Would it matter if ZF was inconsistent?
Frederick Williams
provably 10^(10^10) formulae long. (Nor can it prove other undesirable
things more shortly.) Would you stop using it? Would it disturb you?
(* Or your favourite set theory)
--
Which of the seven heavens / Was responsible her smile /
Wouldn't be sure but attested / That, whoever it was, a god /
Worth kneeling-to for a while / Had tabernacled and rested.
WM
wrote:
> Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
> provably 10^(10^10) formulae long.
The shortest proof of inconsistence is less than 10^5 bits.
http://www.hs-augsburg.de/~mueckenh/GU/Pr%fcfung%20GU0907.pdf
look up number 11.
> (Nor can it prove other undesirable
> things more shortly.) Would you stop using it?
At any rate. Better never take it up.
> Would it disturb you?
No. It is fine as it is. True mathematics has never been connected to
astrology or set theory.
Regards, WM
Jan Burse
> On 16 Jul., 15:02, Frederick Williams <frederick.willia...@tesco.net>
> wrote:
>> Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
>> provably 10^(10^10) formulae long.
>
> The shortest proof of inconsistence is less than 10^5 bits.
>
> http://www.hs-augsburg.de/~mueckenh/GU/Pr%fcfung%20GU0907.pdf
>
> look up number 11.
>
"Der bin�re Baum kann dazu dienen, die Existenz einer aktualen
Unendlichkeit ad absurdum zu f�hren."
"Die Pfadmenge kann n�mlich als gr��er (nach Cantor), oder als
gleichgro� (pro Knoten wird ein weiterer Pfad unterscheidbar) oder als
kleiner als eine abz�hlbar unendliche Menge nachgewiesen werden."
".. ist der Nachweis erbracht, dass die Menge der Knoten gr��er als die
der Pfade ist."
This translates to:
"The binary tree can be used to show the absurdity of an
actual infinity."
"I can be either shown that the set of path is greater (after
Cantor), or equal (for each node there is a new path distinction)
or less than a countable set."
".. we have proved that the set of nodes is bigger than the
set of paths."
Funny indeed.
So what is the logical flaw. The logical flaw is to find the
way somebody steps from 0...n to 0..... Here is a simpler form of
this logical flaw:
Lets assume achilles and the turtle can run same speed, and achilles
is one meter in front of the turtle. Achilles runs on the left
lane and the turtle runs on the right lane. We will now show that
the infinite left lane is longer than the infinite right lane (sic!).
We show that achilles is always at step x+1
when the turtle is at step x.
Induction foundation: The turtle starts at point 0, and achilles
starts at point 1, (achilles is one meter in front of the turtle)
Induction step: The turtle and achilles have the same velocity, so
when achilles is at step x'-1+1=x', and the turtle is at x'-1
(for x'>0), then after that achilles will be at step x'+1 and
the turtle will be at step x'.
What is the flaw here?
Well very easy, there is a jumping to conclusions from some
ordering on the elements to the cardinality of the set
of these elements. But cardinality does not work this way,
it is careless about orderings on the elements. (Thats why
it is called cardinality and not ordinal)
We will now show that the left L and the right lane R have
same cardinality. We assume that L and R come into being
as follows:
L = u Lx, Li = {x+1}
R = u Rx Ri = {x}
Now we proof that f:R->L, with f(x)=x+1 is surjective
and injective. Injectivity is no problem. Surjectivity
goes as follows: Lets assume k is in L. Then there is
some x such that k is in Lx. Hence k=x+1. Hence x in R
and k=f(x).
But since the earth is flat, this is all not true,
and achilles will fall down, at the border of the
disk, the turtle will see this and stop. So turtle
will be walking {0,..,zap} and achilles will be
walking {1,...,zap}, so to the contrary we can conclude:
The right lane is longer than the left lane.
Bye
Zap = Distance to the border of the disc, estimated
to be 9^9^9.
James Burns
> Suppose that ZF(*) could prove 0=/=0, but the shortest proof
> was provably 10^(10^10) formulae long. (Nor can it prove
> other undesirable things more shortly.)
> Would you stop using it? Would it disturb you?
>
> (* Or your favourite set theory)
I think I don't know how I would react. This seems too far
outside my experience.
I can guess, though: I would read about paraconsistent logics
immdiately. I would hope that there are logics only a tiny
bit paraconsistent, that still produce familiar results from
familiar axioms, but justify ignoring the "distant" (in some
sense) inconsistency as irrelevant.
I don't think it would disturb me -- fascinate, more likely.
But I am not a mathematician. Apart from a fairly narrow
range of results, mathematics, logic and philosophy are my
playground.
Perhaps a couple of short stories where something along
these lines happen would be an aid to the imagination:
"Division by Zero", by Ted Chiang
"Dark Integers", by Greg Egan
Jim Burns
"Division by Zero", by Ted Chiang
http://www.fantasticmetropolis.com/i/division/full/
"Dark Integers", by Greg Egan
http://www.asimovs.com/_issue_0805/DarkINtegers.shtml
Review
http://groups.google.com/group/sfabc-nj-science-science-fiction-group/browse_thread/thread/97d1c71957930de8
MoeBlee
wrote:
> Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
> provably 10^(10^10) formulae long. (Nor can it prove other undesirable
> things more shortly.) Would you stop using it? Would it disturb you?
First, I'd want to know what consistent subsets of the axioms there
are. Then get some sense of what is lost by cutting back to a
consistent subset but adding some weaker version of each removed
axiom. Or, if a satisfactorily comprehensive theory could not be found
that way, I'd look for a different theory (I might even do that right
away).
If ZF were found inconsistent, it would fascinate me more than disturb
me, I think. It would be unfortunate; but it would be better to know
than not to know, and I'd be happy to be alive to witness such a
monumental discovery.
MoeBlee
MoeBlee
> On 16 Jul., 15:02, Frederick Williams <frederick.willia...@tesco.net>
> wrote:
>
> > Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
> > provably 10^(10^10) formulae long.
>
> The shortest proof of inconsistence is less than 10^5 bits.
You've never produced a proof that ZF is inconsistent. ('inconsistent'
is not merely meant 'contrary to WM's own mathematical notions'.)
MoeBlee
WM
> WM schrieb:
>
> > On 16 Jul., 15:02, Frederick Williams <frederick.willia...@tesco.net>
> > wrote:
> >> Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
> >> provably 10^(10^10) formulae long.
>
> > The shortest proof of inconsistence is less than 10^5 bits.
>
> >http://www.hs-augsburg.de/~mueckenh/GU/Pr%fcfung%20GU0907.pdf
>
> > look up number 11.
>
> Unendlichkeit ad absurdum zu führen."
> "Die Pfadmenge kann nämlich als größer (nach Cantor), oder als
> gleichgroß (pro Knoten wird ein weiterer Pfad unterscheidbar) oder als
> kleiner als eine abzählbar unendliche Menge nachgewiesen werden."
> ".. ist der Nachweis erbracht, dass die Menge der Knoten größer als die
> der Pfade ist."
>
> This translates to:
> "The binary tree can be used to show the absurdity of an
> actual infinity."
> Cantor), or equal (for each node there is a new path distinction)
> or less than a countable set."
> ".. we have proved that the set of nodes is bigger than the
> set of paths."
>
> Funny indeed.
>
> So what is the logical flaw. The logical flaw is to find the
> way somebody steps from 0...n to 0.....
No, with respect to set theory you are wrong. If every part of the
tree can be conquered, then there is nothing remaining. We have the
whole tree. (In principle you are right, however. Infinity is never
finished.)
>Here is a simpler form of
> this logical flaw:
>
> Lets assume achilles and the turtle can run same speed, and achilles
> is one meter in front of the turtle. Achilles runs on the left
> lane and the turtle runs on the right lane. We will now show that
> the infinite left lane is longer than the infinite right lane (sic!).
>
> We show that achilles is always at step x+1
> when the turtle is at step x.
>
> Induction foundation: The turtle starts at point 0, and achilles
> starts at point 1, (achilles is one meter in front of the turtle)
>
> Induction step: The turtle and achilles have the same velocity, so
> when achilles is at step x'-1+1=x', and the turtle is at x'-1
> (for x'>0), then after that achilles will be at step x'+1 and
> the turtle will be at step x'.
>
> What is the flaw here?
Let us assume a bijection between N and 2N.
1 - 2
2 - 4
3 - 6
...
What is the logical flaw here?
>
> Well very easy, there is a jumping to conclusions from some
> ordering on the elements to the cardinality of the set
> of these elements. But cardinality does not work this way,
> it is careless about orderings on the elements. (Thats why
> it is called cardinality and not ordinal)
Correct. Your explanation covers all those cases, in particular |N| = |
Q| and the like. How should it be "proved" without ordering?
But if the actually infinite tree would exist, then we could conquer
it by defining when which part is conquered and showing that there is
no part remaining unconquered forever.
Regards, WM
Jan Burse
> How should it be "proved" without ordering?
Why do you think a function f:N -> N\0, f(x)=x+1
involves ordering.
Yes, N and N\0 can be ordered, and somehow I
invoked ordering when I proved surjectivity.
But I could also assume that the left lane
came into being by the following process:
L = u L2x u u L2x+1
(first the even ones, and then the odd ones)
So the order would be:
1, 3, 5, ... 2, 4, 6, ...
Still L and R have same cardinality.
> But if the actually infinite tree would exist
But the question is not how many nodes are
in the tree, but how many infinite paths
are in the tree.
A node does not identify an infinite path.
It only identifies a finite path from to
root to the node itself.
So it is kind of fractal. When ever you
have conquered down to the level n,
with your 2^n leaf nodes, then each
leaf node points again to an infinite tree.
So we have the following equation:
c = c^(2^n)+2^n
This equation can be satisfied by a countable
infinity, or an uncountable infinity. So
your construction does not exclude that
the infinite tree exists. It is just
undecided about it.
Further arguments have to be persued
whether the infinite tree exists. And
one further argument might be diagonalization,
which then excludes the countable case.
Lets me think (in a subsequent post) whether
a further argument can be deduced by the
tree construction itself.
Bye
Jan Burse
> What is the logical flaw here?
For the tree: You get cheated by your drawing.
Don't confuse drawing with logical reasoning.
I think there are similar examples in geometry.
