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Distinguishability of paths of the Infinite Binary tree???
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Zuhair
Dec 23, 2012, 3:20:35 PM12/23/12
to
This is just a minor conundrum that I want to discuss about WM's
argument about the complete Infinite binary tree (CIBT). It is really
about Cantor's argument. But the discussion here will be at intuitive
level rather than just formal level.
Let's start with "distinguish-ability" at finite binary trees, and
then make some rough analogy with the infinite binary tree.
Let's take the binary tree with two levels below the root node level
which is the following:
0
/ \
0 1
/ \ | \
0 1 0 1
Now with this three one can say that distinquish-ability is present at
all levels below the root node level, so we have two distinguishable
paths at level 1 that are 0-0 and 0-1. While at level 2 we have four
distinguishable paths that are 0-0-0, 0-0-1, 0-1-0, 0-1-1. However the
reason why we had increased distinguish-ability at level 2 is because
we had differential labeling of nodes at that level! Now if we remove
that differential labeling we'll see that we can only distinguish two
longer paths by the labeling of their nodes, like in the following
tree:
0
/ \
0 1
/ \ | \
0 0 0 0
Now clearly the only distinguishability present in that tree is at
level 1 because all nodes at level 2 are not distinguished by their
labeling. So the result is that there is no increase in the number of
distinguishable paths of the above tree when we move from level 1 to
level 2. See:
Paths at level 1 are: 0-0 , 0-1. Only Two paths.
Paths at level 2 are : 0-0-0, 0-1-0. Only Two paths.
Similarly take the tree:
0
/ \
0 1
/ \ | \
1 1 1 1
Paths at level 1 are: 0-0 , 0-1. Only Two paths.
Paths at level 2 are : 0-0-1, 0-1-1. Only Two paths.
So the increment in number of paths in the original binary tree of
level 2 after the root node, is actually due to having distinct
labeling of nodes at level 2. If we don't have distinct labeling at a
further level the number of distinguished longer paths stops at the
last level where distinguished labeling is present.
This is obviously the case for FINITE binary trees.
Now lets come to the complete infinite binary tree, which is an actual
infinite tree where at each level a node have two path each down to
one node below and each one of those below nodes is having a distinct
label from the other, and labeling of any node of that tree is
restricted to either 1 or 2.
Now at FINITE level of the infinite binary tree, we only have
countably many finite paths down from the root node. i.e. the set of
all distinguishable (by labeling of their nodes) finite paths of the
CIBT is actually COUNTABLE.
BUT the set of all Paths that are distinguishable on finite basis is
UNCOUNTABLE. Furthermore the subtree of the CIBT that have all finite
paths of the CIBT as initial segments of its paths is actually the
CIBT itself according to Cantor, all of those uncountably many
infinitely long paths are distinguishable at finite level! But
distinguishability at finite level is COUNTABLE?
To clarify any two paths P1, P2 of the CIBT are said to be
distinguishable on finite basis iff there exist an n_th far node down
from the root node of P1 that is labeled differently from the n_th far
node down from the root node of P2. Of course P1,P2 might be infinite
paths, or might be finite.
Of course we know that every node in any path of the CIBT is at finite
distance down from the root node, so there is no node of the CIBT that
lies at some infinite distance down from the root node, so there is no
distingushiability of paths of the CIBT beyond that occurring at
finite level, so the total number of infinite paths of the CIBT must
be the same as the total number of distinguished finite paths of the
CIBT. So it must be countable. But that is not the case?
Also the proof of Cantor is actually about uncountability of paths
that are distinguishable on finite basis.
So how come we can have uncountably many distinguishable paths on
finite basis while finite distinguishability itself is countable? what
is the source for the extra distinguishability over the finite level?
we don't have any node at infinite level???
One of course can easily say that at actual infinity level some
results are COUNTER-INTUITIVE, and would say that the above aspect is
among those counter-intuitive aspects, much like a set having an equal
size to some proper superset. But I must confess that the above line
of counter-intuitiveness is too puzzling to me??? There must be
something that I'm missing?
Any insights?
Zuhair
argument about the complete Infinite binary tree (CIBT). It is really
about Cantor's argument. But the discussion here will be at intuitive
level rather than just formal level.
Let's start with "distinguish-ability" at finite binary trees, and
then make some rough analogy with the infinite binary tree.
Let's take the binary tree with two levels below the root node level
which is the following:
0
/ \
0 1
/ \ | \
0 1 0 1
Now with this three one can say that distinquish-ability is present at
all levels below the root node level, so we have two distinguishable
paths at level 1 that are 0-0 and 0-1. While at level 2 we have four
distinguishable paths that are 0-0-0, 0-0-1, 0-1-0, 0-1-1. However the
reason why we had increased distinguish-ability at level 2 is because
we had differential labeling of nodes at that level! Now if we remove
that differential labeling we'll see that we can only distinguish two
longer paths by the labeling of their nodes, like in the following
tree:
0
/ \
0 1
/ \ | \
0 0 0 0
Now clearly the only distinguishability present in that tree is at
level 1 because all nodes at level 2 are not distinguished by their
labeling. So the result is that there is no increase in the number of
distinguishable paths of the above tree when we move from level 1 to
level 2. See:
Paths at level 1 are: 0-0 , 0-1. Only Two paths.
