On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
I mapped all nodes of the allways right going path to paths that go >>>>> left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many successors. >>> Which node does not have inifinely many successors ?
They are dark. We know however, that the complete set of nodes of a
path has no successors. That means the nodes of the path have all been
used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
And you can't show how they affect the validity of your claims or
my proofs.
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths that
go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many
successors.
They are dark. We know however, that the complete set of nodes of a
path has no successors. That means the nodes of the path have all
been used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It is not provable. Earlier generations of mathematicians adhered to potential infinity: Only what can be identifioed exists.
And you can't show how they affect the validity of your claims or
my proofs.
Assuming Cantor's theory leads to a contradiction.
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths that
go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many
successors.
They are dark. We know however, that the complete set of nodes of a
path has no successors. That means the nodes of the path have all
been used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It is not provable. Earlier generations of mathematicians adhered to potential infinity: Only what can be identifioed exists.
Jeff BarnettAnd you can't show how they affect the validity of your claims or
my proofs.
To discuss the complete Binary Tree, we must assume that it is complete. This assumption yields ra|o nodes and, since each node produces one additional sheaf, ra|o paths.
Assuming Cantor's theory leads to a contradiction.--
On 16/04/2026 17:26, WM wrote:
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths that >>>>>>> go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many
successors.
They are dark. We know however, that the complete set of nodes of a
path has no successors. That means the nodes of the path have all
been used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It is
not provable. Earlier generations of mathematicians adhered to
potential infinity: Only what can be identifioed exists.
So you lied when you said that I only mapped nodes with infinitely
many successors.
And you can't show how they affect the validity of your claims or
my proofs.
To discuss the complete Binary Tree, we must assume that it is complete. This assumption yields ra|o nodes and, since each node produces one additional sheaf, ra|o paths.
So all of your talk about paths that cover nodes has benn about
non-nextent things.
Assuming Cantor's theory leads to a contradiction.
Indeed, Cantor's theory contradicts your claims. But that does not
really matter as you have already contradicted your own claims.
But you have not shown any internal contradiction in Cantor's theory.
On 4/16/2026 8:26 AM, WM wrote:
Yes, we must adhere to Cantor's definition of actual infinity. It is
not provable. Earlier generations of mathematicians adhered to
potential infinity: Only what can be identifioed exists.
Are you suggesting that a "definition" must be provable??
Answer please - The curious and ROTFLO crowd need closure on the above question.
Am 17.04.2026 um 09:08 schrieb Mikko:
On 16/04/2026 17:26, WM wrote:
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths that >>>>>>>> go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many
successors.
They are dark. We know however, that the complete set of nodes of a >>>>> path has no successors. That means the nodes of the path have all
been used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It is
not provable. Earlier generations of mathematicians adhered to
potential infinity: Only what can be identifioed exists.
So you lied when you said that I only mapped nodes with infinitely
many successors.
No. When discussing the complete Binary Tree we must accept that it is complete first. That implies that most nodes are dark.
On 17/04/2026 16:10, wm wrote:
Am 17.04.2026 um 09:08 schrieb Mikko:
On 16/04/2026 17:26, WM wrote:
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths >>>>>>>>> that go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many
successors.
They are dark. We know however, that the complete set of nodes of >>>>>> a path has no successors. That means the nodes of the path have
all been used up.
So you can't identifiy a node without infintely may successors and
have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It is
not provable. Earlier generations of mathematicians adhered to
potential infinity: Only what can be identifioed exists.
So you lied when you said that I only mapped nodes with infinitely
many successors.
No. When discussing the complete Binary Tree we must accept that it is
complete first. That implies that most nodes are dark.
Does "complete" imply that every node has two subnodes?
Am 18.04.2026 um 11:52 schrieb Mikko:That is sufficient for my proofs. It also simplifies the numbering of
On 17/04/2026 16:10, wm wrote:
Am 17.04.2026 um 09:08 schrieb Mikko:
On 16/04/2026 17:26, WM wrote:
Am 16.04.2026 um 10:41 schrieb Mikko:
On 15/04/2026 17:20, wm wrote:
Am 15.04.2026 um 09:18 schrieb Mikko:
On 14/04/2026 21:38, WM wrote:
Am 14.04.2026 um 08:16 schrieb Mikko:
Which node does not have inifinely many successors ?I mapped all nodes of the allways right going path to paths >>>>>>>>>> that go left
at some point beause I wanted to do so.
But you did not. You mapped only nodes with infinitely many >>>>>>>>> successors.
They are dark. We know however, that the complete set of nodes of >>>>>>> a path has no successors. That means the nodes of the path have >>>>>>> all been used up.
So you can't identifiy a node without infintely may successors and >>>>>> have no proof that there is any but just assume that there are.
Yes, we must adhere to Cantor's definition of actual infinity. It
is not provable. Earlier generations of mathematicians adhered to
potential infinity: Only what can be identifioed exists.
So you lied when you said that I only mapped nodes with infinitely
many successors.
No. When discussing the complete Binary Tree we must accept that it
is complete first. That implies that most nodes are dark.
Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf into
two sheaves. The node is mapped to one of them. It does not matter to
which one, because the other sheaf carries a node from a higher-a level.
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf
into two sheaves. The node is mapped to one of them. It does not
matter to which one, because the other sheaf carries a node from a
higher-a level.
If the tree is complete in that sense then it also is infinite and its
every subtree is also complete in that sense and infinite.
Am 19.04.2026 um 11:55 schrieb Mikko:
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf
into two sheaves. The node is mapped to one of them. It does not
matter to which one, because the other sheaf carries a node from a
higher-a level.
Then we have found a contradiction-
If the tree is complete in that sense then it also is infinite and its
every subtree is also complete in that sense and infinite.
Here is the proof of equicardinality of nodes and paths:
-a-a |
-a-a |
-a-a |
-a-a v
-a-a O
-a / \
-a/-a-a \
|-a-a-a-a |
v-a-a-a-a v
And even clearer:
|
|
|
|-a-a-a-a-a O
|-a-a-a-a-a |
|-a-a-a-a-a |
v-a-a-a-a-a v
On 19/04/2026 13:55, wm wrote:
Am 19.04.2026 um 11:55 schrieb Mikko:
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf
into two sheaves. The node is mapped to one of them. It does not
matter to which one, because the other sheaf carries a node from a
higher-a level.
Then we have found a contradiction-
Then the next step is to extract from that contracidtion a simpler contradiction, preferably the simplest possible one.
One may even
hope that it could be simplified to something that can be published
as a paradox.
If the tree is complete in that sense then it also is infinite and its
every subtree is also complete in that sense and infinite.
Here is the proof of equicardinality of nodes and paths:
-a-a-a |
-a-a-a |
-a-a-a |
-a-a-a v
-a-a-a O
-a-a / \
-a-a/-a-a \
|-a-a-a-a |
v-a-a-a-a v
And even clearer:
|
|
|
|-a-a-a-a-a O
|-a-a-a-a-a |
|-a-a-a-a-a |
v-a-a-a-a-a v
No, neither is a proof of anything. For a proof of equicardinality
you need to proofs of that every requirement in the definition of equicardinality is satisfied.
Am 20.04.2026 um 11:34 schrieb Mikko:
On 19/04/2026 13:55, wm wrote:
Am 19.04.2026 um 11:55 schrieb Mikko:
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf
into two sheaves. The node is mapped to one of them. It does not
matter to which one, because the other sheaf carries a node from a
higher-a level.
Then we have found a contradiction-
Then the next step is to extract from that contracidtion a simpler
contradiction, preferably the simplest possible one.
See below. I think there is the simplest possible contradiction.
One may even
hope that it could be simplified to something that can be published
as a paradox.
It has been published already:
W. M|+ckenheim: "Transfinity - A Source Book", ELIVA Press, Chisinau
(2024). p. 283
https://www.elivapress.com/en/book/book-9877032691/
The tunnel effect of set theory, reddit/mathematics (12 April 2026) https://www.reddit.com/r/mathematics/comments/1sjdj57/ the_tunnel_effect_of_set_theory/
If the tree is complete in that sense then it also is infinite and its >>>> every subtree is also complete in that sense and infinite.
Here is the proof of equicardinality of nodes and paths:
-a-a-a |
-a-a-a |
-a-a-a |
-a-a-a v
-a-a-a O
-a-a / \
-a-a/-a-a \
|-a-a-a-a |
v-a-a-a-a v
And even clearer:
|
|
|
|-a-a-a-a-a O
|-a-a-a-a-a |
|-a-a-a-a-a |
v-a-a-a-a-a v
No, neither is a proof of anything. For a proof of equicardinality
you need to proofs of that every requirement in the definition of
equicardinality is satisfied.