Take the x-axis, a unit circle, and then a line
passing quadrant I, II and III, such that it
crosses the circle two times.
One intersection point (1) with the circle is
above the x-axis, and the other intersection
point (2) is below the x-axis.
We can draw the following analogy:
Circle/Ruler: Your infinite tree construction.
Point (1): The infinite tree is uncountable.
Point (2): The infinite tree is countable.
Now there are two camps: One camp says my
construction only yields point (1), the
other camp says my construction only yields
point (2).
But in the end picture is wrong, line
goes through quadrant I and IV, so line
is always above x-axis....
Bye
Jan Burse
> So we have the following equation:
>
> c = c^(2^n)+2^n
>
Oops, the following should be enough:
c = c^(2^n)
Jan Burse
> WM schrieb:
>> What is the logical flaw here?
>
> For the tree: You get cheated by your drawing.
> Don't confuse drawing with logical reasoning.
Anyway from L�wenheim Skolem downward theorem
we know, that this problem (multiple solutions)
will always arise when your construction
is "first order".
Some second order spice (quantifying over sets)
has to be injected, so that construction
sticks to the uncountable...
Bye
WM
> WM schrieb:
>
> > How should it be "proved" without ordering?
>
> Why do you think a function f:N -> N\0, f(x)=x+1
> involves ordering.
Have you ever seen the proof of equnumerousity of N and Q?
However, if for any element of a set we can say when it is used, then
the whole set is use (because there is nothing remaining).
>
> Yes, N and N\0 can be ordered, and somehow I
> invoked ordering when I proved surjectivity.
>
> But I could also assume that the left lane
> came into being by the following process:
>
> L = u L2x u u L2x+1
> (first the even ones, and then the odd ones)
>
> So the order would be:
>
> 1, 3, 5, ... 2, 4, 6, ...
>
> Still L and R have same cardinality.
Of course. And still there is order.
>
> > But if the actually infinite tree would exist
>
> But the question is not how many nodes are
> in the tree, but how many infinite paths
> are in the tree.
The question is: How many infinite paths must be conquered so that
nobody is able to find a path that is not conquered. (Of course I do
not publish which subset of paths I have used, in order to show set
theorists that their claim is nonsense.)
>
> A node does not identify an infinite path.
> It only identifies a finite path from to
> root to the node itself.
I construct the complete binary tree by means of a countable set of
nodes. Tell me which nodes have not been constructed.
>
> So it is kind of fractal. When ever you
> have conquered down to the level n,
> with your 2^n leaf nodes, then each
> leaf node points again to an infinite tree.
>
> So we have the following equation:
>
> c = c^(2^n)+2^n
>
> This equation can be satisfied by a countable
> infinity, or an uncountable infinity. So
> your construction does not exclude that
> the infinite tree exists. It is just
> undecided about it.
No. My argument shows that a countable set of paths fills the complete
infinite binary tree such that nobody can tell what further paths are
missing (none is). That proves that the whole blathering about actual
infinity is nonsense. There is not even a single actually infinite
path.
>
> Further arguments have to be persued
> whether the infinite tree exists. And
> one further argument might be diagonalization,
> which then excludes the countable case.
By definition the tree contains every binary sequence. So there is no
diagonalization possible.
Regards, WM
WM
> WM schrieb:
>
> > What is the logical flaw here?
>
> For the tree: You get cheated by your drawing.
> Don't confuse drawing with logical reasoning.
>
> I think there are similar examples in geometry.
> Take the x-axis, a unit circle, and then a line
> passing quadrant I, II and III, such that it
> crosses the circle two times.
>
> One intersection point (1) with the circle is
> above the x-axis, and the other intersection
> point (2) is below the x-axis.
>
> We can draw the following analogy:
Don't draw analogies. Tell me which paths are missing in the tree that
has been constructed by means of a countable number of paths.
Regards, WM
MoeBlee
> On 16 Jul., 21:21, Jan Burse <janbu...@fastmail.fm> wrote:
>
> > WM schrieb:
>
> > > How should it be "proved" without ordering?
>
> > Why do you think a function f:N -> N\0, f(x)=x+1
> > involves ordering.
>
> Have you ever seen the proof of equnumerousity of N and Q?
Yes, what about it?
> However, if for any element of a set we can say when it is used, then
> the whole set is use (because there is nothing remaining).
Maybe what you mean: "If every member of a set S is in the domain of a
function f then S is a subset of the domain of f. Yes, this is quite
trivially true.
> The question is: How many infinite paths must be conquered so that
> nobody is able to find a path that is not conquered.
What is your mathematical definition of 'conquered'?
>(Of course I do
> not publish which subset of paths I have used, in order to show set
> theorists that their claim is nonsense.)
Good idea. You are quite the clever one.
> > A node does not identify an infinite path.
> > It only identifies a finite path from to
> > root to the node itself.
>
> I construct the complete binary tree by means of a countable set of
> nodes. Tell me which nodes have not been constructed.
There are countably many nodes; uncountably many infinite paths.
You've never proven otherwise.
> No. My argument shows that a countable set of paths fills the complete
> infinite binary tree such that nobody can tell what further paths are
> missing (none is).
Whatever the case about "people telling" things, you've never proven
that there are only countably many paths.
> That proves that the whole blathering about actual
> infinity is nonsense. There is not even a single actually infinite
> path.
There are uncountably many. We prove it. You don't prove otherwise.
> By definition the tree contains every binary sequence. So there is no
Of course diagonalization is available. For any enumeration of a set
of paths, diagonalize to get a path not in the enumeration.
MoeBlee
MoeBlee
Do you think there will ever be a day that a peer-review will approve
your purported proof as an actually correct proof that ZF is
inconsistent ZF? Or more, explicitly, a peer-review that will attest
that you have shown that, (correctly) using only first order logic
applied to the axioms of ZF, there is a derivation of a formula P and
~P?
MoeBlee
MoeBlee
P.S. I've challenged you before, and I'm challenging you again:
Provide rigorous definitions, in the language of ZF, of every term you
use (starting even with "the complete binary tree") and then show that
there exists a derivation, using only first order logic with identity
from the axioms of ZF, of some formula P and ~P.
MoeBlee
Virgil
<026c8c8f-ffe8-4fbe...@o6g2000yqj.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> True mathematics has never been connected to
> astrology or set theory.
A lot of true mathematics is quite compatible with many forms of set
theory, but much of WM's MathUnrealism is not, though it is compatible
with astrology.
--
Virgil
Virgil
<f8f19c70-61f7-4f9d...@y19g2000yqy.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 16 Jul., 17:05, Jan Burse <janbu...@fastmail.fm> wrote:
> > WM schrieb:
> >
> > > On 16 Jul., 15:02, Frederick Williams <frederick.willia...@tesco.net>
> > > wrote:
> > >> Suppose that ZF(*) could prove 0=/=0, but the shortest proof was
> > >> provably 10^(10^10) formulae long.
> >
> > > The shortest proof of inconsistence is less than 10^5 bits.
> >
> > >http://www.hs-augsburg.de/~mueckenh/GU/Pr%fcfung%20GU0907.pdf
> >
> > > look up number 11.
> >
> > "Der bin�re Baum kann dazu dienen, die Existenz einer aktualen
> > Unendlichkeit ad absurdum zu f�hren."
> > "Die Pfadmenge kann n�mlich als gr��er (nach Cantor), oder als
> > gleichgro� (pro Knoten wird ein weiterer Pfad unterscheidbar) oder als
> > kleiner als eine abz�hlbar unendliche Menge nachgewiesen werden."
> > ".. ist der Nachweis erbracht, dass die Menge der Knoten gr��er als die
> > der Pfade ist."
> >
> > This translates to:
> > "The binary tree can be used to show the absurdity of an
> > actual infinity."
> > "It can be either shown that the set of path is greater (after
> > Cantor), or equal (for each node there is a new path distinction)
> > or less than a countable set."
> > ".. we have proved that the set of nodes is bigger than the
> > set of paths."
> >
> > Funny indeed.
> >
> > So what is the logical flaw. The logical flaw is to find the
> > way somebody steps from 0...n to 0.....
>
> No, with respect to set theory you are wrong. If every part of the
> tree can be conquered, then there is nothing remaining. We have the
> whole tree.
If one can have a set of all natural numbers, one can make a tree out of
it in which there exist uncountably many paths.
If, as in WM's world, one cannot have a set of all natural numbers, then
one cannot have infinite binary trees either, and the whole issue
becomes irrelevant.
> (In principle you are right, however. Infinity is never
> finished.)
(Then there are no such things as infinite binary trees, nor paths in
them, to argue about)
WM cannot have it both ways. Either the sort of trees he is arguing
about either do not exist at all or they have properties that WM denies.
>
> Let us assume a bijection between N and 2N.
> 1 - 2
> 2 - 4
> 3 - 6
> ...
>
> What is the logical flaw here?
The first flaw is in not defining 2N. If one grants a set of all
(positive) natural numbers, then one also grants immediately a set of
all even natural numbers and a bijection from one to the other.
> >
> > Well very easy, there is a jumping to conclusions from some
> > ordering on the elements to the cardinality of the set
> > of these elements. But cardinality does not work this way,
> > it is careless about orderings on the elements. (Thats why
> > it is called cardinality and not ordinal)
>
> Correct. Your explanation covers all those cases, in particular |N| = |
> Q| and the like. How should it be "proved" without ordering?
By showing injections in both directions.
>
> But if the actually infinite tree would exist, then we could conquer
> it by defining when which part is conquered and showing that there is
> no part remaining unconquered forever.
That sort of tree exists if and only if a set of all and only natural
numbers also exists. Thus in WM's world it does not exist at all, but in
ZF's world it does.
>
> Regards, WM
--
Virgil
Virgil
<0f84a5f4-a41f-4317...@y17g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 16 Jul., 21:21, Jan Burse <janbu...@fastmail.fm> wrote:
> > WM schrieb:
> >
> > > How should it be "proved" without ordering?
> >
> > Why do you think a function f:N -> N\0, f(x)=x+1
> > involves ordering.
>
> Have you ever seen the proof of equnumerousity of N and Q?
What do you mean by "THE" proof? There are at least two versions of
proof.
> However, if for any element of a set we can say when it is used, then
> the whole set is use (because there is nothing remaining).
"When" is time dependent, which vitiates proofs. Do you mean "if for any
element of a set we can say that it is used" ?