Paths at level 2 are : 0-0-0, 0-1-0. Only Two paths.
Similarly take the tree:
0
/ \
0 1
/ \ | \
1 1 1 1
Paths at level 1 are: 0-0 , 0-1. Only Two paths.
Paths at level 2 are : 0-0-1, 0-1-1. Only Two paths.
So the increment in number of paths in the original binary tree of
level 2 after the root node, is actually due to having distinct
labeling of nodes at level 2. If we don't have distinct labeling at a
further level the number of distinguished longer paths stops at the
last level where distinguished labeling is present.
This is obviously the case for FINITE binary trees.
Now lets come to the complete infinite binary tree, which is an actual
infinite tree where at each level a node have two path each down to
one node below and each one of those below nodes is having a distinct
label from the other, and labeling of any node of that tree is
restricted to either 1 or 2.
Now at FINITE level of the infinite binary tree, we only have
countably many finite paths down from the root node. i.e. the set of
all distinguishable (by labeling of their nodes) finite paths of the
CIBT is actually COUNTABLE.
BUT the set of all Paths that are distinguishable on finite basis is
UNCOUNTABLE. Furthermore the subtree of the CIBT that have all finite
paths of the CIBT as initial segments of its paths is actually the
CIBT itself according to Cantor, all of those uncountably many
infinitely long paths are distinguishable at finite level! But
distinguishability at finite level is COUNTABLE?
To clarify any two paths P1, P2 of the CIBT are said to be
distinguishable on finite basis iff there exist an n_th far node down
from the root node of P1 that is labeled differently from the n_th far
node down from the root node of P2. Of course P1,P2 might be infinite
paths, or might be finite.
Of course we know that every node in any path of the CIBT is at finite
distance down from the root node, so there is no node of the CIBT that
lies at some infinite distance down from the root node, so there is no
distingushiability of paths of the CIBT beyond that occurring at
finite level, so the total number of infinite paths of the CIBT must
be the same as the total number of distinguished finite paths of the
CIBT. So it must be countable. But that is not the case?
Also the proof of Cantor is actually about uncountability of paths
that are distinguishable on finite basis.
So how come we can have uncountably many distinguishable paths on
finite basis while finite distinguishability itself is countable? what
is the source for the extra distinguishability over the finite level?
we don't have any node at infinite level???
One of course can easily say that at actual infinity level some
results are COUNTER-INTUITIVE, and would say that the above aspect is
among those counter-intuitive aspects, much like a set having an equal
size to some proper superset. But I must confess that the above line
of counter-intuitiveness is too puzzling to me??? There must be
something that I'm missing?
Any insights?
Zuhair
WM
Dec 24, 2012, 3:47:12 AM12/24/12
to
Binary Tree occurs at a finite level. There is no difference, whether
the digits or nodes are continuing or not. Evereything in mathematics
happens at a finite level. Therefore blathering about infinite paths
is useless.
Regards, WM
WM
Dec 24, 2012, 4:42:35 AM12/24/12
to
> Also the proof of Cantor is actually about uncountability of paths
> that are distinguishable on finite basis.
That is the point!
> that are distinguishable on finite basis.
>
> So how come we can have uncountably many distinguishable paths on
> finite basis while finite distinguishability itself is countable? what
> is the source for the extra distinguishability over the finite level?
> we don't have any node at infinite level???
The source is the belief that a set can be finished (1) without being
> So how come we can have uncountably many distinguishable paths on
> finite basis while finite distinguishability itself is countable? what
> is the source for the extra distinguishability over the finite level?
> we don't have any node at infinite level???
finished (2).
(1) We can be sure that the complete set does not contain a certain
number (the diagonal).
(2) We can be sure that for every line of the following list, there is
another line
0.0
0.1
0.11
0.111
...
such that the diagonal constructed up to line n, 0.111...1, is not in
the list as an entry (the entry of the next line) but is distinct from
every line entry.
>
> One of course can easily say that at actual infinity level some
> results are COUNTER-INTUITIVE, and would say that the above aspect is
> among those counter-intuitive aspects, much like a set having an equal
> size to some proper superset. But I must confess that the above line
> of counter-intuitiveness is too puzzling to me??? There must be
> something that I'm missing?
>
> Any insights?
leading matheologians have acquired during their whole life time.
There is nothing to happen "in the infinite". And it is completely
irrelevant whether the paths after the distinction are finite or
infinite. Everything that happens in a Cantor-list and in a Binary
Tree happens at a finite level.
Wait a few days until the shock will have diminished. Transfinity is
nonsense!
Regards, WM
Virgil
Dec 24, 2012, 2:14:04 PM12/24/12
to
In article
<d85d67ab-aa37-4091...@x10g2000yqx.googlegroups.com>,
(each individual listed entry differs from the "diagonal" at an
"individual finite position") but shows that it can occur infinitely
often.
> >
--
<d85d67ab-aa37-4091...@x10g2000yqx.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > Also the proof of Cantor is actually about uncountability of paths
> > that are distinguishable on finite basis.
>
> That is the point!