It is. Every node produces one and only one new sheaf. It is a bijection.
On 20/04/2026 14:21, wm wrote:
Am 20.04.2026 um 11:34 schrieb Mikko:
On 19/04/2026 13:55, wm wrote:
Am 19.04.2026 um 11:55 schrieb Mikko:
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a sheaf >>>>>> into two sheaves. The node is mapped to one of them. It does not
matter to which one, because the other sheaf carries a node from a >>>>>> higher-a level.
Then we have found a contradiction-
Then the next step is to extract from that contracidtion a simpler
contradiction, preferably the simplest possible one.
See below. I think there is the simplest possible contradiction.
One may even
hope that it could be simplified to something that can be published
as a paradox.
It has been published already:
W. M|+ckenheim: "Transfinity - A Source Book", ELIVA Press, Chisinau
(2024). p. 283
https://www.elivapress.com/en/book/book-9877032691/
The tunnel effect of set theory, reddit/mathematics (12 April 2026)
https://www.reddit.com/r/mathematics/comments/1sjdj57/
the_tunnel_effect_of_set_theory/
In that case your "Then we have found a contradiction" does not mean
a new contradiction.
A contradiction that can't be stated here but takes a book to express
is not really simple.
If the tree is complete in that sense then it also is infinite and its >>>>> every subtree is also complete in that sense and infinite.
Here is the proof of equicardinality of nodes and paths:
-a-a-a |
-a-a-a |
-a-a-a |
-a-a-a v
-a-a-a O
-a-a / \
-a-a/-a-a \
|-a-a-a-a |
v-a-a-a-a v
And even clearer:
|
|
|
|-a-a-a-a-a O
|-a-a-a-a-a |
|-a-a-a-a-a |
v-a-a-a-a-a v
No, neither is a proof of anything. For a proof of equicardinality
you need to proofs of that every requirement in the definition of
equicardinality is satisfied.
It is. Every node produces one and only one new sheaf. It is a bijection.
No, it is not. A proof is a sequence of statements, starting with
clearly identified premises.
Saying "bijection" is not enough. You must prove that there is a
function from paths to nodes that does not map any two paths to
the same node. Without that you have no proof of bijection.
Am 21.04.2026 um 09:25 schrieb Mikko:
On 20/04/2026 14:21, wm wrote:
Am 20.04.2026 um 11:34 schrieb Mikko:
On 19/04/2026 13:55, wm wrote:
Am 19.04.2026 um 11:55 schrieb Mikko:
On 18/04/2026 15:40, wm wrote:
That is sufficient for my proofs.Does "complete" imply that every node has two subnodes?
Of course. This allows a simple mapping: Every node splits a
sheaf into two sheaves. The node is mapped to one of them. It
does not matter to which one, because the other sheaf carries a >>>>>>> node from a higher-a level.
Then we have found a contradiction-
Then the next step is to extract from that contracidtion a simpler
contradiction, preferably the simplest possible one.
See below. I think there is the simplest possible contradiction.
One may even
hope that it could be simplified to something that can be published
as a paradox.
It has been published already:
W. M|+ckenheim: "Transfinity - A Source Book", ELIVA Press, Chisinau
(2024). p. 283
https://www.elivapress.com/en/book/book-9877032691/
The tunnel effect of set theory, reddit/mathematics (12 April 2026)
https://www.reddit.com/r/mathematics/comments/1sjdj57/
the_tunnel_effect_of_set_theory/
In that case your "Then we have found a contradiction" does not mean
a new contradiction.
It is a contradiction not yet realized by many.
A contradiction that can't be stated here but takes a book to express
is not really simple.
It does not take a book but has been written on one page of the book.
If the tree is complete in that sense then it also is infinite and >>>>>> its
every subtree is also complete in that sense and infinite.
Here is the proof of equicardinality of nodes and paths:
-a-a-a |
-a-a-a |
-a-a-a |
-a-a-a v
-a-a-a O
-a-a / \
-a-a/-a-a \
|-a-a-a-a |
v-a-a-a-a v
And even clearer:
|
|
|
|-a-a-a-a-a O
|-a-a-a-a-a |
|-a-a-a-a-a |
v-a-a-a-a-a v
No, neither is a proof of anything. For a proof of equicardinality
you need to proofs of that every requirement in the definition of
equicardinality is satisfied.
It is. Every node produces one and only one new sheaf. It is a
bijection.
No, it is not. A proof is a sequence of statements, starting with
clearly identified premises.
Premise: The set of nodes is countable. Fact: No new path can be
produced without a node. The set of paths is countable. Contradiction.
Insufficient as explained above.Saying "bijection" is not enough. You must prove that there is a
function from paths to nodes that does not map any two paths to
the same node. Without that you have no proof of bijection.
I have proved a one-to-one map from nodes to paths.
On 21/04/2026 13:46, wm wrote:
It is a contradiction not yet realized by many.
If it is already in a book it is not new.
A contradiction that can't be stated here but takes a book to express
is not really simple.
It does not take a book but has been written on one page of the book.
Still not simple wnough if it can't be stated here.
Premise: The set of nodes is countable. Fact: No new path can be
produced without a node. The set of paths is countable. Contradiction.
Your "The set of paths is countable" does not follow from what precedes.
A node can be shared by many paths (e.g., the root node is shared by all paths). Two paths are different if there is one node that is in one of
them but not in the other.
The simplest way to prove countability of paths is to construct a
function that gives every path a number and another function that
gives every number a path and then prove that each is the inverse
function of the other.
But apparently you can't do that.
Insufficient as explained above.Saying "bijection" is not enough. You must prove that there is a
function from paths to nodes that does not map any two paths to
the same node. Without that you have no proof of bijection.
I have proved a one-to-one map from nodes to paths.
Am 22.04.2026 um 10:03 schrieb Mikko:
On 21/04/2026 13:46, wm wrote:
It is a contradiction not yet realized by many.
If it is already in a book it is not new.
In fact, I have already published it in newsgroups about 2005.
A contradiction that can't be stated here but takes a book to express
is not really simple.
It does not take a book but has been written on one page of the book.
Still not simple wnough if it can't be stated here.
It has been stated here:
Premise: The set of nodes is countable. Fact: No new path can be
produced without a node. The set of paths is countable. Contradiction.
Your "The set of paths is countable" does not follow from what precedes.
A node can be shared by many paths (e.g., the root node is shared by all
paths). Two paths are different if there is one node that is in one of
them but not in the other.
A path distinct from a being one can only be created by a node.
If there> are many paths created, they will have to prove theirexistence by
becoming distinct at further nodes.
The simplest way to prove countability of paths is to construct a
function that gives every path a number and another function that
gives every number a path and then prove that each is the inverse
function of the other.
If it was so easily done I would hardly have been the first to find the inconsistency.
But apparently you can't do that.
Therefore I use the countability of the nodes. A proof does not become invalid by leaving the well-trodden paths.
That does not make sense. How should the first phrase be parsed?Insufficient as explained above.Saying "bijection" is not enough. You must prove that there is a
function from paths to nodes that does not map any two paths to
the same node. Without that you have no proof of bijection.
I have proved a one-to-one map from nodes to paths.
Try to find a point where a path can leave the system of paths already produced without a node. Important: Try to overcome your psychological block.
On 22/04/2026 16:21, WM wrote:
Premise: The set of nodes is countable. Fact: No new path can be
produced without a node. The set of paths is countable. Contradiction.
Oh, you meant that? None of these sentences contradict the others.
The sentence "the sent of paths is countable" contradicts other considerations but that does not matter because it is not implied
the preceding sentences.
Your "The set of paths is countable" does not follow from what precedes. >>> A node can be shared by many paths (e.g., the root node is shared by all >>> paths). Two paths are different if there is one node that is in one of
them but not in the other.
A path distinct from a being one can only be created by a node.
No, it can't. A node does not create a path.
If there> are many paths created, they will have to prove theirexistence by
becoming distinct at further nodes.
Yes. For any two paths there are infinitely many nodes that are
only in one of those paths. To show one of those nodes is sufficient
to show that you have two distinct paths.
Leaving logic (even for one step) makes a proof invalid.
Try to find a point where a path can leave the system of paths alreadyThat does not make sense. How should the first phrase be parsed?
produced without a node. Important: Try to overcome your psychological
block.