> >
> > Yes, N and N\0 can be ordered, and somehow I
> > invoked ordering when I proved surjectivity.
> >
> > But I could also assume that the left lane
> > came into being by the following process:
> >
> > � � �L = u L2x u u L2x+1
> > � � (first the even ones, and then the odd ones)
> >
> > So the order would be:
> >
> > � � �1, 3, 5, ... 2, 4, 6, ...
> >
> > Still L and R have same cardinality.
>
> Of course. And still there is order.
> >
> > �> But if the actually infinite tree would exist
> >
> > But the question is not how many nodes are
> > in the tree, but how many infinite paths
> > are in the tree.
>
> The question is: How many infinite paths must be conquered so that
> nobody is able to find a path that is not conquered. (Of course I do
> not publish which subset of paths I have used, in order to show set
> theorists that their claim is nonsense.)
Only a German would find it useful to "conquer" paths. The rest of us
merely want to include them in the appropriate tree.
Besides, whatever provably countable set of paths WM has "conquered" but
refused to publish, Cantor's diagonal argument shows that the very proof
of its countability also proves incompleteness.
> >
> > A node does not identify an infinite path.
> > It only identifies a finite path from to
> > root to the node itself.
>
> I construct the complete binary tree by means of a countable set of
> nodes. Tell me which nodes have not been constructed.
Whoever does WM claim has been saying that any nodes have not been
"constructed"? It is that not all sets of nodes cannot be covered by any
countable set of sets of nodes. Do try to get it right, WM.
> >
> > So it is kind of fractal. When ever you
> > have conquered down to the level n,
> > with your 2^n leaf nodes, then each
> > leaf node points again to an infinite tree.
> >
> > So we have the following equation:
> >
> > � � �c = c^(2^n)+2^n
> >
> > This equation can be satisfied by a countable
> > infinity, or an uncountable infinity. So
> > your construction does not exclude that
> > the infinite tree exists. It is just
> > undecided about it.
>
> No. My argument shows that a countable set of paths fills the complete
> infinite binary tree such that nobody can tell what further paths are
> missing (none is).
Your argument does no such thing, WM. Your arguments shows that every
node can be covered by a countable set of sets of nodes, but do NOT show
that every set of nodes can be covered by a countable set of sets of
nodes.
> That proves that the whole blathering about actual
> infinity is nonsense. There is not even a single actually infinite
> path.
Then there is no tree at all, and all of WM's arguments about paths in
such nonexistent trees are merely hot air.
Unless one can have actually infinite sets, such as the set of all
naturals, none of WM's arguments about maximal infinite binary trees are
arguments about anything.
> >
> > Further arguments have to be persued
> > whether the infinite tree exists. And
> > one further argument might be diagonalization,
> > which then excludes the countable case.
>
> By definition the tree contains every binary sequence. So there is no
> diagonalization possible.
By definition, existence of any actual maximal infinite binary tree
depends on the existence of at least one actually infinite set.
Until WM cedes the existence of at least one actually infinite set, he
denies the very existence of the trees whose properties he is eternally
arguing about.
And once WM cedes that existence, he loses, since one can then prove
that such trees have the properties he denies but not prove the
properties he claims.
>
> Regards, WM
--
Virgil
Virgil
<34f1627c-ecca-4500...@c29g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
No paths are missing from the tree, but most paths are missing from the
set of those paths used in the construction.
In fact, the very proof of the countability of any set of paths sets up
the proof of the incompleteness of that set of paths.
--
Virgil
Jan Burse
> By definition the tree contains every binary sequence. So there is no
> diagonalization possible.
Actually the infinite tree can indeed be conquered, by the
countable. This is mathematically true. But still there are
uncountable infinite paths.
See: Achill and the turtle can also conquere the infinite
lane, but they cannot finish it. And thats the same problem
with your infinite tree. You conquere it, but you do not
finish it.
Assume there are stones layed out in the lane. These stones
are black and white. Now Achill is asked whether all the
stones on his lane are white. He will not be able to answer
this question, because although he can conquere he cannot
finish.
Bye
WM
> WM schrieb:
>
> > By definition the tree contains every binary sequence. So there is no
> > diagonalization possible.
>
> Actually the infinite tree can indeed be conquered, by the
> countable. This is mathematically true.
Congratulations, you have mastered the first step.
> But still there are
> uncountable infinite paths.
That cannot be.nYou will se this when constructing the the tree from a
list by simply dropping some bits along these lines:
0.111000...
0.1111000...
will be unioned to
000...
0.111
1000...
Got it? Then you see that no addtional paths have a chance to creap
into the tree. They simply do not exist. But if you deny sober
thinking, then you can also believe that additional lines creap into
Cantor's list.
Regards, WM
LudovicoVan
> WM schrieb:
>
> > By definition the tree contains every binary sequence. So there is no
> > diagonalization possible.
>
> Actually the infinite tree can indeed be conquered, by the
> countable. This is mathematically true. But still there are
> uncountable infinite paths.
My 2c: Since a bijection can be given between paths and leaf nodes
over the transfinite tree, how can there be uncountable paths if the
nodes are countable? I.e., I believe the fact you mention is properly
a contradiction: either the set of nodes is itself uncountable, or the
diagonal argument is nonsense: you cannot have a consistent theory
where both are true.
-LV
LudovicoVan
> There are uncountably many [paths]. We prove it. You don't prove otherwise.
My 2c: Since a bijection can be given between paths and leaf nodes
over the transfinite tree, how can there be uncountable paths if the
nodes are countable? I.e., I believe the fact you mention is properly
a contradiction: either the set of nodes is itself uncountable, or the
diagonal argument is nonsense: you cannot have a consistent theory
where both are true.
BTW, that's informal, yet it's a proof.
-LV
Aatu Koskensilta
> My 2c: Since a bijection can be given between paths and leaf nodes
> over the transfinite tree, how can there be uncountable paths if the
> nodes are countable?
But surely you know by now how the next round will go? Someone will
ask you, "What leaf nodes?" (adding perhaps a helpful "Idiot!"). And
so it goes, unfolding with a tedious inevitability and utter
predictability. Such is the predictability (and inevitability) of the
ensuing "debate" there's no need for anyone to actually go through the
motions.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
LudovicoVan
> LudovicoVan <ju...@diegidio.name> writes:
> > My 2c: Since a bijection can be given between paths and leaf nodes
> > over the transfinite tree, how can there be uncountable paths if the
> > nodes are countable?
>
> But surely you know by now how the next round will go? Someone will
> ask you, "What leaf nodes?" (adding perhaps a helpful "Idiot!"). And
> so it goes, unfolding with a tedious inevitability and utter
> predictability. Such is the predictability (and inevitability) of the
> ensuing "debate" there's no need for anyone to actually go through the
> motions.
I can appreciate your sentiment: but I am a hopeless optimist, and
even quite insensible to name calling. So I'll keep trying, until that
memorable day when someone will give a proper mathematical answer/
explanation to the above.
-LV
WM
> In article
> <026c8c8f-ffe8-4fbe-9c16-4a010d683...@o6g2000yqj.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > True mathematics has never been connected to
> > astrology or set theory.
>
> A lot of true mathematics is quite compatible with many forms of set
> theory,
Stop blustering.
Set theory shows there are uncountably many paths in the complete
infinite binary tree. I have shown (and every child not yet deformed
by matheology can see it): It is impossible that more different paths
can be there than points of diversification, because below every
point (i.e. below each node) only one more path can be distinguished
than before.
And don't come up with "uncountable many" run through that node.
Exactly one more path can be distinguished below the node than above
the node. If there are more than this additional one, then they must
become distinct at later nodes, but that shall not bother us here.
It is really not hard to see. Only Fools Of Matheology will suffer
from blindness.
Regards, WM
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 16 Jul., 21:45, Jan Burse <janbu...@fastmail.fm> wrote:
> > > WM schrieb:
>
> > > > What is the logical flaw here?
>
> > > For the tree: You get cheated by your drawing.
> > > Don't confuse drawing with logical reasoning.
>
> > > I think there are similar examples in geometry.
> > > Take the x-axis, a unit circle, and then a line
> > > passing quadrant I, II and III, such that it
> > > crosses the circle two times.
>
> > > One intersection point (1) with the circle is
> > > above the x-axis, and the other intersection
> > > point (2) is below the x-axis.
>
> > > We can draw the following analogy:
>
> > Don't draw analogies. Tell me which paths are missing in the tree that
> > has been constructed by means of a countable number of paths.
>
> No paths are missing from the tree, but most paths are missing from the
> set of those paths used in the construction.
There is no chance for paths to creap into the tree during
construction. But if so, then deconstruct the tree and get a list of
all paths.
No, the tree contains all paths that have been used to construct it
--- and no single one more. Otherwise math becomes magic --- and we
cannot be sure whether paths are lurking in Cantor's list too.
Regards, WM
WM
> If, as in WM's world, one cannot have a set of all natural numbers, then
> one cannot have infinite binary trees either, and the whole issue
> becomes irrelevant.
But one can assume that an actual infinity of nodes exists. And then
one can show that this assumption is was wrong.
>
> > (In principle you are right, however. Infinity is never
> > finished.)
>
> (Then there are no such things as infinite binary trees, nor paths in
> them, to argue about)
Correct, but most easily to see when their exiatence is assumed first.
>
> WM cannot have it both ways. Either the sort of trees he is arguing
> about either do not exist at all or they have properties that WM denies.
>
>
>
> > Let us assume a bijection between N and 2N.
> > 1 - 2
> > 2 - 4
> > 3 - 6
> > ...
>
> > What is the logical flaw here?
>
> The first flaw is in not defining 2N. If one grants a set of all
> (positive) natural numbers, then one also grants immediately a set of
> all even natural numbers and a bijection from one to the other.
I wanted to show only that ordering is necessary for proving
equinumerousity.
>
>
>
> > > Well very easy, there is a jumping to conclusions from some
> > > ordering on the elements to the cardinality of the set
> > > of these elements. But cardinality does not work this way,
> > > it is careless about orderings on the elements. (Thats why
> > > it is called cardinality and not ordinal)
>
> > Correct. Your explanation covers all those cases, in particular |N| = |
> > Q| and the like. How should it be "proved" without ordering?
>
> By showing injections in both directions.