The Cantor argument only deals with distinguishability on a finite basis
> On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > Also the proof of Cantor is actually about uncountability of paths
> > that are distinguishable on finite basis.
>
> That is the point!
(each individual listed entry differs from the "diagonal" at an
"individual finite position") but shows that it can occur infinitely
often.
> >
--
Virgil
Dec 24, 2012, 2:18:03 PM12/24/12
to
In article
<2f2588c2-ca27-4586...@b8g2000yqh.googlegroups.com>,
EVERY "level" of an infinite path is finite, so there is no problem with
finiteness of levels.
The only problem with WM are the possible self-imposed limitations on
his mind.
--
<2f2588c2-ca27-4586...@b8g2000yqh.googlegroups.com>,
finiteness of levels.
The only problem with WM are the possible self-imposed limitations on
his mind.
--
George Greene
Dec 24, 2012, 3:31:59 PM12/24/12
to
On Dec 23, 3:20 pm, Zuhair <zaljo...@gmail.com> wrote:
> BUT the set of all Paths that are distinguishable on finite basis is
> UNCOUNTABLE.
Well, every path is distinguishable at some finite index from every
> BUT the set of all Paths that are distinguishable on finite basis is
> UNCOUNTABLE.
other path,
since all the paths consist of ONLY finite indices.
And the number of paths is uncountable. So this is neither surprising
nor insightful.
> Furthermore the subtree of the CIBT that have all finite
> paths of the CIBT as initial segments of its paths is actually the
> CIBT itself according to Cantor,
The subtree of the CIBT beginning at any node of the CIBT *IS* the
CIBT.
The whole CIBT is arguably nothing but a triple of (left-copy-of-
itself, root-node, right-copy-of-itself).
George Greene
Dec 24, 2012, 3:32:59 PM12/24/12
to
> Let's take the binary tree with two levels below the root node level
> which is the following:
>
> 0
> / \
> 0 1
> / \ | \
> 0 1 0 1
Oh, SHUT UP. You are labeling the NODES again.
> which is the following:
>
> 0
> / \
> 0 1
> / \ | \
> 0 1 0 1
I TOLD YOU NOT TO DO THAT.
The EDGES are what are labeled 0 and 1.
The node is just labeled with the path that ends at it.
WM
Dec 25, 2012, 6:33:54 AM12/25/12
to
On 24 Dez., 20:14, Virgil <vir...@ligriv.com> wrote:
> In article
> <d85d67ab-aa37-4091-9474-a089288c3...@x10g2000yqx.googlegroups.com>,
nothing but all finite paths.
Regards, WM
> In article
> <d85d67ab-aa37-4091-9474-a089288c3...@x10g2000yqx.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Also the proof of Cantor is actually about uncountability of paths
> > > that are distinguishable on finite basis.
>
> > That is the point!
>
> The Cantor argument only deals with distinguishability on a finite basis
> (each individual listed entry differs from the "diagonal" at an
> "individual finite position") but shows that it can occur infinitely
> often
without leaving the finite domain, i.e., the Binary Tree that contains
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Also the proof of Cantor is actually about uncountability of paths
> > > that are distinguishable on finite basis.
>
> > That is the point!
>
> The Cantor argument only deals with distinguishability on a finite basis
> (each individual listed entry differs from the "diagonal" at an
> "individual finite position") but shows that it can occur infinitely
> often
nothing but all finite paths.
Regards, WM
WM
Dec 25, 2012, 6:36:57 AM12/25/12
to
On 24 Dez., 21:31, George Greene <gree...@email.unc.edu> wrote:
> On Dec 23, 3:20 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > BUT the set of all Paths that are distinguishable on finite basis is
> > UNCOUNTABLE.
>
> Well, every path is distinguishable at some finite index from every
> other path,
> since all the paths consist of ONLY finite indices.
> And the number of paths is uncountable. So this is neither surprising
> nor insightful.
It is surprising for the majority of matheologians who continue to
> On Dec 23, 3:20 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > BUT the set of all Paths that are distinguishable on finite basis is
> > UNCOUNTABLE.
>
> Well, every path is distinguishable at some finite index from every
> other path,
> since all the paths consist of ONLY finite indices.
> And the number of paths is uncountable. So this is neither surprising
> nor insightful.
claim that "for infinite paths" there are other laws to be applied
than in the case of finite paths.
>
> > Furthermore the subtree of the CIBT that have all finite
> > paths of the CIBT as initial segments of its paths is actually the
> > CIBT itself according to Cantor,
>
> And according to everybody else as well.
Not according to the majority of matheologians as I can assure you
> > Furthermore the subtree of the CIBT that have all finite
> > paths of the CIBT as initial segments of its paths is actually the
> > CIBT itself according to Cantor,
>
> And according to everybody else as well.
from the experience of many years - not 10, but nearly.
>
> The subtree of the CIBT beginning at any node of the CIBT *IS* the
> CIBT.
> The whole CIBT is arguably nothing but a triple of (left-copy-of-
> itself, root-node, right-copy-of-itself).
And therefore everything in the CIBT is countable.
> The subtree of the CIBT beginning at any node of the CIBT *IS* the
> CIBT.
> The whole CIBT is arguably nothing but a triple of (left-copy-of-
> itself, root-node, right-copy-of-itself).