What would "find without a node" or "leave wighout a node" or "produced wihout a node" (or whatever you tried to say) mean?
Am 23.04.2026 um 09:40 schrieb Mikko:
On 22/04/2026 16:21, WM wrote:
Premise: The set of nodes is countable. Fact: No new path can be
produced without a node. The set of paths is countable. Contradiction.
Oh, you meant that? None of these sentences contradict the others.
The sentence "the sent of paths is countable" contradicts other
considerations but that does not matter because it is not implied
the preceding sentences.
It is implied by No new path can be produced without a node.
Your "The set of paths is countable" does not follow from what
precedes.
A node can be shared by many paths (e.g., the root node is shared by
all
paths). Two paths are different if there is one node that is in one of >>>> them but not in the other.
A path distinct from a being one can only be created by a node.
No, it can't. A node does not create a path.
A node makes it distinct from the incoming one.
That is creating a new path.
Am 24.04.2026 um 08:45 schrieb Mikko:
On 23/04/2026 16:01, WM wrote:
Am 23.04.2026 um 09:40 schrieb Mikko:
On 22/04/2026 16:21, WM wrote:
Premise: The set of nodes is countable. Fact: No new path can be >>>>>>> produced without a node. The set of paths is countable.
Contradiction.
Oh, you meant that? None of these sentences contradict the others.
The sentence "the sent of paths is countable" contradicts other
considerations but that does not matter because it is not implied
the preceding sentences.
It is implied by No new path can be produced without a node.
No, is not.
It is the definition of node: A node is a point where a path is
separated from another path.
Your "The set of paths is countable" does not follow from what
precedes.
If you want to claim otherwise you need a proof.
It is the definition of node: Every node is a point of separation, and
every point of separation is a node. If you dislike this definition, try
to find another one.
A node makes it distinct from the incoming one.
What does "the incoming one" mean?
The incoming path is the connection between the node and the root. It is
a path that above the node contained the paths which below the node run different ways.
That is creating a new path.>What does it mean to "create" a path?
Above the node there is one, below the node there are two.
The treee is what it is. You
cannot ad a node or a path to it and you can't remove a node or a
path from it.
The Binary Tree is fixed, but we can describe how it evolves.
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is
separated from another path.
Witch coes not imply that the set of paths is countable.
No need of any definition of node. The nature of nodes is irrelvant,
only the realtions between them matter.
A node makes it distinct from the incoming one.
What does "the incoming one" mean?
The incoming path is the connection between the node and the root. It
is a path that above the node contained the paths which below the node
run different ways.
So it is a finite path from root to node.
That is creating a new path.>What does it mean to "create" a path?
Above the node there is one, below the node there are two.
So your set of paths include those that don't start from the root?
Anyway, there are also paths that neither end at the node nor
start from it but go though the node.
The treee is what it is. You
cannot ad a node or a path to it and you can't remove a node or a
path from it.
The Binary Tree is fixed, but we can describe how it evolves.
Tnat it is fixed means that it does not evolve.
Am 25.04.2026 um 11:14 schrieb Mikko:
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is
separated from another path.
Witch coes not imply that the set of paths is countable.
There is a simple step from:
Every node splits the incoming path into two paths
to
There are as many paths as nodes.
Below I have explained it in more detail.>
No need of any definition of node. The nature of nodes is irrelvant,
only the realtions between them matter.
Every bode increases the number of distinct path by 1. Try this:
2 nodes produce 2 paths
3 nodes produce 3 paths.
n nodes produce n paths.
ra|reC nodes produce ra|reC paths.
A node makes it distinct from the incoming one.
What does "the incoming one" mean?
The incoming path is the connection between the node and the root. It
is a path that above the node contained the paths which below the
node run different ways.
So it is a finite path from root to node
That part is finite, but the incoming path continues infinitely - only without the other path.
That is creating a new path.>What does it mean to "create" a path?
Above the node there is one, below the node there are two.
So your set of paths include those that don't start from the root?
No, all paths start from the root, but they cannot be distinguished
above the node.
Anyway, there are also paths that neither end at the node nor
start from it but go though the node.
All paths are infinite starting at the root.
It can't because it is fixed.The treee is what it is. You
cannot ad a node or a path to it and you can't remove a node or a
path from it.
The Binary Tree is fixed, but we can describe how it evolves.
Tnat it is fixed means that it does not evolve.
It evolves with respect to the points we consider
On 25/04/2026 16:36, WM wrote:
Am 25.04.2026 um 11:14 schrieb Mikko:
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is
separated from another path.
Witch coes not imply that the set of paths is countable.
There is a simple step from:
Every node splits the incoming path into two paths
to
There are as many paths as nodes.
Below I have explained it in more detail.>
No need of any definition of node. The nature of nodes is irrelvant,
only the realtions between them matter.
Every node increases the number of distinct path by 1. Try this:
2 nodes produce 2 paths
3 nodes produce 3 paths.
n nodes produce n paths.
ra|reC nodes produce ra|reC paths.
Have you read Galilei's Two New Sciences? There he gives sevaral
examples of getting contradictions if one assumes that what is
true about finite is applicable to infinite.
So there are infinitely many nodes[paths] through a node. Infinitely many
of them continue to the left subnode and infinitely many to the
right subnode.
No, all paths start from the root, but they cannot be distinguished
above the node.
In that case there are as many above the node as below because they
are the same paths.
Anyway, there are also paths that neither end at the node nor
start from it but go though the node.
All paths are infinite starting at the root.
And there are infinitely many of them.
It evolves with respect to the points we considerIt can't because it is fixed.
Am 26.04.2026 um 10:30 schrieb Mikko:
On 25/04/2026 16:36, WM wrote:
Am 25.04.2026 um 11:14 schrieb Mikko:
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is
separated from another path.
Witch coes not imply that the set of paths is countable.
There is a simple step from:
Every node splits the incoming path into two paths
to
There are as many paths as nodes.
Below I have explained it in more detail.>
No need of any definition of node. The nature of nodes is irrelvant,
only the realtions between them matter.
Every node increases the number of distinct path by 1. Try this:
2 nodes produce 2 paths
3 nodes produce 3 paths.
n nodes produce n paths.
ra|reC nodes produce ra|reC paths.
Have you read Galilei's Two New Sciences? There he gives sevaral
examples of getting contradictions if one assumes that what is
true about finite is applicable to infinite.
Have you read Cantor's works: There he explains that infinite sets can
be put in bijection.
Every node splits the incoming sheaf. One of its paths has a node from above. Another one gets this node. One-to-one map.
So there are infinitely many nodes[paths] through a node. Infinitely many
of them continue to the left subnode and infinitely many to the
right subnode.
Yes. But we can invent a pre-root-node and choose one path which this
the pre-root-node is mapped to:
-ao
-a|
-ao
/ \
When a sheaf is split at node N, it contains a path carrying a node
already. This path goes to the left or right. And we can choose another
path that goes in the other direction and gets the node N mapped to. And
for every split at every we can choose another path which this node is mapped to. (The incoming path has got its node above already.)
No, all paths start from the root, but they cannot be distinguished
above the node.
In that case there are as many above the node as below because they
are the same paths.
They are not the same paths, because they get separated below -
countably many.
Yes. The bijection described above shows the equicardinality.
It evolves with respect to the points we considerIt can't because it is fixed.
Can a sequence increase?
On 26/04/2026 22:33, WM wrote:
Am 26.04.2026 um 10:30 schrieb Mikko:
On 25/04/2026 16:36, WM wrote:
Am 25.04.2026 um 11:14 schrieb Mikko:
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is
separated from another path.
Witch coes not imply that the set of paths is countable.
There is a simple step from:
Every node splits the incoming path into two paths
to
There are as many paths as nodes.
Below I have explained it in more detail.>
No need of any definition of node. The nature of nodes is irrelvant, >>>>> only the realtions between them matter.
Every node increases the number of distinct path by 1. Try this:
2 nodes produce 2 paths
3 nodes produce 3 paths.
n nodes produce n paths.
ra|reC nodes produce ra|reC paths.
Have you read Galilei's Two New Sciences? There he gives sevaral
examples of getting contradictions if one assumes that what is
true about finite is applicable to infinite.
Have you read Cantor's works: There he explains that infinite sets can
be put in bijection.
I havn't read all of them but I have read the most relevant one.
He explains that in some cases infinite sets cannot be put in
bijection.
Every node splits the incoming sheaf. One of its paths has a node from
above. Another one gets this node. One-to-one map.