For instance every node of the tree can be bijected with one of the
paths whcih I used for construction. But wait, if during construction
(of the bijection) additional element are allowed to appear, how then
does a bijection prove anything?
Regards, WM
Daryl McCullough
>
>LudovicoVan <ju...@diegidio.name> writes:
>
>> My 2c: Since a bijection can be given between paths and leaf nodes
>> over the transfinite tree, how can there be uncountable paths if the
>> nodes are countable?
>
>But surely you know by now how the next round will go? Someone will
>ask you, "What leaf nodes?" (adding perhaps a helpful "Idiot!"). And
>so it goes, unfolding with a tedious inevitability and utter
>predictability. Such is the predictability (and inevitability) of the
>ensuing "debate" there's no need for anyone to actually go through the
>motions.
Sure there is. It's tradition. It's culture. People keep performing
"Romeo and Juliet" and "Hamlet" even though everyone knows how they
will end.
--
Daryl McCullough
Ithaca, NY
Jan Burse
> will be unioned to
>
> 000...
> 0.111
> 1000...
>
> Got it? Then you see that no addtional paths have a chance to creap
> into the tree. They simply do not exist. But if you deny sober
> thinking, then you can also believe that additional lines creap into
> Cantor's list.
You will never be able to finish the path for
pi in binary representation for example. You
can conquere it, that everybody knows. But you
cannot finish it.
Thats why pi can be computed to 1000 digits, or
even a million digits, what ever. We can
arbitrarily approximate it. But how many irrational
paths like pi are there?
Bye
WM
And they keep researching set theory even though everyone knows how
that will end.
Regards, WM
WM
> WM schrieb:
>
> > will be unioned to
>
> > 000...
> > 0.111
> > 1000...
>
> > Got it? Then you see that no addtional paths have a chance to creap
> > into the tree. They simply do not exist. But if you deny sober
> > thinking, then you can also believe that additional lines creap into
> > Cantor's list.
>
> You will never be able to finish the path for
> pi in binary representation for example. You
> can conquere it, that everybody knows.
I choose for concquering the tree all finite paths ending at the
nodes. Then I extend these paths by tails to taste. These tails can
consist of the binary expansion of pi. Then pi is in the tree.
> But you cannot finish it.
I need not finish it. I take it from the shelf.
>
> Thats why pi can be computed to 1000 digits, or
> even a million digits, what ever. We can
> arbitrarily approximate it. But how many irrational
> paths like pi are there?
According to your exposition, no path like pi is there.
According to the inability of set theorists to find missing paths in
my tree, there are no paths like pi at all. The paths are the same as
the emperors new cloths. All fit into an empty set. (That's mainly why
set theorists need it.)
Regards, WM
MoeBlee
> On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > There are uncountably many [paths]. We prove it. You don't prove otherwise.
>
> My 2c: Since a bijection can be given between paths and leaf nodes
> over the transfinite tree,
There ARE NO leaf nodes.
> how can there be uncountable paths if the
> nodes are countable?
There are countably many nodes but uncountably many paths. (Though
each path itself is a denumerable sequence).
>I.e., I believe the fact you mention is properly
> a contradiction: either the set of nodes is itself uncountable, or the
> diagonal argument is nonsense: you cannot have a consistent theory
> where both are true.
>
> BTW, that's informal, yet it's a proof.
Then all you have to do is show a derivation of a sentence P and ~P to
show that the theory is inconsistent. Let me know when you've done
that.
MoeBlee
MoeBlee
wrote:
> It's tradition. It's culture. People keep performing
> "Romeo and Juliet" and "Hamlet" even though everyone knows how they
> will end.
This is more like 'Punch and Judy'.
MoeBlee
Herbert Newman
> On Jul 17, 4:04�am, LudovicoVan <ju...@diegidio.name> wrote:
>>
>> bla bla bla
Moe, you know that LV is (high level) troll, right?
Herb
Herbert Newman
> On Jul 17, 4:04�am, LudovicoVan <ju...@diegidio.name> wrote:
>
MoeBlee
> (high level) troll
What do you mean by 'high level' here?
MoeBlee
Herbert Newman
That he's more sophisticated than the usual ones (i.e. deceiving). :-)
Herb
Jan Burse
> I choose for concquering the tree all finite paths ending at the
> nodes. Then I extend these paths by tails to taste. These tails can
> consist of the binary expansion of pi. Then pi is in the tree.
Yes, pi is in the infinite tree, even in your tree, since your
tree is my tree. But it is not in your paths from the root to
a node, and also not in my paths from the root to a node, since
this paths are the same for me and you.
It will never be, since pi is irrational.
> The paths are the same as the emperors new cloths. All fit
> into an empty set.
Then pi does not exist.
Bye
WM
> WM schrieb:
>
> > I choose for concquering the tree all finite paths ending at the
> > nodes. Then I extend these paths by tails to taste. These tails can
> > consist of the binary expansion of pi. Then pi is in the tree.
>
> Yes, pi is in the infinite tree, even in your tree, since your
> tree is my tree. But it is not in your paths from the root to
> a node, and also not in my paths from the root to a node, since
> this paths are the same for me and you.
>
> It will never be, since pi is irrational.
If we assume its complete existence, then we must believe that we can
use it.
>
> > The paths are the same as the emperors new cloths. All fit
> > into an empty set.
>
> Then pi does not exist.
That topic is a little bit more sophisticated. (I have to confess
that, some years ago, I was of your opinion.) But that seems not to be
the right track. I think that something exists, if we can talk about
and understand one and the same by it.
There are many finite definitions (remember the many, many sequences
and series and formulae) involving what Jones once baptized pi. So pi
exists. But it does not exist as a complete binary sequence. By use of
some definition, we can calculate (in principle, neglecting all
physical constraints) every bit of pi, but always there will be
infinitely many bits remaining uncalculated. So we have to
distinguish: There exists a number called pi, but this number,
contrary to the creed of set theorists, does not have a complete
sequence of bits (or digits).
Therefore nobody can find it or not find it in the tree. Set theorists
in general seem to be not intelligent enough (or may have some other
handicap that hinders them) to recognize, that "forall n we can find
pi's digit(n)" does not imply "we can find all of pi's digits".
Regards, WM
Jan Burse
> Therefore nobody can find it or not find it in the tree. Set theorists
> in general seem to be not intelligent enough (or may have some other
> handicap that hinders them) to recognize, that "forall n we can find
> pi's digit(n)" does not imply "we can find all of pi's digits".
What is this handicap exactly? Are there digits of pi,
that do not have an index n?
I am very curious.
Bye
LudovicoVan
> On Jul 17, 4:04 am, LudovicoVan <ju...@diegidio.name> wrote:
>
> > On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > There are uncountably many [paths]. We prove it. You don't prove otherwise.
>
> > My 2c: Since a bijection can be given between paths and leaf nodes
> > over the transfinite tree,
>
> There ARE NO leaf nodes.
Yes, of course there are.
> > how can there be uncountable paths if the
> > nodes are countable?
>
> There are countably many nodes but uncountably many paths. (Though
> each path itself is a denumerable sequence).
A splendid example of your "explanations".
> >I.e., I believe the fact you mention is properly
> > a contradiction: either the set of nodes is itself uncountable, or the
> > diagonal argument is nonsense: you cannot have a consistent theory
> > where both are true.
>
> > BTW, that's informal, yet it's a proof.
>
> Then all you have to do is show a derivation of a sentence P and ~P to
> show that the theory is inconsistent. Let me know when you've done
> that.
Let me know when you get anything else.
-LV
WM
Apparently not. But what about all indexes?
>
> I am very curious.
Then give the complete sequence of pi, or better: try it. Try to write
the digit sequence of pi or of 1/2 such that not infinitely many
digits remain unwritten. You will see the difference between
"proving" (in the matheological sense) that something exists and
demonstrating it.
Regards, WM
LudovicoVan
> > On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > There are uncountably many [paths]. We prove it. You don't prove otherwise.
>
> > My 2c: Since a bijection can be given between paths and leaf nodes
> > over the transfinite tree,
>
> There ARE NO leaf nodes.
Thanks for the question!
Below I have tried some transfinite-inductive definitions, leveraging
the infinite cartesian product to define the infinite case. For
illustration, there is also an implementation in Prolog that works for
the finite cases, with some sample output.
The predicate nodes(n) defines the set (actually, the list) of nodes
at level (depth) n in the binary tree, i.e. those nodes with n-1
parents up to (but excluding) the root. In particular, each node
encodes the whole path from the root to it, i.e. paths and nodes are
equivalent. For instance, an element of nodes(3) is [1, [0, [1, []]]],
corresponding to the path R-1-0-1.
The predicate tree(n) defines the set (actually, the list) of nodes in
a tree of level (depth) n. In particular, we will call *leaf nodes*
that subset of tree(n) that is the set nodes(n). And the set paths(n)
in tree(n) (i.e. the set of paths of length n in the tree of depth n)
is equivalent to the set nodes(n), i.e. the set of leaf nodes in tree
(n).
Extending the definitions to the transfinite, we have in particular
that the set of infinite paths in tree(w) corresponds to the set nodes
(w): i.e., again, the set of leaf nodes.
Let mathematics triumph!!
-LV
/*
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% http://en.wikipedia.org/wiki/Cartesian_product#Infinite_products
%%% list of nodes at lev. n:
%%% nodes(n) = [0, 1]^n
nodes(0) = [[]]
nodes(n) = [0, 1] x nodes(n-1) = % cartesian product
= Prod_{i=0}^n [0, 1] =
= [0, 1]^n % cartesian exponentiation
nodes(w) = [0, 1]^w
% nodes(0) = [[]]
% nodes(1) = [[0, []], [1, []]]
% nodes(2) = [[0, [0, []]], [0, [1, []]], [1, [0, []]], [1, [1, []]]]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% list of paths in a tree of lev. n:
%%% paths(n) = nodes(n) = [0, 1]^n
paths(n) = nodes(n)
paths(w) = nodes(w)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% list of nodes in a tree of lev. n:
%%% tree(n) = [0, 1]"n
tree(0) = nodes(0)
tree(n) = tree(n-1).nodes(n) = % list-append
= tree(n-1).([0, 1]^n) =
= ([0, 1]^0). ... .([0, 1]^n) =
= Append_{i=0}^n [0, 1]^i =
= [0, 1]"n % list-append exp. (ad-hoc)
tree(w) = [0, 1]"w
% tree(0) = [[]]
% tree(1) = [[], [0, []], [1, []]]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
*/
%%% Implementation in the finite-case (SWI-Prolog):
nodes(Nodes, Lev) :-
findall(Node, node(Node, Lev), Nodes).
paths(Paths, Lev) :-
findall(Path, path(Path, Lev), Paths).
tree(Tree, Lev) :-
findall(Node, tree_node(Node, Lev), Tree).
node([], 0).
node([0, Node0], Lev) :-
Lev > 0,
Lev0 is Lev - 1,
node(Node0, Lev0).
node([1, Node0], Lev) :-
Lev > 0,
Lev0 is Lev - 1,
node(Node0, Lev0).
path(Path, Lev) :-
node(Path, Lev).
tree_node(Node, 0) :-
node(Node, 0).
tree_node(Node, Lev) :-
Lev > 0,
( Lev0 is Lev - 1,
tree_node(Node, Lev0)
; node(Node, Lev)
).