Regards, WM
WM
Dec 25, 2012, 6:40:17 AM12/25/12
to
On 24 Dez., 21:32, George Greene <gree...@email.unc.edu> wrote:
> On Dec 23, 3:20 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > Let's take the binary tree with two levels below the root node level
> > which is the following:
>
> > 0
> > / \
> > 0 1
> > / \ | \
> > 0 1 0 1
>
> Oh, SHUT UP. You are labeling the NODES again.
> I TOLD YOU NOT TO DO THAT.
Yes, you told him that it is best not to think over that problem at
> On Dec 23, 3:20 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > Let's take the binary tree with two levels below the root node level
> > which is the following:
>
> > 0
> > / \
> > 0 1
> > / \ | \
> > 0 1 0 1
>
> Oh, SHUT UP. You are labeling the NODES again.
> I TOLD YOU NOT TO DO THAT.
all, at least not to talk about the results.
> The EDGES are what are labeled 0 and 1.
> The node is just labeled with the path that ends at it.
Both sets are ordered and are isomorphic. Therefore it is completeley
itrrelevant whether nodes or edges are considered. So much you should
have recognized during the 20 years or so that you think about set
theory and crank-arguments.
Regards, WM
Virgil
Dec 25, 2012, 3:59:36 PM12/25/12
to
In article
<76daea40-3902-4687...@b11g2000yqh.googlegroups.com>,
Any such tree is incomplete as a binary tree, as it necessarily contains
paths of all sorts of different lengths (different numbers of nodes or
numbers of branches), while in a complete infinite binary tree all paths
are of exactly the same length.
So that WM is WRONG!
AGAIN!!
AS USUAL!!!
--
<76daea40-3902-4687...@b11g2000yqh.googlegroups.com>,
paths of all sorts of different lengths (different numbers of nodes or
numbers of branches), while in a complete infinite binary tree all paths
are of exactly the same length.
So that WM is WRONG!
AGAIN!!
AS USUAL!!!
--
fom
Dec 25, 2012, 11:47:19 PM12/25/12
to
This is topological completeness.
The open sets of a tree are determined by its finite strings.
It is not that the finite strings are singular. It is that
they "collect" those infinite strings together that have the
same initial segment.
The infinite paths that are eventually constant form a
dense set. That is, they are just like the rational numbers
in respect to the real numbers.
The standard metric for these topologies is given by
d(x,y)= 0 for x=y
d(x,y)= 1/(2^-n) where n is the least n such that -(x_n=y_n)
You may think of your Dedekind cuts as occurring at
strings such as
111111100000000000.....
111111011111111111.....
which are not unlike
1.00000000...
0.99999999...
At any finite length, one can find finite strings
of greater length that lie between. But, there
are no infinite strings between these representations.
As for the "uncountability" argument, consider a
slightly different diagonal,
o,S(o),S(S(o)),S(S(S(o))),....
oo,S(oo),S(S(oo)),S(S(S(oo))),....
ooo,S(ooo),S(S(ooo)),S(S(S(ooo))),....
(let o=0, oo=omega, ooo=omega+omega=2omega, etc.)
There is no uncountability argument here, but there is
a diagonal argument.
If one accepts the "objecthood" or unity of any
horizontal infinity, then one ought not be permitted
to deny the "objecthood" or unity of the diagonal
infinity.
The sequence on the diagonal,
o,S(oo),S(ooo),...
cannot be of the logical category with which
any of the horizontal infinities are concerned.
Thus, there is a logical category of objects
for which we will call the first object
omega^2
Countability vs uncountability arguments are
intended to be counterexamples. If one claims
that there is only one infinity -- the infinity --
a singular notion of infinity as a unity -- an
interpretation of ALL as a singular term encompassing
the finite -- then, the diagonal argument produces
the counterexample.
In the face of such counterexamples, logical hierarchies
can be used to relate the logical types and establish
identity criteria for the higher level based on the
lower level. That is why there is a rational between
every pair of irrationals and an irrational between
every pair of rationals. This is what it means
to be a dense subset.
There are a great many assumptions in these matters --
one being that a completed infinity is always a
presupposition of populating these higher logical
types. Since one can do a great deal of mathematics
without accepting a completed infinity, one can
also accept the necessity of logical types without
accepting the necessity of a population of objects
satisfying the definition of those types.
I hope this helps. I told myself I was not
going to post anymore because the last time
I was out here I was in a bad mood and wrote
things I did not like. Good luck with your
various interests. I need to try to stick
with my resolution better.
WM
Dec 26, 2012, 3:33:22 AM12/26/12
to
> <76daea40-3902-4687-ae1c-53fe5356b...@b11g2000yqh.googlegroups.com>,
countably many infinite paths, all of same length. Try to find and
identify by nodes only one further path. Then your claim may be
considered by rational and sober thinkers. Everything else may be
considered by drunk tinkers.
Regards, WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 24 Dez., 20:14, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <d85d67ab-aa37-4091-9474-a089288c3...@x10g2000yqx.googlegroups.com>,
>
> > > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > > Also the proof of Cantor is actually about uncountability of paths
> > > > > that are distinguishable on finite basis.