Sheafs and mapping from or to sheafs are irrelevant to questions
about paths. There are infinitely may paths in every sheaf.
Anyway, you havn't shown any one-to-one mapping.
So there are infinitely many nodes[paths] through a node. Infinitely
many
of them continue to the left subnode and infinitely many to the
right subnode.
Yes. But we can invent a pre-root-node and choose one path which this
the pre-root-node is mapped to:
-a-ao
-a-a|
-a-ao
/ \
That pre-root node is not a part of the tree and not useful. And if
you don't actually choose a path then your claim that you could
choose is not credible.
When a sheaf is split at node N, it contains a path carrying a node
already. This path goes to the left or right. And we can choose
another path that goes in the other direction and gets the node N
mapped to. And for every split at every we can choose another path
which this node is mapped to. (The incoming path has got its node
above already.)
Again you say "we could choose" but don't choose. Not credible.
No, all paths start from the root, but they cannot be distinguished
above the node.
In that case there are as many above the node as below because they
are the same paths.
They are not the same paths, because they get separated below -
countably many.
Yes, they are. Unless the node is the root node there is no path
above it that is not under it and no path under it that is not
above it.
Yes. The bijection described above shows the equicardinality.
A bijection shows nothging as long as the bijection itself is
not shown.
It evolves with respect to the points we considerIt can't because it is fixed.
Can a sequence increase?
A sequence does not change and does not do anything. It only is.
Sometimes a sequence is said to "increase" but that must not be
interpreted literally. It only means that no element is preceded
by a bigger element.
Expressions like that are OK when everyone knows what is really meant.
But they should be avoided when there is a possibility that they may
be interpreted literally. In particular, you can't know who will read
these messages so you can't trust everybody understands the unusual
meanings of usual words.
Am 27.04.2026 um 12:03 schrieb Mikko:
On 26/04/2026 22:33, WM wrote:
Am 26.04.2026 um 10:30 schrieb Mikko:
On 25/04/2026 16:36, WM wrote:
Am 25.04.2026 um 11:14 schrieb Mikko:
On 24/04/2026 15:59, WM wrote:
It is the definition of node: A node is a point where a path is >>>>>>> separated from another path.
Witch coes not imply that the set of paths is countable.
There is a simple step from:
Every node splits the incoming path into two paths
to
There are as many paths as nodes.
Below I have explained it in more detail.>
No need of any definition of node. The nature of nodes is irrelvant, >>>>>> only the realtions between them matter.
Every node increases the number of distinct path by 1. Try this:
2 nodes produce 2 paths
3 nodes produce 3 paths.
n nodes produce n paths.
ra|reC nodes produce ra|reC paths.
Have you read Galilei's Two New Sciences? There he gives sevaral
examples of getting contradictions if one assumes that what is
true about finite is applicable to infinite.
Have you read Cantor's works: There he explains that infinite sets
can be put in bijection.
I havn't read all of them but I have read the most relevant one.
He explains that in some cases infinite sets cannot be put in
bijection.
Every node splits the incoming sheaf. One of its paths has a node
from above. Another one gets this node. One-to-one map.
Sheafs and mapping from or to sheafs are irrelevant to questions
about paths. There are infinitely may paths in every sheaf.
Anyway, you havn't shown any one-to-one mapping.
So there are infinitely many nodes[paths] through a node. Infinitely
many
of them continue to the left subnode and infinitely many to the
right subnode.
Yes. But we can invent a pre-root-node and choose one path which this
the pre-root-node is mapped to:
-a-ao
-a-a|
-a-ao
/ \
That pre-root node is not a part of the tree and not useful. And if
you don't actually choose a path then your claim that you could
choose is not credible.
Choose any path you like that passes the node. Never you will
distinguish more paths than nodes.
When a sheaf is split at node N, it contains a path carrying a node
already. This path goes to the left or right. And we can choose
another path that goes in the other direction and gets the node N
mapped to. And for every split at every we can choose another path
which this node is mapped to. (The incoming path has got its node
above already.)
Again you say "we could choose" but don't choose. Not credible.
Here is a possibility: Choose the path that passes the right child node
and then goes always right. Choose the path that passes the left child
node and then always goes left.
On 27/04/2026 13:19, WM wrote:
Yes. But we can invent a pre-root-node and choose one path which
this the pre-root-node is mapped to:
-a-ao
-a-a|
-a-ao
/ \
That pre-root node is not a part of the tree and not useful. And if
you don't actually choose a path then your claim that you could
choose is not credible.
Choose any path you like that passes the node. Never you will
distinguish more paths than nodes.
I have already shown that it is possible to find two sets of paths
so that no path is in both sets and every node is in at least one
path in each set.
So far you have proven nothing.Find any level of the Binary Tree where more paths than nodes can be distinguished.
Here is a possibility: Choose the path that passes the right child
node and then goes always right. Choose the path that passes the left
child node and then always goes left.
That is indeed one possibility. In that case every chosen path either
turns left only a finite number of times or turns right only a finite
number of times. A path that turns every second time to the left and
every seconc time to the right is never chosen.
You have not shown a bijection.
Am 28.04.2026 um 10:22 schrieb Mikko:
On 27/04/2026 13:19, WM wrote:
Yes. But we can invent a pre-root-node and choose one path which
this the pre-root-node is mapped to:
-a-ao
-a-a|
-a-ao
/ \
That pre-root node is not a part of the tree and not useful. And if
you don't actually choose a path then your claim that you could
choose is not credible.
Choose any path you like that passes the node. Never you will
distinguish more paths than nodes.
I have already shown that it is possible to find two sets of paths
so that no path is in both sets and every node is in at least one
path in each set.
You have failed. To cover every node of RRR... by paths going left after some node is as possible as to add even numbers with an odd result.
You have not shown a bijection.
Of course every path of the Binary Tree gets chosen and gets a node
mapped to.
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left
after some node is as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
Infinitely many
of those paths are different from RRR, i.e., has farther doesn some
node that RRR has not.
Without proofs your claims are just unproven wild guesses.
You have not shown a bijection.
Of course every path of the Binary Tree gets chosen and gets a node
mapped to.
Sayin "of course" is not proof. Without a proof that every path is
mapped to a different node you have no bijection.
Am 29.04.2026 um 09:42 schrieb Mikko:
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left after some node is
as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
No. That is provable only for nodes which have more successors than predecessors.
Certaily you claim that no others exist. But that is
impossible if all do exist.
Infinitely many
of those paths are different from RRR, i.e., has farther doesn some
node that RRR has not.
But also RRR... has nodes at the same levels further down. Therefore it
is distinct from all other paths.
Without proofs your claims are just unproven wild guesses.
My proof is simple: Every node except the root node separates only one
more path.
More paths cannot become separated than nodes are existing.
Otherwise at at least one node at least two paths must become separated.
What is too difficult for you to understand?
On 29/04/2026 16:02, WM wrote:
Am 29.04.2026 um 09:42 schrieb Mikko:
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left after some node is
as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
No. That is provable only for nodes which have more successors than
predecessors.
It is true about the root node. All other nodes in the tree have a predecessor per the dfinition of "tree". All nodes have two successors
(in this particular tree).
Infinitely many
of those paths are different from RRR, i.e., has farther doesn some
node that RRR has not.
But also RRR... has nodes at the same levels further down. Therefore
it is distinct from all other paths.
Indeed.
My proof is simple: Every node except the root node separates only one
more path.
It does not. Infinitely many of the paths to the root continue to
the left and infinitely many to the right. No more separation is
introduced by that node. Nodes further down separate further but
only to infinite subsets.
More paths cannot become separated than nodes are existing.
So the rest reamins unseparated.
Otherwise at at least one node at least two paths must become separated.
That is not "otherwise", that is anyway. But that is an understatement
as acutally infinitely many paths are separted from infintely-a many
other paths.
What is too difficult for you to understand?
You presented an almost complete proof that every set of nodes
that contain at least one path throuch ecah node of a tree covers
the entire tree. While not exactly a proof it is obvous how to
complete it to a easily verifiable correct proof.
Am 30.04.2026 um 10:54 schrieb Mikko:
On 29/04/2026 16:02, WM wrote:
Am 29.04.2026 um 09:42 schrieb Mikko:
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left after some node
is as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
No. That is provable only for nodes which have more successors than
predecessors.
It is true about the root node. All other nodes in the tree have a
predecessor per the dfinition of "tree". All nodes have two successors
(in this particular tree).
All nodes that you can find or choose have more successors (namely infinitely many) than predecessors (namely finitely many).
