The complete infinite binary tree has only countably many infinite paths.
WM
paths.
0.
/ \
0 1
/ \ / \
0 10 1
...
A) Construct the binary tree starting from a "tree" that has only one
path, say p_0 = 0.000... 0.
|
0
|
0
|
0
...
Add all paths that end by infinitely many zeros. Every path that you
add must start at a node of p_0 or at a node of a path already
constructed. The number of paths in the tree grows by not more and not
less than 1 when 1 path is added. However, after completing that
procedure all nodes and every infinite sequence of bits (including the
path 0.111...) is present, namely represented in the infinite binary
tree.
B) The complete infinite binary tree can be constructed by an infinite
agglomeration of elements of the form
|
o
/ \
The number of distinct lines is increased by 1 by 1 node.
(lines going out - lines coming in - nodes) = (2 - 1 - 1) = 0
This procedure, even when applied infinitely often, cannot but yield
the result 0 respectively a countable number of lines. The set of all
lines limits the set of all distinct paths.
The construction is possible in (B) as well as in (A) because the
elements used for construction is countable infinite.
C) Consider the edges of the complete binary tree (an edge connects
two subsequent nodes of a path). Take all edges and put them on one
and the same level of the tree, side by side, such that the "tree" now
is an array of parallel edges: |||||||... This array limits the number
of possible paths of the tree. It is an upper limit, because every
path there has only one edge. And there is no further edge remaining
to distinguish any further paths.
Remark: Of course a set of n edges can be put in n! different
sequences. But the edges of the binary tree are not subject to
arbitrary ordering. Each one has one and only one fixed place in the
tree. Therefore the number of edges limits the number of paths.
Remark: The tree contains all possible sequences of bits including
0.111... . Nevertheless the tree contains only a countable number of
paths, as we see by each of the arguments (A - C). This shows that
"most" of the real numbers cannot exist as independent bit sequences.
This shows that most of the real numbers are not subject to being put
in a list or resulting from a list as anti-diagonal number.
Regards, WM
William Elliot
> The complete infinite binary tree has only countably many infinite
> paths.
>
> 0.
> / \
> 0 1
> / \ / \
> 0 10 1
> ...
>
It has uncountably many paths, one for each binary sequence.
> A) Construct the binary tree starting from a "tree" that has only one
> path, say p_0 = 0.000... 0.
>
> |
> 0
> |
> 0
> |
> 0
> ...
>
> Add all paths that end by infinitely many zeros. Every path that you
> add must start at a node of p_0 or at a node of a path already
> constructed. The number of paths in the tree grows by not more and not
> less than 1 when 1 path is added. However, after completing that
> procedure all nodes and every infinite sequence of bits (including the
> path 0.111...) is present, namely represented in the infinite binary
> tree.
>
Extreme vagueness. If at each node without a branch, you added another
infinite series you will have countable many series. You repeat this
again, infinitely.
> B) The complete infinite binary tree can be constructed by an infinite
> agglomeration of elements of the form
> |
> o
> / \
> The number of distinct lines is increased by 1 by 1 node.
> (lines going out - lines coming in - nodes) = (2 - 1 - 1) = 0
> This procedure, even when applied infinitely often, cannot but yield
> the result 0 respectively a countable number of lines. The set of all
> lines limits the set of all distinct paths.
> The construction is possible in (B) as well as in (A) because the
> elements used for construction is countable infinite.
>
Are you writing mathematics or science fiction?
> C) Consider the edges of the complete binary tree (an edge connects
> two subsequent nodes of a path). Take all edges and put them on one
> and the same level of the tree, side by side, such that the "tree" now
> is an array of parallel edges: |||||||... This array limits the number
> of possible paths of the tree. It is an upper limit, because every
> path there has only one edge. And there is no further edge remaining
> to distinguish any further paths.
>
You're writing science fiction.
> Remark: Of course a set of n edges can be put in n! different
> sequences. But the edges of the binary tree are not subject to
> arbitrary ordering. Each one has one and only one fixed place in the
> tree. Therefore the number of edges limits the number of paths.
>
Remark: apply at philosophy where such sophistry is philosophical.
> Remark: The tree contains all possible sequences of bits including
> 0.111... . Nevertheless the tree contains only a countable number of
> paths, as we see by each of the arguments (A - C). This shows that
> "most" of the real numbers cannot exist as independent bit sequences.
> This shows that most of the real numbers are not subject to being put
> in a list or resulting from a list as anti-diagonal number.
>
Remark: your application for mathematics has been rejected.
WM
> On Tue, 24 Mar 2009, WM wrote:
> > The complete infinite binary tree has only countably many infinite
> > paths.
>
> > 0.
> > / \
> > 0 1
> > / \ / \
> > 0 10 1
> > ...
> > path, say p_0 = 0.000...
Error corrected (0. belongs to the following figure, not to the number
above):
0.
|
0
|
0
|
0
...
>
> > Add all paths that end by infinitely many zeros. Every path that you
> > add must start at a node of p_0 or at a node of a path already
> > constructed. The number of paths in the tree grows by not more and not
> > less than 1 when 1 path is added. However, after completing that
> > procedure all nodes and every infinite sequence of bits (including the
> > path 0.111...) is present, namely represented in the infinite binary
> > tree.
>
> Extreme vagueness. If at each node without a branch, you added another
> infinite series you will have countable many series. You repeat this
> again, infinitely.
The nodes can be enumerated:
0
1 2
3 4 5 6
...
The connecting line between the root node 0 and node number n is
uniquely determined in the tree. Now proceed as follows for n = 1, 2,
3, ...:
Connect the root node and node number n and after having passed node
number n continue by zeros only. Then you have constructed a path
starting at 0 and passing trough node n and subsequently having
infinitely many zeros. For every n this path is defined, therefore we
have a construction of all paths that end by infinitely many zeros.
(This construction contains many paths more than once. For instance
there are many paths passing trough all nodes 0 at the left hand side
of the tree. Nevertheless all paths belong to a countable set. All
nodes of the tree are in the construction. No sequence that can be
written by digits in Cantor's list is lacking a representation as a
path in the binary tree. Every sequence of bits is in the constructed
tree.)
>
> > B) The complete infinite binary tree can be constructed by an infinite
> > agglomeration of elements of the form
> > |
> > o
> > / \
> > The number of distinct lines is increased by 1 by 1 node.
> > (lines going out - lines coming in - nodes) = (2 - 1 - 1) = 0
> > This procedure, even when applied infinitely often, cannot but yield
> > the result 0 respectively a countable number of lines. The set of all
> > lines limits the set of all distinct paths.
> > The construction is possible in (B) as well as in (A) because the
> > elements used for construction is countable infinite.
>
> Are you writing mathematics or science fiction?
I leave science fiction like finished I leave to set theorists. A
construction is done when all elemnts of a countable set have their
fixed position in a sequence. An example is the construction of an
enumeration of all algebraic numbers.
>
> > C) Consider the edges of the complete binary tree (an edge connects
> > two subsequent nodes of a path). Take all edges and put them on one
> > and the same level of the tree, side by side, such that the "tree" now
> > is an array of parallel edges: |||||||... This array limits the number
> > of possible paths of the tree. It is an upper limit, because every
> > path there has only one edge. And there is no further edge remaining
> > to distinguish any further paths.
>
> You're writing science fiction.
Do you object to infinite sequences for principle reasons? Then you
have my sympathy. But here in sci.logic we must pretend that we
believe in infinite sequences. Every countable set can be put in an
infinite sequence. Therefore the edges can line up according to the
given order..
Regards, WM
George Greene
> A) Construct the binary tree starting from a "tree" that has only one
> path,
Dumbass, YOU CANNOT *construct* "the infinite binary" tree in this
way!
Yes, the thing YOU are constructed will only have countably many
paths,
BUT THAT IS NOT the infinite binary tree!
You mean, by the way, the "denumerably deep" binary tree.
George Greene
> add must start at a node of p_0 or at a node of a path already
> constructed. The number of paths in the tree grows by not more and not
> less than 1 when 1 path is added. However, after completing that
> procedure all nodes and every infinite sequence of bits (including the
> path 0.111...) is present,
THIS IS BULLSHIT!
The path 0.111... IS NOT present! NO path with a FINITE number of 0's
is present! ONLY paths that end wit all 0's have been added!
And these aren't the only paths that are missing, either.
calvin...@gmail.com
> ... all paths belong to a countable set. ...
Okay. Just one simple question here. Let P(x)
mean "x is a path". Let CS(x) mean "x is a
countable set".
WM (not anyone else please!), will you please write
out, formally in first order logic, the statement you have
made above? Thank you.
Virgil
> Consider the edges of the complete binary tree (an edge connects
> two subsequent nodes of a path). Take all edges and put them on one
> and the same level of the tree, side by side, such that the "tree" now
> is an array of parallel edges: |||||||... This array limits the number
> of possible paths of the tree. It is an upper limit, because every
> path there has only one edge.
Actually, a path having only one edge there must be a very short path,
consisting of only one edge.
Actually every infinite path will contain an infinite set of such edges.
Virgil
<97c58f1e-4091-49d0...@v19g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> The nodes can be enumerated:
>
> 0
> 1 2
> 3 4 5 6
> ...
>
> The connecting line between the root node 0 and node number n is
> uniquely determined in the tree. Now proceed as follows for n = 1, 2,
> 3, ...:
> Connect the root node and node number n and after having passed node
> number n continue by zeros only. Then you have constructed a path
> starting at 0 and passing trough node n and subsequently having
> infinitely many zeros. For every n this path is defined, therefore we
> have a construction of all paths that end by infinitely many zeros.
If you mean ending with infinitely many zeros and no ones, correct, but
this omits that uncountably many that end otherwise.
> (This construction contains many paths more than once. For instance
> there are many paths passing trough all nodes 0 at the left hand side
> of the tree. Nevertheless all paths belong to a countable set. All
> nodes of the tree are in the construction. No sequence that can be
> written by digits in Cantor's list is lacking a representation as a
> path in the binary tree. Every sequence of bits is in the constructed
> tree.)
Wrong!!! Your construction never produces any string with more than a
finite number of ones in it, but *most* strings in the complete tree are
of that form.
> >
> > Are you writing mathematics or science fiction?
>
> I leave science fiction like finished I leave to set theorists.
What WM writes is bad science and bad fiction. What WM leaves to set
theorists is any understanding of the structure of a complete infinite
binary tree.
Virgil
<97c58f1e-4091-49d0...@v19g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
While I agree with Mr. Eliot that what you write is mostly fictional, it
is not of the quality that I would call science fiction.
A
> construction is done when all elemnts of a countable set have their
> fixed position in a sequence. An example is the construction of an
> enumeration of all algebraic numbers.
> >
> > > C) Consider the edges of the complete binary tree (an edge connects
> > > two subsequent nodes of a path). Take all edges and put them on one
> > > and the same level of the tree, side by side, such that the "tree" now
> > > is an array of parallel edges: |||||||... This array limits the number
> > > of possible paths of the tree. It is an upper limit, because every
> > > path there has only one edge. And there is no further edge remaining
> > > to distinguish any further paths.
> >
> > You're writing science fiction.
>
> Do you object to infinite sequences for principle reasons?
He objects on principle to the uncountably many of them that you
deliberately leave out.
Then you
> have my sympathy. But here in sci.logic we must pretend that we
> believe in infinite sequences. Every countable set can be put in an
> infinite sequence. Therefore the edges can line up according to the
> given order..
Which has nothing to do with the issue at hand: the uncountability of
the set of all binary sequences (as proved by Cantor and as yet to be
contradicted successfully by anyone).
David C. Ullrich
<muec...@rz.fh-augsburg.de> wrote:
>The complete infinite binary tree has only countably many infinite
>paths.
Why do you continue this? Even if you were right,
surely it's clear by now you're not going to convince
anyone.
Not that you're right, of course - proving that that
tree has uncountably many paths is very easy.
What _could_ be right is a statment of the form
"If we replace all the standard axioms with
the following: ___ then the complete binary tree has
only countably many paths".
But you've never specified those new axioms - it
seems that you continue to insist that there are
only countably many paths in the standard setup.
That's awesomely stupid.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
WM
> On Tue, 24 Mar 2009 04:56:33 -0700 (PDT), WM
>
> >The complete infinite binary tree has only countably many infinite
> >paths.
>
> Why do you continue this? Even if you were right,
> surely it's clear by now you're not going to convince
> anyone.
You are in error. Everybody not blinded by the blinkers of transfinite
set theory will understand very easily that I am right. And here I
have given two slightly different version sof the proof that may
contribute to convince some more readers.
But even those who already have been brain-washed by finished
transfinity do not agree. Some say that my proof is wrong --- of
course without specifying why, only arguing that Cantor's diagonal
proof says the contrary. (But that is no argument when the
applicability of logic to infinite sets is under investigation as here
is the case*). Others say that my arguments are not new. But, as
mentioned above, everybody with undeformed brain understands my
arguments.
*As Brouwer claimed and as Weyl approved: "Classical logic was
abstracted from the mathematics of finite sets and their subsets ....
Forgetful of this limited origin, one afterwards mistook that logic
for something above and prior to all mathematics, and finally applied
it, without justification, to the mathematics of infinite sets. ... As
Brouwer pointed out this is a fallacy, the Fall and Original sin of
set theory even if no paradoxes result from it." [H. Weyl,
"Mathematics and logic: A brief survey serving as a preface to a
review of The Philosophy of Bertrand Russell", American Mathematical
Monthly 53: 2–13, 1946.]
Regards, WM
Aatu Koskensilta
> *As Brouwer claimed and as Weyl approved:
An unfortunate source for support since the diagonal argument is
intuitionistically valid.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
WM
wrote:
If you need this eduction I will refer you to
http://de2.metamath.org/index.html
Your results then can even be approved by automata.
I am not going to prove 1+1 = 2 or anything else by that language. And
if you think that a statement in that language, approved by a machine,
guarantees truth, then you are wrong.
The direct statement that you want to have formalized is: The set of
path can be put in a bijection with N. I do not pretend that this can
be done.
What I have shown is that the set of paths cannot be larger than a
countable set, namely the set of all nodes respectively all lines in
the tree.
Presumably this proof would need some hundred pages in FOPL+ZFC. Or
maybe it cannot be formalized at all. I don't know and I am not
interested in that topic
What I know is that the set of all infinite sequences of bits (or
digits) has not more elements than the set of all natural numbers,
when proved in the way I did it here. Of course it has infinitely more
elements when considered from another standpoint, for instance
considering the fact that between every two rational numbers there are
infinitely many rational numbers and, if they exist, irrational
numbers.
Regards, WM
WM
> WM <mueck...@rz.fh-augsburg.de> writes:
> > *As Brouwer claimed and as Weyl approved:
>
> An unfortunate source for support since the diagonal argument is
> intuitionistically valid.
To be valid means it can be derived using logic. It does not mean the
contrary cannot be derived. This comes in when "finished" infinity is
concerned.
Without finished infinity the diagonal argument does not show that the
diagonal number is not in the sequence, because the sequence is never
complete.
>
> --
> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>
> "Wovon man nicht sprechen kann, darüber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
An unfortunate source for support:
Set theory is wrong
- Ludwig Wittgenstein
WM
> You mean, by the way, the "denumerably deep" binary tree.
It is of no interest whether it has already a name. I mean that binary
tree that has all nodes that a tree can have. And therefore is has all
paths that can be present in a tree.
Regards, WM
WM
No path is missing because no node is missing. This is the magic of
the tree. You construct it from rational paths only. But after
infinity has been finished, it contains all paths.
Could it be that there is no finished infinity and that there are no
complete sequences for irrational numbers (and all others too) and not
Cantor-list-proof --- no valid one, I mean?
Regards, WM
WM
> On Tue, 24 Mar 2009, WM wrote:
> > Consider the edges of the complete binary tree (an edge connects
> > two subsequent nodes of a path). Take all edges and put them on one
> > and the same level of the tree, side by side, such that the "tree" now
> > is an array of parallel edges: |||||||... This array limits the number
> > of possible paths of the tree. It is an upper limit, because every
> > path there has only one edge.
>
> Actually, a path having only one edge there must be a very short path,
> consisting of only one edge.
It is an upper estimation. Compare the proof that e is an irrational
number. There you find at the end that there must be a natural number
between 0 and 1. Contradiction. Here you find that there must be more
edges than can be cast into a seqeunce. Contradiction. Both proof are
equally easy.
Regards, WM
WM
> First thing to note is that: 0.000... is not a "path",
> is it? It appears to be a binary representation of a real
> number.
A path in the binary tree is a binary represwentation of a real number
of the unit interval [0, 1]. There is no difference. Therefore I
insist that any real that can appear in Cantor's list also can appear
in the binary tree. Nevertheless, the tree conatins all
representations of those real numbers, some even twice. There is no
chance to construct an antidiagonal number. Everyone is there. And
every such path runs within one line. Two different paths cannot share
the same line.
> I guess at what it is, as a tree, and let
> us try to construct the tree the way you say. I assume
> that the same strategy is being employed as the one I
> employed in an attempt to make sense of your writing.
> In other words, we defined a tree by starting with some
> tree, and step by step add nodes and edges in a definite
> manner. The tree that results "in the limit" is then
> tree we are talking about. Agreed?
Yes.
>
> To be precise, to say where edges are being added,
> we need to name the nodes. The way I did it we could
> easily name them as elements of N. With yours, perhaps
> not so easily. So something like this is in order. I
> will name the nodes as elements of the set {n->{0,1}: n in N}.
> Each node in other words is named by the path sequence
> leading up to it. This way we don't need edge labels:
> they can be defined by the last component of the node
> they lead into.
>
> So, with this definition, your initial tree T0 is I think
> { fi | fi in n->{0,1} , fi(k) = 0 for k in n}
I propose an easier method:
Count the nodes as I did it
0
1,2
3,4,5,6
7,8,...
or, if you prefer by
0
(1,1), (1,2)
(2,1), (2,2), (2,3), (2,4)
...
(n,1), ..., (n, 2^n)
...
Every node with even second co-ordinate has value 1, every node with
odd second co-ordinate has value 0.
Construct the complete tree (from the path p_0 = 0.000...) by using
all paths that lead to a node with value 1 and afterward having only
nodes with value zero.
Then all the paths (except p_0) can be labelled by their last node
that has value 1.
The order can be lexically, i.e., by first co-ordinate, second co-
ordinate.
> I did not see the entire construction clearly defined.
> With this framework, please go ahead and do so.
>
> I should point out, though, that this tree thing really
> seems to be a red herring. I am convinced that your problem
> can be isolated and identified in a much simpler context,
> including perhaps one that does not even include infinities.
I am open to all arguments other than
0 + 0 + 0 + ... = 2^aleph.
Regards, WM
WM
Continued from the old thread.
On 24 Mrz., 23:42, Virgil <Vir...@gmale.com> wrote:
> None of your three alleged proofs, either separately or in combination,
> speaks to the COMPLETE infinite binary tree, whose paths include all of
> the uncountably many required by Cantor's proof of the uncountability of
> the set of all binary sequences.
I give a construction that yields the complete binary tree, i.e., all
its nodes, such that no further node can be added, Hence no further
edge can be added, hence no further path can be added that is distinct
from all paths whcih are already present.
Regards, WM
lwa...@lausd.net
The one thing I don't understand about these WM binary tree
threads is how WM, as an ultrafinitist who believes that
numbers as large as a googolplex don't exist, can believe
that an infinite tree even exists in the first place, much less
debate over whether it's countable or not.