>
> > > > That is the point!
>
> > > The Cantor argument only deals with distinguishability on a finite basis
> > > (each individual listed entry differs from the "diagonal" at an
> > > "individual finite position") but shows that it can occur infinitely
> > > often
>
> > without leaving the finite domain, i.e., the Binary Tree that contains
> > nothing but all finite paths.
>
> Any such tree is incomplete as a binary tree, as it necessarily contains
> paths of all sorts of different lengths (different numbers of nodes or
> numbers of branches), while in a complete infinite binary tree all paths
> are of exactly the same length.
>
I construct the complete Binary Tree, i.e., all its nodes, by
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 24 Dez., 20:14, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <d85d67ab-aa37-4091-9474-a089288c3...@x10g2000yqx.googlegroups.com>,
>
> > > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > On 23 Dez., 21:20, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > > Also the proof of Cantor is actually about uncountability of paths
> > > > > that are distinguishable on finite basis.
>
> > > > That is the point!
>
> > > The Cantor argument only deals with distinguishability on a finite basis
> > > (each individual listed entry differs from the "diagonal" at an
> > > "individual finite position") but shows that it can occur infinitely
> > > often
>
> > without leaving the finite domain, i.e., the Binary Tree that contains
> > nothing but all finite paths.
>
> Any such tree is incomplete as a binary tree, as it necessarily contains
> paths of all sorts of different lengths (different numbers of nodes or
> numbers of branches), while in a complete infinite binary tree all paths
> are of exactly the same length.
>
countably many infinite paths, all of same length. Try to find and
identify by nodes only one further path. Then your claim may be
considered by rational and sober thinkers. Everything else may be
considered by drunk tinkers.
Regards, WM
Zuhair
Dec 26, 2012, 3:43:38 AM12/26/12
to
On Dec 24, 12:42 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> There is nothing to happen "in the infinite". And it is completely
> irrelevant whether the paths after the distinction are finite or
> infinite. Everything that happens in a Cantor-list and in a Binary
> Tree happens at a finite level.
>
Up till now nobody have answered my question, anyhow. I still find it
>
> There is nothing to happen "in the infinite". And it is completely
> irrelevant whether the paths after the distinction are finite or
> infinite. Everything that happens in a Cantor-list and in a Binary
> Tree happens at a finite level.
>
puzzling really, Cantor has formally proved that there are more
distinguishable reals than are distinguishable finite initial segments
of them, I find that strange since the reals are only distinguishable
by those initial segments, so how they can be more than what makes
them distinguishable? This is too counter-intuitive!?
Probably this counter-intuitive issue is similar to the conflict
between distinguishability and the number of elements of a proper
subset and its set at infinite level, where the set would have
strictly more distinguishable elements than a proper subset of it and
yet they both have the SAME number of elements. So it appears to me
that the number of elements of infinite sets departs from the notion
of distinguishability.
I want to note that I'm not claiming to have paradox in the formal
sense, but there is a kind of extreme counter-intuitiveness involved
here with the notion of uncountability. Indeed this might drive some
to reject being involved with such concepts that would mess about our
intuitive faculaties and they would maintain that such slippery areas
of ideation are better avoided than engaged since they might be too
misleading. Anyhow
Zuhair
Zuhair
Dec 26, 2012, 3:46:14 AM12/26/12
to
equivalent approaches, the argument is already beyond that trivial
point.
Zuhair
George Greene
Dec 26, 2012, 3:49:22 AM12/26/12
to
On Dec 26, 3:46 am, Zuhair <zaljo...@gmail.com> wrote:
> It doesn't matter really whether nodes or edges are labeled,
It DOES SO TOO matter, otherwise we wouldn't be arguing about it.
> It doesn't matter really whether nodes or edges are labeled,
> they are equivalent approaches,
> the argument is already beyond that trivial
> point.
previous ones.
Thank you sincerely.
Virgil
Dec 26, 2012, 4:12:18 AM12/26/12
to
In article
<98ac2b0a-a54a-4b05...@b16g2000vbh.googlegroups.com>,
Then why do you say above that "the Binary Tree that contains nothing
but all finite paths."?
> Try to find and
> identify by nodes only one further path.
AS soon as you tell us what paths you have included in your tree, we
will be able to tell you at least some of the ones that you have left
out. But until you show us your paths we have no way of knowing which
ones are yours and which ones are not yours.
> Then your claim may be
> considered by rational and sober thinkers.
It already has been considered by rational and sober thinkers and
accepted by them and it is only drunk 'tinkers' like yourself who object.
> Everything else may be
> considered by drunk tinkers.
Even as a drunk 'tinker' WM is no good.
No matter how he tinkers with things they do not work.
--
<98ac2b0a-a54a-4b05...@b16g2000vbh.googlegroups.com>,
but all finite paths."?
> Try to find and
> identify by nodes only one further path.
will be able to tell you at least some of the ones that you have left
out. But until you show us your paths we have no way of knowing which
ones are yours and which ones are not yours.
> Then your claim may be
> considered by rational and sober thinkers.
accepted by them and it is only drunk 'tinkers' like yourself who object.
> Everything else may be
> considered by drunk tinkers.