Infinitely many
of those paths are different from RRR, i.e., has farther doesn some
node that RRR has not.
But also RRR... has nodes at the same levels further down. Therefore
it is distinct from all other paths.
Indeed.
Therefore it cannot be covered by other paths, not even by infinitely many.
My proof is simple: Every node except the root node separates only
one more path.
It does not. Infinitely many of the paths to the root continue to
the left and infinitely many to the right. No more separation is
introduced by that node. Nodes further down separate further but
only to infinite subsets.
So you deny the existence of single paths in the Binary Tree?
On 30/04/2026 15:49, WM wrote:
Am 30.04.2026 um 10:54 schrieb Mikko:
On 29/04/2026 16:02, WM wrote:
Am 29.04.2026 um 09:42 schrieb Mikko:
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left after some node >>>>>> is as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
No. That is provable only for nodes which have more successors than
predecessors.
It is true about the root node. All other nodes in the tree have a
predecessor per the dfinition of "tree". All nodes have two successors
(in this particular tree).
All nodes that you can find or choose have more successors (namely
infinitely many) than predecessors (namely finitely many).
If your "successor" includes successors of the successors and your "predecessor" includes predecessors of the predecessors then yes.
But it is true about all nodes.
Therefore it cannot be covered by other paths, not even by infinitely
many.
No path can be covered by another path.
My proof is simple: Every node except the root node separates only
one more path.
It does not. Infinitely many of the paths to the root continue to
the left and infinitely many to the right. No more separation is
introduced by that node. Nodes further down separate further but
only to infinite subsets.
So you deny the existence of single paths in the Binary Tree?
What does "single" mean above?
I do deny existence of nodes that don't have more than one path through
them.
Am 01.05.2026 um 10:33 schrieb Mikko:
On 30/04/2026 15:49, WM wrote:
Am 30.04.2026 um 10:54 schrieb Mikko:
On 29/04/2026 16:02, WM wrote:
Am 29.04.2026 um 09:42 schrieb Mikko:
On 28/04/2026 16:18, WM wrote:
To cover every node of RRR... by paths going left after some node >>>>>>> is as possible as to add even numbers with an odd result.
There are infinitely many paths to every node of RRR.
No. That is provable only for nodes which have more successors than >>>>> predecessors.
It is true about the root node. All other nodes in the tree have a
predecessor per the dfinition of "tree". All nodes have two successors >>>> (in this particular tree).
All nodes that you can find or choose have more successors (namely
infinitely many) than predecessors (namely finitely many).
If your "successor" includes successors of the successors and your
"predecessor" includes predecessors of the predecessors then yes.
Of course they all are included.
But it is true about all nodes.
That is not possible if "all" is a meaningful expression here. If all
nodes are mapped, then none remains.
That is not possible if only those
nodes are mapped which have successors (because the successors remain).
Therefore it cannot be covered by other paths, not even by infinitely
many.
No path can be covered by another path.
Yes. Nevertheless it is often claimed that a path like RRR... can be
covered by an infinite sequence of other paths ending with LLL.... Since
all paths of this sequence are useless, the claim is wrong, of course.>
My proof is simple: Every node except the root node separates only
one more path.
It does not. Infinitely many of the paths to the root continue to
the left and infinitely many to the right. No more separation is
introduced by that node. Nodes further down separate further but
only to infinite subsets.
So you deny the existence of single paths in the Binary Tree?
What does "single" mean above?
It means a sheave or separated subset containing only one path.
I do deny existence of nodes that don't have more than one path through
them.
Then you deny single paths.
On 01/05/2026 16:14, wm wrote:
Am 01.05.2026 um 10:33 schrieb Mikko:
All nodes that you can find or choose have more successors (namely
infinitely many) than predecessors (namely finitely many).
If your "successor" includes successors of the successors and your
"predecessor" includes predecessors of the predecessors then yes.
Of course they all are included.
That is not "of course". The words are used both ways.
But it is true about all nodes.
That is not possible if "all" is a meaningful expression here. If all
nodes are mapped, then none remains.
If you say that all nodes are not mapped
What does "single" mean above?
It means a sheave or separated subset containing only one path.
That's bad language. The expression "single path" should denote a
path, not a sheave nor a subset.
I do deny existence of nodes that don't have more than one path through
them.
Then you deny single paths.
Perhaps if you interprete "single paths" in your way but not if
it is interpreted in a sensible way.
Am 03.05.2026 um 10:09 schrieb Mikko:
On 02/05/2026 17:02, wm wrote:
That is not possible if "all" is a meaningful expression here. If
all nodes are mapped, then none remains.
If you say that all nodes are not mapped
I say all nodes are mapped.
That you say "if all nodes are mapped" means that you don't believe
that all nodes are mapped.
The meaning is: After all nodes are mapped, then none remains.
That paths can be indexed with real numbers is irrelevant to theSingle path needs no interpretation.
Your use of meaningless expressions is a sign of your intent to deceive.
A single path or a sheaf containing only one path are expressions
describing a real number. That should be basic knkwledge.
On 03/05/2026 14:30, wm wrote:
Am 03.05.2026 um 10:09 schrieb Mikko:
On 02/05/2026 17:02, wm wrote:
That is not possible if "all" is a meaningful expression here. If >>>>>> all nodes are mapped, then none remains.
If you say that all nodes are not mapped
I say all nodes are mapped.
That you say "if all nodes are mapped" means that you don't believe
that all nodes are mapped.
The meaning is: After all nodes are mapped, then none remains.
Maybe you meant that but you said otherwise.
That paths can be indexed with real numbers is irrelevantSingle path needs no interpretation.
Your use of meaningless expressions is a sign of your intent to deceive.
A single path or a sheaf containing only one path are expressions
describing a real number. That should be basic knkwledge.
Though it seems that you can't rememberwaht the topic of this discussion is.
Am 04.05.2026 um 08:37 schrieb Mikko:
On 03/05/2026 14:30, wm wrote:
Am 03.05.2026 um 10:09 schrieb Mikko:
On 02/05/2026 17:02, wm wrote:
That is not possible if "all" is a meaningful expression here. If >>>>>>> all nodes are mapped, then none remains.
If you say that all nodes are not mapped
I say all nodes are mapped.
That you say "if all nodes are mapped" means that you don't believe
that all nodes are mapped.
The meaning is: After all nodes are mapped, then none remains.
Maybe you meant that but you said otherwise.
It is a conditional clause (a "first conditional" structure) indicating
that the speaker is uncertain. It is a possibility, but not a certainty.
It does not mean the contrary.
That paths can be indexed with real numbers is irrelevantSingle path needs no interpretation.
Your use of meaningless expressions is a sign of your intent to
deceive.
A single path or a sheaf containing only one path are expressions
describing a real number. That should be basic knkwledge.
The path are not indexed but *are* representing real numbers.
Though it seems that you can't rememberwaht the topic of this discussion is.
The topic is that AI has unerstood that only countably many sheaves can
be distinguished in the Binary Tree.
This proves that only countably many real numbers can be distinguished.
Have you understood it too?
On 04/05/2026 13:47, WM wrote:
It seems that you now have that certainty, so now you can reformulate
your claim without that condition.
A single path or a sheaf containing only one path are expressionsThat paths can be indexed with real numbers is irrelevant
describing a real number. That should be basic knowledge.
The path are not indexed but *are* representing real numbers.
The word "resimbling" is too vague for mathematical discussion.
This proves that only countably many real numbers can be distinguished.
Have you understood it too?
Real numbers are not in the scople of the subject line or OP.
Perhaps you can confirm?
Am 05.05.2026 um 11:16 schrieb Mikko:
On 04/05/2026 13:47, WM wrote:
It seems that you now have that certainty, so now you can reformulate
your claim without that condition.
There are dark numbers. When you choose a natural number, then it has
more successors than predecessors. When you choose the collection "all natural" numbers then none remains.That proves that you cannot choose
every natural number as an individual. Same is true for nodes.
On 05/05/2026 18:03, WM wrote:
There are dark numbers. When you choose a natural number, then it has
more successors than predecessors. When you choose the collection "all
natural" numbers then none remains.That proves that you cannot choose
every natural number as an individual. Same is true for nodes.
Whdn we need to choose an individual node or an individual natural
number we don't care about all of them. When we need to say something
about every node or every natural number we con't care about individial
ones.