I can try to come up with a theory in which infinite trees
can exist yet are countable, but that would be pointless
unless I can see how infinite trees can even exist in an
ultrafinitist theory.
For example, ZF-Powerset+~Powerset can prove that
countably infinite sets exist, but can't prove that any
uncountable sets exist. Perhaps in such a theory, we
can try to define an infinite tree in such a way that it
would have only countably many paths. But then it
would betray WM's ultrafinitism, since there would be
nodes such that the number of edges between it and
the root node is exactly a googolplex.
An ultrafinitist ought to believe that infinite trees don't
even exist in the first place. Infinite trees wouldn't have
countably infinitely many paths, but _zero_ paths.
The only way I can harmonize ultrafinitism with the
existence of an infinite tree would be to declare that
some big number such as googolplex or AP's 242!+1
(which is approximately googol^5, more or less) to
be already infinity. But even this seems to go against
what WM has said about googolplex -- he seems to
say that it doesn't exist at all, not merely that it's an
infinite number larger than all finite natural numbers.
Perhaps WM can settle this once and for all. So now
I ask of WM, since you believe that the infinite tree
has countably infinitely many paths (some of which
terminate at nodes, others don't), how many of those
countably infinitely many paths (terminate at nodes
and) have exactly googolplex (10^10^100) edges?
Gus Gassmann
You don't actually believe you can get a coherent answer out of WM on
anything, do you? Let's see. The number of paths in the binary tree is
countably infinite because the paths map bijectively onto the nodes.
And there are as many nodes as there are natural numbers. But the set
of natural numbers is not actually infinite, it is potentially
infinite, which is to say, it is finite and hence a FISON, but it is
not clear which FISON, because that changes over time. Also, since
there are at most 10^100 natural numbers in existence and people have
obviously been using natural numbers greater than 10^100 (such as the
googolplex), there must be holes in the natural numbers, that is,
there must exist numbers that have no successor, but, lo and behold, a
little while later there all of a sudden is another natural number
because someone has thought of it, and if that person ceases to think
of the number, the number obviously ceases to exist. Add to that that
N is really a little bit more than a FISON, anyway, and you'll come
close to what WM thinks about natural numbers. But today is a new
moon, so everything is possible.
The only thing in all this that is crystal clear is that set theory is
inconsistent, and there is no way to make it consistent, because it
obviously clashes with the good professor's particular delusion.
Virgil
<44f6c2b3-90c3-4a8f...@e38g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Where do you find that?
I find that an "array"of countably many edges, which is what you have,
can be cast into a sequence quite easily, as can every non-empty subset
of that array.
> Contradiction. Both proof are
> equally easy.
But yours is flawed.
Virgil
<fb197784-091e-49dd...@h28g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 10:21, David C. Ullrich <dullr...@sprynet.com> wrote:
> > On Tue, 24 Mar 2009 04:56:33 -0700 (PDT), WM
> >
> > <mueck...@rz.fh-augsburg.de> wrote:
> > >The complete infinite binary tree has only countably many infinite
> > >paths.
> >
> > Why do you continue this? Even if you were right,
> > surely it's clear by now you're not going to convince
> > anyone.
>
> You are in error. Everybody not blinded by the blinkers of transfinite
> set theory will understand very easily that I am right.
Whom, excluding the captive audience of your students, have you ever
convinced of the truth of your foolishness?
> And here I
> have given two slightly different version sof the proof that may
> contribute to convince some more readers.
Not bloody likely!
> But even those who already have been brain-washed by finished
> transfinity do not agree. Some say that my proof is wrong --- of
> course without specifying why, only arguing that Cantor's diagonal
> proof says the contrary.
Since you still cannot successfully fault the Cantor argument, no
counterargument of your own can override it.
> (But that is no argument when the
> applicability of logic to infinite sets is under investigation as here
> is the case*). Others say that my arguments are not new. But, as
> mentioned above, everybody with undeformed brain understands my
> arguments.
We also understand their faults, but then WM ignores those faults when
the are pointed out to him.
>
> *As Brouwer claimed and as Weyl approved: "Classical logic was
> abstracted from the mathematics of finite sets and their subsets ....
> Forgetful of this limited origin, one afterwards mistook that logic
> for something above and prior to all mathematics, and finally applied
> it, without justification, to the mathematics of infinite sets. ... As
> Brouwer pointed out this is a fallacy, the Fall and Original sin of
> set theory even if no paradoxes result from it." [H. Weyl,
> "Mathematics and logic: A brief survey serving as a preface to a
> review of The Philosophy of Bertrand Russell", American Mathematical
> Monthly 53: 213, 1946.]
Sin alone, other than the sin of introducing self-contradictory
assumptions, is not a mathematical concept.
The only contradictions WM can find are between infinite sets and WM's
mathUnrealism, not between infinite sets and anything mathematical.
calvin...@gmail.com
> > WM (not anyone else please!), will you please write
> > out, formally in first order logic, the statement you have
> > made above? Thank you.
>
> If you need this eduction I will refer you tohttp://de2.metamath.org/index.html
> Your results then can even be approved by automata.
How could my results in translating an English statement
into formal logic be "approved" of by automata?
> I am not going to prove 1+1 = 2 or anything else by that language. And
> if you think that a statement in that language, approved by a machine,
> guarantees truth, then you are wrong.
I did not ask you to *prove* anything. Simply to *state* the English
statement I have in formal logic.
> The direct statement that you want to have formalized is: The set of
> path can be put in a bijection with N. I do not pretend that this can
> be done.
It certainly is not. I did not ask anything of the sort. I gave you
the
formalization of countable set as the unary predicate "CS". I did
not mention N, I did not mention bijections. And the statement
I asked about, in any case, is in no way similar to that statement.
But wait. Are you now saying that the very statement "The set of
paths can be put in a bijection with N" cannot even be *formalized*?
What? What?
Anyway, could you please answer the simple question I asked you.
It takes only about 20 ascii characters or so, The statement again:
Virgil
<1930378e-cd11-4c0a...@w35g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mrz., 23:54, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On 24 Mrz., 16:07, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > ... all paths belong to a countable set. ...
> >
> > Okay. Just one simple question here. Let P(x)
> > mean "x is a path". Let CS(x) mean "x is a
> > countable set".
> >
> > WM (not anyone else please!), will you please write
> > out, formally in first order logic, the statement you have
> > made above? Thank you.
>
> If you need this eduction I will refer you to
> http://de2.metamath.org/index.html
As usual, WM misses the point.
That reference does not translate English, or German, into first order
logic, and it is only such a translation which was requested.
> Your results then can even be approved by automata.
They wold be YOUR results, and would not at all be approved within any
of the axiom sets available.
>
> I am not going to prove 1+1 = 2 or anything else by that language. And
> if you think that a statement in that language, approved by a machine,
> guarantees truth, then you are wrong.
Things approved by that machinery have a good deal better chance at
truth than anything approved in WM's world of MathUnrealism.
>
> The direct statement that you want to have formalized is: The set of
> path can be put in a bijection with N. I do not pretend that this can
> be done.
> What I have shown is that the set of paths cannot be larger than a
> countable set, namely the set of all nodes respectively all lines in
> the tree.
Except that your alleged proof is base on hidden assumptions contrary to
standard mathematics.
> Presumably this proof would need some hundred pages in FOPL+ZFC. Or
> maybe it cannot be formalized at all. I don't know and I am not
> interested in that topic
Your claim could easily be disproved since that metamath analyzer
presumes ZFC.
> What I know is that the set of all infinite sequences of bits (or
> digits) has not more elements than the set of all natural numbers,
It's not so much what you don't know that does you in, its what you
know that just ain't so.
calvin...@gmail.com
> > You are in error. Everybody not blinded by the blinkers of transfinite
> > set theory will understand very easily that I am right.
>
> Whom, excluding the captive audience of your students, have you ever
> convinced of the truth of your foolishness?
How do we even know he convinced any students?
We need some such student, who has for his teacher
a great figure, if he is right, to stand up and say he
was convinced. Is no such student reading these
heroic epics of his master? (Unforunately, it may
be hard to rule out sock-puppets should such a
student show up here).
> > But even those who already have been brain-washed by finished
> > transfinity do not agree. Some say that my proof is wrong --- of
> > course without specifying why, only arguing that Cantor's diagonal
> > proof says the contrary.
>
> Since you still cannot successfully fault the Cantor argument, no
> counterargument of your own can override it.
Well, perhaps no such argument can *override* Cantor's,
but it can at least meet it on an equal footing.
It seems that (from time to time anyway) WM's
argument is that set theory is inconsistent. It seems
that at these time he accepts Cantor's argument, yet
also accepts an argument establishing the opposite
conclusion.
If this is the case, he should clearly say so, and then
suggest either a new, hopefully consistent axiomatization
for set theory (or something similar, appropriate for
doing mathematics), or instead, a way of dealing with
inconsistency so that the logic does not "explode'
(presumably by outlining the nature of some
para-consistent logic to replace regular first order
logic)
Virgil
<2263ae2b-52fb-427b...@o11g2000yql.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 11:02, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > *As Brouwer claimed and as Weyl approved:
> >
> > An unfortunate source for support since the diagonal argument is
> > intuitionistically valid.
>
> To be valid means it can be derived using logic. It does not mean the
> contrary cannot be derived. This comes in when "finished" infinity is
> concerned.
>
> Without finished infinity the diagonal argument does not show that the
> diagonal number is not in the sequence, because the sequence is never
> complete.
Neither the completion of the listing of the binary sequences, nor the
completion of any listing of any binary sequence, including the
anti-diagonal, is required in that proof, so that WM's argument is, as
usual, faulty.
Virgil
<ab68b10e-b425-4422...@d19g2000yqb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
A countable set of nodes will generate a unary tree as well as a binary
tree, so that merely having enough nodes for a complete infinite binary
tree is insufficient to guarantee completeness of an infinite binary
tree.
Virgil
<d8385cbb-eb46-4922...@w9g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mrz., 21:32, George Greene <gree...@email.unc.edu> wrote:
> > On Mar 24, 7:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > Add all paths that end by infinitely many zeros. Every path that you
> > > add must start at a node of p_0 or at a node of a path already
> > > constructed. The number of paths in the tree grows by not more and not
> > > less than 1 when 1 path is added. However, after completing that
> > > procedure all nodes and every infinite sequence of bits (including the
> > > path 0.111...) is present,
> >
> > THIS IS BULLSHIT!
> > The path 0.111... IS NOT present! NO path with a FINITE number of 0's
> > is present! ONLY paths that end wit all 0's have been added!
> > And these aren't the only paths that are missing, either.
The set of all paths having only finitely many left branchings passes
through every node in the tree but excludes all paths having infinitely
many left branchings.
So that, despite WM's claim to the contrary,
'all nodes' is not enough to guarantee 'all paths'.
And WM has again clearly demonstrated his mathematical incompetence.
>
> No path is missing because no node is missing.
WRONG! The set of all paths having only finitely many left branchings
passes through every node in the tree but excludes all paths having
infinitely many left branchings.
> This is the magic of
> the tree. You construct it from rational paths only. But after
> infinity has been finished, it contains all paths.
In mathematics, we do not rely on magic, as it often, as in this case,
proves to be mathematically unrealiable.
>
> Could it be that there is no finished infinity and that there are no
> complete sequences for irrational numbers (and all others too) and not
> Cantor-list-proof --- no valid one, I mean?
Or could it be that WM's MathUnrealistic world relies too much on magic
and not enough on sound mathematical methods.
Virgil
<3cc6cf3c-9ead-4068...@p11g2000yqe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> From the old thread:
>
> > First thing to note is that: 0.000... is not a "path",
> > is it? It appears to be a binary representation of a real
> > number.
>
> A path in the binary tree is a binary represwentation of a real number
> of the unit interval [0, 1]. There is no difference. Therefore I
> insist that any real that can appear in Cantor's list also can appear
> in the binary tree. Nevertheless, the tree conatins all
> representations of those real numbers, some even twice. There is no
> chance to construct an antidiagonal number. Everyone is there.
Then, as Cantor proved, there must be too many to be countable.
> And
> every such path runs within one line. Two different paths cannot share
> the same line.
What in blazes are "lines" in a tree. If they are to be straight lines,
at least in the standard diagramatic form, there are at most two of
them, between which all other paths are constrained..
>
> I am open to all arguments other than
> 0 + 0 + 0 + ... = 2^aleph.
ON the contrary, you are closed to all arguments that in any way do not
conform to your MathUnrealism creed.
And the beliefs within that creed are in no way logically coherent.
Virgil
<35ae61bc-9deb-4fc2...@e38g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
But there are paths in that tree that spring into existence without
constriction.
Your construction is the equivalent of one in which every constructed
path has at most finitely many right branches.
But when "finished" that tree must have through each node a path with
infinitely many right branches.
Either that or infinitely many nodes with no right branch.
So that the "completion" creates paths not constructed before it.
Uncountably many of them.
Which again shows how the completion of a infinite process may differ
wildly from every finite stage prior to that completion.
Virgil
<4a272e6f-4498-450c...@d38g2000prn.googlegroups.com>,
lwa...@lausd.net wrote:
> On Mar 24, 3:14 pm, Virgil <Vir...@gmale.com> wrote:
> > On Tue, 24 Mar 2009, WM wrote:
> > > Consider the edges of the complete binary tree (an edge connects
> > > two subsequent nodes of a path). Take all edges and put them on one
> > > and the same level of the tree, side by side, such that the "tree" now
> > > is an array of parallel edges: |||||||... This array limits the number
> > > of possible paths of the tree. It is an upper limit, because every
> > > path there has only one edge.
> > Actually, a path having only one edge there must be a very short path,
> > consisting of only one edge.
> > Actually every infinite path will contain an infinite set of such edges.
>
> The one thing I don't understand about these WM binary tree
> threads is how WM, as an ultrafinitist who believes that
> numbers as large as a googolplex don't exist, can believe
> that an infinite tree even exists in the first place, much less
> debate over whether it's countable or not.
WM is one of those, like Ralph Waldo Emerson, who believes that a
foolish consistency is the hobgoblin of little minds.
And so embraces inconsistency too enthusiastically.
Virgil
<17d878bf-3d4d-4797...@q18g2000vbn.googlegroups.com>,
"calvin...@gmail.com" <calvin...@gmail.com> wrote:
> On Mar 25, 3:33 pm, Virgil <Vir...@gmale.com> wrote:
>
> > > You are in error. Everybody not blinded by the blinkers of transfinite
> > > set theory will understand very easily that I am right.
> >
> > Whom, excluding the captive audience of your students, have you ever
> > convinced of the truth of your foolishness?
>
> How do we even know he convinced any students?
We don't, though WM claims to have browbeaten them enough to get them
to regurgitate bits of his credo on exams.
> We need some such student, who has for his teacher
> a great figure, if he is right, to stand up and say he
> was convinced. Is no such student reading these
> heroic epics of his master? (Unforunately, it may
> be hard to rule out sock-puppets should such a
> student show up here).
>
> > > But even those who already have been brain-washed by finished
> > > transfinity do not agree. Some say that my proof is wrong --- of
> > > course without specifying why, only arguing that Cantor's diagonal
> > > proof says the contrary.
> >
> > Since you still cannot successfully fault the Cantor argument, no
> > counterargument of your own can override it.
>
> Well, perhaps no such argument can *override* Cantor's,
> but it can at least meet it on an equal footing.
> It seems that (from time to time anyway) WM's
> argument is that set theory is inconsistent. It seems
> that at these time he accepts Cantor's argument, yet
> also accepts an argument establishing the opposite
> conclusion.
If in his "system", he accepts simultaneously both such a P and such a
not P, I don't see why he should object to anything ever.
calvin...@gmail.com
> 0 [say instead (0,1)
> (1,1), (1,2)
> (2,1), (2,2), (2,3), (2,4)
> ...
> (n,1), ..., (n, 2^n)
So those are the nodes. {(n,m)| n,m in N, m<=2^n}
> Every node with even second co-ordinate has value 1, every node with
> odd second co-ordinate has value 0.
Labeling function L( (n,m) ) = m+1 mod 2
> Construct the complete tree (from the path p_0 = 0.000...) by using
> all paths that lead to a node with value 1 and afterward having only
> nodes with value zero.
Okay, well, a tree is a set of edges. Let us instead add the edges.
That is more definite and precise. Paths result by supervening on
the pattern of edges.
Can you rigorously specify the edges that are being added, in
a manner such as I have done above (instead of just with pictures
and informal words).
When you do that, you will see there are more paths that you
attempted to limit the set of paths to.
> Then all the paths (except p_0) can be labelled by their last node
> that has value 1.
Here, I think, is your problem. You are not noticing that when
you add a single path, you are actually adding many paths.
The net result in the end is that after all your paths have been
added, there are many other paths that exist, because of the
edges that have been added to make the paths, which you
have not noticed.
Perhaps the problem arises because of using the idea of one
tree as a limit of the other. I don't think that is necessary here.
As above for the nodes, we should be able to define the final
tree explicitly without using a limit.
Here is the problem you may have with using a limit. The
limit tree is a union of all the edges in the sequence of trees.
That is indeed true. If there is an edge in the limit tree, that
edge must exist in some tree in the sequence. HOWEVER,
the same is not true about paths. Paths exist in the limit
tree that do *not* exist in any tree in the sequence. This
is because the paths "interact" with one another in the way
edges do not.
Every path you add is of the form a finite sequence of 0's
and 1's, following by infinite sequence of 0's. However,
there are paths in the final tree that are not like this. That is
because the paths interact with one another. You must look
at the individual edges and how then can be made into
paths in the final tree. You do not add paths, you add edges.
Paths come FOR FREE. You cannot pick and choose the
paths you add. They sneak in as sets of joined edges and
there is nothing you can do about it.
calvin...@gmail.com
> > How do we even know he convinced any students?
>
> We don't, though WM claims to have browbeaten them enough to get them
> to regurgitate bits of his credo on exams.
Actually, if I recall, he explicitly claims *not* to have
browbeaten them in this manner.
> > It seems
> > that at these time he accepts Cantor's argument, yet
> > also accepts an argument establishing the opposite
> > conclusion.
>
> If in his "system", he accepts simultaneously both such a P and such a
> not P, I don't see why he should object to anything ever.
Why not?
In any case, presuming he accepts both some P and also ~P,
maybe you don't see why he objects to other things
occasionally, but should we first not establish whether
or not he does accept both some P and also ~P (not necessarily
absolutely simultaneously, by the way!)
WM
> On Mar 24, 3:14 pm, Virgil <Vir...@gmale.com> wrote:
>
> > On Tue, 24 Mar 2009, WM wrote:
> > > Consider the edges of the complete binary tree (an edge connects
> > > two subsequent nodes of a path). Take all edges and put them on one
> > > and the same level of the tree, side by side, such that the "tree" now
> > > is an array of parallel edges: |||||||... This array limits the number
> > > of possible paths of the tree. It is an upper limit, because every
> > > path there has only one edge.
> > Actually, a path having only one edge there must be a very short path,
> > consisting of only one edge.
> > Actually every infinite path will contain an infinite set of such edges.
>
> The one thing I don't understand about these WM binary tree
> threads is how WM, as an ultrafinitist who believes that
> numbers as large as a googolplex don't exist, can believe
> that an infinite tree even exists in the first place, much less
> debate over whether it's countable or not.
There are two lines of arguing.
1) Of course there is no actually infinite set. But this would not
convince set theorists that their theories are useless.
2) If there were actually infinite sets like the digits of an
irrational number or the nodes of the binary tree, then this would
yield a contradiction. So for case (2) I accept actually infinite
sets, but show that this assumption leads to contradictions.