/*
%%% Sample output:
221 ?- nodes(Nodes, 3).
Nodes = [[0, [0, [0, []]]], [0, [0, [1, []]]], [0, [1, [0, []]]], [0,
[1, [1, []]]], [1, [0, [0, []]]], [1, [0, [1, []]]], [1, [1, [0,
[]]]], [1, [1, [1, []]]]].
222 ?- paths(Paths, 3).
Paths = [[0, [0, [0, []]]], [0, [0, [1, []]]], [0, [1, [0, []]]], [0,
[1, [1, []]]], [1, [0, [0, []]]], [1, [0, [1, []]]], [1, [1, [0,
[]]]], [1, [1, [1, []]]]].
223 ?- tree(Tree, 3).
Tree = [[], [0, []], [1, []], [0, [0, []]], [0, [1, []]], [1, [0,
[]]], [1, [1, []]], [0, [0, [0, []]]], [0, [0, [1, []]]], [0, [1, [0,
[]]]], [0, [1, [1, []]]], [1, [0, [0, []]]], [1, [0, [1, []]]], [1,
[1, [0, []]]], [1, [1, [1, []]]]].
224 ?- node(Node, 3).
Node = [0, [0, [0, []]]] ;
Node = [0, [0, [1, []]]] ;
Node = [0, [1, [0, []]]] ;
Node = [0, [1, [1, []]]] ;
Node = [1, [0, [0, []]]] ;
Node = [1, [0, [1, []]]] ;
Node = [1, [1, [0, []]]] ;
Node = [1, [1, [1, []]]] ;
false.
225 ?- path(Path, 3).
Path = [0, [0, [0, []]]] ;
Path = [0, [0, [1, []]]] ;
Path = [0, [1, [0, []]]] ;
Path = [0, [1, [1, []]]] ;
Path = [1, [0, [0, []]]] ;
Path = [1, [0, [1, []]]] ;
Path = [1, [1, [0, []]]] ;
Path = [1, [1, [1, []]]] ;
false.
226 ?- tree_node(Node, 3).
Node = [] ;
Node = [0, []] ;
Node = [1, []] ;
Node = [0, [0, []]] ;
Node = [0, [1, []]] ;
Node = [1, [0, []]] ;
Node = [1, [1, []]] ;
Node = [0, [0, [0, []]]] ;
Node = [0, [0, [1, []]]] ;
Node = [0, [1, [0, []]]] ;
Node = [0, [1, [1, []]]] ;
Node = [1, [0, [0, []]]] ;
Node = [1, [0, [1, []]]] ;
Node = [1, [1, [0, []]]] ;
Node = [1, [1, [1, []]]] ;
false.
*/
MoeBlee
> > There ARE NO leaf nodes.
Then we're not talking about the same thing.
> > There are countably many nodes but uncountably many paths. (Though
> > each path itself is a denumerable sequence).
>
> A splendid example of your "explanations".
I didn't claim it is an explanation.
> > Then all you have to do is show a derivation of a sentence P and ~P to
> > show that the theory is inconsistent. Let me know when you've done
> > that.
>
> Let me know when you get anything else.
What else should I get? When you've proven some P and ~P from the
axioms of set theory, using first order logic, I'll buy you a steak
dinner.
MoeBlee
Jan Burse
> Then give the complete sequence of pi, or better: try it. Try to write
> the digit sequence of pi or of 1/2 such that not infinitely many
> digits remain unwritten. You will see the difference between
> "proving" (in the matheological sense) that something exists and
> demonstrating it.
>
> Regards, WM
No, I cannot display all the concrete decimal digits of pi. Why should
I be able to do that?
Why do you think proving should only be restricted to displayable
objects. What would be the advantage?
Bye
Jan Burse
> Extending the definitions to the transfinite, we have in particular
> that the set of infinite paths in tree(w) corresponds to the set nodes
> (w): i.e., again, the set of leaf nodes.
You only get all the prefixes of the infinite paths. But you
do not get the infinite paths itself.
Here is an infinite path:
x1 x2 x3 x4 ...
Here is a prefix:
x1 x2 x3
In general, an infinite path is:
f: N -> 2
A prefix of f is:
g: {0..n-1} -> 2, such that
f(i)=g(i) for all i<n
Bye
Jan Burse
There are countable many prefixes.
But uncountable many infinite paths.
Jan Burse
The prefixes can be viewed as finite words over 2.
The words of length n are denoted by:
2^n, where 2={0,1}
All finite words over 2 are denoted as follows:
2* = u_n<w 2^n
The infinite paths can be viewed as infinite words over 2.
The infinite words over 2 are denoted by:
2^w
2* is countable.
2^w is uncountable.
MoeBlee
> Below I have tried some transfinite-inductive definitions, leveraging
> the infinite cartesian product to define the infinite case. For
> illustration, there is also an implementation in Prolog that works for
> the finite cases, with some sample output.
>
> The predicate nodes(n) defines the set (actually, the list) of nodes
> at level (depth) n in the binary tree, i.e. those nodes with n-1
> parents up to (but excluding) the root. In particular, each node
> encodes the whole path from the root to it, i.e. paths and nodes are
> equivalent. For instance, an element of nodes(3) is [1, [0, [1, []]]],
> corresponding to the path R-1-0-1.
>
> The predicate tree(n) defines the set (actually, the list) of nodes in
> a tree of level (depth) n. In particular, we will call *leaf nodes*
> that subset of tree(n) that is the set nodes(n).
If I understand your specifications, your leaf nodes are not leaf
nodes of the complete (infinite) binary tree but rather of certain
subsets of the complete (infinite) binary tree and resultingly leaf
nodes, under your definition, turn out just to be nodes of the
complete (infinite) binary tree. Of course, whether a node is a leaf
node depends on what tree we're talking about. In this context, the
tree under discussion has been the complete (infinite) binary tree,
which has no leaf nodes, though, of course, certain subtrees have leaf
nodes (as, again, 'leaf node' is defined PER a given tree).
> And the set paths(n)
> in tree(n) (i.e. the set of paths of length n in the tree of depth n)
> is equivalent to the set nodes(n), i.e. the set of leaf nodes in tree
> (n).
If I understand your specifications, each tree(n) is a finite tree.
The matter of previous discussion had not been as to finite trees. But
perhaps that is what you are getting at with the following:
> Extending the definitions to the transfinite, we have in particular
> that the set of infinite paths in tree(w) corresponds to the set nodes
> (w): i.e., again, the set of leaf nodes.
Perhaps your conviction is based on whatever your method is of
"extending the definitions to the transfinite", which may not be
equivalent with the actual set theoretical definition of the infinite
binary tree.
If you ever wish to see a rigorous treatment and definition in set
theory, just let me know, and I'll recommend some books for you
(anyone of which is adequate for the task).
MoeBlee
LudovicoVan
> LudovicoVan schrieb:
>
> > Extending the definitions to the transfinite, we have in particular
> > that the set of infinite paths in tree(w) corresponds to the set nodes
> > (w): i.e., again, the set of leaf nodes.
>
> You only get all the prefixes of the infinite paths. But you
> do not get the infinite paths itself.
BS. Do I get the set nodes(w) or not? Otherwise, is there any error in
the transfinite definitions or what?
-LV
Jan Burse
No there is no error. Only imprecise terminology.
LudovicoVan
> If I understand your specifications, each tree(n) is a finite tree.
> The matter of previous discussion had not been as to finite trees.
Then you don't get it: I have given transfinite-inductive definitions,
they work in the infinite.
-LV
LudovicoVan
> LudovicoVan schrieb:
> > On 17 July, 20:29, Jan Burse <janbu...@fastmail.fm> wrote:
> >> LudovicoVan schrieb:
>
> >>> Extending the definitions to the transfinite, we have in particular
> >>> that the set of infinite paths in tree(w) corresponds to the set nodes
> >>> (w): i.e., again, the set of leaf nodes.
> >> You only get all the prefixes of the infinite paths. But you
> >> do not get the infinite paths itself.
>
> > BS. Do I get the set nodes(w) or not? Otherwise, is there any error in
> > the transfinite definitions or what?
>
Thanks, I am touched and enlightened.
-LV
Jan Burse
> The prefixes can be viewed as finite words over 2.
> The words of length n are denoted by:
>
> 2^n, where 2={0,1}
>
> All finite words over 2 are denoted as follows:
>
> 2* = u_n<w 2^n
>
> The infinite paths can be viewed as infinite words over 2.
> The infinite words over 2 are denoted by:
>
> 2^w
>
> 2* is countable.
> 2^w is uncountable.
But this is not the end of the world. For example
there are interesting subsets of 2^w, which we
can very easily deal with.
For example all words of the form:
a b^w
Correspond to periodic binary expansions, and thus
are the rationals in effect.
This subset is countable, because it is isomorphic
to 2* x 2*.
Check this out:
http://www.liafa.jussieu.fr/~jep/Resumes/InfiniteWords.html
MoeBlee
> Do I get the set nodes(w)
I don't know your notation, but just in case I'm in the ballpark
regarding your notation, in the tree we're talking about, there are no
nodes whose level is indexed by w (omega) itself. Rather, the only
levels are those that are indexed by a natural number, and for every
natural number there is a level indexed by that natural number.