Am 06.05.2026 um 10:22 schrieb Mikko:It is well known that there are uncaountably many reals between any two different reals but only cauntably many names. An obvious conseqence is
On 05/05/2026 18:03, WM wrote:
There are dark numbers. When you choose a natural number, then it has
more successors than predecessors. When you choose the collection
"all natural" numbers then none remains.That proves that you cannot
choose every natural number as an individual. Same is true for nodes.
Whdn we need to choose an individual node or an individual natural
number we don't care about all of them. When we need to say something
about every node or every natural number we con't care about individial
ones.
con't means can't? Then you said a truth. And in particular with reals
we see the idea of dark numbers even better: Between two real numbers
there are infinitely many real numbers, infinitely many of which you
cannot name whatever you try.
On 06/05/2026 15:11, WM wrote:
Am 06.05.2026 um 10:22 schrieb Mikko:It is well known that there are uncaountably many reals between any two different reals but only cauntably many names. An obvious conseqence is
On 05/05/2026 18:03, WM wrote:
There are dark numbers. When you choose a natural number, then it
has more successors than predecessors. When you choose the
collection "all natural" numbers then none remains.That proves that
you cannot choose every natural number as an individual. Same is
true for nodes.
Whdn we need to choose an individual node or an individual natural
number we don't care about all of them. When we need to say something
about every node or every natural number we con't care about individial
ones.
con't means can't? Then you said a truth. And in particular with reals
we see the idea of dark numbers even better: Between two real numbers
there are infinitely many real numbers, infinitely many of which you
cannot name whatever you try.
that most of the reals have no name.
In constructive mathematics, where nothing unnamed exists, it is
provable that reals are not constructively countable.
Am 07.05.2026 um 09:32 schrieb Mikko:
On 06/05/2026 15:11, WM wrote:
Am 06.05.2026 um 10:22 schrieb Mikko:It is well known that there are uncaountably many reals between any two
On 05/05/2026 18:03, WM wrote:
There are dark numbers. When you choose a natural number, then it
has more successors than predecessors. When you choose the
collection "all natural" numbers then none remains.That proves that >>>>> you cannot choose every natural number as an individual. Same is
true for nodes.
Whdn we need to choose an individual node or an individual natural
number we don't care about all of them. When we need to say
something about every node or every natural number we con't care
about individial
ones.
con't means can't? Then you said a truth. And in particular with
reals we see the idea of dark numbers even better: Between two real
numbers there are infinitely many real numbers, infinitely many of
which you cannot name whatever you try.
different reals but only cauntably many names. An obvious conseqence is
that most of the reals have no name.
Nice that you understand this. But many do not. A preprint by J.D.
Hamkins et al. contains the following phrases, starting smugly: "One occasionally hears the argument rCo let us call it the math-tea argument, for perhaps it is heard at a good math tea rCo that there must be real numbers that we cannot describe or define, because there are only
countably many definitions, but uncountably many reals. Does it
withstand scrutiny? [...]
-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory that every real is definable in the language of set theory without parameters?
-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC axioms of set theory are consistent, then there are models of ZFC in which every object, including every real number, every function on the reals, every
set of reals, every topological space, every ordinal and so on, is
uniquely definable without parameters. [J.D. Hamkins et al.: "Pointwise definable models of set theory", arXiv (2012)]
Obviously he has not understood that he disproved ZFC.
But uncountability is not the reason for the existence of undefinable
numbers because between two rational numbers on the real line there
are infinitely many rational numbers, infinitely many of which you
cannot name whatever you try.
In constructive mathematics, where nothing unnamed exists, it is
provable that reals are not constructively countable.
Even there we have most rational numbers between two given rational
numbers can never be found although they must be there in actual infinity.
At least in the Binary Tree all should be present unless darkness veils them.
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory that >> every real is definable in the language of set theory without parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC axioms >> of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal and
so on, is uniquely definable without parameters. [J.D. Hamkins et al.:
"Pointwise definable models of set theory", arXiv (2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
It does not even make sense to say "disproved ZFC".
But uncountability is not the reason for the existence of undefinable
numbers because between two rational numbers on the real line there
are infinitely many rational numbers, infinitely many of which you
cannot name whatever you try.
Every integer can be named. Every rational number can be named with
two integers. Nothing else satisfies the definition of "rational
number".
In constructive mathematics, where nothing unnamed exists, it is
provable that reals are not constructively countable.
Even there we have most rational numbers between two given rational
numbers can never be found although they must be there in actual
infinity.
In constructive mathematics what cannot be constructed (of "found") does
not exist.
At least in the Binary Tree all should be present unless darkness
veils them.
It seems that you don't have coherent opinions abot the binary tree.
Am 08.05.2026 um 09:40 schrieb Mikko:
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory that >>> every real is definable in the language of set theory without
parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC axioms >>> of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal and
so on, is uniquely definable without parameters. [J.D. Hamkins et
al.: "Pointwise definable models of set theory", arXiv (2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
He did! He claimed that uncountably many names must exist if ZFC is
correct,
It does not even make sense to say "disproved ZFC".
If uncountable sets are countable in ZFC, then it is disproved.
Then the notion of countability is nonsensed.
But uncountability is not the reason for the existence of undefinable
numbers because between two rational numbers on the real line there
are infinitely many rational numbers, infinitely many of which you
cannot name whatever you try.
Every integer can be named. Every rational number can be named with
two integers. Nothing else satisfies the definition of "rational
number".
Every two named rationals have infinitely many not named rationals
between them, infinitely many of which cannot be named. Do you disagree?
In constructive mathematics, where nothing unnamed exists, it is
provable that reals are not constructively countable.
Even there we have most rational numbers between two given rational
numbers can never be found although they must be there in actual
infinity.
In constructive mathematics what cannot be constructed (of "found") does
not exist.
I said "in actual infinity".
If we believe that actual infinity is true, then there are numbers
which cannot be constructed.
At least in the Binary Tree all should be present unless darkness
veils them.
It seems that you don't have coherent opinions abot the binary tree.
It seems that you don't understand. Every node adds another sheaf
branching off - and not more!
On 08/05/2026 15:26, wm wrote:
Am 08.05.2026 um 09:40 schrieb Mikko:
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory that >>>> every real is definable in the language of set theory without
parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC axioms >>>> of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal
and so on, is uniquely definable without parameters. [J.D. Hamkins
et al.: "Pointwise definable models of set theory", arXiv (2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
He did! He claimed that uncountably many names must exist if ZFC is
correct,
What did he mean with "name"? Apparently not the same as we do.
It does not even make sense to say "disproved ZFC".
If uncountable sets are countable in ZFC, then it is disproved.
It does not mean that if it means nothing.
Then the notion of countability is nonsensed.
That does not mean anything.
But uncountability is not the reason for the existence of undefinable
numbers because between two rational numbers on the real line there
are infinitely many rational numbers, infinitely many of which you
cannot name whatever you try.
Every integer can be named. Every rational number can be named with
two integers. Nothing else satisfies the definition of "rational
number".
Every two named rationals have infinitely many not named rationals
between them, infinitely many of which cannot be named. Do you disagree?
No, they are all named. All unnamed numbers are irrational.
If we believe that actual infinity is true, then there are numbers
which cannot be constructed.
Only a claim can be true. An actual infinity is not a claim so it
cannot be true.
At least in the Binary Tree all should be present unless darkness
veils them.
It seems that you don't have coherent opinions abot the binary tree.
It seems that you don't understand. Every node adds another sheaf
branching off - and not more!
The only thing to be understood of an incoherent presentation is that
the presentation is incoherent.
Am 09.05.2026 um 09:39 schrieb Mikko:
On 08/05/2026 15:26, wm wrote:
Am 08.05.2026 um 09:40 schrieb Mikko:
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory >>>>> that every real is definable in the language of set theory without
parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC
axioms of set theory are consistent, then there are models of ZFC
in which every object, including every real number, every function
on the reals, every set of reals, every topological space, every
ordinal and so on, is uniquely definable without parameters. [J.D.
Hamkins et al.: "Pointwise definable models of set theory", arXiv
(2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
He did! He claimed that uncountably many names must exist if ZFC is
correct,
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
It does not even make sense to say "disproved ZFC".
If uncountable sets are countable in ZFC, then it is disproved.
It does not mean that if it means nothing.
It means something. Otherwise he would not have published it.
Then the notion of countability is nonsensed.
That does not mean anything.
That means that most mathematicians adhere to nonsense.
But uncountability is not the reason for the existence of undefinable >>>> -a> numbers because between two rational numbers on the real line there >>>> -a> are infinitely many rational numbers, infinitely many of which you >>>> -a> cannot name whatever you try.