>
> I can try to come up with a theory in which infinite trees
> can exist yet are countable, but that would be pointless
> unless I can see how infinite trees can even exist in an
> ultrafinitist theory.
>
> For example, ZF-Powerset+~Powerset can prove that
> countably infinite sets exist, but can't prove that any
> uncountable sets exist. Perhaps in such a theory, we
> can try to define an infinite tree in such a way that it
> would have only countably many paths. But then it
> would betray WM's ultrafinitism, since there would be
> nodes such that the number of edges between it and
> the root node is exactly a googolplex.
>
> An ultrafinitist ought to believe that infinite trees don't
> even exist in the first place. Infinite trees wouldn't have
> countably infinitely many paths, but _zero_ paths.
>
> The only way I can harmonize ultrafinitism with the
> existence of an infinite tree would be to declare that
> some big number such as googolplex or AP's 242!+1
> (which is approximately googol^5, more or less) to
> be already infinity. But even this seems to go against
> what WM has said about googolplex -- he seems to
> say that it doesn't exist at all, not merely that it's an
> infinite number larger than all finite natural numbers.
I do not object to the existence of any number you can define.
10^100^100^100^100 does exist. We can talk about it.
>
> Perhaps WM can settle this once and for all. So now
> I ask of WM, since you believe that the infinite tree
> has countably infinitely many paths (some of which
> terminate at nodes, others don't),
None of them terminates.
> how many of those
> countably infinitely many paths (terminate at nodes
> and) have exactly googolplex (10^10^100) edges?
All are infinite. Each one has aleph_0 nodes. This is assumed for the
tree. Otherwise the discussion would be boring.
Regards, WM
MoeBlee
> 2) If there were actually infinite sets like the digits of an
> irrational number or the nodes of the binary tree, then this would
> yield a contradiction. So for case (2) I accept actually infinite
> sets, but show that this assumption leads to contradictions.
A contradiction with WHAT?
A contradiction with certain of your own nebulous principles (if they
aren't already contradictory among themselves).
But you've never shown that there is a sentence P such that both P and
~P are theorems of ZFC. Specifically, you've never shown that it is a
theorem of ZFC that the infinite binary tree has only countably many
paths.
MoeBlee
P.S. Note to lwal: It is WM's failure to recognize such distinctions
as I just mentioned that contribute to an appraisal that he is a crank.
calvin...@gmail.com
> But you've never shown that there is a sentence P such that both P and
> ~P are theorems of ZFC. Specifically, you've never shown that it is a
> theorem of ZFC that the infinite binary tree has only countably many
> paths.
Unfortunately, he can wiggle out of this, by saying simply
that ZFC is inadequate because it does not permit the
formulation of his argument for ~P. But then the onus
should be upon him to give a much more clear version of
his argument. We are all used of ZFC being very
expressive, and also powerful enough to prove about
all we want in normal mathematics. But that does not
necessarily mean that it is. (However unlikely it is that
someone like WM (see below for what this means)
could realize this while virtually no technically competent
mathematican can).
> P.S. Note to lwal: It is WM's failure to recognize such distinctions
> as I just mentioned that contribute to an appraisal that he is a crank.
I think another good contribution comes from his response,
for example, to the question I posed that he write out
formally the statement "all paths are contained in a
countable set", given predicate symbols for "path" and
"countable set".
He confused this with (1) a different statement entirely
(2) proving something rather than just stating it and (3)
a question that could be evaluated by a formal theorem
proving or checking system.
MoeBlee
<calvin.ost...@gmail.com> wrote:
> On Mar 25, 5:17 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > But you've never shown that there is a sentence P such that both P and
> > ~P are theorems of ZFC. Specifically, you've never shown that it is a
> > theorem of ZFC that the infinite binary tree has only countably many
> > paths.
>
> Unfortunately, he can wiggle out of this, by saying simply
> that ZFC is inadequate because it does not permit the
> formulation of his argument for ~P. But then the onus
> should be upon him to give a much more clear version of
> his argument. We are all used of ZFC being very
> expressive, and also powerful enough to prove about
> all we want in normal mathematics. But that does not
> necessarily mean that it is. (However unlikely it is that
> someone like WM (see below for what this means)
> could realize this while virtually no technically competent
> mathematican can).
It's been offered to him a hundred times that he is welcome to state
whatever alternative syntax, primitives, axioms, definitions, and
rules of inference he wishes to present. His ordinary response is that
he is not interested in formal theories.
> > P.S. Note to lwal: It is WM's failure to recognize such distinctions
> > as I just mentioned that contribute to an appraisal that he is a crank.
>
> I think another good contribution comes from his response,
> for example, to the question I posed that he write out
> formally the statement "all paths are contained in a
> countable set", given predicate symbols for "path" and
> "countable set".
I offered to work with him to formalize his argument. I even posted
pages worth of precise definitions of the various terminology that was
being slung by him. He was welcomed to to accept or to modify any of
these defintions for whatever his purposes. The result was him
breaking down into yet further incoherence, including intractable use/
mention confusions and even an argument by him that BLATANTLY violated
the the transitivity of identity.
> He confused this with (1) a different statement entirely
> (2) proving something rather than just stating it and (3)
> a question that could be evaluated by a formal theorem
> proving or checking system.
I hope you do understand that these rounds of confusion generated by
him, along with his remarkable intractability, have been going on for
YEARS, over dozens and dozens of threads, some of them thousands and
thousands of posts long.
You want to hear a REALLY funny one? Recently he posted to a
professional logic forum. He claimed in his post that it is
statistically evident that there is a movement away from acceptance of
infinite sets. Another poster asked for statistical evidence. WM's
reply started with (I'm paraphrasing), "In my experience..." and went
on to give ANECDOTAL evidence. And what is even MORE funny is that his
ANECDOTAL evidence is all of the form that his STUDENTS (at, what I
understand to be, a kind of trade school college) are acquiescent to
accept his claims about such things. In other words, not only is he
foolish enough, right in front of a forum of professional logicians,
to take anecdotal evidence as statistical evidence, but even his
anecdotal evidence is that there is a LACK of objection from his own
students, who we may suppose are not educated in set theory OTHER than
by WM's own idiosyncratic approach.
In sum, the dude's bad news. Don't hope that he'll actually REASON
with you.
MoeBlee
calvin...@gmail.com
> It's been offered to him a hundred times that he is welcome to state
> whatever alternative syntax, primitives, axioms, definitions, and
> rules of inference he wishes to present. His ordinary response is that
> he is not interested in formal theories.
I've seen that. Another sign of crankhood. Maybe he
should be pushed on this point, instead. If we all "gang up"
on him (in a most friendly way) about this...?
I am specifically referring to the clarity of expression that
formalization allows, not the rigor that it adds to argument
itself.
"If your thoughts are rubbish merely,
don't express yourself too clearly"
---Piet Hein
> I hope you do understand that these rounds of confusion generated by
> him, along with his remarkable intractability, have been going on for
> YEARS, over dozens and dozens of threads, some of them thousands and
> thousands of posts long.
I did look back into the history a little bit, a couple of years at
least.
> You want to hear a REALLY funny one? Recently he posted to a
> professional logic forum. He claimed in his post that it is
> statistically evident that there is a movement away from acceptance of
> infinite sets.
His post in FOM to which you refer is in fact how
I came to meet his acquaintance for the first time. I
emailed him to complain, and after a lengthy go-round in
email found all these other threads in other fora.
> And what is even MORE funny is that his
> ANECDOTAL evidence is all of the form that his STUDENTS (at, what I
> understand to be, a kind of trade school college) are acquiescent to
> accept his claims about such things. In other words, not only is he
> foolish enough, right in front of a forum of professional logicians,
> to take anecdotal evidence as statistical evidence,
I don't know that this is so bad. In the lack of statistical
evidence collected by methodologically sound techniques,
anecdotal evidence is maybe the best one has to go on
to provide hints. I don't want to disparage "anecdotal
evidence" entirely. But this particular evidence is not very
good of course.
I'd like to hear from some of these students! They are
apparently being maltreated, if only because they
are getting little idea of what legitimate mathematical
reasoning, and its necessary prerequisite of adequate
clarity, are like.
If you visit his website you can see that he is ranked
very highly by his students!
> but even his
> anecdotal evidence is that there is a LACK of objection from his own
> students, who we may suppose are not educated in set theory OTHER than
> by WM's own idiosyncratic approach.
Well, his claim is that all the others (those in the "old generation"
and those under their influence) have been brainwashed for a
long period of time. We don't believe that (particularly since
there are all sorts of technically competent logicians who hold
severely restricted metaphysical views about infinity, but who
at least express themselves clearly in their technical work), but
there is at least some semblance of reasoning involved there.
Ralf Bader
> On 25 Mrz., 11:02, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > *As Brouwer claimed and as Weyl approved:
>>
>> An unfortunate source for support since the diagonal argument is
>> intuitionistically valid.
>
> To be valid means it can be derived using logic. It does not mean the
> contrary cannot be derived.
If the contrary can also be derived intuitionistically then something is
wrong with intuitionism.
> This comes in when "finished" infinity is
> concerned.
>
> Without finished infinity the diagonal argument does not show that the
> diagonal number is not in the sequence, because the sequence is never
> complete.
Of course it can be shown, under appropriate circumstances, that some number
does not appear in a given sequence. Intuitionism would be in a really
hopeless position if he couldn't do that.
>> --
>> Aatu Koskensilta (aatu.koskensi...@uta.fi)
>>
>> "Wovon man nicht sprechen kann, darüber muss man schweigen"
>> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
>
> An unfortunate source for support:
> Set theory is wrong
> - Ludwig Wittgenstein
Oh yes, that is of course convincing beyond any doubt. Btw, from
Wittgenstein's "Bemerkungen" one can cull out almost infintely many
quotations that show what Wittgenstein would have thought
about "Matherealism", if he had known about that braindead idiocy.
Oh, and to say something about the subject of this thread, on the frontpage
of the German translation of Conway's "On numbers and games" a really big
binary tree is shown. It is used to describe the same structure about which
Knuth has written a booklet on "Surreal numbers". Two authors writing in
booklength about a subject which Mueckenheim asserts can be seen as
wretched by some three lines of reasoning. The inevitable conclusion is
that either John Conway and Donald Knuth both are dunces or Mueckenheim is
one.
Oh, and Brouwer was able to write: "From the present point of view of
intuitionism therefore all mathematical sets of
units which are entitled to the name can be developed out of the basal
intuition, and this can only be done by combining a finite number of times
the two operations: “to create a finite ordinal number” and “to create the
infinite ordinal number ω”;", p. 58 in
http://www.ams.org/bull/2000-37-01/S0273-0979-99-00802-2/S0273-0979-99-00802-2.pdf
What is at stake is not the "existence" of that infinite ordinal, but
certain ways of reasoning that involve omega (for example arguments relying
on the smallest counterexample when proving an assertion about natural
numbers). But Mueckenheimian pseudoarguments do not touch that
intuitionistically forbidden realm (or more generally the realm where
infinity could have the capacity to generate problems), because they
already start as chaotic nonsense and then deteriorate quickly.
--
How lucky we are that Cantor introduced curly brackets! But it was no[t]
he who introduced the silly distinction between a and {a} that enables
so called mathematicians to build card houses on nothing.
(Prof. Dr. W. Mückenheim, mathematical mastermind of "Augsburg University of
Applied Science" , in sci.math)
Owen Jacobson
> On 24 Mrz., 23:54, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
>> On 24 Mrz., 16:07, WM <mueck...@rz.fh-augsburg.de> wrote:
>>
>>> ... all paths belong to a countable set. ...
>>
>> Okay. Just one simple question here. Let P(x)
>> mean "x is a path". Let CS(x) mean "x is a
>> countable set".
>>
>> WM (not anyone else please!), will you please write
>> out, formally in first order logic, the statement you have
>> made above? Thank you.
>
> If you need this eduction I will refer you to
> http://de2.metamath.org/index.html
> Your results then can even be approved by automata.
>
> I am not going to prove 1+1 = 2 or anything else by that language. And
> if you think that a statement in that language, approved by a machine,
> guarantees truth, then you are wrong.
>
> The direct statement that you want to have formalized is: The set of
> path can be put in a bijection with N. I do not pretend that this can
> be done.
Discussion over, then. This is a necessary property for a "countably
infinite" set.
> What I have shown is that the set of paths cannot be larger than a
> countable set, namely the set of all nodes respectively all lines in
> the tree.
Leaving aside whether you've shown that or not, proving that a set is
no larger than N is not the same as proving it to be the same size as
N. Proving it to be the same size as N would involve either invoking
the continuum hypothesis (which would mean acknowledging that modern
infinite set theory is sound) or constructing an injection from N to
the set of paths in an infinite binary tree and an injection from the
set of paths in an infinite binary tree to N.
> Presumably this proof would need some hundred pages in FOPL+ZFC. Or
> maybe it cannot be formalized at all. I don't know and I am not
> interested in that topic
Definitions:
1: R is a node.
2: ancestors(R) is {}.
3: For every node x, y and z are nodes such that:
a) x != y and x != z
b) y != z
c) y ∉ ancestors(x) and z ∉ ancestors(x)
d) ancestors(y) = ancestors(z) = ancestors(x) ∪ {x}
4: For a set of nodes P, P is a path if and only if P contains R and
for every pair of distinct nodes x ≠ y in P, y ∈ ancestors(x) or x ∈
ancestors(y).
5: For path P, P is an infinite path if and only if for every node x ∈
P there is a node y ∈ P such that x ∈ ancestors(y).
(There may be a simpler set of definitions. The goal for me was to
take, say, R = 0 and construct the binary tree (0->1,2; 1->3,4; 2->5,6;
...) by induction, and to provide a hopefully-correct and
hopefully-formal definition of an infinite path. {0, 1, 3, ...} should
be an infinite path in the above tree, and {0, 1} and {0, 1, 3} are
both finite paths in it.)
Conjecture:
There exists a bijection between the set of all infinite paths and the
set of natural numbers N.
Would you accept a formal, computer-checkable proof that this
conjecture is false in ZFC as proving that your conjecture is false in
ZFC? The above conjecture and definitions are what everyone else in
this thread hears when you say "The complete infinite binary tree has
only countably many infinite paths."
> What I know is that the set of all infinite sequences of bits (or
> digits) has not more elements than the set of all natural numbers,
> when proved in the way I did it here.
Depressingly, there's a kernel of a worthwhile discussion in this
sentence. However, you don't seem to want to discuss alterate set
theories where this might be true. It's provably false in ZFC.
-o
Tim Little
> The one thing I don't understand about these WM binary tree threads
> is how WM, as an ultrafinitist who believes that numbers as large as
> a googolplex don't exist, can believe that an infinite tree even
> exists in the first place, much less debate over whether it's
> countable or not.
That has been asked before. If I recall correctly, his answer was
along the lines of it merely being a premise for a reductio ad
absurdum argument.
He is not merely an ultrafinitist: he argues that the very concept of
infinity itself has been *proven inconsistent*. As one example, he
claims that he has proven ZF inconsistent.
> I can try to come up with a theory in which infinite trees can exist
> yet are countable, but that would be pointless unless I can see how
> infinite trees can even exist in an ultrafinitist theory.
Since his claim concerns ZF, inventing a new system is useless for
modelling his beliefs.
> For example, ZF-Powerset+~Powerset can prove that countably infinite
> sets exist, but can't prove that any uncountable sets exist.
How certain are you of that? I have an idea of how I might go about
proving the opposite. I'll see how it turns out when I have a bit of
spare time.
- Tim
lwa...@lausd.net
> On 2009-03-25, lwal...@lausd.net <lwal...@lausd.net> wrote:
> > The one thing I don't understand about these WM binary tree threads
> > is how WM, as an ultrafinitist who believes that numbers as large as
> > a googolplex don't exist, can believe that an infinite tree even
> > exists in the first place, much less debate over whether it's
> > countable or not.
> That has been asked before. If I recall correctly, his answer was
> along the lines of it merely being a premise for a reductio ad
> absurdum argument.
OK, I see. So it's sort of like "assume infinite sets exist. Then the
binary tree is both countable and uncountable -- contradiction. Hence
only finite sets exist."
> > For example, ZF-Powerset+~Powerset can prove that countably infinite
> > sets exist, but can't prove that any uncountable sets exist.
> How certain are you of that? I have an idea of how I might go about
> proving the opposite. I'll see how it turns out when I have a bit of
> spare time.
That might be interesting to see. I wonder whether ~Powerset will be
really used in Little's proof, or would it work in ZF-Powerset alone.
If the former, then it seems interesting that, in ZF-Powerset, we'd
have
proofs of both "Powerset -> uncountable sets exist" as well as
"(~Powerset) -> uncountable sets exist," yet not a proof of just
"uncountable sets exist," which seems a bit odd in classical logic.
If the latter, then we see that Powerset isn't responsible for the
existence of uncountable sets after all. This might ruin WM's theory,
but it might help out tommy1729's theory in another thread (a thread
of which Little is already aware, I believe). For tommy1729 has
desired a mereological theory in which there's no evident way to
form a Powerset, yet he still wants uncountable sets, including the
set R of real numbers.
Speaking of mereology, it's a bit difficult to tell whether WM would
support a mereological theory. Notice that from Ralf Bader's sig:
"How lucky we are that Cantor introduced curly brackets! But it was no
[t]
he who introduced the silly distinction between a and {a} that
enables
so called mathematicians to build card houses on nothing.
(Prof. Dr. W. Mückenheim, mathematical mastermind of "Augsburg
University of
Applied Science" , in sci.math) "
which sounds mereological. I think it was Herbert Newman who
also mentioned mereology in a WM thread. So perhaps if
mereology doesn't work to give us R in tommy1729's theory,
maybe we could try mereology in WM's theory instead (in
which we don't even want infinite N, much less uncountable R).
Virgil
<dfd24aae-8d33-44b4...@y9g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Except that the only contradictions are between WM's hidden assumptions
and the actual propesrties of an actual complete infinite binary tree.
There are no contradictions without WM's hidden assumptions.
According to WM's claims elsewhere,10^100^100^100^100 cannot exist until
a unary representation of it also exists.
Which clearly rules it out.
So that now, at least in Wm's mathUnrealism, we have numbers which
simultaneously exist and do not exist.
And since one can produce perfectly reasonable definitions of both the
cardinal number aleph_0 and the ordinal number omega, WM has now
conceded their existence as well.
> >
> > Perhaps WM can settle this once and for all. So now
> > I ask of WM, since you believe that the infinite tree
> > has countably infinitely many paths (some of which
> > terminate at nodes, others don't),
>
> None of them terminates.
>
> > how many of those
> > countably infinitely many paths (terminate at nodes
> > and) have exactly googolplex (10^10^100) edges?
>
> All are infinite. Each one has aleph_0 nodes. This is assumed for the
> tree. Otherwise the discussion would be boring.
How many paths in a compete infinite binary tree have infinitely many of
both left branching edges and right branching edges? According to
Cantor, there should be uncountably many of them.
Virgil
<3e6158c6-2dea-4fb9...@f11g2000vbf.googlegroups.com>,
"calvin...@gmail.com" <calvin...@gmail.com> wrote:
> On Mar 25, 4:47 pm, Virgil <Vir...@gmale.com> wrote:
> > > How do we even know he convinced any students?
> >
> > We don't, though WM claims to have browbeaten them enough to get them
> > to regurgitate bits of his credo on exams.
>
> Actually, if I recall, he explicitly claims *not* to have
> browbeaten them in this manner.