No matter how he tinkers with things they do not work.
--
Virgil
Dec 26, 2012, 4:21:03 AM12/26/12
to
In article
<58fbc0c5-854b-4c63...@d4g2000vbw.googlegroups.com>,
infinite binary sequence that distinguishes it from ALL others.
For any finite set of such strings, finite initial segments suffice to
distinguish all of them from each oterhbut for at least some infinite
set, no finite set of finite initial segments suffices.
--
<58fbc0c5-854b-4c63...@d4g2000vbw.googlegroups.com>,
Zuhair <zalj...@gmail.com> wrote:
> On Dec 24, 12:42�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > There is nothing to happen "in the infinite". And it is completely
> > irrelevant whether the paths after the distinction are finite or
> > infinite. Everything that happens in a Cantor-list and in a Binary
> > Tree happens at a finite level.
> >
>
> Up till now nobody have answered my question, anyhow. I still find it
> puzzling really, Cantor has formally proved that there are more
> distinguishable reals than are distinguishable finite initial segments
> of them, I find that strange since the reals are only distinguishable
> by those initial segments, so how they can be more than what makes
> them distinguishable? This is too counter-intuitive!?
Note, however, that there is no finite initial segment of any one
> On Dec 24, 12:42�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > There is nothing to happen "in the infinite". And it is completely
> > irrelevant whether the paths after the distinction are finite or
> > infinite. Everything that happens in a Cantor-list and in a Binary
> > Tree happens at a finite level.
> >
>
> Up till now nobody have answered my question, anyhow. I still find it
> puzzling really, Cantor has formally proved that there are more
> distinguishable reals than are distinguishable finite initial segments
> of them, I find that strange since the reals are only distinguishable
> by those initial segments, so how they can be more than what makes
> them distinguishable? This is too counter-intuitive!?
infinite binary sequence that distinguishes it from ALL others.
>
> Probably this counter-intuitive issue is similar to the conflict
> between distinguishability and the number of elements of a proper
> subset and its set at infinite level, where the set would have
> strictly more distinguishable elements than a proper subset of it and
> yet they both have the SAME number of elements. So it appears to me
> that the number of elements of infinite sets departs from the notion
> of distinguishability.
Depends on the level of distinguishability at issue.
> Probably this counter-intuitive issue is similar to the conflict
> between distinguishability and the number of elements of a proper
> subset and its set at infinite level, where the set would have
> strictly more distinguishable elements than a proper subset of it and
> yet they both have the SAME number of elements. So it appears to me
> that the number of elements of infinite sets departs from the notion
> of distinguishability.
For any finite set of such strings, finite initial segments suffice to
distinguish all of them from each oterhbut for at least some infinite
set, no finite set of finite initial segments suffices.
>
> I want to note that I'm not claiming to have paradox in the formal
> sense, but there is a kind of extreme counter-intuitiveness involved
> here with the notion of uncountability. Indeed this might drive some
> to reject being involved with such concepts that would mess about our
> intuitive faculaties and they would maintain that such slippery areas
> of ideation are better avoided than engaged since they might be too
> misleading. Anyhow
What drives WM is shear orneryness.
> I want to note that I'm not claiming to have paradox in the formal
> sense, but there is a kind of extreme counter-intuitiveness involved
> here with the notion of uncountability. Indeed this might drive some
> to reject being involved with such concepts that would mess about our
> intuitive faculaties and they would maintain that such slippery areas
> of ideation are better avoided than engaged since they might be too
> misleading. Anyhow
--
Zuhair
Dec 26, 2012, 6:29:04 AM12/26/12
to
> <58fbc0c5-854b-4c63-bd5f-58faa3908...@d4g2000vbw.googlegroups.com>,
Yes but a countable set of them suffices! no?
>
>
> > I want to note that I'm not claiming to have paradox in the formal
> > sense, but there is a kind of extreme counter-intuitiveness involved
> > here with the notion of uncountability. Indeed this might drive some
> > to reject being involved with such concepts that would mess about our
> > intuitive faculaties and they would maintain that such slippery areas
> > of ideation are better avoided than engaged since they might be too
> > misleading. Anyhow
>
> What drives WM is shear orneryness.
Possibly I don't know, but there is some Intuitive issue that WM is
addressing. Anyhow those kinds of discussion are not really easy to
run because they are discussions at Truth level which is in a sense
higher than just formal level. One can always still keep insisting
that all sets are countable and that uncountability is just a form of
a Pseudo-argument as far as reality of the matters is concerned like
saying that the quantifiers in Cantor's argument can only be suitably
understood to be first order, i.e. ranging over "elements" of the
universe of discourse, and so doesn't cover ALL functions in reality,
because some functions (which are subsets of the universe of
discourse) might not be elements of the universe of discourse! that is
usually the basis for it being possible to have a countable model of a
theory that proves existence of uncountably many objects, and in this
scenario uncountability pops up as an artifact due to a defect in the
theory's ability to define all functions and not due to something that
reflects some issue that is present in the real world of sets. Those
kinds of arguments might really be motivated by presenting strong
intuitive similes against Cantor like that one present here, albeit
I'm not so sure if the point that I've presented here is sufficient
for such a drastic alternative move. Anyhow that doesn't mean that
Uncountability is not interesting, in reality it is, even if it is
just a kind of internal (intra-theory) manifestation, but by then it
would only be interesting at formal level. It won't have any
philosophical significance. Anyhow.