Every integer can be named. Every rational number can be named with
two integers. Nothing else satisfies the definition of "rational
number".
Every two named rationals have infinitely many not named rationals
between them, infinitely many of which cannot be named. Do you disagree?
No, they are all named. All unnamed numbers are irrational.
Try to name all rationals between 0 and eps.
If we believe that actual infinity is true, then there are numbers
which cannot be constructed.
Only a claim can be true. An actual infinity is not a claim so it
cannot be true.
Actual infinity can only be a claim.
On 10/05/2026 16:47, WM wrote:
Am 09.05.2026 um 09:39 schrieb Mikko:
On 08/05/2026 15:26, wm wrote:
Am 08.05.2026 um 09:40 schrieb Mikko:
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory >>>>>> that every real is definable in the language of set theory without >>>>>> parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC >>>>>> axioms of set theory are consistent, then there are models of ZFC >>>>>> in which every object, including every real number, every function >>>>>> on the reals, every set of reals, every topological space, every
ordinal and so on, is uniquely definable without parameters. [J.D. >>>>>> Hamkins et al.: "Pointwise definable models of set theory", arXiv >>>>>> (2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
He did! He claimed that uncountably many names must exist if ZFC is
correct,
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more
general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
Or pehaps he accpets
uncountable character sets.
No, they are all named. All unnamed numbers are irrational.
Try to name all rationals between 0 and eps.
Too much work that would serve no purpose.
It is sufficient to note that
you have not proven about one rational that it cannot be named, let
alone about infinitely many.
Actual infinity can only be a claim.
No. Actual infinity is not a sentence. Only a sentence can be a claim.
Am 11.05.2026 um 09:39 schrieb Mikko:
On 10/05/2026 16:47, WM wrote:
Am 09.05.2026 um 09:39 schrieb Mikko:
On 08/05/2026 15:26, wm wrote:
Am 08.05.2026 um 09:40 schrieb Mikko:
On 07/05/2026 13:35, wm wrote:
-a-a-a-a-aQuestion 1. Is it consistent with the axioms of set theory >>>>>>> that every real is definable in the language of set theory
without parameters?
-a-a-a-a-aThe answer is Yes. Indeed, much more is true: if the ZFC >>>>>>> axioms of set theory are consistent, then there are models of ZFC >>>>>>> in which every object, including every real number, every
function on the reals, every set of reals, every topological
space, every ordinal and so on, is uniquely definable without
parameters. [J.D. Hamkins et al.: "Pointwise definable models of >>>>>>> set theory", arXiv (2012)]
Obviously he has not understood that he disproved ZFC.
Because he did not.
He did! He claimed that uncountably many names must exist if ZFC is >>>>> correct,
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more
general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
No. He makes fun of the math tea argument which says that there are only countably many (necessarily finite) names. Nobody denies that there are uncountably many infinite strings representing real numbers - at least
not at a math tea.
Or pehaps he accpets
uncountable character sets.
Same nonsense. All character sets which can be written or imagined are finite. Even if the universe was filled with characters of smallest size
the set would be countable, in the accessible universe even finite.
No, they are all named. All unnamed numbers are irrational.
Try to name all rationals between 0 and eps.
Too much work that would serve no purpose.
The purpose is to inform you that most are dark.
It is sufficient to note that
you have not proven about one rational that it cannot be named, let
alone about infinitely many.
Between all rationals that can be named there are infinitely many
unnamed, most of which cannot be named. This situation cannot be
changed. That is proof enough.
Actual infinity can only be a claim.
No. Actual infinity is not a sentence. Only a sentence can be a claim.
The claim is: actual infinity exists.
On 11/05/2026 13:56, WM wrote:
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more
general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
No. He makes fun of the math tea argument which says that there are
only countably many (necessarily finite) names. Nobody denies that
there are uncountably many infinite strings representing real numbers
- at least not at a math tea.
You mean, he doesn't care about mathematical truth?
The purpose is to inform you that most are dark.
Other than every is not most.
It means accepted as the basis of set theory.The claim is: actual infinity exists.
Depends on what "exists" means.
Am 12.05.2026 um 09:19 schrieb Mikko:
On 11/05/2026 13:56, WM wrote:
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more
general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
No. He makes fun of the math tea argument which says that there are
only countably many (necessarily finite) names. Nobody denies that
there are uncountably many infinite strings representing real numbers
- at least not at a math tea.
You mean, he doesn't care about mathematical truth?
What he said is wrong.
The purpose is to inform you that most are dark.
Other than every is not most.
Define or name rational numbers between two different given ones, may
they be chosen as close as you can accomplish, such that not infinitely
many remain not named. Dark numbers exist because it is clear that you
would fail.
On 12/05/2026 13:52, wm wrote:
Am 12.05.2026 um 09:19 schrieb Mikko:
On 11/05/2026 13:56, WM wrote:
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more
general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
No. He makes fun of the math tea argument which says that there are
only countably many (necessarily finite) names. Nobody denies that
there are uncountably many infinite strings representing real
numbers - at least not at a math tea.
You mean, he doesn't care about mathematical truth?
What he said is wrong.
Who knows what he really said?
Define or name rational numbers between two different given ones, may
they be chosen as close as you can accomplish, such that not
infinitely many remain not named. Dark numbers exist because it is
clear that you would fail.
If rreC and rreU are two different named number then (rreC + rreU) / 2 is a named number between them. Therefore for every natural number k the
number rreureeree = (rreu + rreUreereU) / 2 is a named natural number between rreC and
rreU and there are infinitely many of them, one for each k.
Am 13.05.2026 um 11:59 schrieb Mikko:
On 12/05/2026 13:52, wm wrote:
Am 12.05.2026 um 09:19 schrieb Mikko:
On 11/05/2026 13:56, WM wrote:
What did he mean with "name"? Apparently not the same as we do.
He said uniquely definable. That giving an individual name.
That is compatible with the idea that hes meaning of "name" is more >>>>>> general than the usual meaning. Apparently he accpets inifinitely
long strings or uncaountable sets as "names".
No. He makes fun of the math tea argument which says that there are >>>>> only countably many (necessarily finite) names. Nobody denies that
there are uncountably many infinite strings representing real
numbers - at least not at a math tea.
You mean, he doesn't care about mathematical truth?
What he said is wrong.
Who knows what he really said?
I told you.
Therefore you shoulsd know. But in case you have a short
memory, here it is again:
"One occasionally hears the argument rCo let us call it the math-tea argument, for perhaps it is heard at a good math tea rCo that there must
be real numbers that we cannot describe or define, because there are
only countably many definitions, but uncountably many reals. Does it withstand scrutiny? [...]
-a-a-a Question 1. Is it consistent with the axioms of set theory that every real is definable in the language of set theory without parameters?
-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC axioms of set theory are consistent, then there are models of ZFC in which every object, including every real number, every function on the reals, every
set of reals, every topological space, every ordinal and so on, is
uniquely definable without parameters. [J.D. Hamkins et al.: "Pointwise definable models of set theory", arXiv (2012)]
Complete nonsense.
Define or name rational numbers between two different given ones, may
they be chosen as close as you can accomplish, such that not
infinitely many remain not named. Dark numbers exist because it is
clear that you would fail.
If rreC and rreU are two different named number then (rreC + rreU) / 2 is a >> named number between them. Therefore for every natural number k the
number rreureeree = (rreu + rreUreereU) / 2 is a named natural number between rreC and
rreU and there are infinitely many of them, one for each k.
Potentially infinitely many. Actually infinitely many remain unnamed.
And here is a statement shaped according to your standards:
The function of a node in the Binary Tree: One sheaf goes in from above,
two sheaves go out downwards. The node adds 2 - 1 = 1 sheaf.
On 13/05/2026 23:50, wm wrote:
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory, here
it is again:
"One occasionally hears the argument rCo let us call it the math-tea
argument, for perhaps it is heard at a good math tea rCo that there must
be real numbers that we cannot describe or define, because there are
only countably many definitions, but uncountably many reals. Does it
withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory that
every real is definable in the language of set theory without parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC axioms
of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal and
so on, is uniquely definable without parameters. [J.D. Hamkins et al.:
"Pointwise definable models of set theory", arXiv (2012)]
Complete nonsense.
Potentially infinitely many. Actually infinitely many remain unnamed.
They all actually have a name.
That you have never seen or heard all
of those names is irrelevant.