From the manner WM exhibits here, I take leave to doubt that he would
not downgrade a student for disagreeing with him.
>
> > > It seems
> > > that at these time he accepts Cantor's argument, yet
> > > also accepts an argument establishing the opposite
> > > conclusion.
> >
> > If in his "system", he accepts simultaneously both such a P and such a
> > not P, I don't see why he should object to anything ever.
>
> Why not?
Given both P and not-P, in any reasonably comprehensive logical system,
one can prove anything and its negation.
>
> In any case, presuming he accepts both some P and also ~P,
> maybe you don't see why he objects to other things
> occasionally, but should we first not establish whether
> or not he does accept both some P and also ~P (not necessarily
> absolutely simultaneously, by the way!)
My "simultaneously" only meant "in the same logical system".
Sorry if that was not clear.
WM
wrote:
> On Mar 25, 7:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > WM (not anyone else please!), will you please write
> > > out, formally in first order logic, the statement you have
> > > made above? Thank you.
>
> > If you need this eduction I will refer you tohttp://de2.metamath.org/index.html
> > Your results then can even be approved by automata.
>
> How could my results in translating an English statement
> into formal logic be "approved" of by automata?
If the statement was wrong, the machine would detect it (in some
cases).
>
> > I am not going to prove 1+1 = 2 or anything else by that language. And
> > if you think that a statement in that language, approved by a machine,
> > guarantees truth, then you are wrong.
>
> I did not ask you to *prove* anything. Simply to *state* the English
> statement I have in formal logic.
So you want to see that thereis a countable set y such that forall x :
p(x) is an element of y? Or would you prefer when I said that forall x
thereis an y ...? (Certainly you would prefer when I said the latter,
but I don't.)
For some education on the use of quantifiers I could refer you to my
lesson
http://www.hs-augsburg.de/~mueckenh/Mathematerial/M01.PPT#299,5,Folie
5
but I am afraid without explaining words it does not teach very good.
>
> > The direct statement that you want to have formalized is: The set of
> > path can be put in a bijection with N. I do not pretend that this can
> > be done.
>
> It certainly is not. I did not ask anything of the sort. I gave you
> the
> formalization of countable set as the unary predicate "CS". I did
> not mention N, I did not mention bijections.
You mentioned countable. That implies N and bijections because it is
defined as possibility to put the set in bijection with N.
>
> But wait. Are you now saying that the very statement "The set of
> paths can be put in a bijection with N" cannot even be *formalized*?
> What? What?
The set of paths cannot be put in bijection with N. The paths cannot
be enumerated. The proof is as follows:
The nodes can be enumerated. The number of paths can be shown to be
limited by the number of nodes (via the number of lines).
Regards, WM
WM
wrote:
> On Mar 25, 3:33 pm, Virgil <Vir...@gmale.com> wrote:
>
> > > You are in error. Everybody not blinded by the blinkers of transfinite
> > > set theory will understand very easily that I am right.
>
> > Whom, excluding the captive audience of your students, have you ever
> > convinced of the truth of your foolishness?
>
> How do we even know he convinced any students?
Recently I got a mail from a professional mathematician:
Ich teile uebrigens mit einem Ihrer Studenten zeitweilig das Buero und
Sie haben recht:
Ihm leuchtet alles ein, was Sie sagen.
Nun, warum haben gerade Berufsmathematiker da Probleme?
Not a very good statistic, I agree. But if you could look at the tree
before having been spoilt by set theory then you would not have any
problems. It is merely a very small group of matheologists who "want
spread the word".
> We need some such student, who has for his teacher
> a great figure, if he is right, to stand up and say he
> was convinced. Is no such student reading these
> heroic epics of his master? (Unforunately, it may
> be hard to rule out sock-puppets should such a
> student show up here).
Of course. But why don't you come out of your ivory tower and ask
people who can think logically without calling them logicians. You
would be surprised.
Regards, WM
WM
> > Since you still cannot successfully fault the Cantor argument, no
> > counterargument of your own can override it.
>
> Well, perhaps no such argument can *override* Cantor's,
> but it can at least meet it on an equal footing.
> It seems that (from time to time anyway) WM's
> argument is that set theory is inconsistent. It seems
> that at these time he accepts Cantor's argument, yet
> also accepts an argument establishing the opposite
> conclusion.
If we accept Cantor's argument, then we get to the concusion that the
binary tree has uncountable many paths. I have shown in the original
post that the binary tree cannot have uncountably many paths. This
dilemma is caused by the acceptance of finished infinity. Nothing else
could be expected by such a muddled concept.
My conclusion is: There is no finished infinity.
>
> If this is the case, he should clearly say so, and then
> suggest either a new, hopefully consistent axiomatization
> for set theory (or something similar, appropriate for
> doing mathematics),
There is no axiomatization required for mathematics. I hesitate to use
this word that has become disreputable by the work of modern
"logicians". But if you like to cast the basic truth in the form of
axioms, then include the axiom:
I s a number.
If you have n strokes, then you can make n + 1 strokes. In his theory,
basically due to Paul Lorenzen, you will not have a set of all
numbers. But you will have all numbers that ae required. I think it
was this concept that Abraham Robinson (inventor of non-standard
analysis and pupil of Fraenkel, a front man of set theory) addressed
when he said: (i) Infinite totalities do not exist in any sense of the
word (i.e., either really or ideally). More precisely, any mention, or
purported mention, of infinite totalities is, literally, meaningless.
(ii) Nevertheless, we should continue the business of Mathematics 'as
usual', i.e., we should act as if infinite totalities really
existed." (In: Formalism 64, auch abgedruckt in Robinson 1979, p.
507.)
Exactly in this way we can proceed.
> or instead, a way of dealing with
> inconsistency so that the logic does not "explode'
Drop finished infinities. Then logics can be used for all remaining
branches of mathematics. (In other sciences it never lost it
accuracy, because none of them uses finished infinities. Have you
ever wondered why no science is interested in the results of modern
logics?)
Regards, WM
calvin...@gmail.com
> > > We don't, though WM claims to have browbeaten them enough to get them
> > > to regurgitate bits of his credo on exams.
>
> > Actually, if I recall, he explicitly claims *not* to have
> > browbeaten them in this manner.
>
> From the manner WM exhibits here, I take leave to doubt that he would
> not downgrade a student for disagreeing with him.
Maybe your doubt is reasonable, but what you said
above is not that you suspected that he has browbeaten
his students, etc., but rather that he *claims* to have browbeaten
them, etc. I assume you mean that he claims to have down
something to his students (which he describes as properly
educating them) but which you describe as browbeating. I
think you have to be very careful with this way of offering
opinions while performing indirect quotation.
(Regarding quotation: you never answered about how the
"cup" symbol disappeared in your quotation from Wikipedia.
Was it a browser problem?)
> > > If in his "system", he accepts simultaneously both such a P and such a
> > > not P, I don't see why he should object to anything ever.
>
> > Why not?
>
> Given both P and not-P, in any reasonably comprehensive logical system,
> one can prove anything and its negation.
Are you saying that this is currently the case, or that this
must necessarily be the case? How familiar are you with
the development of the various forms of paraconsistent
logic? Maybe none of them are "reasonably comprehensive"
as of yet?
Although I myself don't believe in true contradictions as
Graham Priest claims to, I think I have to agree that it
would be a good thing if a logical system did not explode
as soon as it was able to prove one contradiction.
WM
wrote:
Agreed. But no path ends there. So it is not important.
> > (1,1), (1,2)
> > (2,1), (2,2), (2,3), (2,4)
> > ...
> > (n,1), ..., (n, 2^n)
>
> So those are the nodes. {(n,m)| n,m in N, m<=2^n}
>
> > Every node with even second co-ordinate has value 1, every node with
> > odd second co-ordinate has value 0.
>
> Labeling function L( (n,m) ) = m+1 mod 2
>
> > Construct the complete tree (from the path p_0 = 0.000...) by using
> > all paths that lead to a node with value 1 and afterward having only
> > nodes with value zero.
>
> Okay, well, a tree is a set of edges. Let us instead add the edges.
> That is more definite and precise. Paths result by supervening on
> the pattern of edges.
That is possible. But it will refer to my proof (B).
>
> Can you rigorously specify the edges that are being added, in
> a manner such as I have done above (instead of just with pictures
> and informal words).
As well as labelling the nodes you can also label the edges leading to
the nodes by the same scheme.
(1,1), (1,2)
(2,1), (2,2), (2,3), (2,4)
...
(n,1), ..., (n, 2^n)
>
So use just
the edges {(n,m)| n,m in N, m<=2^n}
> When you do that, you will see there are more paths that you
> attempted to limit the set of paths to.
I do not see that. I see that there is an nfinite but countable set of
lines. I wold be nice if you said whether you could agree.
>
> > Then all the paths (except p_0) can be labelled by their last node
> > that has value 1.
>
> Here, I think, is your problem. You are not noticing that when
> you add a single path, you are actually adding many paths.
What many of paths do I add when I add 0.1000...?
> The net result in the end is that after all your paths have been
> added, there are many other paths that exist, because of the
> edges that have been added to make the paths, which you
> have not noticed.
When you consider my proof (A), then you see that I add always exatly
one path.
You can even consider all paths that end on some node (regardless of
its value) and then always turn left. I this way many paths are added
infinitely often, like
0.1|000..., 0.10|000..., 0.100|000
where "|" appears behind the node of destination.
>
> Perhaps the problem arises because of using the idea of one
> tree as a limit of the other. I don't think that is necessary here.
> As above for the nodes, we should be able to define the final
> tree explicitly without using a limit.
The tree consists of all lines that are constructed by paths that end
by infinitely many nodes. That is not a limit. Simply take the set of
all those paths.
>
> Here is the problem you may have with using a limit. The
> limit tree is a union of all the edges in the sequence of trees.
> That is indeed true. If there is an edge in the limit tree, that
> edge must exist in some tree in the sequence. HOWEVER,
> the same is not true about paths. Paths exist in the limit
> tree that do *not* exist in any tree in the sequence. This
> is because the paths "interact" with one another in the way
> edges do not.
For that argument I deviced my proof (A). Paths may be imagined to do
this or that. They cannot but follow one line in the tree. And there
ae all lines in the tree. There is no infinite sequence of bits that
is lacking. Neverteless all this is constructed by a countable number
of paths.
>
> Every path you add is of the form a finite sequence of 0's
> and 1's, following by infinite sequence of 0's. However,
> there are paths in the final tree that are not like this. That is
> because the paths interact with one another. You must look
> at the individual edges and how then can be made into
> paths in the final tree. You do not add paths, you add edges.
> Paths come FOR FREE. You cannot pick and choose the
> paths you add. They sneak in as sets of joined edges and
> there is nothing you can do about it.
Please be more clear. And, as you criticize the edge construction,
please refer to proof (A).
You want to prove that there are more paths in the tree than lines
paved by a countable set of paths.
Regards, WM
WM
wrote:
> On Mar 25, 6:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > It's been offered to him a hundred times that he is welcome to state
> > whatever alternative syntax, primitives, axioms, definitions, and
> > rules of inference he wishes to present. His ordinary response is that
> > he is not interested in formal theories.
>
> I've seen that.
That is nonsense. MoeBlee continued to utter insults. In those cases I
decline to continue the discussion.
Most theorems of mathematics have not been formalized. In my opinion
such a formalization is without any value, because many proofs about
absolute nonsense like inaccesible cadinals have been formalized.
Therefore I am not motivated to start that business. But if ZFC + FOPL
is the "fundament of mathematics" then it should be possible to treat
the binary tree. Evereybody who is capable of doing so is invited to
do.
Regards, WM
WM
> > The direct statement that you want to have formalized is: The set of
> > path can be put in a bijection with N. I do not pretend that this can
> > be done.
>
> Discussion over, then. This is a necessary property for a "countably
> infinite" set.
And if set theory is free of contradictions, then every other set is
either finite or larger than a countable set. But just this latter is
contradicted by my proof in the first posting of this thread. What do
you conclude?
>
> > What I have shown is that the set of paths cannot be larger than a
> > countable set, namely the set of all nodes respectively all lines in
> > the tree.
>
> Leaving aside whether you've shown that or not, proving that a set is
> no larger than N is not the same as proving it to be the same size as
> N.
It is proving that the set does not contain more elements than N
contains. In the present case it is proving that most real numbers do
not exist as distinct infinite sequences of bits.
Further we know that most real numbes have no finite definition: The
set of finite words over a finite alphabet is countable. The set of
meanings of these words, i.e., the set of languages, is countable. The
set of finite alphabets is countable. The cartesian product of these,
and possibly some further features, is countable.
Therefore we know that most of what set theorists call the real
numbers do not exist at all, neither as infinite bit sequences nor as
finite definitions like pi, e, or Liouvilles numbers.
> Proving it to be the same size as N would involve either invoking
> the continuum hypothesis (which would mean acknowledging that modern
> infinite set theory is sound) or constructing an injection from N to
> the set of paths in an infinite binary tree and an injection from the
> set of paths in an infinite binary tree to N.
You can see my proofs. There is no continuum hypothesis necessary nor
a bijectionwith N. All we need is faith in the truth that an infinite
sum of zeros is not larger than countable infinity.
As I said above: There is no bijection. Nevertheless the set of paths
is not arger than the set of natural numbers.
>
> Would you accept a formal, computer-checkable proof that this
> conjecture is false in ZFC as proving that your conjecture is false in
> ZFC? The above conjecture and definitions are what everyone else in
> this thread hears when you say "The complete infinite binary tree has
> only countably many infinite paths."
>
> > What I know is that the set of all infinite sequences of bits (or
> > digits) has not more elements than the set of all natural numbers,
> > when proved in the way I did it here.
>
> Depressingly, there's a kernel of a worthwhile discussion in this
> sentence. However, you don't seem to want to discuss alterate set
> theories where this might be true. It's provably false in ZFC.
ZFC is of no interest to me. I am interested in the contradiction.
Of course, if we take another stand point, then we can prove tha there
are far more real unvers than natural numbers. Consider only the
harmonic sequence (1/n) and the rationals between the points of the
sequence.
It is impossible to consistently enumerate the infinite. That is my
point.
Regards WM
calvin...@gmail.com
> > How could my results in translating an English statement
> > into formal logic be "approved" of by automata?
>
> If the statement was wrong, the machine would detect it (in some
> cases).
I suppose it is possible that it could correct my translation
from English to formal logic, but not according
to any rigorous rules I am aware of. In any case I could
not find the place to type in the English for translation
to formal logic: can you please provide the precise url
of a page that has a text box I can use?
Anyway, I was obviously asking for your translation, not
a machine's.
> > I did not ask you to *prove* anything. Simply to *state* the English
> > statement I have in formal logic.
>
> So you want to see that thereis a countable set y such that forall x :
> p(x) is an element of y? Or would you prefer when I said that forall x
> thereis an y ...? (Certainly you would prefer when I said the latter,
> but I don't.)
Well, for the time being I just wanted to see if you could
say anything at all clearly. I guess I'll just give up for now.
Still haven't got it, since "p" as I gave it to you was a predicate
symbol, not a function symbol.
> > It certainly is not. I did not ask anything of the sort. I gave you
> > the
> > formalization of countable set as the unary predicate "CS". I did
> > not mention N, I did not mention bijections.
>
> You mentioned countable. That implies N and bijections because it is
> defined as possibility to put the set in bijection with N.
No, that was all finessed because I gave you a the formalization
of countable set as the unary predicate "CS".
> Recently I got a mail from a professional mathematician:
> Ich teile uebrigens mit einem Ihrer Studenten zeitweilig das Buero und
> Sie haben recht:
> Ihm leuchtet alles ein, was Sie sagen.
> Nun, warum haben gerade Berufsmathematiker da Probleme?
Could you please translate? Not that it will prove anything
of course.
> Not a very good statistic, I agree. But if you could look at the tree
> before having been spoilt by set theory then you would not have any
> problems. It is merely a very small group of matheologists who "want
> spread the word".
Statistically, it's not a small group that disagrees with you. It
is
virtually everyone who understands the basic mathematics involved.
> But why don't you come out of your ivory tower and ask
> people who can think logically without calling them logicians. You
> would be surprised.
I can't very well do this, because I can't make any sense
out of your "construction" of the tree. But certainly when
I describe to them what I consider to be a "complete
infinite binary tree", such people immediately agree (once
they understand what "uncountable" means), that the
set of all paths in it is uncountable.
Many of them do tend to take what would be called an
"anti-realist" attitude toward infinities, however. Not because
there are contradictions, but because it seems excessively
fanciful and useless, and also in some cases because of
epistemological problems (along the lines of what is commonly
referred to as Benacerraf's Dilemma (see his paper
"Mathematical Truth")). How much of this was
"brow-beating", I do not know.
> If we accept Cantor's argument, then we get to the concusion that the
> binary tree has uncountable many paths. I have shown in the original
> post that the binary tree cannot have uncountably many paths. This
> dilemma is caused by the acceptance of finished infinity. Nothing else
> could be expected by such a muddled concept.
>
> My conclusion is: There is no finished infinity.
Too bad, it appears my speculation was wrong about the
possibility you were a dialetheist. Of course I guess I
really knew that was not so, but it would have been
more interesting, and also probably more logical.
I assume you are now saying that you *do* accept Cantor's
argument, and thus therefore the axioms he (or his
modern equivalents) argues from must be inconsistent,
since, yes, you derive the opposite conclusion as well.
However, this does not entitle you to throw out, in particular,
the axiom of infinity from ZFC. If you have an inconsistent
set (which none of us feels that you have established, of
course), there can be plenty of different ways to restore
the consistency (if that is what you feel to be necessary).
So your argument is still so far very incomplete (as well
as being wrong anyway as far as it goes).
But forget about even going all the way and showing
infinity is bad stuff. You do realize how utterly famous
you would be if you could merely prove clearly
that ZF was inconsistent!? Wow. You should really
try to write this up *clearly* instead of
wasting your time in these silly newsgroups. Who would
have the last laugh then, and also, more importantly,
what a great service to mankind this would be!!
The fact that you don't seem to care about doing this,
offhandedly suggesting that perhaps your proof would
be way too long in formal ZFC, is very strange and
disheartening. I mean, you think your proof is okay
and its like only a few lines long. Do a little more work
and put it into ZFC. You will be in the Pantheon of
great mathematical logicians.
You don't want to bother? Okay, it starts to look like
you are just a big troll, I am afraid.
> Drop finished infinities. Then logics can be used for all remaining
> branches of mathematics. (In other sciences it never lost it
> accuracy, because none of them uses finished infinities. Have you
> ever wondered why no science is interested in the results of modern
> logics?)
I don't know for sure that this is true, but most science does
not apear to *need* very much mathematics, at least not very
much infinity, for sure. But this is not because infinity
is incoherent, and it does need some infinity. It needs the real
and complex numbers, various function spaces built up from
these. Perhaps it could get by with somewhat less, the
constructive or computable reals for example, but it would
likely be very difficult and unpleasant. There are lots of
people looking into this sort of thing today: people like Stephen
Simpson and Solomon Feferman.
David C. Ullrich
<muec...@rz.fh-augsburg.de> wrote:
>On 25 Mrz., 10:21, David C. Ullrich <dullr...@sprynet.com> wrote:
>> On Tue, 24 Mar 2009 04:56:33 -0700 (PDT), WM
>>
>> <mueck...@rz.fh-augsburg.de> wrote:
>> >The complete infinite binary tree has only countably many infinite
>> >paths.