> --
>
>
> > I want to note that I'm not claiming to have paradox in the formal
> > sense, but there is a kind of extreme counter-intuitiveness involved
> > here with the notion of uncountability. Indeed this might drive some
> > to reject being involved with such concepts that would mess about our
> > intuitive faculaties and they would maintain that such slippery areas
> > of ideation are better avoided than engaged since they might be too
> > misleading. Anyhow
>
> What drives WM is shear orneryness.
addressing. Anyhow those kinds of discussion are not really easy to
run because they are discussions at Truth level which is in a sense
higher than just formal level. One can always still keep insisting
that all sets are countable and that uncountability is just a form of
a Pseudo-argument as far as reality of the matters is concerned like
saying that the quantifiers in Cantor's argument can only be suitably
understood to be first order, i.e. ranging over "elements" of the
universe of discourse, and so doesn't cover ALL functions in reality,
because some functions (which are subsets of the universe of
discourse) might not be elements of the universe of discourse! that is
usually the basis for it being possible to have a countable model of a
theory that proves existence of uncountably many objects, and in this
scenario uncountability pops up as an artifact due to a defect in the
theory's ability to define all functions and not due to something that
reflects some issue that is present in the real world of sets. Those
kinds of arguments might really be motivated by presenting strong
intuitive similes against Cantor like that one present here, albeit
I'm not so sure if the point that I've presented here is sufficient
for such a drastic alternative move. Anyhow that doesn't mean that
Uncountability is not interesting, in reality it is, even if it is
just a kind of internal (intra-theory) manifestation, but by then it
would only be interesting at formal level. It won't have any
philosophical significance. Anyhow.
> --
WM
Dec 26, 2012, 7:09:10 AM12/26/12
to
On 26 Dez., 05:47, fom <fomJ...@nyms.net> wrote:
> I hope this helps.
No. In mathematics we have real numbers that are defined exclusively
by their digits (or nodes), without any mythological labelling and
without being written in different colours of ink, when belonging to
"different logical types". We are interested only in the single
logiocal type of real number that make up the entries of a Cantor list
as well as its diagonal. There is not the slightest difference and
therefore no reference to logic in analysis.
> I told myself I was not
> going to post anymore because the last time
> I was out here I was in a bad mood and wrote
> things I did not like.
And you will not like what you wrote here (therefore I deleted it),
when you will have understood that the complete Binary Tree can be
constructed within countably many steps (node by node), i.e., there is
nothing uncountable that could be distinguished by nodes. .
Regards, WM
> I hope this helps.
No. In mathematics we have real numbers that are defined exclusively
by their digits (or nodes), without any mythological labelling and
without being written in different colours of ink, when belonging to
"different logical types". We are interested only in the single
logiocal type of real number that make up the entries of a Cantor list
as well as its diagonal. There is not the slightest difference and
therefore no reference to logic in analysis.
> I told myself I was not
> going to post anymore because the last time
> I was out here I was in a bad mood and wrote
> things I did not like.
when you will have understood that the complete Binary Tree can be
constructed within countably many steps (node by node), i.e., there is
nothing uncountable that could be distinguished by nodes. .
Regards, WM
WM
Dec 26, 2012, 7:12:18 AM12/26/12
to
On 26 Dez., 09:43, Zuhair <zaljo...@gmail.com> wrote:
> I want to note that I'm not claiming to have paradox in the formal
> sense, but there is a kind of extreme counter-intuitiveness involved
There is nothing depending on any intuition. The CIBT can be
> I want to note that I'm not claiming to have paradox in the formal
> sense, but there is a kind of extreme counter-intuitiveness involved
constructed in countably many steps, adding one node in every step and
never removing anything. That means thare are not more than countable
many different configurations. Therefore not more than countably many
different things can be distinguished by nodes. Where is any necessity
for "intuition" in this clear mathematical argument?
Regards, WM
WM
Dec 26, 2012, 7:14:46 AM12/26/12
to
matheologians like George Greene have no real insight, although he,
for one, has been active in this field for more than 20 years,
according to his own testimony.
Regards, WM
gus gassmann
Dec 26, 2012, 7:24:59 AM12/26/12
to
that there are uncountably many.
There are at least two counter-intuitive notions when dealing with
infinities: There is an infinite set, each of whose elements are finite
(viz. the sequence of initial segments {{1}, {1,2}, {1,2,3}, ...}; and
the set of all countable subsets of a countable set is uncountable. The
only thing this shows is that intuition is sometimes insufficient to
grasp complex things.
WM
Dec 26, 2012, 10:30:41 AM12/26/12
to
On 26 Dez., 10:21, Virgil <vir...@ligriv.com> wrote:
> Note, however, that there is no finite initial segment of any one
> infinite binary sequence that distinguishes it from ALL others.
Note that there is no sequence of digits that alone defines a number.
> Note, however, that there is no finite initial segment of any one
> infinite binary sequence that distinguishes it from ALL others.
You need always a finite definition like 0.111 and so in infinity.