And here is a statement shaped according to your standards:
The function of a node in the Binary Tree: One sheaf goes in from
above, two sheaves go out downwards. The node adds 2 - 1 = 1 sheaf.
That means that sheaves are infinitely divisible.
But the sheaves are
not interesting. Paths are.
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:
No, you did not. You presented some claims about what he said butCan't you read? Below is the quote and its source.
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory, here
it is again:
"One occasionally hears the argument rCo let us call it the math-tea
argument, for perhaps it is heard at a good math tea rCo that there
must be real numbers that we cannot describe or define, because there
are only countably many definitions, but uncountably many reals. Does
it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory that >>> every real is definable in the language of set theory without
parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC axioms >>> of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal and
so on, is uniquely definable without parameters. [J.D. Hamkins et
al.: "Pointwise definable models of set theory", arXiv (2012)]
Complete nonsense.
Potentially infinitely many. Actually infinitely many remain unnamed.
They all actually have a name.
That is belief. Anyhow almost all "names" cannot be reproduced.
That you have never seen or heard all
of those names is irrelevant.
In my opinion your belief is irrelevant. Anyhow this question is irrelevant.>
And here is a statement shaped according to your standards:
The function of a node in the Binary Tree: One sheaf goes in from
above, two sheaves go out downwards. The node adds 2 - 1 = 1 sheaf.
That means that sheaves are infinitely divisible.
That is the same kind of belief. There are countably many sheaves. They
may or may not contain infinitely many paths. This question is
irrelevant because these paths cannot be distinguished by nodes.
But the sheaves areOnly path which can be distinguished by nodes.
not interesting. Paths are.
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory,
here it is again:
"One occasionally hears the argument rCo let us call it the math-tea
argument, for perhaps it is heard at a good math tea rCo that there
must be real numbers that we cannot describe or define, because
there are only countably many definitions, but uncountably many
reals. Does it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory that >>>> every real is definable in the language of set theory without
parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC axioms >>>> of set theory are consistent, then there are models of ZFC in which
every object, including every real number, every function on the
reals, every set of reals, every topological space, every ordinal
and so on, is uniquely definable without parameters. [J.D. Hamkins
et al.: "Pointwise definable models of set theory", arXiv (2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
Not by different nodes. By what else?And here is a statement shaped according to your standards:
The function of a node in the Binary Tree: One sheaf goes in from
above, two sheaves go out downwards. The node adds 2 - 1 = 1 sheaf.
That means that sheaves are infinitely divisible.
But the sheaves areOnly path which can be distinguished by nodes.
not interesting. Paths are.
Paths that can't be would be even more interesting.
Am 15.05.2026 um 08:03 schrieb Mikko:
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory,
here it is again:
"One occasionally hears the argument rCo let us call it the math-tea >>>>> argument, for perhaps it is heard at a good math tea rCo that there >>>>> must be real numbers that we cannot describe or define, because
there are only countably many definitions, but uncountably many
reals. Does it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory >>>>> that every real is definable in the language of set theory without
parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC
axioms of set theory are consistent, then there are models of ZFC
in which every object, including every real number, every function
on the reals, every set of reals, every topological space, every
ordinal and so on, is uniquely definable without parameters. [J.D.
Hamkins et al.: "Pointwise definable models of set theory", arXiv
(2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
It is nonsense since there are only countably many definitions.
On 15/05/2026 19:44, WM wrote:
Am 15.05.2026 um 08:03 schrieb Mikko:
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory,
here it is again:
"One occasionally hears the argument rCo let us call it the math-tea >>>>>> argument, for perhaps it is heard at a good math tea rCo that there >>>>>> must be real numbers that we cannot describe or define, because
there are only countably many definitions, but uncountably many
reals. Does it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory >>>>>> that every real is definable in the language of set theory without >>>>>> parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC >>>>>> axioms of set theory are consistent, then there are models of ZFC >>>>>> in which every object, including every real number, every function >>>>>> on the reals, every set of reals, every topological space, every
ordinal and so on, is uniquely definable without parameters. [J.D. >>>>>> Hamkins et al.: "Pointwise definable models of set theory", arXiv >>>>>> (2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
It is nonsense since there are only countably many definitions.
How is the claim that there are only countably many definitions made
nonsense by the fact that there are only countably many dfinitions?
Am 16.05.2026 um 11:47 schrieb Mikko:
On 15/05/2026 19:44, WM wrote:Hamkins' claim that uncountably many reals are nameable is nonsense.
Am 15.05.2026 um 08:03 schrieb Mikko:
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory, >>>>>>> here it is again:
"One occasionally hears the argument rCo let us call it the
math-tea argument, for perhaps it is heard at a good math tea rCo >>>>>>> that there must be real numbers that we cannot describe or
define, because there are only countably many definitions, but
uncountably many reals. Does it withstand scrutiny? [...]
Question 1. Is it consistent with the axioms of set theory
that every real is definable in the language of set theory
without parameters?
The answer is Yes. Indeed, much more is true: if the ZFC
axioms of set theory are consistent, then there are models of ZFC >>>>>>> in which every object, including every real number, every
function on the reals, every set of reals, every topological
space, every ordinal and so on, is uniquely definable without
parameters. [J.D. Hamkins et al.: "Pointwise definable models of >>>>>>> set theory", arXiv (2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
It is nonsense since there are only countably many definitions.
How is the claim that there are only countably many definitions made
nonsense by the fact that there are only countably many dfinitions?
REgards, WM
Am 16.05.2026 um 11:47 schrieb Mikko:First you must prove that he made such claim.
On 15/05/2026 19:44, WM wrote:Hamkins' claim that uncountably many reals are nameable is nonsense.
Am 15.05.2026 um 08:03 schrieb Mikko:
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but
you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory, >>>>>>> here it is again:
"One occasionally hears the argument rCo let us call it the math- >>>>>>> tea argument, for perhaps it is heard at a good math tea rCo that >>>>>>> there must be real numbers that we cannot describe or define,
because there are only countably many definitions, but
uncountably many reals. Does it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory >>>>>>> that every real is definable in the language of set theory
without parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC >>>>>>> axioms of set theory are consistent, then there are models of ZFC >>>>>>> in which every object, including every real number, every
function on the reals, every set of reals, every topological
space, every ordinal and so on, is uniquely definable without
parameters. [J.D. Hamkins et al.: "Pointwise definable models of >>>>>>> set theory", arXiv (2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
It is nonsense since there are only countably many definitions.
How is the claim that there are only countably many definitions made
nonsense by the fact that there are only countably many dfinitions?
On 17/05/2026 17:17, wm wrote:
Am 16.05.2026 um 11:47 schrieb Mikko:First you must prove that he made such claim.
On 15/05/2026 19:44, WM wrote:Hamkins' claim that uncountably many reals are nameable is nonsense.
Am 15.05.2026 um 08:03 schrieb Mikko:
On 14/05/2026 18:00, WM wrote:
Am 14.05.2026 um 10:54 schrieb Mikko:
On 13/05/2026 23:50, wm wrote:Can't you read? Below is the quote and its source.
No, you did not. You presented some claims about what he said but >>>>>>> you did not quote anytihing.
Therefore you shoulsd know. But in case you have a short memory, >>>>>>>> here it is again:
"One occasionally hears the argument rCo let us call it the math- >>>>>>>> tea argument, for perhaps it is heard at a good math tea rCo that >>>>>>>> there must be real numbers that we cannot describe or define, >>>>>>>> because there are only countably many definitions, but
uncountably many reals. Does it withstand scrutiny? [...]
-a-a-a-a Question 1. Is it consistent with the axioms of set theory >>>>>>>> that every real is definable in the language of set theory
without parameters?
-a-a-a-a The answer is Yes. Indeed, much more is true: if the ZFC >>>>>>>> axioms of set theory are consistent, then there are models of >>>>>>>> ZFC in which every object, including every real number, every >>>>>>>> function on the reals, every set of reals, every topological
space, every ordinal and so on, is uniquely definable without >>>>>>>> parameters. [J.D. Hamkins et al.: "Pointwise definable models of >>>>>>>> set theory", arXiv (2012)]
Complete nonsense.
What he says here is perfectly correct and uses the words in their
usual meanings.
It is nonsense since there are only countably many definitions.
How is the claim that there are only countably many definitions made
nonsense by the fact that there are only countably many dfinitions?
However, there is a problem with the quoted text. The proof that a
consistent theory has a finite or countable model is restricted to
first order theories. But the concepts of infinity and countability
are second order concepts. Therefore one must be careful when these
are discussed at the same time.
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