>>
>> Why do you continue this? Even if you were right,
>> surely it's clear by now you're not going to convince
>> anyone.
>
>You are in error. Everybody not blinded by the blinkers of transfinite
>set theory will understand very easily that I am right.
Huh? I claim it should be clear you're not going to convince
anyone. Saying that whoever _will_ understand you're right
is not much of a response. Who _has_ _been_ convinced you're
right?
Not including students in your classes or former students.
They don't count, for obvious reasons.
>And here I
>have given two slightly different version sof the proof that may
>contribute to convince some more readers.
>But even those who already have been brain-washed by finished
>transfinity do not agree. Some say that my proof is wrong --- of
>course without specifying why, only arguing that Cantor's diagonal
>proof says the contrary. (But that is no argument when the
>applicability of logic to infinite sets is under investigation as here
>is the case*). Others say that my arguments are not new.
It's curious that you don't mention the large number of "others"
who explain exactly where the _error_ in your arguments is.
Or rather the _gap_. You say
"This procedure, even when applied infinitely often, cannot but yield
the result 0 respectively a countable number of lines."
over and over, but it's simply not true. Your asserting it does
not make it true. The fact that it seems clearly true to you
does not make it true. You need to give a _proof_ of this,
and you've never done so - in each of your attempts there's
always a similar unjustified assertion.
>But, as
>mentioned above, everybody with undeformed brain understands my
>arguments.
Guffaw. That's very convenient. Of course it follows that every
competent mathemtician on the planet has a deformed brain.
No reason anyone would find _that_ assertion hard to swallow.
>*As Brouwer claimed and as Weyl approved: "Classical logic was
>abstracted from the mathematics of finite sets and their subsets ....
>Forgetful of this limited origin, one afterwards mistook that logic
>for something above and prior to all mathematics, and finally applied
>it, without justification, to the mathematics of infinite sets. ... As
>Brouwer pointed out this is a fallacy, the Fall and Original sin of
>set theory even if no paradoxes result from it." [H. Weyl,
>"Mathematics and logic: A brief survey serving as a preface to a
>review of The Philosophy of Bertrand Russell", American Mathematical
>Monthly 53: 2–13, 1946.]
That's fascinating, but of no relevance whatever. A quote
where one of those guys stated that that tree had only countably
many paths would be relevant - got one of those?
>Regards, WM
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
WM
> On Mar 25, 9:01 pm, Tim Little <t...@little-possums.net> wrote:
>
> > On 2009-03-25, lwal...@lausd.net <lwal...@lausd.net> wrote:
> > > The one thing I don't understand about these WM binary tree threads
> > > is how WM, as an ultrafinitist who believes that numbers as large as
> > > a googolplex don't exist, can believe that an infinite tree even
> > > exists in the first place, much less debate over whether it's
> > > countable or not.
> > That has been asked before. If I recall correctly, his answer was
> > along the lines of it merely being a premise for a reductio ad
> > absurdum argument.
>
> OK, I see. So it's sort of like "assume infinite sets exist. Then the
> binary tree is both countable and uncountable -- contradiction. Hence
> only finite sets exist."
Yes, you got it. But instead of saying "only finite sets exist" which
would most mathematicians immediately cause to stop reading, one
should say sets can be potentially infinite in the sense infinity was
applied over thousands of years: oo. (Unless MatheRealism is invoked.
Then sets of numbers are finite but values of numbers are not
limited.)
Regards, WM
calvin...@gmail.com
> > > 0 [say instead (0,
> Agreed. But no path ends there. So it is not important.
It provides a uniform naming scheme that makes it
much easier to make general statements about
the tree.
> > Can you rigorously specify the edges that are being added, in
> > a manner such as I have done above (instead of just with pictures
> > and informal words).
>
> As well as labelling the nodes you can also label the edges leading to
> the nodes by the same scheme.
> (1,1), (1,2)
> (2,1), (2,2), (2,3), (2,4)
> ...
> (n,1), ..., (n, 2^n)
>
> So use just
> the edges {(n,m)| n,m in N, m<=2^n}
No...
You know, a tree consists of a set of nodes, and a set of
edges that join the nodes. We only name the nodes so
that we can separate them out from one another in order
to specify the edges. So, to start with, I tried to get you
to name the nodes in a way that would be convenient for
later reference when the edges were added. So far so
good. However, those names strictly speaking said
nothing about the structure of the tree. All the real work
comes in giving the edges. That is what ultimately
defines the tree.
But have you done that? No. You merely *named*
the edges. That is not what you must do. You must
say what the edges are in terms of what nodes they
connect. Edges are a subset of the cartesian product
of the set of nodes with itself.
So, please continue, and specify what your tree is by
saying what the set of edges are, as a subset of NxN,
where N is the set of nodes defined above.
> > > Then all the paths (except p_0) can be labelled by their last node
> > > that has value 1.
>
> > Here, I think, is your problem. You are not noticing that when
> > you add a single path, you are actually adding many paths.
>
> What many of paths do I add when I add 0.1000...?
What I said here was based on trying to understand your
account, and I did not. It is true that if
you add paths only, you are adding only one
path at a time. But you will not add all paths
by doing that. The only way to add all paths is to
add edges one at a time in the right way. You will never get
all paths by trying to add a countable number of infinite
paths. You cannot construct the tree by adding paths.
You must construct it by adding edges. And in the right
way. You will never add a single infinite
path at any time. Only in the final tree will there be
infinite paths. It all sounds very paradoxical, must
drive you crazy. But the point is this: the final tree
has infinite paths in it because these paths are sets
edges that such that each edge is added at some
*different* finite stage, not along with all its pathmates
all at once. They are added one edge at each of stage
S(i), for some strictly increasing function S of i. So notice:
these infinite paths do not occur in any of the finite trees made
during the construction, because no such tree has all the edges
required! Only the final tree has all the edges required.
Infinity is tricky.
In any case, it is probably quite possible to define the
final tree not as a limit, but directly, by clearly specifying
the set of nodes as I have asked you to do. Once we
have all tree defined, by saying what its edges are,
we can then look to see which paths are in it, or not.
WM
wrote:
> On Mar 26, 2:36 am, Virgil <Vir...@gmale.com> wrote:
>
> > > > We don't, though WM claims to have browbeaten them enough to get them
> > > > to regurgitate bits of his credo on exams.
>
> > > Actually, if I recall, he explicitly claims *not* to have
> > > browbeaten them in this manner.
>
> > From the manner WM exhibits here, I take leave to doubt that he would
> > not downgrade a student for disagreeing with him.
>
> Maybe your doubt is reasonable, but what you said
> above is not that you suspected that he has browbeaten
> his students,
Look into the last proposed answer to question 12 in
http://www.hs-augsburg.de/~mueckenh/GU/Pr%fcfung%20GU0401.pdf
Everyone can have his own opinion! Could I punish students of
engineering sciences because they are not cleverer than most
contemporary mathematicians?
Regards, WM
calvin...@gmail.com
> Most theorems of mathematics have not been formalized. In my opinion
> such a formalization is without any value, because many proofs about
> absolute nonsense like inaccesible cadinals have been formalized.
> Therefore I am not motivated to start that business. But if ZFC + FOPL
> is the "fundament of mathematics" then it should be possible to treat
> the binary tree. Evereybody who is capable of doing so is invited to
> do.
Okay, I think I have had it. Sorry. First of all, absolute nonsense
that has been proven, I think you will find, has only been proven
from absolute nonsense as premisses.
Second of all, it is totally irrelevant that "most theorems of
mathematics have not been formalized". A very large
number of theorems (more importantly their proofs)
*have* been formalized, and many of these are far, far,
more complex that your alleged theorem.
If your theorem is truly a theorem, it would not be very hard to
formalize it. You will then have a rigorous, known down,
irrefutable,
argument that will establish your mind-blowing, game-changing,
world-shaking, conclusion.
The fact that these considerations do not in the least affect your
behavior on this issue *strongly* suggests that you are just some
kind of strange joker at best. Surely you must see this. So if you
are not, why not get busy, and make a formal proof of this alleged
contradiction?
An analogy to Hume's (?) dictum that "spectacular claims require
spectacular evidence" applies here, moreso than to the large
number of other proofs that have been formalized. Largely
because people think your non-formal proof is *nonsense*.
They cannot think that anymore if you provide a formal proof
(or even merely a *largely* formal proof). Don't you see this?
So what explains your lack of action?
Even if *you* don't like them in order to convince yourself, you
should see that such a formalization possesses *immense* value,
contrary to your remark above.
Unless you can come up with a plausible reply to these
considerations, I'm giving up on this whole thing.
WM
> > Recently I got a mail from a professional mathematician:
> > Ich teile uebrigens mit einem Ihrer Studenten zeitweilig das Buero und
> > Sie haben recht:
> > Ihm leuchtet alles ein, was Sie sagen.
> > Nun, warum haben gerade Berufsmathematiker da Probleme?
>
> Could you please translate? Not that it will prove anything
> of course.
We had some correspondence about my findings. He wrote: "I am sharing
the office from time to time with one of your [former] students [they
did some work in programming together, as far as I remember] and you
are right: he accepts everthing you say. Well, why have just
professional mathematicians problems there?"
>
> > Not a very good statistic, I agree. But if you could look at the tree
> > before having been spoilt by set theory then you would not have any
> > problems. It is merely a very small group of matheologists who "want
> > spread the word".
>
> Statistically, it's not a small group that disagrees with you. It
> is
> virtually everyone who understands the basic mathematics involved.
... who understands Cantor's proof (that is not very difficult) and
bases his intellectual life on its truth. That is the danger.
>
> > But why don't you come out of your ivory tower and ask
> > people who can think logically without calling them logicians. You
> > would be surprised.
>
> I can't very well do this, because I can't make any sense
> out of your "construction" of the tree. But certainly when
> I describe to them what I consider to be a "complete
> infinite binary tree", such people immediately agree (once
> they understand what "uncountable" means), that the
> set of all paths in it is uncountable.
Never! They cannot understand it because it is not there.
>
> Many of them do tend to take what would be called an
> "anti-realist" attitude toward infinities, however. Not because
> there are contradictions, but because it seems excessively
> fanciful and useless, and also in some cases because of
> epistemological problems (along the lines of what is commonly
> referred to as Benacerraf's Dilemma (see his paper
> "Mathematical Truth")).
I would very much like to do so. I it available in the net?
>
> > If we accept Cantor's argument, then we get to the concusion that the
> > binary tree has uncountable many paths. I have shown in the original
> > post that the binary tree cannot have uncountably many paths. This
> > dilemma is caused by the acceptance of finished infinity. Nothing else
> > could be expected by such a muddled concept.
>
> > My conclusion is: There is no finished infinity.
>
> Too bad, it appears my speculation was wrong about the
> possibility you were a dialetheist. Of course I guess I
> really knew that was not so, but it would have been
> more interesting, and also probably more logical.
No. I think logic can be used to discover the truth. But if finished
infinity exists, then it is clearly a matter of dialetheism, simply by
the wording.
>
> I assume you are now saying that you *do* accept Cantor's
> argument, and thus therefore the axioms he (or his
> modern equivalents) argues from must be inconsistent,
> since, yes, you derive the opposite conclusion as well.
>
> However, this does not entitle you to throw out, in particular,
> the axiom of infinity from ZFC. If you have an inconsistent
> set (which none of us feels that you have established, of
> course), there can be plenty of different ways to restore
> the consistency (if that is what you feel to be necessary).
> So your argument is still so far very incomplete
Of course you can always find some other condition to blame. But it is
not necessary to search so far. It is suficient to abolish finished
infinity --- and everything is safe and sound and in mathematics there
is no disagreement between intelligent enough people.
> (as well
> as being wrong anyway as far as it goes).
Here you enter the usual attitude. Please try to find a way of
arguing:
Either the tree cannot be construced by countably many paths.
Or there are paths that are not bound to the nodes constructed.
Or ... whatever
Only the argument that Cantor proved the uncountability is not
admitted when alkig about inconsistency.
>
> But forget about even going all the way and showing
> infinity is bad stuff. You do realize how utterly famous
> you would be if you could merely prove clearly
> that ZF was inconsistent!? Wow. You should really
> try to write this up *clearly* instead of
> wasting your time in these silly newsgroups.
I thought that you wanted to disproe me. Now you are shouting.
> Who would
> have the last laugh then, and also, more importantly,
> what a great service to mankind this would be!!
>
> The fact that you don't seem to care about doing this,
> offhandedly suggesting that perhaps your proof would
> be way too long in formal ZFC, is very strange and
> disheartening.
Do you know a way to translate it? Perhaps during this process we
could see where the problems come in?
> I mean, you think your proof is okay
> and its like only a few lines long. Do a little more work
> and put it into ZFC. You will be in the Pantheon of
> great mathematical logicians.
>
> You don't want to bother? Okay, it starts to look like
> you are just a big troll, I am afraid.
Once you said that you probably even in the finite case could show me
an error. Did you change your mind?
Regards, WM
WM
> It's curious that you don't mention the large number of "others"
> who explain exactly where the _error_ in your arguments is.
That is a typical reply. Where is the error?
Oh, I see, you belong to the very few who give an argument.
>
> Or rather the _gap_. You say
>
> "This procedure, even when applied infinitely often, cannot but yield
> the result 0 respectively a countable number of lines."
>
> over and over, but it's simply not true. Your asserting it does
> not make it true. The fact that it seems clearly true to you
> does not make it true. You need to give a _proof_ of this,
> and you've never done so - in each of your attempts there's
> always a similar unjustified assertion.
If you assume that 0 + 0 + 0 + ... > aleph_0 is acceptable, then you
are right. Then my proof fails. It is my wish to convince people that
this cannot be assumed.
>
> That's fascinating, but of no relevance whatever. A quote
> where one of those guys stated that that tree had only countably
> many paths would be relevant - got one of those?
No. But now we have clear positions. If actual infinity exists, then a
sum of aleph_0 zeros yields 2^aleph_0, or at least can do so under
certain circumstances.
Of course there is no further discussion between us necessary or
useful. But be sure, you are the first to analyze my error in such a
clear way. I am really grateful to you.
Regards, WM
WM
wrote:
Every edge leads to one and only one node. If we call the edge by E(x,
y) where (x, y) is the node, then the edge is uniquely defined. There
is not much work to do. Further if we work with edges, then we need
not work with nodes at all. Therefore even the E can be dropped. Why
do you see problems where they are not?
>
>
> > > > Then all the paths (except p_0) can be labelled by their last node
> > > > that has value 1.
>
> > > Here, I think, is your problem. You are not noticing that when
> > > you add a single path, you are actually adding many paths.
>
> > What many of paths do I add when I add 0.1000...?
>
> What I said here was based on trying to understand your
> account, and I did not. It is true that if
> you add paths only, you are adding only one
> path at a time. But you will not add all paths
> by doing that.
Before goning on, please anwer the question: Do you agree that I
construct all the nodes respectively all the edges of the tree by my
prescription?
> The only way to add all paths is to
> add edges one at a time in the right way. You will never get
> all paths by trying to add a countable number of infinite
> paths. You cannot construct the tree by adding paths.
Why not?
> You must construct it by adding edges. And in the right
> way. You will never add a single infinite
> path at any time. Only in the final tree will there be
> infinite paths. It all sounds very paradoxical, must
> drive you crazy. But the point is this: the final tree
> has infinite paths in it because these paths are sets
> edges that such that each edge is added at some
> *different* finite stage, not along with all its pathmates
> all at once. They are added one edge at each of stage
> S(i), for some strictly increasing function S of i. So notice:
> these infinite paths do not occur in any of the finite trees made
> during the construction, because no such tree has all the edges
> required! Only the final tree has all the edges required.
Therefore I do not add edges but infinite paths in my proof (A)
>
> Infinity is tricky.
Yes. You can construct the complete tree by all path that end by
zeros. And you can deconstruct it by all paths that end by ones.
You can construct it by all paths that end by the sequence of pi. Or
you can construct it by path that end by whatever you choose from time
to time.
But you cannot construct it by using all paths that represent all real
numbers of the unit interval unless you use some paths more than once.
All paths simply do not fit into the tree. Nevertheless they are in
it, after infinity has been finished (and given they all exist).
>
> In any case, it is probably quite possible to define the
> final tree not as a limit, but directly, by clearly specifying
> the set of nodes as I have asked you to do. Once we
> have all tree defined, by saying what its edges are,
> we can then look to see which paths are in it, or not.
If the tree has all nodes, after the end of construction, then there
are all paths.
Regards, WM
WM
wrote:
> Unless you can come up with a plausible reply to these
> considerations, I'm giving up on this whole thing.
That is the usual ending. My problem is: I have to invest many letters
to get mathematicians convinced that they cannot find an error. (If
they could find an error there was no formalization required for such
an easy argument to name the error.) But instead of confessing that
they cannot find any error, they declare, that there is an error and
that I should find it myself or should provide a formal proof. Then
they leave the discussion and another one shows up, and I have to
punch the keys again.
Regards, WM
calvin...@gmail.com
> Every edge leads to one and only one node. If we call the edge by E(x,
> y) where (x, y) is the node, then the edge is uniquely defined. There
> is not much work to do. Further if we work with edges, then we need
> not work with nodes at all. Therefore even the E can be dropped. Why
> do you see problems where they are not?
It is not what you *call* the edges, it is what they *are*. An
edge is essentially something between two nodes. So, to
specify *what it is*, you must give the two nodes. You don't
do that, do you? You just say, E(x,y) is what we call the edge
between x and y. Fine, but what are x and y? They are just
formal names. You must give the set of {(x,y)}, a subset
of NxN, in order to even define the tree. The problem I see
is that you haven't defined the tree at all here.
(This is assuming, again, what we should be able to do,
define the tree T in one shot, as a set of edges between
nodes, rather than building it up as a limit, which you have
been trying to do all along, but which is leading
into confusion)
> Before goning on, please anwer the question: Do you agree that I
> construct all the nodes respectively all the edges of the tree by my
> prescription?
Above, you have said *nothing* about the tree. Let us
look yet again at the limit method. You start with
an empty tree, or maybe a tree containing the
edges for the path you call 0.0000..., and at each step
you add another infinite path to tree T(n) to get
tree T(n+1) (you do not say clearly
which path). Then, I believe, you say that T =
the union of the T(i), i in N. (Or more simply,
if the infinite set of edges added at step i is p(i),
the tree then is union p(i), in in N.
This is not a complete specification until you specify
precisely what each p(i) is.
> > The only way to add all paths is to
> > add edges one at a time in the right way. You will never get
> > all paths by trying to add a countable number of infinite
> > paths. You cannot construct the tree by adding paths.
>
> Why not?
Because there are an uncountable number of paths, and
the process has only a countable number of steps, one
for each i in N.
However, if you add them correctly, the key point is
that you are also adding edges. And it is these edges,
not the paths, that end up making up all the other paths
that you have not added. So the edges do the job in
the end. And all these other paths sneak in although
you did not add them (at any given step).
> > S(i), for some strictly increasing function S of i. So notice:
> > these infinite paths do not occur in any of the finite trees made
> > during the construction, because no such tree has all the edges
> > required! Only the final tree has all the edges required.
>
> Therefore I do not add edges but infinite paths in my proof (A)
That is your problem. You must add edges. You do, by
adding paths, and the edges, not the paths, do the job.
> All paths simply do not fit into the tree. Nevertheless they are in
> it, after infinity has been finished (and given they all exist).
What? They don't fit in, yet they are there? Well, that
makes no sense. But you appear to be coming close
to saying the right thing. You do not PUT all the paths
into the tree yourself, one at a time. You only put in
a countable number of paths in total. But in the end
the uncountable number are there.
