguished by nodes/bits.
Whether there are more paths than sheaves is uniteresting because the
sheaves are uninteresting.
The interesting queston is whether there is
a mapping from nodes to paths that covers every path. If not the set
of paths is uncountable.
Therefore ther is a contradiction in set theory.
Perhaps in your set theory but not in Cantor's or ZF.
Cantor "proved" uncountably many reals/paths that can be distinguished
by bits/nodes.
Cantor proved what he proved. You can't show any error in the proofs.
Others have proved the same with more formal proofs.
The important point is that every real
number is represented by at least one path.
Yes. And there are only countably many paths that differ by nodes.
So you say but you can't prove that there is a mapping from paths
to nodes that does not map any path to a node that another path
is mapped to. Without such mapping paths are not countable.
Without such mapping nodes are countable, and distinguishable paths
are as many as nodes.
That does not follow.
I map every node on a paths containing it.
Syntax error: should be "a path" or "paths" without "a".
If your mapping does not map every node to exactly one path you don't
get a bijection needed for countability.
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any proof
of the existece or non-existence unnameable numbers. However, from the
usual axioms follow that unnameable rationals exist if and only if
unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are actually infinitely many, because you can name only finitely many. Infinitely
many remain forever unnamed.
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any proof >>> of the existece or non-existence unnameable numbers. However, from the
usual axioms follow that unnameable rationals exist if and only if
unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are actually
infinitely many, because you can name only finitely many. Infinitely
many remain forever unnamed.
If only finitely many natural numbers can be named then there is the
smallest natural number that cannot be named.
Am 13.05.2026 um 11:39 schrieb Mikko:Then the term "named" should not be used in any mathematical context but
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any
proof
of the existece or non-existence unnameable numbers. However, from the >>>> usual axioms follow that unnameable rationals exist if and only if
unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are actually
infinitely many, because you can name only finitely many. Infinitely
many remain forever unnamed.
If only finitely many natural numbers can be named then there is the
smallest natural number that cannot be named.
This is a widespread common error. The collection of named numbers is potentially infinite, that is: not fixed, without upper limit but always finite.
On 13/05/2026 23:39, wm wrote:
Am 13.05.2026 um 11:39 schrieb Mikko:Then the term "named" should not be used in any mathematical context but
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any
proof
of the existece or non-existence unnameable numbers. However, from the >>>>> usual axioms follow that unnameable rationals exist if and only if
unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are
actually infinitely many, because you can name only finitely many.
Infinitely many remain forever unnamed.
If only finitely many natural numbers can be named then there is the
smallest natural number that cannot be named.
This is a widespread common error. The collection of named numbers is
potentially infinite, that is: not fixed, without upper limit but
always finite.
only the term "nameable" meaning whatever can be named.
Am 14.05.2026 um 10:46 schrieb Mikko:No, but if there is a set of nameable numbers then all natural numbers
On 13/05/2026 23:39, wm wrote:Even "nameable" would not include all natural numbers.
Am 13.05.2026 um 11:39 schrieb Mikko:Then the term "named" should not be used in any mathematical context but
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for any >>>>>> proof
of the existece or non-existence unnameable numbers. However, from >>>>>> the
usual axioms follow that unnameable rationals exist if and only if >>>>>> unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are
actually infinitely many, because you can name only finitely many.
Infinitely many remain forever unnamed.
If only finitely many natural numbers can be named then there is the
smallest natural number that cannot be named.
This is a widespread common error. The collection of named numbers is
potentially infinite, that is: not fixed, without upper limit but
always finite.
only the term "nameable" meaning whatever can be named.
On 14/05/2026 17:52, WM wrote:
Am 14.05.2026 um 10:46 schrieb Mikko:No, but if there is a set of nameable numbers then all natural numbers
On 13/05/2026 23:39, wm wrote:Even "nameable" would not include all natural numbers.
Am 13.05.2026 um 11:39 schrieb Mikko:Then the term "named" should not be used in any mathematical context but >>> only the term "nameable" meaning whatever can be named.
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for
any proof
of the existece or non-existence unnameable numbers. However,
from the
usual axioms follow that unnameable rationals exist if and only if >>>>>>> unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are
actually infinitely many, because you can name only finitely many. >>>>>> Infinitely many remain forever unnamed.
If only finitely many natural numbers can be named then there is the >>>>> smallest natural number that cannot be named.
This is a widespread common error. The collection of named numbers
is potentially infinite, that is: not fixed, without upper limit but
always finite.
are in it per the induction axiom.
Am 15.05.2026 um 07:50 schrieb Mikko:
On 14/05/2026 17:52, WM wrote:
Am 14.05.2026 um 10:46 schrieb Mikko:No, but if there is a set of nameable numbers then all natural numbers
On 13/05/2026 23:39, wm wrote:Even "nameable" would not include all natural numbers.
Am 13.05.2026 um 11:39 schrieb Mikko:Then the term "named" should not be used in any mathematical context
On 12/05/2026 14:36, wm wrote:
Am 12.05.2026 um 09:50 schrieb Mikko:
On 11/05/2026 14:42, WM wrote:
In axiomatic mathematics the usual axioms are insufficient for >>>>>>>> any proof
of the existece or non-existence unnameable numbers. However, >>>>>>>> from the
usual axioms follow that unnameable rationals exist if and only if >>>>>>>> unnameable natural numbers exist.
Of course. Almost all natural numbers are dark, if there are
actually infinitely many, because you can name only finitely
many. Infinitely many remain forever unnamed.
If only finitely many natural numbers can be named then there is the >>>>>> smallest natural number that cannot be named.
This is a widespread common error. The collection of named numbers
is potentially infinite, that is: not fixed, without upper limit
but always finite.
but
only the term "nameable" meaning whatever can be named.
are in it per the induction axiom.
But there is no set of nameable numbers.
This collection is subject to
induction and as such potentially infinite. Note that by induction you cannot create the set rao.
Always almost all natural numbers are greater > than any n resultingfrom induction.
On 15/05/2026 19:39, WM wrote:
But there is no set of nameable numbers.
You can't prove that.
Always almost all natural numbers are greater than any n resulting
from induction.
That's right: for every n there is only finitely many natural numbers
that are smaller than n but infinitely many natural numbers that are
greater than n.
Am 16.05.2026 um 11:43 schrieb Mikko:In mathematics there is no distinction between potential and acutal. The distinction only arises when mathiematics is applied to situations where
On 15/05/2026 19:39, WM wrote:
But there is no set of nameable numbers.
You can't prove that.
Here ix the proof:
Always almost all natural numbers are greater than any n resulting
from induction.
That's right: for every n there is only finitely many natural numbers
that are smaller than n but infinitely many natural numbers that are
greater than n.
And the nameable n are not fixed. The collection is potentially infinite
- not a set.
On 17/05/2026 17:15, wm wrote:
Am 16.05.2026 um 11:43 schrieb Mikko:In mathematics there is no distinction between potential and acutal.
On 15/05/2026 19:39, WM wrote:
But there is no set of nameable numbers.
You can't prove that.
Here ix the proof:
Always almost all natural numbers are greater than any n resulting
from induction.
That's right: for every n there is only finitely many natural numbers
that are smaller than n but infinitely many natural numbers that are
greater than n.
And the nameable n are not fixed. The collection is potentially
infinite - not a set.
Anyway, in the usual set theories there are infinite sets.
For example,
in ZF the axiom of infinity ensures the existence of an infinite set.
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