Moeblee
LudovicoVan
The you must be stupid. First, read the post to know what the bloody
notation means. Second, the only levels you conceive are those within
the limits of your incompetence only.
-LV
MoeBlee
The complete (infinite) binary tree requires no "transfinite inductive
definition". Meanwhile, what other thing you have in mind is not clear
to me. Perhaps you would start with some commonly known set theoretic
notations and use them to give precise definitions of each of your
terms. But, meanwhile, even more fundamentally, perhaps you should
check that you understand a set theoretic definition of the particular
tree we're talking about.
MoeBlee
LudovicoVan
> > 2* is countable.
> > 2^w is uncountable.
>
> But this is not the end of the world.
Above all because it is completely irrelevant.
I'm not going to call you a liar, just thanks for the lough.
-LV
Jan Burse
Do you "all above" or "above all".
What is irrelevant? This:
> 2* is countable.
> 2^w is uncountable.
Or this?
Virgil
<a1964cdd-ac8c-4156...@c2g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 08:13, Jan Burse <janbu...@fastmail.fm> wrote:
> > WM schrieb:
> >
> > > By definition the tree contains every binary sequence. So there is no
> > > diagonalization possible.
> >
> > Actually the infinite tree can indeed be conquered, by the
> > countable. This is mathematically true.
>
>
> Congratulations, you have mastered the first step.
Not without defining "conquered".
>
> > But still there are
> > uncountable infinite paths.
>
> That cannot be.
Perhaps not in WM's world, but in any world in which one has a set of
all natural numbers, it is an inevitability.
> nYou will se this when constructing the the tree from a
> list by simply dropping some bits along these lines:
>
> 0.111000...
>
> 0.1111000...
>
> will be unioned to
>
> 000...
> 0.111
> 1000...
>
> Got it? Then you see that no addtional paths have a chance to creap
> into the tree.
They will be there as soon as the tree itself is "there".
That WM claims they are not "there" means that the tree itself is not
"there".
> They simply do not exist.
Then neither does the tree.
> But if you deny sober
> thinking,
We certainly deny WM's considerably less than sober thinking.
> but then you can also believe that additional lines creap into
> Cantor's list.
In Cantor's diagonal proof, 'the' list is never Cantor's.
--
Virgil
LudovicoVan
> What is irrelevant?
All you have said is irrelevant. Do I get the set nodes(w) or not?
I.e. you are not saying what is the problem with my definitions.
-LV
Virgil
<9d3e59d5-f3ec-47e7...@32g2000yqj.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > There are uncountably many [paths]. We prove it. You don't prove otherwise.
>
> My 2c: Since a bijection can be given between paths and leaf nodes
> over the transfinite tree, how can there be uncountable paths if the
> nodes are countable? I.e., I believe the fact you mention is properly
> a contradiction: either the set of nodes is itself uncountable, or the
> diagonal argument is nonsense: you cannot have a consistent theory
> where both are true.
>
> BTW, that's informal, yet it's a proof.
>
> -LV
In the complete infinite binary tree as standardly constructed, every
node has 2 child nodes so that there are no such things as leaf nodes.
So whatever sort of infinite tree LudovicoVan is talking about, it is
not the one everyone else is considering.
--
Virgil
Virgil
<f5054ed0-3472-4e25...@k6g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 01:15, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <026c8c8f-ffe8-4fbe-9c16-4a010d683...@o6g2000yqj.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > �True mathematics has never been connected to
> > > astrology or set theory.
> >
> > A lot of true mathematics is quite compatible with many forms of set
> > theory,
>
> Stop blustering.
> Set theory shows there are uncountably many paths in the complete
> infinite binary tree. I have shown (and every child not yet deformed
> by matheology can see it): It is impossible that more different paths
> can be there than points of diversification, because below every
> point (i.e. below each node) only one more path can be distinguished
> than before.
Wm mistakes distinguishing one path from another, which can be done by a
single node, with distinguishing one path from all others, which
requires infinitely many nodes.
>
> And don't come up with "uncountable many" run through that node.
Why not, since it is quite true that just as many pass through any
other node as pass through the root node.
> Exactly one more path can be distinguished below the node than above
> the node. If there are more than this additional one, then they must
> become distinct at later nodes, but that shall not bother us here.
It should bother us here as it takes infinitely many nodes to isolate
any one path from all others.
>
> It is really not hard to see. Only Fools Of Matheology will suffer
> from blindness.
That sort of ''blindness' sees a good better than the sort of perverted
vision WM has.
--
Virgil
Virgil
<4e07c06e-4a86-484e...@y7g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 02:07, Virgil <virg...@nowhere.com> wrote:
> > > Don't draw analogies. Tell me which paths are missing in the tree that
> > > has been constructed by means of a countable number of paths.
> >
> > No paths are missing from the tree, but most paths are missing from the
> > set of those paths used in the construction.
>
> There is no chance for paths to creap into the tree during
> construction. But if so, then deconstruct the tree and get a list of
> all paths.
When the tree is "deconstructed" it has no paths at all. Paths are
artifacts of completed trees, and without a tree to exist in are not
paths at all (they may exist as something else, but not as paths).
>
> No, the tree contains all paths that have been used to construct it
> --- and no single one more.
Those so called 'paths' do not exist as paths until the tree is
complete. Until then they are merely actually infinite binary sequences.
The very definition of pathhood is as a set of nodes in a tree.
> Otherwise math becomes magic
Anything sufficiently beyond the comprehension of a person is magic to
that person, just as actual mathematics is so often magic to WM.
> --- and we
> cannot be sure whether paths are lurking in Cantor's list too.
Cantor does not have such a list of his own, but demonstrated for all
time the incompleteness of anyone else's list.
--
Virgil
Virgil
<a8a6b8f9-c3fc-4a31...@r33g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 01:29, Virgil <virg...@nowhere.com> wrote:
>
> > If, as in WM's world, one cannot have a set of all natural numbers, then
> > one cannot have infinite binary trees either, and the whole issue
> > becomes irrelevant.
>
> But one can assume that an actual infinity of nodes exists. And then
> one can show that this assumption is was wrong.
Only when one also assumes, as WM always does, hat no such actual
infinite can exist. So WM's arguments here are always circular.
> >
> > > (In principle you are right, however. Infinity is never
> > > finished.)
> >
> > (Then there are no such things as infinite binary trees, nor paths in
> > them, to argue about)
>
> Correct, but most easily to see when their exiatence is assumed first.
When one also assumes, as WM always does, both some P and some ~P, one
can prove, and disprove, anything one wants.
> >
> > WM cannot have it both ways. Either the sort of trees he is arguing
> > about either do not exist at all or they have properties that WM denies.
> >
> >
> >
> > > Let us assume a bijection between N and 2N.
> > > 1 - 2
> > > 2 - 4
> > > 3 - 6
> > > ...
> >
> > > What is the logical flaw here?
> >
> > The first flaw is in not defining 2N. If one grants a set of all
> > (positive) natural numbers, then one also grants immediately a set of
> > all even natural numbers and a bijection from one to the other.
>
> I wanted to show only that ordering is necessary for proving
> equinumerousity.
Such a proof is well beyond WM's capabilities.
> >
> >
> >
> > > > Well very easy, there is a jumping to conclusions from some
> > > > ordering on the elements to the cardinality of the set
> > > > of these elements. But cardinality does not work this way,
> > > > it is careless about orderings on the elements. (Thats why
> > > > it is called cardinality and not ordinal)
> >
> > > Correct. Your explanation covers all those cases, in particular |N| = |
> > > Q| and the like. How should it be "proved" without ordering?
> >
> > By showing injections in both directions.
>
> For instance every node of the tree can be bijected with one of the
> paths whcih I used for construction. But wait, if during construction
> (of the bijection) additional element are allowed to appear, how then
> does a bijection prove anything?
If one applies the Cantor diagonal argument to the list of paths used in
the construction (and they can be listed because the nodes can be
listed), then one sees that there are as many paths in that tree
uncounted as counted.
At least is one does not blind oneself.
--
Virgil
LudovicoVan
> LudovicoVan <ju...@diegidio.name> wrote:
> > On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > There are uncountably many [paths]. We prove it. You don't prove otherwise.
>
> > My 2c: Since a bijection can be given between paths and leaf nodes
> > over the transfinite tree, how can there be uncountable paths if the
> > nodes are countable? I.e., I believe the fact you mention is properly
> > a contradiction: either the set of nodes is itself uncountable, or the
> > diagonal argument is nonsense: you cannot have a consistent theory
> > where both are true.
>
> > BTW, that's informal, yet it's a proof.
>
> node has 2 child nodes so that there are no such things as leaf nodes.
The same property trivially holds in my tree too. OTOH, the definition
I have given of "leaf nodes" is unambiguous, salient, and does *not*
mention child nodes at all: the set of leaf nodes of tree(x) is simply
the set nodes(x).
> So whatever sort of infinite tree LudovicoVan is talking about, it is
> not the one everyone else is considering.
BS. Otherwise give proper references.
-LV
Virgil
<75b35957-8689-4946...@y17g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 15:10, stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
> > Aatu Koskensilta says...
> >
> >
> >
> > >LudovicoVan <ju...@diegidio.name> writes:
> >
> > >> My 2c: Since a bijection can be given between paths and leaf nodes
> > >> over the transfinite tree, how can there be uncountable paths if the
> > >> nodes are countable?
> >
> > >But surely you know by now how the next round will go? Someone will
> > >ask you, "What leaf nodes?" (adding perhaps a helpful "Idiot!"). And
> > >so it goes, unfolding with a tedious inevitability and utter
> > >predictability. Such is the predictability (and inevitability) of the
> > >ensuing "debate" there's no need for anyone to actually go through the
> > >motions.
> >
> > Sure there is. It's tradition. It's culture. People keep performing
> > "Romeo and Juliet" and "Hamlet" even though everyone knows how they
> > will end.
>
> And they keep researching set theory even though everyone knows how
> that will end.
WM does not know ow it will end, since he deliberately blinds himself to
its logic.
--
Virgil
Herbert Newman
> On Jul 17, 12:40�pm, LudovicoVan <ju...@diegidio.name> wrote:
>
>> Do I get the set nodes(w)
>>
> I don't know your notation ...
Maybe you should try to learn it. I'm sure it's worthwhile!