How is that possible. It is because of the edges. When
you take an infinite set of paths that you put in, and take
one edge only from each of them, what you get is a new
path that you NEVER put in, among the COUNTABLE paths
you put in. That path is in the final tree, contributing
to the UNCOUNTABLE number in the tree.
Consider the path you denote .111111111....
Well, you never put this path in, of course. But it
is there ONLY in the infinite tree. It is there because
it takes one edge from
.10000000.....
one edge from
.11000000.....
one edge from
.11100000....
etc.
It is the single edges that do the work, not the puny number
of infinite paths that you added one by one.
> If the tree has all nodes, after the end of construction, then there
> are all paths.
Right. You added explicitly only a countable number of
paths, but there are an uncountable number that sneak
their way in by the end, as a result of them picking one
edge from each of an infinite number of the puny
countable infinity of paths that you explicitly put in.
So, there is no contradiction is there? You add
a countable number of paths, but they break into
edges and the edges recombine (only in the final
tree) to become many many more paths.
calvin...@gmail.com
> On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
>
> > Unless you can come up with a plausible reply to these
> > considerations, I'm giving up on this whole thing.
>
> That is the usual ending.
Why should it be?? Are you saying you cannot come
up with a plausible reply to the considerations I gave?
What explains your lack of motivation to develop a clear
formal proof of your theorem, and instead your motivation
to write many letters to many mathematicians who
dismiss you because you do not have a formal proof??
Now I am fully with them, that I think you will fail to
develop such a proof, because what you have now is
not any kind of proof at all. But forget about that view
that I and these others have, for Gods Sake! If you
are right, HOW HARD do you think it can possibly be
to develop a formal proof of this that they cannot reject.
IT WOULD NOT BE THAT HARD!!
> Then
> they leave the discussion and another one shows up, and I have to
> punch the keys again.
Yes, can't you see how irrational your behavior is here?
I mean, by your own lights. You just keep shooting
yourself in the foot like this. It is as if you WANT nobody
to believe you! Maybe that IS what you want. But think
how much better it would be to be believed, to be the one
who finally cracked this gigantic world wide conspiracy!
I believe you will find when you start to do this that you
cannot. You will see where there are errors, non-sequiturs,
equivocations, that cannot be repaired. Perhaps you are
afraid really to try. That would be one explanation.
WM
wrote:
> On Mar 26, 7:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Every edge leads to one and only one node. If we call the edge by E(x,
> > y) where (x, y) is the node, then the edge is uniquely defined. There
> > is not much work to do. Further if we work with edges, then we need
> > not work with nodes at all. Therefore even the E can be dropped. Why
> > do you see problems where they are not?
>
> It is not what you *call* the edges, it is what they *are*. An
> edge is essentially something between two nodes. So, to
> specify *what it is*, you must give the two nodes.
No. The goal node is sufficient.
> You don't
> do that, do you? You just say, E(x,y) is what we call the edge
> between x and y. Fine, but what are x and y? They are just
> formal names. You must give the set of {(x,y)}, a subset
> of NxN, in order to even define the tree. The problem I see
> is that you haven't defined the tree at all here.
x is the level number. It runs from the root level 0 through the
naturl numbers.
y = 2^x. Where is the problem?
>
> (This is assuming, again, what we should be able to do,
> define the tree T in one shot, as a set of edges between
> nodes, rather than building it up as a limit, which you have
> been trying to do all along, but which is leading
> into confusion)
>
> > Before goning on, please anwer the question: Do you agree that I
> > construct all the nodes respectively all the edges of the tree by my
> > prescription?
>
> Above, you have said *nothing* about the tree. Let us
> look yet again at the limit method. You start with
> an empty tree, or maybe a tree containing the
> edges for the path you call 0.0000..., and at each step
> you add another infinite path to tree T(n) to get
> tree T(n+1) (you do not say clearly
> which path). Then, I believe, you say that T =
> the union of the T(i), i in N. (Or more simply,
> if the infinite set of edges added at step i is p(i),
> the tree then is union p(i), in in N.
>
> This is not a complete specification until you specify
> precisely what each p(i) is.
p(x, y) = is the path that starts at the root node and goes to the yth
node in the xth level and subsequently ends with zeros.
>
> > > The only way to add all paths is to
> > > add edges one at a time in the right way. You will never get
> > > all paths by trying to add a countable number of infinite
> > > paths. You cannot construct the tree by adding paths.
>
> > Why not?
>
> Because there are an uncountable number of paths, and
> the process has only a countable number of steps, one
> for each i in N.
So all nodes will be covered.
>
> > If the tree has all nodes, after the end of construction, then there
> > are all paths.
>
> Right. You added explicitly only a countable number of
> paths, but there are an uncountable number that sneak
> their way in by the end,
If you believe that this can happen in mathematics, why then don't you
believe that in Cantor's list all real numbers can sneak in after it
has been completed? Of course every line number n that you search does
not contain the diagonal number. But after completion it sneaks its
way.
Concluding: If you allow legerdemain in the tree, then you must also
allow it in Cantor's list. Then uncountability is unsubstantiated
nevertheless.
Regards, WM
Virgil
<877821f0-5f14-44ef...@h28g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 23:43, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 25, 6:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> >
> > > It's been offered to him a hundred times that he is welcome to state
> > > whatever alternative syntax, primitives, axioms, definitions, and
> > > rules of inference he wishes to present. His ordinary response is that
> > > he is not interested in formal theories.
> >
> > I've seen that.
>
> That is nonsense. MoeBlee continued to utter insults. In those cases I
> decline to continue the discussion.
Wm only declines because he is because he is too incompetent to comply.
>
> Most theorems of mathematics have not been formalized. In my opinion
> such a formalization is without any value, because many proofs about
> absolute nonsense like inaccesible cadinals have been formalized.
> Therefore I am not motivated to start that business. But if ZFC + FOPL
> is the "fundament of mathematics" then it should be possible to treat
> the binary tree. Evereybody who is capable of doing so is invited to
> do.
Several of us have done so with sufficient formality to convince those
not enslaved by their preconceptions, but WM has carefully ignored each
and every such presentation.
Virgil
<ea334777-9379-47aa...@r33g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> But instead of saying "only finite sets exist" which
> would most mathematicians immediately cause to stop reading, one
> should say sets can be potentially infinite
The intuitionist approach does not say that no infinite sets exist nor
that there do exist any of those ambiguities WM calls "potentially
infinite sets", so that WM offends everyone by his idiocies.
Virgil
<b90d1ad7-5c00-47cd...@r29g2000vbp.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
>
> > Unless you can come up with a plausible reply to these
> > considerations, I'm giving up on this whole thing.
>
> That is the usual ending. My problem is: I have to invest many letters
> to get mathematicians convinced that they cannot find an error.
Many of us have pointed out at least some of the many errors in WM's
claimed proofs, but WM is immune to such clearness and responds to such
points only with obfuscation if at all.
> (If
> they could find an error there was no formalization required for such
> an easy argument to name the error.)
Almost anyone with a modicum of mathematics can find multiple errors in
all WM's attempts to prove his nonsense. And those that try, do.
WM
wrote:
> On Mar 26, 8:11 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> > wrote:
>
> > > Unless you can come up with a plausible reply to these
> > > considerations, I'm giving up on this whole thing.
>
> > That is the usual ending.
>
> Why should it be?? Are you saying you cannot come
> up with a plausible reply to the considerations I gave?
> What explains your lack of motivation to develop a clear
> formal proof of your theorem, and instead your motivation
> to write many letters to many mathematicians who
> dismiss you because you do not have a formal proof??
The first posting of this thread gives three clear proofs, much more
lucid than any formal proof can be.
>
> Now I am fully with them, that I think you will fail to
> develop such a proof, because what you have now is
> not any kind of proof at all. But forget about that view
> that I and these others have, for Gods Sake! If you
> are right, HOW HARD do you think it can possibly be
> to develop a formal proof of this that they cannot reject.
> IT WOULD NOT BE THAT HARD!!
I have talked to an expert who is really good in that field. He said
that he was not able to formalize my tree argument.
>
> > Then
> > they leave the discussion and another one shows up, and I have to
> > punch the keys again.
>
> Yes, can't you see how irrational your behavior is here?
> I mean, by your own lights. You just keep shooting
> yourself in the foot like this. It is as if you WANT nobody
> to believe you!
I consider that I have been very successful here. I have, for the
first time, received substantial arguments.
Usually the responses of swell-headed logicians read like: "I leave
finding the errors in your reasoning as two exercises for you." (And I
must ask, how can such fools be employed as teachers at a university?
Our job is to explain what is unclear, and to explain again, when
questions remain, and to explain as long as there is demand.)
Now you are talking about sneaking in after infinity has been
finished. Well that is an answer. But it would also apply to Cantor's
list.
David Ullrich gives the argument that an infinite sum of zeros can
surpass aleph_0.
Well that would also invalidate Cantor's list proof.
That are arguments I can apply. I can show them to others and ask what
they find more convincing, finished infinite sets that sneak in, or
the honest confession that infinity does not come in different
quantities. In my opinion, the choice is easy, very easy --- even
without AC.
Regards, WM
WM
> In article
> <b90d1ad7-5c00-47cd-b623-4884c2ced...@r29g2000vbp.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> > wrote:
>
> > > Unless you can come up with a plausible reply to these
> > > considerations, I'm giving up on this whole thing.
>
> > That is the usual ending. My problem is: I have to invest many letters
> > to get mathematicians convinced that they cannot find an error.
>
> Many of us have pointed out at least some of the many errors in WM's
> claimed proofs, but WM is immune to such clearness and responds to such
> points only with obfuscation if at all.
Could you please repeat one of these arguments? Perhaps that one that
is the clearest in your opinion? I must have overlooked them
completely. But I would be grateful to see them.
Regards, WM
Gus Gassmann
And this is WM's usual cop-out: "I have to re-iterate my mistakes so
many times, all the while ignoring the mathematical arguments provided
to me. When the mathematicians finally give up at my obtuseness and
walk away, another one shows up and I have to trot out my ignorance
all over again."
It's a shame, really, that WM has not, by his own admission, been able
to convince _one_ mathematician that his position has merit. The only
ones to whom he can apparently peddle his nonsense is the students in
his course, who depend on him for grades and are asked to regurgitate
arguments that are patently false.
Virgil
<b081e596-72c9-49a1...@g38g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 20:37, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 25, 7:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > WM (not anyone else please!), will you please write
> > > > out, formally in first order logic, the statement you have
> > > > made above? Thank you.
> >
> > > If you need this eduction I will refer you
> > > tohttp://de2.metamath.org/index.html
> > > Your results then can even be approved by automata.
> >
> > How could my results in translating an English statement
> > into formal logic be "approved" of by automata?
>
> If the statement was wrong, the machine would detect it (in some
> cases).
But how could the automata determine whether the machine language
statement was an appropriate rendering of the ordinary language
statement?
Translations, especially machine translations, from one language to
another are notoriously tricky.
> >
> > > I am not going to prove 1+1 = 2 or anything else by that language. And
> > > if you think that a statement in that language, approved by a machine,
> > > guarantees truth, then you are wrong.
> >
> > I did not ask you to *prove* anything. Simply to *state* the English
> > statement I have in formal logic.
>
> So you want to see that thereis a countable set y such that forall x :
> p(x) is an element of y? Or would you prefer when I said that forall x
> thereis an y ...? (Certainly you would prefer when I said the latter,
> but I don't.)
>
> For some education on the use of quantifiers I could refer you to my
> lesson
> http://www.hs-augsburg.de/~mueckenh/Mathematerial/M01.PPT#299,5,Folie
> 5
> but I am afraid without explaining words it does not teach very good.
Considering WM's perpetual quantifier dyslexia, the thought of him
trying to educate others on the workings of quantifiers is frightening.
> >
> > > The direct statement that you want to have formalized is: The set of
> > > path can be put in a bijection with N. I do not pretend that this can
> > > be done.
> >
> > It certainly is not. I did not ask anything of the sort. I gave you
> > the
> > formalization of countable set as the unary predicate "CS". I did
> > not mention N, I did not mention bijections.
>
> You mentioned countable. That implies N and bijections because it is
> defined as possibility to put the set in bijection with N.
> >
> > But wait. Are you now saying that the very statement "The set of
> > paths can be put in a bijection with N" cannot even be *formalized*?
> > What? What?
>
> The set of paths cannot be put in bijection with N. The paths cannot
> be enumerated. The proof is as follows:
> The nodes can be enumerated. The number of paths can be shown to be
> limited by the number of nodes (via the number of lines).
>
> Regards, WM
Virgil
<d3dcf25f-7e5d-4e86...@d19g2000yqb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 20:48, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 25, 3:33 pm, Virgil <Vir...@gmale.com> wrote:
> >
> > > > You are in error. Everybody not blinded by the blinkers of transfinite
> > > > set theory will understand very easily that I am right.
> >
> > > Whom, excluding the captive audience of your students, have you ever
> > > convinced of the truth of your foolishness?
> >
> > How do we even know he convinced any students?
>
> Recently I got a mail from a professional mathematician:
> Ich teile uebrigens mit einem Ihrer Studenten zeitweilig das Buero und
> Sie haben recht:
> Ihm leuchtet alles ein, was Sie sagen.
> Nun, warum haben gerade Berufsmathematiker da Probleme?
>
> Not a very good statistic, I agree. But if you could look at the tree
> before having been spoilt by set theory then you would not have any
> problems. It is merely a very small group of matheologists who "want
> spread the word".
The quality of mathematics generated by those that WM unjustifiably
disparages by calling them "matheologists" puts to shame the
pseudo-mathematics that WM himself generates.
>
> > We need some such student, who has for his teacher
> > a great figure, if he is right, to stand up and say he
> > was convinced. Is no such student reading these
> > heroic epics of his master? (Unforunately, it may
> > be hard to rule out sock-puppets should such a
> > student show up here).
>
> Of course. But why don't you come out of your ivory tower and ask
> people who can think logically without calling them logicians. You
> would be surprised.
I am continually surprised by the illogic of WM who can not, or at least
does not, see anything that shown how foolishly inconsistent and even
self-contradictory his "system" is.
A most unsystematic system indeed.
Virgil
<46996c55-6a4f-4a81...@p11g2000yqe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 20:48, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 25, 3:33 pm, Virgil <Vir...@gmale.com> wrote:
>
> > > Since you still cannot successfully fault the Cantor argument, no
> > > counterargument of your own can override it.
> >
> > Well, perhaps no such argument can *override* Cantor's,
> > but it can at least meet it on an equal footing.
> > It seems that (from time to time anyway) WM's
> > argument is that set theory is inconsistent. It seems
> > that at these time he accepts Cantor's argument, yet
> > also accepts an argument establishing the opposite
> > conclusion.
>
> If we accept Cantor's argument, then we get to the concusion that the
> binary tree has uncountable many paths. I have shown in the original
> post that the binary tree cannot have uncountably many paths.
And that argument has been shown to be flawed.
> This
> dilemma is caused by the acceptance of finished infinity.
The dilemma exists only in WM's anti-mathematics, and is caused by his
not having a coherent and consistent system of axioms and definitions to
work from.
> Nothing else
> could be expected by such a muddled concept.
In actual mathematics, there is no such muddle,
but in WM's anti-mathematics there is little else.
>
> My conclusion is: There is no finished infinity.
Our conclusion is: WM's anti-mathematics is in a muddle.
> >
> > If this is the case, he should clearly say so, and then
> > suggest either a new, hopefully consistent axiomatization
> > for set theory (or something similar, appropriate for
> > doing mathematics),
>
> There is no axiomatization required for mathematics. I hesitate to use
> this word that has become disreputable by the work of modern
> "logicians". But if you like to cast the basic truth in the form of
> axioms, then include the axiom:
> I s a number.
> If you have n strokes, then you can make n + 1 strokes. In his theory,
> basically due to Paul Lorenzen, you will not have a set of all
> numbers. But you will have all numbers that ae required. I think it
> was this concept that Abraham Robinson (inventor of non-standard
> analysis and pupil of Fraenkel, a front man of set theory) addressed
> when he said: (i) Infinite totalities do not exist in any sense of the
> word (i.e., either really or ideally). More precisely, any mention, or
> purported mention, of infinite totalities is, literally, meaningless.
> (ii) Nevertheless, we should continue the business of Mathematics 'as
> usual', i.e., we should act as if infinite totalities really
> existed." (In: Formalism 64, auch abgedruckt in Robinson 1979, p.
> 507.)
>
> Exactly in this way we can proceed.
But that is exactly the way in which WM refuses to proceed.
>
> > or instead, a way of dealing with
> > inconsistency so that the logic does not "explode'
>
> Drop finished infinities.
Robinson didn't himself, and advised not doing so,
so why should anyone following his advice do so?
And for Robinson, the complete infinite binary tree will have
uncountably many paths.
MoeBlee
> I think it
> was this concept that Abraham Robinson (inventor of non-standard
> analysis and pupil of Fraenkel, a front man of set theory) addressed
> when he said: (i) Infinite totalities do not exist in any sense of the
> word (i.e., either really or ideally). More precisely, any mention, or
> purported mention, of infinite totalities is, literally, meaningless.
> (ii) Nevertheless, we should continue the business of Mathematics 'as
> usual', i.e., we should act as if infinite totalities really
> existed." (In: Formalism 64, auch abgedruckt in Robinson 1979, p.
> 507.)
>
> Exactly in this way we can proceed.
WRONG. Sure, you typed the Robinson quote okay. But Robinson proceeded
by working in ZFC along with his own philosophical stipulations yet
still with an interest in and conviction of the purpose of acting "as
if" infinite sets exist. You, on the other hand, reject ZFC and your
notion of "potentially infinte" is NOT the same as acting as if there
are infinite sets.
MoeBlee
MoeBlee
> On 25 Mrz., 23:43, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
>
> > On Mar 25, 6:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > It's been offered to him a hundred times that he is welcome to state
> > > whatever alternative syntax, primitives, axioms, definitions, and
> > > rules of inference he wishes to present. His ordinary response is that
> > > he is not interested in formal theories.
>
> > I've seen that.
>
> That is nonsense. MoeBlee continued to utter insults. In those cases I
> decline to continue the discussion.
I made DESERVED insults of you AFTER you had gone well past showing
yourself to be intractable and incapable of even basic communication
regarding mathematical definitions.
Also, you have stated that you are not interested in formal theories.
It is not nonsense to say that you have said so.
> Most theorems of mathematics have not been formalized. In my opinion
> such a formalization is without any value, because many proofs about
> absolute nonsense like inaccesible cadinals have been formalized.
You ALSO decline to formalize your OWN notions or framework (or
whatever it should be called).
And your argument above makes little sense. That you feel that
formalization has been used for incorrect purposes (formalization of
theories regarding inaccessible cardinals) doesn't entail that
formalization could be used for what you consider to be a correct
purpose (your own notions).
> Therefore I am not motivated to start that business.
Your motivations, of course, are entirely your own prerogative, but it
does not follow that since formalization has been used in ways you
deplore that formalization can't also be used in ways that are
consistent with your agenda.
> But if ZFC + FOPL
The "+" there is redundant. The pure theory of predicate logic is a
subset of ZFC.
> is the "fundament of mathematics" then it should be possible to treat
> the binary tree. Evereybody who is capable of doing so is invited to
> do.
Many of us have taken the invitation, only to find that your arguments
are fallacious where they make sense and otherwise incoherent.
MoeBlee
Virgil
<d3c81ac6-a6a0-4c6f...@g38g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mrz., 21:55, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 25, 11:50 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > 0 [say instead (0,1)
>
> Agreed. But no path ends there. So it is not important.
>
> > > (1,1), (1,2)
> > > (2,1), (2,2), (2,3), (2,4)
> > > ...
> > > (n,1), ..., (n, 2^n)
> >
> > So those are the nodes. {(n,m)| n,m in N, m<=2^n}
> >
> > > Every node with even second co-ordinate has value 1, every node with
> > > odd second co-ordinate has value 0.
> >
> > Labeling function L( (n,m) ) = m+1 mod 2
> >
> > > Construct the complete tree (from the path p_0 = 0.000...) by using
> > > all paths that lead to a node with value 1 and afterward having only
> > > nodes with value zero.
> >
> > Okay, well, a tree is a set of edges. Let us instead add the edges.
> > That is more definite and precise. Paths result by supervening on
> > the pattern of edges.
>
> That is possible. But it will refer to my proof (B).
Only to disprove it.
> >
> > Can you rigorously specify the edges that are being added, in
> > a manner such as I have done above (instead of just with pictures
> > and informal words).
>
> As well as labelling the nodes you can also label the edges leading to
> the nodes by the same scheme.
> (1,1), (1,2)
> (2,1), (2,2), (2,3), (2,4)
> ...
> (n,1), ..., (n, 2^n)
> >
> So use just
> the edges {(n,m)| n,m in N, m<=2^n}
>
> > When you do that, you will see there are more paths that you
> > attempted to limit the set of paths to.
>
> I do not see that.
Try new glasses.
> I see that there is an nfinite but countable set of
> lines. I wold be nice if you said whether you could agree.
As I do not know what you mean by a "line", I cannot say.
> When you consider my proof (A), then you see that I add always exatly
> one path.
The way you do it prohibits any path from having a path with infinitely
many branches in the opposite direction from the ones you add.
> You can even consider all paths that end on some node (regardless of
> its value) and then always turn left. I this way many paths are added
> infinitely often, like
> 0.1|000..., 0.10|000..., 0.100|000
> where "|" appears behind the node of destination.
Then no path in YOUR tree has infinitely many right turns, even though
the complete tree demands them.
> >
> > Perhaps the problem arises because of using the idea of one
> > tree as a limit of the other. I don't think that is necessary here.
> > As above for the nodes, we should be able to define the final
> > tree explicitly without using a limit.
>
> The tree consists of all lines that are constructed by paths that end
> by infinitely many nodes. That is not a limit. Simply take the set of
> all those paths.
Your 'constructive' generation of such a tree always omits most paths.
You can constructively generate complete trees to any finite depth, but
not a complete infinite tree.
> >
> > Here is the problem you may have with using a limit. The
> > limit tree is a union of all the edges in the sequence of trees.
> > That is indeed true. If there is an edge in the limit tree, that
> > edge must exist in some tree in the sequence. HOWEVER,
> > the same is not true about paths. Paths exist in the limit
> > tree that do *not* exist in any tree in the sequence. This
> > is because the paths "interact" with one another in the way
> > edges do not.
>
> For that argument I deviced my proof (A). Paths may be imagined to do
> this or that. They cannot but follow one line in the tree. And there
> ae all lines in the tree. There is no infinite sequence of bits that
> is lacking. Neverteless all this is constructed by a countable number
> of paths.
Claimed but never proven, and all constructions so far presented have
been shown to fail.
The Cantor binary proof says uncountable.
None of WM's allegedly countable constructions has yet proved to produce
the complete tree, at least outside of WM's MathUnerealism.
So Cantor still rules.
> >
> > Every path you add is of the form a finite sequence of 0's
> > and 1's, following by infinite sequence of 0's. However,
> > there are paths in the final tree that are not like this. That is
> > because the paths interact with one another. You must look
> > at the individual edges and how then can be made into
> > paths in the final tree. You do not add paths, you add edges.
> > Paths come FOR FREE. You cannot pick and choose the
> > paths you add. They sneak in as sets of joined edges and
> > there is nothing you can do about it.
>
> Please be more clear. And, as you criticize the edge construction,
> please refer to proof (A).
> You want to prove that there are more paths in the tree than lines
> paved by a countable set of paths.
On the contrary, we claim nothing at all about "lines" but we do claim,
and have proved, that no construction method you have yet produced
produces all possible binary sequences as paths.
Virgil
<4bcc7eed-c70d-44dc...@n20g2000vba.googlegroups.com>,
"calvin...@gmail.com" <calvin...@gmail.com> wrote:
> (Regarding quotation: you never answered about how the
> "cup" symbol disappeared in your quotation from Wikipedia.
> Was it a browser problem?)
Just sloppy proofreading.
>
> > > > If in his "system", he accepts simultaneously both such a P and such a
> > > > not P, I don't see why he should object to anything ever.
> >
> > > Why not?
> >
> > Given both P and not-P, in any reasonably comprehensive logical system,
> > one can prove anything and its negation.
>
> Are you saying that this is currently the case, or that this
> must necessarily be the case?
I said "If ..., then..."
Virgil
<f0a9f084-c660-4463...@37g2000yqp.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> And if set theory is free of contradictions, then every other set is
> either finite or larger than a countable set.
False! "Every other set is either finite or larger than a countable
set" would require that set theory HAS contradictions.
> But just this latter is
> contradicted by my proof in the first posting of this thread. What do
> you conclude?
That WM is playing the fool again. AS usual.
Virgil
<cbfa70b6-6a3e-4115...@p11g2000yqe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> ... who understands Cantor's proof (that is not very difficult) and
> bases his intellectual life on its truth. That is the danger.
To base one's intellectual life on its falsehood is at least equally
dangerous.
> >
> > > But why don't you come out of your ivory tower and ask
> > > people who can think logically without calling them logicians. You
> > > would be surprised.
> >
> > I can't very well do this, because I can't make any sense
> > out of your "construction" of the tree. But certainly when
> > I describe to them what I consider to be a "complete
> > infinite binary tree", such people immediately agree (once
> > they understand what "uncountable" means), that the
> > set of all paths in it is uncountable.
>
> Never! They cannot understand it because it is not there.
It is there to people who see it there.
> >
It is suficient to abolish finished
> infinity --- and everything is safe and sound and in mathematics there
> is no disagreement between intelligent enough people.
Prohibiting thinking only works with dumb enough people
> Either the tree cannot be construced by countably many paths.
> Or there are paths that are not bound to the nodes constructed.
> Or ... whatever
The "whatever" is that in completing the tree, so that through every
node there is at least one left branching path and one right branching
path, one creates more paths than can be counted through every node.
> Only the argument that Cantor proved the uncountability is not
> admitted when alkig about inconsistency.
it is asked and answered in mathematics.
> >
> > But forget about even going all the way and showing
> > infinity is bad stuff. You do realize how utterly famous
> > you would be if you could merely prove clearly
> > that ZF was inconsistent!? Wow. You should really
> > try to write this up *clearly* instead of
> > wasting your time in these silly newsgroups.
>
> I thought that you wanted to disproe me. Now you are shouting.
Shouting in newsgroups requires posting in all upper case.
Virgil
<306c4b18-ce31-48b0...@e38g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mrz., 11:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > It's curious that you don't mention the large number of "others"
> > who explain exactly where the _error_ in your arguments is.
>
> That is a typical reply. Where is the error?
Give us an argument of yours whose conclusion clashes with standard
mathematics, including standard set theories, and several of us will
find at least one error in it.
>
> If you assume that 0 + 0 + 0 + ... > aleph_0 is acceptable
WE do not assume things so outre.
> >
> > That's fascinating, but of no relevance whatever. A quote
> > where one of those guys stated that that tree had only countably
> > many paths would be relevant - got one of those?
>
> No. But now we have clear positions. If actual infinity exists, then a
> sum of aleph_0 zeros yields 2^aleph_0, or at least can do so under
> certain circumstances.
That may be the case in WM's anti-mathematics, but not in any actual
mathematics.
Virgil
<57912645-3985-40ce...@l10g2000vba.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mrz., 11:58, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > But have you done that? No. You merely *named*
> > the edges. That is not what you must do. You must
> > say what the edges are in terms of what nodes they
> > connect. Edges are a subset of the cartesian product
> > of the set of nodes with itself.
> > So, please continue, and specify what your tree is by
> > saying what the set of edges are, as a subset of NxN,
> > where N is the set of nodes defined above.
>
> Every edge leads to one and only one node. If we call the edge by E(x,
> y) where (x, y) is the node, then the edge is uniquely defined. There
> is not much work to do. Further if we work with edges, then we need
> not work with nodes at all. Therefore even the E can be dropped. Why
> do you see problems where they are not?
Label each node with a 1-origin natural assigned in row succession:
1
2 3
4 5 6 7
8 9 10 11 12 13 14 15
.........................
Then the left child of node n is node 2*n + 0
and the right child of node n is node 2*n + 1
Then a left branching edge is a pair of form (n, 2*n + 0)
and a right branching edge is a pair of form (n, 2*n + 1)
And, in general left branches are even and right branches are odd, and a
chain of edges depends only on its root node and the sequence of 0's for
evens and 1's for odds identifying the direction of each successive
branching.
Denote the leading node of any sequence of nodes by its number followed
by a decimal point and the following edges by sequence 0's and 1's
representing directins of brfanching, and one has a unique way of
representing any chain of parent to child links.
Note the leftmost path is 1,2,4,8,...
And the rightmost path is 1,3,7,15,...
In that infinite tree (meaning with no terminal nodes), all PATH
designators in that tree begin with '1.' followed by an endless sequence
of 0's and 1's, and each different sequence denotes a different path.
> >
> >
> > > > > Then all the paths (except p_0) can be labelled by their last node
> > > > > that has value 1.
> >
> > > > Here, I think, is your problem. You are not noticing that when
> > > > you add a single path, you are actually adding many paths.
> >
> > > What many of paths do I add when I add 0.1000...?
> >
> > What I said here was based on trying to understand your
> > account, and I did not. It is true that if
> > you add paths only, you are adding only one
> > path at a time. But you will not add all paths
> > by doing that.
>
> Before goning on, please anwer the question: Do you agree that I
> construct all the nodes respectively all the edges of the tree by my
> prescription?
But that is like only having all members of a set and then trying to
count subsets. But for any set the number of members is strictly less
than the number of subsets. This is certainly true for all finite sets,
so that WM cannot logically object to it for infinite sets.
>
> Yes. You can construct the complete tree by all path that end by
> zeros. And you can deconstruct it by all paths that end by ones.
> You can construct it by all paths that end by the sequence of pi. Or
> you can construct it by path that end by whatever you choose from time
> to time.
> But you cannot construct it by using all paths that represent all real
> numbers of the unit interval unless you use some paths more than once.
In the complete infinite binary tree, which paths does WM claim must
appear more than once?
> All paths simply do not fit into the tree.
And where do those alleged paths that don't fit come from, WM?
Since all paths are suitable sequences of edges, they are all
necessarily IN the tree.
calvin...@gmail.com
On Mar 26, 12:45 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > It is not what you *call* the edges, it is what they *are*. An
> > edge is essentially something between two nodes. So, to
> > specify *what it is*, you must give the two nodes.
>
> No. The goal node is sufficient.
No, a tree is a set of edges. To specify the tree, you must
specify the edges. An edge is essentially a set of two nodes.
You must specify both nodes. To specify a tree, you must
give a set of nodes N and a set of edges E in NxN. That is
just what a tree is.
> > You don't
> > do that, do you? You just say, E(x,y) is what we call the edge
> > between x and y. Fine, but what are x and y? They are just
> > formal names. You must give the set of {(x,y)}, a subset
> > of NxN, in order to even define the tree. The problem I see
> > is that you haven't defined the tree at all here.
>
> x is the level number. It runs from the root level 0 through the
> naturl numbers.
> y = 2^x. Where is the problem?
These are not nodes. To specify a tree, you must give the
edges. An edge is a pair of nodes. The nodes are from the
set {(n,m} : n in N, m in N. m<=n}. You must give E as
a subset of this set. (Sorry, I see the typographic limitations
have caused confusion, since I called this set N, and the
natural numbers are also denoted with an "N", but a fancy
one I cannot depict here. But if you had any understanding
of what you were doing you would have understood that.)
> > This is not a complete specification until you specify
> > precisely what each p(i) is.
>
> p(x, y) = is the path that starts at the root node and goes to the yth
> node in the xth level and subsequently ends with zeros.
No, p(i) is the path you add at stage i. What is this path,
precisely, you have never said and still don't say and just
say nonsense instead.
> > Right. You added explicitly only a countable number of
> > paths, but there are an uncountable number that sneak
> > their way in by the end,
>
> If you believe that this can happen in mathematics, why then don't you
> believe that in Cantor's list all real numbers can sneak in after it
> has been completed? Of course every line number n that you search does
> not contain the diagonal number. But after completion it sneaks its
> way.
NO. Geez. Because the digital strings in each number in Cantor's
list do not interact in the way your path's do. The diagonal number
is the only one that is made this way, and it is not in the list. But
in your tree, most of the paths are made by selecting one edge
from each of a number of the paths you have added. In effect
the process itself automatically creates an uncountable number of
something like diagonal numbers, and sticks them in the tree
without you being able to do anything about it.
> Concluding: If you allow legerdemain in the tree, then you must also
> allow it in Cantor's list. Then uncountability is unsubstantiated
> nevertheless.
You omitted an account of how the path you denote
as .11111.... gets in the tree
If you add your strange infinite paths ending in all 0's correctly,
it (and all other paths) will be in the tree, WITHOUT YOU HAVING
TO ADD THEM.
You add
.100000...
.110000...
.111000...
so .111111.... gets put in ONLY AT THE END, BUT IT IS THERE.
It and every other combination of 0's and 1's is there, despite
the fact you added only a puny countable number of paths
Its because you add edges when you add paths, and the edges
are reused in new paths you did not add at any time.
Excuse me folks if there are more typos or other reparable
errors that usual I am losing patience in proofreading these things.
George Greene
> On 24 Mrz., 21:32, George Greene <gree...@email.unc.edu> wrote:
> > The path 0.111... IS NOT present! NO path with a FINITE number of 0's
> > is present! ONLY paths that end wit all 0's have been added!
> > And these aren't the only paths that are missing, either.
On Mar 25, 7:21 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> No path is missing because no node is missing.
This DOES NOT FOLLOW.
OBVIOUSLY, since there are only a countable number of nodes
in the tree, you can cover every node with a countable number of
paths.
Indeed, you can cover every node USING ONLY FINITE paths.
YET OBVIOUSLY, a collection of ONLY FINITE paths is missing
EVERY infinite path!
There is no need for you to complicate matters by trying to add
a path ending in an infinite number of 0's.
You could just say that since your collection of finite paths covers
every node, there cannot be any nodes left over to be in any infinite
path.
George Greene
> the tree.
No, this is just ignorant-ass bullshit that DOES NOT LOGICALLY FOLLOW.
But then, you don't know shit about proving anything logically to
begin with,
SO WHAT ELSE IS NEW.
George Greene
> It is of no interest whether it has already a name. I mean that binary
> tree that has all nodes that a tree can have. And therefore is has all
> paths that can be present in a tree.
"is has" is not grammatical.
"Has" IS PURELY METAPHORICAL if you are going to use natural
language to talk about this. THAT IS WHY math REQUIRES you to
use formal language INSTEAD.
Formally, the terms mean WHAT THE AXIOMS SAY they mean,
AS OPPOSED to whatever ignorant bullshit you personally MADE UP.
George Greene
> No path is missing because no node is missing.
I'm going to say this one more time:
IF YOU ARE RIGHT about this, IF it logically follows that
when no node is missing, no path is missing either,
THEN THE SET OF ALL FINITE PATHS (or even HALF of it,
the set of all finite paths ENDING IN 0)
has the property that EVERY node in the INFINITE tree occurs on SOME
such path!
Therefore, you are not only saying -- YOU , NOT WE , are saying this
--
that no infinite paths exist, AND that no finite paths ending in 1
exist!
The only problem with this is, the exact same symmetric argument
applies to THE CLASS OF ALL FINITE PATHS ENDING IN 1 -- since
every node in the tree ALSO occurs on some path in THAT collection,
YOUR AXIOM IMPLIES that NO finite path ending in 0 EXISTS EITHER!
But since EVERY path is either inifinite, finite ending in 0, or
finite ending in 1,
IT FOLLOWS FROM YOUR AXIOM THAT *NO PATHS EXIST AT ALL*.
THIS IS WHAT *YOU* have averred.
calvin...@gmail.com
> > When you consider my proof (A), then you see that I add always exatly
> > one path.
>
> The way you do it prohibits any path from having a path with infinitely
> many branches in the opposite direction from the ones you add.
I don't think we know this because he is not being clear.
However, I think now that under the most natural way
of understanding what he is saying, this is incorrect.
All paths will occur in the final tree (the union of what you get
by unioning all the infinite paths he adds).
> > The tree consists of all lines that are constructed by paths that end
> > by infinitely many nodes. That is not a limit. Simply take the set of
> > all those paths.
>
> Your 'constructive' generation of such a tree always omits most paths.
> You can constructively generate complete trees to any finite depth, but
> not a complete infinite tree.
No, it need not. There is a countable set of paths, such that
its union yields the entire tree, and thus all paths.
> > For that argument I deviced my proof (A). Paths may be imagined to do
> > this or that. They cannot but follow one line in the tree. And there
> > ae all lines in the tree. There is no infinite sequence of bits that
> > is lacking. Neverteless all this is constructed by a countable number
> > of paths.
>
> Claimed but never proven, and all constructions so far presented have
> been shown to fail.
No, I don't think so. He has not rigorously provided the
construction, but it is quite trivial. It adds only a countable
number of paths, and there is no infinite sequence of bits
that is lacking. All of this is correct. The point is that
the paths interact. Again, suppose he adds no path
corresponding to .1111....
He does add .100000....
.110000...
.111000...
So when they are all unioned, the path .111111... will also be
there along with an uncountable number of other paths
that he never *explicitly* added in his countable list.
calvin...@gmail.com
> > > Given both P and not-P, in any reasonably comprehensive logical system,
> > > one can prove anything and its negation.
>
> > Are you saying that this is currently the case, or that this
> > must necessarily be the case?
>
> I said "If ..., then..."
I meant, are you saying it is currently the case, or that it must be
the case, that "Given both P and not-P, in any reasonably
comprehensive logical system, one can prove anything and its
negation."?
Certainly there are many logical systems now in which given
both P and ~P, one cannot thereby prove anything (and
its negation, of course. That is redundant). So, either you
have not heard of these at all, or you are saying that none of
them is "reasonably comprehensive" (why not?) or that
none of them could ever develop into any "reasonably
comprehensive" (why not?)
calvin...@gmail.com
> The first posting of this thread gives three clear proofs, much more
> lucid than any formal proof can be.
That is not at all so, as has been repeatedly shown. However,
it is irrelevant. The conspiracy brainwashers extraordinare
want a simple easy formal proof. So provide it and CHANGE
THE WORLD ALREADY.
Your intellectual irresponsibility here is SHOCKING. You
could ENLIGHTEN THE WORLD and dethrone the naked
emperors, but you refuse to do so.
> I have talked to an expert who is really good in that field. He said
> that he was not able to formalize my tree argument.
No doubt a kind dismissal. As I said, your argument could
only be easily formalized if it is makes sense and is correct,
at most one of which is actually true. By saying "he cannot
formalize" it no doubt he meant that it is either nonsensical
or incorrect.