Herb
Herbert Newman
> On Jul 17, 12:16�pm, LudovicoVan <ju...@diegidio.name> wrote:
>>
>> bla bla bla
>>
> blub blub blub
>
> MoeBlee
You really don't get it, do you? You are "discussing" with a troll.
(You know the saying: Don't wrestle with a pig, because you both get dirty
and the pig enjoys it!)
It's really funny how this guy is hooking you up.
Herb
Herbert Newman
>> ... you don't get it:
Right, Moe is dumb like a doornail! He NEVER get's it!
>> I have given transfinite-inductive definitions,
>> they work in the infinite.
Oh right, you have given "transfinite-inductive definitions": they work in
the infinite, right! (Maybe you should try to give head?)
> The complete (infinite) binary tree requires no "transfinite inductive
> definition".
Really?! Tell me more, Moe!
> Meanwhile, what other thing you have in mind is not clear to me.
You might ask! I'm sure he would be glad to explain it to you!
Herb
Virgil
<1200d071-eca5-4db0...@k6g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 17:20, Jan Burse <janbu...@fastmail.fm> wrote:
> > WM schrieb:
> >
> > > will be unioned to
> >
> > > � � � � � 000...
> > > 0.111
> > > � � � � � 1000...
> >
> > > Got it? Then you see that no addtional paths have a chance to creap
> > > into the tree. They simply do not exist. But if you deny sober
> > > thinking, then you can also believe that additional lines creap into
> > > Cantor's list.
> >
> > You will never be able to finish the path for
> > pi in binary representation for example. You
> > can conquere it, that everybody knows.
>
> I choose for concquering the tree all finite paths ending at the
> nodes. Then I extend these paths by tails to taste. These tails can
> consist of the binary expansion of pi. Then pi is in the tree.
>
> > But you cannot finish it.
>
> I need not finish it. I take it from the shelf.
Wm's shelf does not contain it.
WM's shelf does not contain any actually infinite sets at all, not even
a set of all naturals, without which no such trees can even be imagined.
>
> According to the inability of set theorists to find missing paths in
> my tree, there are no paths like pi at all.
If WM's "tree" actually satisfies the requirements of a maximal infinite
binary tree, then it has unlistably many paths, and if is does not
satisfy those requirements, it is irrelevant in any discussion of trees
which do.
> The paths are the same as
> the emperors new cloths. All fit into an empty set. (That's mainly why
> set theorists need it.)
WM is off in his fairy tale land again. it is his own delusion of trees
with which he is attempting to cloth the emperor.
--
Virgil
Virgil
<7962583b-651b-44d9...@k30g2000yqf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Set theorists
> in general seem to be not intelligent enough (or may have some other
> handicap that hinders them) to recognize, that "forall n we can find
> pi's digit(n)" does not imply "we can find all of pi's digits".
That is the very fallacy on which so many of WM's own false claims are
based, that what is true for each n individually must necessarily be
true for the set of all n.
--
Virgil
Virgil
<0304187a-5eb8-4bf2...@18g2000yqa.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 18:07, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 17, 4:04�am, LudovicoVan <ju...@diegidio.name> wrote:
> >
> > > On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
> >
> > > > There are uncountably many [paths]. We prove it. You don't prove
> > > > otherwise.
> >
> > > My 2c: Since a bijection can be given between paths and leaf nodes
> > > over the transfinite tree,
> >
> > There ARE NO leaf nodes.
>
> Yes, of course there are.
Name one!
>
> > > how can there be uncountable paths if the
> > > nodes are countable?
> >
> > There are countably many nodes but uncountably many paths. (Though
> > each path itself is a denumerable sequence).
>
> A splendid example of your "explanations".
It is a good deal better than yours.
>
> > >I.e., I believe the fact you mention is properly
> > > a contradiction: either the set of nodes is itself uncountable, or the
> > > diagonal argument is nonsense: you cannot have a consistent theory
> > > where both are true.
> >
> > > BTW, that's informal, yet it's a proof.
> >
> > Then all you have to do is show a derivation of a sentence P and ~P to
> > show that the theory is inconsistent. Let me know when you've done
> > that.
>
> Let me know when you get anything else.
People have been trying to deduce "P and ~P" in ZF for years without
ever succeeding to the satisfaction of anyone but themselves.
If LudovicoVan has actually succeeded, he could make a huge name for
himself by publishing his derivation.
Since he has not done so, one suspects he is merely blowing hot air.
--
Virgil
Virgil
<29d43009-6502-457a...@v20g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Jul., 21:02, Jan Burse <janbu...@fastmail.fm> wrote:
> > WM schrieb:
> >
> > > Therefore nobody can find it or not find it in the tree. Set theorists
> > > in general seem to be not intelligent enough (or may have some other
> > > handicap that hinders them) to recognize, that "forall n we can find
> > > pi's digit(n)" does not imply "we can find all of pi's digits".
> >
> > What is this handicap exactly? Are there digits of pi,
> > that do not have an index n?
>
> Apparently not. But what about all indexes?
> >
> > I am very curious.
>
> Then give the complete sequence of pi, or better: try it. Try to write
> the digit sequence of pi or of 1/2 such that not infinitely many
> digits remain unwritten. You will see the difference between
> "proving" (in the matheological sense) that something exists and
> demonstrating it.
A number can be perfectly well defined and known to exist without its
having a know binary or decimal expansion.
The currrent form of decimal notation is, after all, a fairly recent
development in the total history of mathematics.
--
Virgil
Ross A. Finlayson
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
< ...
> > --- and we
> > cannot be sure whether paths are lurking in Cantor's list too.
>
> Cantor does not have such a list of his own, but demonstrated for all
> time the incompleteness of anyone else's list.
>
> --
> Virgil
Not a list like the natural/unit equivalency function, as so modeled
by real functions, in terms of nested intervals or the antidiagonal
argument, that is not so. (For standard theorists Goedel showed your
theory is incomplete so there are true statements about the objects of
interest that aren't theorems of standard i.e. well-founded theories,
if consistent. Quite modernly Paris and Kirby show nonstandard
countable integers as a result.) In fact a reasonable person can see
in a reinterpretation of Cantor's results, in an appropriate (and
utilitarian) numerical framework, that actually the numerical
continuum of the unit interval's real numbers increase monotonically.
That wouldn't be a novel notion to those such as, for example, Newton
or Leibniz, whose analytical methods see widespread application. (EF
is a CDF.) So, the canon over time offers reasonable alternatives,
including those which see daily application, contrasted to transfinite
cardinals, which don't.
To reintroduce another notion that doesn't meet intuition, about the
rationals and irrationals _each_ being dense in the reals, a countable
array of nodes is dense in these paths of the infinite binary tree.
If there are uncountably many ways to make lists then consider
Cantor's methods and when they don't stop.
Ross
Jan Burse
There is no problem with your definition.
Nodes are nodes, and infinite paths are infinite paths.
So what?
Virgil
<e70aec30-c798-4a65...@h31g2000yqd.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 18:07, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 17, 4:04�am, LudovicoVan <ju...@diegidio.name> wrote:
> > > On 16 July, 22:55, MoeBlee <jazzm...@hotmail.com> wrote:
> >
> > > > There are uncountably many [paths]. We prove it. You don't prove
> > > > otherwise.
> >
> > > My 2c: Since a bijection can be given between paths and leaf nodes
> > > over the transfinite tree,
> >
> > There ARE NO leaf nodes.
>
> Thanks for the question!
>
> Below I have tried some transfinite-inductive definitions, leveraging
> the infinite cartesian product to define the infinite case. For
> illustration, there is also an implementation in Prolog that works for
> the finite cases, with some sample output.
>
> The predicate nodes(n) defines the set (actually, the list) of nodes
> at level (depth) n in the binary tree, i.e. those nodes with n-1
> parents up to (but excluding) the root. In particular, each node
> encodes the whole path from the root to it, i.e. paths and nodes are
> equivalent. For instance, an element of nodes(3) is [1, [0, [1, []]]],
> corresponding to the path R-1-0-1.
>
> The predicate tree(n) defines the set (actually, the list) of nodes in
> a tree of level (depth) n. In particular, we will call *leaf nodes*
> that subset of tree(n) that is the set nodes(n). And the set paths(n)
> in tree(n) (i.e. the set of paths of length n in the tree of depth n)
> is equivalent to the set nodes(n), i.e. the set of leaf nodes in tree
> (n).
>
> Extending the definitions to the transfinite, we have in particular
> that the set of infinite paths in tree(w) corresponds to the set nodes
> (w): i.e., again, the set of leaf nodes.
>
> Let mathematics triumph!!
According to your definition, there is no difference between nodes and
paths, but in the definitions everyone else is using, there is an
essential difference: a node is not a set of nodes, but a path is a set
of nodes of a certain type.
Thus what may be the case in your trees need not be the case in ours and
vice versa.
And according to our definitions, e.g., see below, the paths in a
maximal complete infinite binary tree cannot have leaf nodes.
Model for a maximal infinite complete binary tree:
The set of nodes is the (1-origin) set of naturals, N = {1,2,3,...} with
the usual arithmetic.
1 is the root node.
For each n in N, 2*n_+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
1 is in P and
for each n in P one and only one of 2*n+0 and 2*n+1 is in P.
In THIS model, paths do not have leaf nodes. Further, no node is a path
nor any path a node.
--
Virgil
Jan Burse
> The same property trivially holds in my tree too. OTOH, the definition
> I have given of "leaf nodes" is unambiguous, salient, and does *not*
> mention child nodes at all: the set of leaf nodes of tree(x) is simply
> the set nodes(x).
Did you ever look at a real tree? Usually
one calls a leaf node, which is where the
tree has its leafs, namely there where the
branching stops.
Bye
LudovicoVan
> LudovicoVan schrieb:
>
> > On 17 July, 20:59, Jan Burse <janbu...@fastmail.fm> wrote:
>
> >> What is irrelevant?
>
> > All you have said is irrelevant. Do I get the set nodes(w) or not?
> > I.e. you are not saying what is the problem with my definitions.
>
> Nodes are nodes, and infinite paths are infinite paths.
> So what?
Just come back from holidays? So, given that the set paths(w) isn't
but the set nodes(w), either the set of paths is countable, or the set
of nodes is uncountable. QED. Corollary: since the set of nodes is
clearly countable, the diagonal argument is invalid.
-LV
MoeBlee
> the set of leaf nodes of tree(x) is simply
> the set nodes(x).
And that is not what mathematicians ordinarily understand by the term
'leaf node'.
You're free to make whatever definitions you like, but you should
understand that your definition is not what people ordinarily mean.
> > So whatever sort of infinite tree LudovicoVan is talking about, it is
> > not the one everyone else is considering.
>
> BS. Otherwise give proper references.
Moschovakis 'Notes On Set Theory'; Monk 'Introduction To Set Theory';
Levy 'Basic Set Theory' (as I recall); others available.
MoeBlee
LudovicoVan
> LudovicoVan schrieb:
>
> > The same property trivially holds in my tree too. OTOH, the definition
> > I have given of "leaf nodes" is unambiguous, salient, and does *not*
> > mention child nodes at all: the set of leaf nodes of tree(x) is simply
> > the set nodes(x).
>
> Did you ever look at a real tree?
Did you? What is it that so much offends your holy intuition? That the
diagonal argument is invalid, maybe??
-LV
Virgil
<ce104fc7-35de-425a...@d4g2000yqa.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 20:29, Jan Burse <janbu...@fastmail.fm> wrote:
> > LudovicoVan schrieb:
> >
> > > Extending the definitions to the transfinite, we have in particular
> > > that the set of infinite paths in tree(w) corresponds to the set nodes
> > > (w): i.e., again, the set of leaf nodes.
> >
> > You only get all the prefixes of the infinite paths. But you
> > do not get the infinite paths itself.
>
> BS. Do I get the set nodes(w) or not? Otherwise, is there any error in
> the transfinite definitions or what?
As you are using a different definition of your trees, what you say
about your trees is of little relevance to what we say about ours.
--
Virgil
Virgil
<5b6b52c7-2562-42ab...@b14g2000yqd.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 20:38, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > If I understand your specifications, each tree(n) is a finite tree.
> > The matter of previous discussion had not been as to finite trees.
>
> Then you don't get it: I have given transfinite-inductive definitions,
> they work in the infinite.
>
> -LV
But differ from our definitions, so are not relevant to out trees.
--
Virgil
Virgil
<3e8387e1-83a0-4c06...@b15g2000yqd.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 20:51, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 17, 12:40�pm, LudovicoVan <ju...@diegidio.name> wrote:
> >
> > > Do I get the set nodes(w)
> >
> > I don't know your notation, but just in case I'm in the ballpark
> > regarding your notation, in the tree we're talking about, there are no
> > nodes whose level is indexed by w (omega) itself. Rather, the only
> > levels are those that are indexed by a natural number, and for every
> > natural number there is a level indexed by that natural number.
>
> The you must be stupid. First, read the post to know what the bloody
> notation means. Second, the only levels you conceive are those within
> the limits of your incompetence only.
>
> -LV
We have define our trees one way and you have defined yours differently,
so what holds in yours can be, and sometimes is, false in ours.
--
Virgil
Virgil
<97dc91e7-facb-4a04...@o6g2000yqj.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
That yours are different from the ones we are using is enough to make
some of your conclusions false in our trees
--
Virgil
Virgil
<0a06aca9-11dd-468e...@y7g2000yqa.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
Here is a concrete example of our sort of maximal infinite binary tree:
The set of nodes is the (one origin) set of naturals n = {1,2,3,...}.
1 is the root node.
For every n in N, 2*n+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
(a) 1 is a member of P, and
(b) for all n in P one and only one of 2*n+0 and 2*n+1 is in P
In this sort of tree, there are no leaf nodes.
--
Virgil
LudovicoVan
> LudovicoVan <ju...@diegidio.name> wrote:
> > On 17 July, 20:59, Jan Burse <janbu...@fastmail.fm> wrote:
>
> > > What is irrelevant?
>
> > All you have said is irrelevant. Do I get the set nodes(w) or not?
> > I.e. you are not saying what is the problem with my definitions.
>
> some of your conclusions false in our trees
You are a shameless bullshitter. The only conclusions that I prove
false, if any, are your holy diagonal argument and its holy
consequences.
-LV
Virgil
<f94a0c17-190b-4d20...@h31g2000yqd.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
Whether the set of nodes is countable depends on one's definition of
nodes. By our definition they are countable, by yours they are not.
Virgil
<fdd687e3-855c-42ef...@k1g2000yqf.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 17 July, 23:03, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <97dc91e7-facb-4a04-82dd-f3a09df74...@o6g2000yqj.googlegroups.com>,
> > �LudovicoVan <ju...@diegidio.name> wrote:
> > > On 17 July, 20:59, Jan Burse <janbu...@fastmail.fm> wrote:
> >
> > > > What is irrelevant?
> >
> > > All you have said is irrelevant. Do I get the set nodes(w) or not?
> > > I.e. you are not saying what is the problem with my definitions.
> >
> > That yours are different from the ones we are using is enough to make
> > some of your conclusions false in our trees
>
> You are a shameless bullshitter. The only conclusions that I prove
> false, if any, are your holy diagonal argument and its holy
> consequences.
What you claim to have proved about trees does not apply to trees that
do not satisfy your own definition, and ours do not.
Here is a concrete example of our sort of maximal infinite binary tree:
The set of nodes is the (one origin) set of naturals n = {1,2,3,...}.
1 is the root node.
For every n in N, 2*n+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
(a) 1 is a member of P, and
(b) for all n in P one and only one of 2*n+0 and 2*n+1 is in P
In this sort of tree, there are no leaf nodes, as there is no natural n
for which there are not naturals of form 2*n+0 and 2*n+1.
For it to be otherwise, there must be at least one natural, n, for which
neither 2*n+0 or 2*n+1 are naturals.
--
Virgil
Virgil
<4496e5e5-73ee-49b2...@18g2000yqa.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
It is perfectly valid for the paths in OUR trees.
Here is a concrete example of our sort of maximal infinite binary tree:
The set of nodes is the (one origin) set of naturals n = {1,2,3,...}.
1 is the root node.
For every n in N, 2*n+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
(a) 1 is a member of P, and
(b) for all n in P one and only one of 2*n+0 and 2*n+1 is in P
In this sort of tree, there are no leaf nodes.
--
Virgil
LudovicoVan
You are a shameless bullshitter. First, there is no problem in my
definition of leaf nodes, of any kind, and you make a fuss on this
that just tells your anti-mathematical attitude. Second, your
definition is finite, mine is transfinite, that's the only difference
between what I have shown and what you claim to be the only sensible
definition ever. Third, a transfinite definition is actually one that
makes sense w.r.t. the diagonal argument, because with your finite
definitions (a la Cantor's indeed) there is no sensible ground to talk
about complete infinite paths over a complete infinite tree.
And with this I have finished once again with your spam,
-LV
LudovicoVan
> It is perfectly valid for the paths in OUR trees.
Spammer.
-LV
Herbert Newman
>>> So whatever sort of infinite tree Ludovico Van (LV) is talking about,
>>> it is not the one everyone else is considering.
>>A
>> BS. Otherwise give proper references. [LV]
>>
> Moschovakis 'Notes On Set Theory'; Monk 'Introduction To Set Theory';
> Levy 'Basic Set Theory' (as I recall); others available.
I'm sure LV will acknowledge that! :-)
Herb
Virgil
<1b76f1be-b153-4844...@h21g2000yqa.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
The problem with your definitions is that they differ from the the
definition everyone else is using which are compatible with the
following:
Here is a concrete example of our sort of maximal infinite binary tree:
The set of nodes is the (one origin) set of naturals n = {1,2,3,...}.
1 is the root node.
For every n in N, 2*n+0 is its left child and 2*n+1 is its right child.
A path is a subset, P, of N such that
(a) 1 is a member of P, and
(b) for all n in P one and only one of 2*n+0 and 2*n+1 is in P
In this sort of tree, there are no leaf nodes.
> Second, your
> definition is finite, mine is transfinite, that's the only difference
> between what I have shown and what you claim to be the only sensible
> definition ever.
It is not a matter of sensibility, but of actuality.
What is true of one definition need not be true of a different one. And
when everyone but you is using one definition and you use a different
one, what you deduce from your definition may well be irrelevant to what
holds in the everyone else's.
In the concrete definition above,. Which accords with the trees that
everyone else except you and WM have been talking about, there are no
leaf nodes.
> Third, a transfinite definition is actually one that
> makes sense w.r.t. the diagonal argument, because with your finite
> definitions (a la Cantor's indeed) there is no sensible ground to talk
> about complete infinite paths over a complete infinite tree.
On the contrary, one can easily biject the set of (complete) paths in
our trees with the set of all mappings from N to {0,1}, which is a good
deal more sensible than your garbage.
>
> And with this I have finished once again with your spam,
>
> -LV
--
Virgil
MoeBlee
I am supremely confident that he will rush to buy them, then to study
them as if his life depended on it, thus to master them in every
detail, and then to hunger continually for more.
MoeBlee
Virgil
<05907597-f021-4de9...@o6g2000yqj.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
That is not a legitimate refutation of anything, and certainly not of
the truth that infinite trees, at least as we define them, do not have
leaf nodes..
If you want to play around with different sorts of trees, feel free to
do so, but do not claim that we must accede to your definition when your
results conflict with our own.
--
Virgil
Peter Webb
"LudovicoVan" <ju...@diegidio.name> wrote in message
news:fdd687e3-855c-42ef...@k1g2000yqf.googlegroups.com...
-LV
======================
You have redefined the term "leaf node". In common usage, a leaf node can
only be connected to one other node. This is not true of your usage of "leaf
node". Using the same word to refer to two completely different concepts is
bound to cause problems. Try reformulating your argument so that the terms
you use reflect common use. Feel free to invent new words (as long as you
define them fully), rather than using words which already mean something
different (such as "leaf node"). Your error will become more obvious.
HTH
Peter Webb
>> > diagonalization possible.
>>
>> Actually the infinite tree can indeed be conquered, by the
>> countable. This is mathematically true. But still there are
>> uncountable infinite paths.
>
> My 2c: Since a bijection can be given between paths and leaf nodes
> over the transfinite tree,
Please define the "transfinite tree". What numbers (or whatever) comprise
the nodes, and what exactly is the rule which determines if two nodes are
directly connected by a path?