Please provide his name, let us get him here to see what
he said.
By the way, you don't need an expert in any field. This is
a very very simple piece of mathematics we have here.
> Now you are talking about sneaking in after infinity has been
> finished. Well that is an answer. But it would also apply to Cantor's
> list.
I explained clearly why it does not. The difference is in that adding
every node in a breadth first fashion, for example (add finite paths
in this way, even extend with all 0's to make them infinite if you
want but it is not necessary), you in fact add successive
approximations
to every infinite path. In the end, ALTHOUGH YOU HAVE NOT ADDED
them explicitly yourself, all the paths are there.
This is not how it works with Cantor's list. No infinite set of
numbers
in his list, when added, automatically add another number to the
list! Can't you see that? Geez.
> David Ullrich gives the argument that an infinite sum of zeros can
> surpass aleph_0.
He did? I doubt that. It's not true. Unless he meant that
the cardinality of a set containing an infinite number of objects
labelled with zero can surpass aleph_0. No doubt you
misunderstood him.
Virgil
<568b6bdd-aa51-4e36...@n20g2000vba.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mrz., 14:07, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> wrote:
> > On Mar 26, 8:11 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> > > wrote:
> >
> > > > Unless you can come up with a plausible reply to these
> > > > considerations, I'm giving up on this whole thing.
> >
> > > That is the usual ending.
> >
> > Why should it be?? Are you saying you cannot come
> > up with a plausible reply to the considerations I gave?
> > What explains your lack of motivation to develop a clear
> > formal proof of your theorem, and instead your motivation
> > to write many letters to many mathematicians who
> > dismiss you because you do not have a formal proof??
>
> The first posting of this thread gives three clear proofs, much more
> lucid than any formal proof can be.
However clear or lucid WM claims his proofs may be, they are not valid
proofs of what he clim they prove, which is of considerably greater
mathematical import, regardless.
> >
> > Now I am fully with them, that I think you will fail to
> > develop such a proof, because what you have now is
> > not any kind of proof at all. But forget about that view
> > that I and these others have, for Gods Sake! If you
> > are right, HOW HARD do you think it can possibly be
> > to develop a formal proof of this that they cannot reject.
> > IT WOULD NOT BE THAT HARD!!
>
> I have talked to an expert who is really good in that field. He said
> that he was not able to formalize my tree argument.
No one is able to formalize it in such a way as validly to conclude that
a complete infinite binary tree has only countably many paths.
> >
> > > Then
> > > they leave the discussion and another one shows up, and I have to
> > > punch the keys again.
> >
> > Yes, can't you see how irrational your behavior is here?
> > I mean, by your own lights. You just keep shooting
> > yourself in the foot like this. It is as if you WANT nobody
> > to believe you!
>
> I consider that I have been very successful here.
> I have, for the
> first time, received substantial arguments.
You have for the first time acknowedged them.
> Usually the responses of swell-headed logicians read like: "I leave
> finding the errors in your reasoning as two exercises for you."
I, and a number of others, have, on numerous occasions, been quite
concrete and specific about the errors in your work.
But you chose to ignore us.
> (And I
> must ask, how can such fools be employed as teachers at a university?
As most of them are a good deal less foolish than WM, one wonders how he
came to be employed at any level of education.
> Our job is to explain what is unclear, and to explain again, when
> questions remain, and to explain as long as there is demand.
That presumes that the explainer is sufficiently competent to be able to
explain, which WM is not.
> Now you are talking about sneaking in after infinity has been
> finished. Well that is an answer. But it would also apply to Cantor's
> list.
Cantor's proof does not require any of the infinities involved to be
"completed", but merely shows the necessary incompleteness of any list
of binary sequences.
Ralf Bader
> On Mar 26, 7:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>> Every edge leads to one and only one node. If we call the edge by E(x,
>> y) where (x, y) is the node, then the edge is uniquely defined. There
>> is not much work to do. Further if we work with edges, then we need
>> not work with nodes at all. Therefore even the E can be dropped. Why
>> do you see problems where they are not?
>
> It is not what you *call* the edges, it is what they *are*. An
> edge is essentially something between two nodes. So, to
> specify *what it is*, you must give the two nodes. You don't
> do that, do you? You just say, E(x,y) is what we call the edge
> between x and y. Fine, but what are x and y? They are just
> formal names. You must give the set of {(x,y)}, a subset
> of NxN, in order to even define the tree. The problem I see
> is that you haven't defined the tree at all here.
Neither is this the problem, nor will you be able to change Mueckenheim's
mindset by an iota.
There is a common definition of tree in set theory: A tree is a partially
ordered set (A,<) such that for any a e A, the set p_a = {b e A|b<a} of
predecessors of a is well-ordered by (the restriction to p_a of) <. In the
case at hand, we can take for A the set of words of finite length on the
alphabet {0,1}, and for two such words a, b we put a<b iff there is a
non-void word c such that b = ac (the concatenation of the words a and c).
A path then is a subset of A which is totally ordered by (the restriction
of) <. It is not too difficult (exercise level) to set up a bijection
between the set of paths of (A,<) and the set of mappings from N to {0,1}.
All this can happen within ZFC (apart from the last sentence it could also
happen intuitionistically)
Mueckenheim is not reasoning in the above style. He wants to figure things
out by looking at some kind of mental picture, and this is doomed to fail.
We do not have senses to discover anything infinite in the empirical world
(whether there is anything of that kind or not) and we can not "look" at
anything infinite by inner vision. We can only reason abstractly about the
infinite. Mueckenheim, by his method, is perplexed by the fact that this
infinite binary tree can be covered by a countable set S of paths, in the
sense that any node of the tree is on at least one path from S. How then
should there be more paths? Where should a path not from S run through the
tree where all nodes and edges are already occupied by paths from S? The
conviction that such path simply can't exist is understandable and may even
be true, in some sense, for that mentally imagined tree (but the mental
picture of that tree is dim, and its inspection doesn't reveal anything
useful). But this is not the tree as defined above within ZFC. That one can
indeed reason about the infinite in the abstract sense and not based on
inner vision (which leads to all those antinomies described by Bolzano -
that is the place where Mueckenheim's discovery, if it is one, fits in) is
the great discovery of Cantor.
lwa...@lausd.net
> On Mar 26, 2:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > Most theorems of mathematics have not been formalized. In my opinion
> > such a formalization is without any value, because many proofs about
> > absolute nonsense like inaccesible cadinals have been formalized.
> Your motivations, of course, are entirely your own prerogative, but it
> does not follow that since formalization has been used in ways you
> deplore that formalization can't also be used in ways that are
> consistent with your agenda.
Hmmm. This debate between MoeBlee and WM reminds me of
what I've been wondering all along -- whether it's possible to
give a rigorous theory in which WM is right, and whether either
WM or the standard set theorist even want to come up with such
a theory.
I still disagree with those like WM and AP who believe that,
since one rigorously axiomatized theory (such as the theory
ZFC+Inaccessible mentioned by WM) proves undesirable
theorems, therefore axiomatization is itself to begin. If one
doesn't like the theorems proved by certain axioms, then just
choose different axioms -- which is what I've been trying to do
all along!
On the other hand, I notice the irony that WM mentions
inaccessible cardinals. Most standard set theorists work in
ZFC, in which one can't prove the existence of any of the
inaccessible cardinals. So inaccessibles in ZFC have the
same status as omega in ZF-Infinity. And so I compare how
WM feels about Infinity to how standard set theorists feel
about inaccessible cardinals. If the standard set theorists
wish to force WM to accept Infinity and only use theorists in
which the existence of omega is provable, then we must also
force the standard set theorists to accept inaccessibles and
only use theories in which their existence is provable --
which would rule out ZFC as well!
Here we go with Robinson again. Sometimes I wonder whether, if
only Robinson were alive today and posting to sci.math, would he
side with WM or the standard set theorists. Both sides seem to
claim that he'd side with them rather than their opponents.
lwa...@lausd.net
> On 26 Mrz., 06:40, lwal...@lausd.net wrote:
> > OK, I see. So it's sort of like "assume infinite sets exist. Then the
> > binary tree is both countable and uncountable -- contradiction. Hence
> > only finite sets exist."
> Yes, you got it. But instead of saying "only finite sets exist" which
> would most mathematicians immediately cause to stop reading, one
> should say sets can be potentially infinite
But unfortunately, saying "potentially infinite" causes most standard
set theorists to stop reading as well!
calvin...@gmail.com
> >> Why
> >> do you see problems where they are not?
>
> > The problem I see
> > is that you haven't defined the tree at all here.
>
> Neither is this the problem, nor will you be able to change Mueckenheim's
> mindset by an iota.
>
> There is a common definition of tree in set theory: A tree is a partially
> ordered set (A,<) such that for any a e A, the set p_a = {b e A|b<a} of
> predecessors of a is well-ordered by (the restriction to p_a of) <.
It *is* the problem in that one paragraph WM is referring to. I
was talking about a problem in the small here: that he has not
wielded any formal machinery adequately in order to define anything
precisely. Our thread for a while has been considering trees as
certain sorts of graphs (talking about roots, nodes, edges, paths)
and not as certain kinds of partial orders although of course one
can conceive of it either way. I just wanted *one* way.
> Mueckenheim is not reasoning in the above style. He wants to figure things
> out by looking at some kind of mental picture, and this is doomed to fail.
> We do not have senses to discover anything infinite in the empirical world
> (whether there is anything of that kind or not) and we can not "look" at
> anything infinite by inner vision.
I am not so convinced of that (the last clause that is) although
such attempts would for most people, if not all at least
some of time, be inherently risky. It WM's case, the risk
factor has went way outside the red and broken the dial.
> Where should a path not from S run through the
> tree where all nodes and edges are already occupied by paths from S? The
> conviction that such path simply can't exist is understandable
I agree. Strangely, though, this does not appear to be
what what WM is saying. He does occasionally say that
all paths are in the tree (he says at least "all binary
sequences are" -- same thing in this context).
The problem is that he does not seem to see that a path can
exist in the tree even though he has not explicitly added it as
one of the paths that exhaustively produces all of the edges of
the tree. So he thinks there is a contradiction: "all of the
supposedly uncountable paths are there, yet I added only
countable many. " All we need to get him to see for
the refutation of this "proof" is that paths that are
not explicitly added to the tree by him as complete
paths are nevertheless in the tree.
> and may even
> be true, in some sense, for that mentally imagined tree
That is hard to imagine however. Maybe the image is
sort of like an Escher woodprint with stairs going up
forever in a finite space?
MoeBlee
What "status" is that? Different set theorists have different regard
for the notion of the existence of inaccessible cardinals.
> And so I compare how
> WM feels about Infinity to how standard set theorists feel
> about inaccessible cardinals.
Why do you do that? Why don't you let these "standard set theorists"
SPEAK FOR THEMSELVES on the subject rather than put words in their
mouths. You do know that people publish articles and books on this
subject, right?
> If the standard set theorists
> wish to force WM to accept Infinity and only use theorists in
> which the existence of omega is provable,
Would you please name just one "standard set theorist" who wishes to
force WM to accept the existence of infinite sets?
> then we must also
> force the standard set theorists to accept inaccessibles and
> only use theories in which their existence is provable --
> which would rule out ZFC as well!
What a ridiculous argument! Do you stop to THINK for a moment about
what you type? (1) You've not even shown that any "standard set
theorists" wish to force anyone to adopt any axioms at all. (2) Even
if one adopts the axiom of infinity, one is not therefore logically
required to adopt all other infinitistic axioms. One may have a whole
range of possible reasons for not adopting every possible infinitistic
axiom. One may feel that the axiom of infinity is adequate for one's
purposes and that other higher axioms are not needed. Or one may feel
that the axiom of infinity is intuitively satisfactory but other
higher axioms are not or are not compelled by ones most basic
intuitions about mathematics, sets, or the iterative conception or
whatever other conception. Moreover, a great amount of research in set
theory is devoted to search and justification for new axioms that may
themselves imply such things as the existence of inaccessible
cardinals but only as the new axioms appeal to more compelling
intuitions than merely the bald claim of the existence of inaccessible
cardinals. Etc.
> > I think it
> > was this concept that Abraham Robinson (inventor of non-standard
> > analysis and pupil of Fraenkel, a front man of set theory) addressed
> > when he said: (i) Infinite totalities do not exist in any sense of the
> > word (i.e., either really or ideally). More precisely, any mention, or
> > purported mention, of infinite totalities is, literally, meaningless.
> > (ii) Nevertheless, we should continue the business of Mathematics 'as
> > usual', i.e., we should act as if infinite totalities really
> > existed." (In: Formalism 64, auch abgedruckt in Robinson 1979, p.
> > 507.)
> > Exactly in this way we can proceed.
>
> WRONG. Sure, you typed the Robinson quote okay. But Robinson
> proceeded
> by working in ZFC along with his own philosophical stipulations yet
> still with an interest in and conviction of the purpose of acting "as
> if" infinite sets exist. You, on the other hand, reject ZFC and your
> notion of "potentially infinte" is NOT the same as acting as if there
> are infinite sets.
>
> Here we go with Robinson again. Sometimes I wonder whether, if
> only Robinson were alive today and posting to sci.math, would he
> side with WM or the standard set theorists. Both sides seem to
> claim that he'd side with them rather than their opponents.
I don't know what exact proposition you have in mind for drawing
"sides". But Robinson spent his life proving theorems in ZFC and gave
his argument for doing that despite his rejection of the meaninfulness
of claiming the existence of infinite sets. I don't know why you even
expect that there is a "side" for Robinson to take other than his OWN
as he himself explains it. Why are you so preoccupied with what "side"
people are on when it is so much more productive, so much less
unproductively reductive, not to describe this subject like it were a
reinstallment of the Battle of Gettysburg.
MoeBlee
Tim Little
> On 2009-03-25, lwa...@lausd.net <lwa...@lausd.net> wrote:
>> For example, ZF-Powerset+~Powerset can prove that countably infinite
>> sets exist, but can't prove that any uncountable sets exist.
>
> How certain are you of that? I have an idea of how I might go about
> proving the opposite. I'll see how it turns out when I have a bit of
> spare time.
It didn't turn out well. I thought that there might be a loophole in
"all sets are countable" along the lines of provable absence of some
bijection. In the usual treatment, functions are built up from
subsets of cartesian products, and the powerset axiom is used to prove
the existence of cartesian products. However, countable sets don't
need it - NxN is easily constructible using the other axioms.
So in ZF-Powerset+"All sets are countable", all finite sets would have
powersets, and all infinite sets wouldn't. There would be no
uncountable ordinals, and only one infinite cardinal. There would be
infinitely many permutations of N, and even sets of infinitely many
permutations, but no set of all of them.
Amusingly, the Dedekind construction of reals can be applied - but the
class of reals is then provably not a set. Nor is any nontrivial
interval, and obviously there are no functions over any real
intervals. This is not unexpected of course, since one does not even
need set theory to prove that any complete ordered field is
uncountable.
(As a side note, I find it very amusing that Archimedes Plutonium's
numbers are not Archimedean)
I'm not entirely certain, but I suspect that any model of ZF can be
made into a model of ZF - Powerset + ~Powerset by restricting sets to
any maximum infinite cardinality alpha such that alpha^2 = alpha.
- Tim
Ralf Bader
> On Mar 26, 7:21 pm, Ralf Bader <ba...@nefkom.net> wrote:
>
>> >> Why
>> >> do you see problems where they are not?
>>
>> > The problem I see
>> > is that you haven't defined the tree at all here.
>>
>> Neither is this the problem, nor will you be able to change Mueckenheim's
>> mindset by an iota.
>>
>> There is a common definition of tree in set theory: A tree is a partially
>> ordered set (A,<) such that for any a e A, the set p_a = {b e A|b<a} of
>> predecessors of a is well-ordered by (the restriction to p_a of) <.
>
> It *is* the problem in that one paragraph WM is referring to. I
> was talking about a problem in the small here: that he has not
> wielded any formal machinery adequately in order to define anything
> precisely.
And Mueckenheim has repeatedly declared that he is not interested in formal
machinery. He also has asserted somewhere that mathematics would be a part
physics. So the notion "three" is for him of the same category as, say,
"proton". Mathematical propositions can be verified (according to
Mueckenheim) by observation and experiment, not by formal machinery.
> Our thread for a while has been considering trees as
> certain sorts of graphs (talking about roots, nodes, edges, paths)
> and not as certain kinds of partial orders although of course one
> can conceive of it either way. I just wanted *one* way.
The nodes would be the elements of A, in my definition. A root may not
exist; it would be the minimal element of the intersection of all
predecessor sets, if that intersection is nonempty. An "edge" ending in
node a exists only if p_a has a maximal element, which need not be the
case. And here comes the point why here the set-theoretical trees are more
appropriate than the graph-theoretical ones: Take for A the set of all
words of length <= omega on the alphabet {0,1}, and extend < by requiring
that a word of length omega is larger than all of its finite initial
segments. This adds endpoints to all the paths in the binary tree
considered so far in this thread. Of course, this will not impress
Mueckenheim, but now any path has an exclusive element of his own in the
tree. Generally, for any ordinal a, the full set-theoretical binary tree of
height a can be formed. And even the tree whose height is the class of all
ordinals. And the endpoints of its paths can be viewed as filling up a line
which is the real line in the case of height omega+1, and a "surreal" line
with much more points for larger height. Cf.
http://oak.cats.ohiou.edu/~ehrlich/AAC(Draft).pdf
I'd require people who assert that set-theory is just some nonsense to
explain why such well-formed structures can grow on the basis of that
nonsense, while the alleged contradictions steadfastly refuse
manifestation.
It is not my intention to make Mueckenheim see anything. As long as his path
addition process runs, there are always only finitely many paths - exactly
those which he has added explicitly. Mueckenheim then will concoct some
weird ideas about limits and say that the "set theorists" require that in
the "last step" of the path addition process the number of paths would jump
from finite to uncountably infinite, leaving out countable infinity. And
that set theorists must assert such nonsense to cover up the
inconsistencies in their theories. That is how he handled this in the past.
>> and may even
>> be true, in some sense, for that mentally imagined tree
>
> That is hard to imagine however. Maybe the image is
> sort of like an Escher woodprint with stairs going up
> forever in a finite space?
This is a nice analogy.
Virgil
<0aef09d2-1a15-42db...@r3g2000vbp.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mrz., 18:09, Virgil <Vir...@gmale.com> wrote:
> > In article
> > <b90d1ad7-5c00-47cd-b623-4884c2ced...@r29g2000vbp.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 26 Mrz., 12:22, "calvin.ost...@gmail.com" <calvin.ost...@gmail.com>
> > > wrote:
> >
> > > > Unless you can come up with a plausible reply to these
> > > > considerations, I'm giving up on this whole thing.
> >
> > > That is the usual ending. My problem is: I have to invest many letters
> > > to get mathematicians convinced that they cannot find an error.
> >
> > Many of us have pointed out at least some of the many errors in WM's
> > claimed proofs, but WM is immune to such clearness and responds to such
> > points only with obfuscation if at all.
>
> Could you please repeat one of these arguments? Perhaps that one that
> is the clearest in your opinion? I must have overlooked them
> completely. But I would be grateful to see them.
Present what you allege to be a proof, and if I disagree, I will
critique it.
Among others you have claimed in your arguments re complete infinite
binary trees and Cantor's binary proof, that when including only
infinite binary strings of 0's and 1's with finitely many 1's in them
that somehow you have included with them all the strings with infinitely
many 1's in them. You have not!