You cannot use a finite sequence of digits alone to define a number,
because something unexpected could follow. And you cannot an infinite
sequence of digits to define a number, because the definition would
never get ready. All you can do is to use a finite definition like 1/3
or 0.111... or 0.25 and the rest is silence or zeros.
> For any finite set of such strings, finite initial segments suffice to
> distinguish all of them from each oterhbut for at least some infinite
> set, no finite set of finite initial segments suffices.
be infinite. But without finite definition, the Cantor-list is as
undefined as its diagonal.
>
> What drives WM is shear orneryness.
What is that? In doubt I contradict.
> What drives WM is shear orneryness.
Regards, WM
WM
Dec 26, 2012, 10:33:23 AM12/26/12
to
On 26 Dez., 10:12, Virgil <vir...@ligriv.com> wrote:
> > Try to find and
> > identify by nodes only one further path.
>
> AS soon as you tell us what paths you have included in your tree,
So you do no longer claim that the Cantor-argument works by digits or
> > Try to find and
> > identify by nodes only one further path.
>
> AS soon as you tell us what paths you have included in your tree,
nodes?
> we
> will be able to tell you at least some of the ones that you have left
> out. But until you show us your paths we have no way of knowing which
> ones are yours and which ones are not yours.
Binary Tree) and no nodes of your paths (namely the empty set of
nodes), you cannot distinguish my paths and your paths. But according
to Cantor's argument, you should be able to distinguish the diagonal
by nodes only.
> > Then your claim may be
> > considered by rational and sober thinkers.
>
> It already has been considered by rational and sober thinkers
> and
> accepted by them and it is only drunk 'tinkers' like yourself who object.
paths by nodes? Didn't you just say this few lines above?
Regards, WM
WM
Dec 26, 2012, 10:49:00 AM12/26/12
to
On 26 Dez., 13:24, gus gassmann <g...@nospam.com> wrote:
> On 26/12/2012 7:29 AM, Zuhair wrote:
>
>
>
> >> Depends on the level of distinguishability at issue.
>
> >> For any finite set of such strings, finite initial segments suffice to
> >> distinguish all of them from each oterhbut for at least some infinite
> >> set, no finite set of finite initial segments suffices.
>
> > Yes but a countable set of them suffices! no?
>
> Of course. And how many such countable sets are there? Cantor showed
> that there are uncountably many.
Cantor showed that by digits or nodes.
> On 26/12/2012 7:29 AM, Zuhair wrote:
>
>
>
> >> Depends on the level of distinguishability at issue.
>
> >> For any finite set of such strings, finite initial segments suffice to
> >> distinguish all of them from each oterhbut for at least some infinite
> >> set, no finite set of finite initial segments suffices.
>
> > Yes but a countable set of them suffices! no?
>
> Of course. And how many such countable sets are there? Cantor showed
> that there are uncountably many.
And I showed that they cannot be distinguished by nodes.
>
> There are at least two counter-intuitive notions when dealing with
> infinities: There is an infinite set, each of whose elements are finite
> (viz. the sequence of initial segments {{1}, {1,2}, {1,2,3}, ...}; and
> the set of all countable subsets of a countable set is uncountable. The
> only thing this shows is that intuition is sometimes insufficient to
> grasp complex things.
No, it shows that there is no nonsense great enough for matheologians
> There are at least two counter-intuitive notions when dealing with
> infinities: There is an infinite set, each of whose elements are finite
> (viz. the sequence of initial segments {{1}, {1,2}, {1,2,3}, ...}; and
> the set of all countable subsets of a countable set is uncountable. The
> only thing this shows is that intuition is sometimes insufficient to
> grasp complex things.
not to believe in (and to call their thinking "complex" and a simple
and clear contradiction "intuition"). No set of finite subsets of |N
exists, that was uncountable. Only a subset containing uncountably
many infinite subsets is uncountable. However, it is impossible to
define infinite subsets by themselves. You need always a finite
definition. The set of all finite definitions however is countable. No
intuition requred.
Cantor's and Hessenberg's "proofs" simply show that infinity is never
finished and a complete infinite set is not part of sober thinking.
You could be excused perhaps if the CIBT was the only contradiction of
your belief. But there is a lot more, for instance this one:
http://planetmath.org/?op=getobj&from=objects&id=12607
Regards, WM
Zuhair
Dec 26, 2012, 2:14:44 PM12/26/12
to
On Dec 26, 6:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
> Cantor's and Hessenberg's "proofs" simply show that infinity is never
> finished and a complete infinite set is not part of sober thinking.
>
> Regards, WM
>
>
> Cantor's and Hessenberg's "proofs" simply show that infinity is never
> finished and a complete infinite set is not part of sober thinking.
>
To make the discussion fruitful, lets take all possibilities available
and see what is the response to each.
(1) To say that the formal proof of Cantor is clear and exact in
formal terms, but the distinguishability argument is clear on
intuitive level but has not been verified in formal terms, so
accordingly we have the option of saying that Infinity do not copy
intuitions derived from the finite world, and deem the result as just
counter-intuitive but not paradoxical. I think this is the standard
approach.
(2) To say that the distinguishability argument is so clear and to
accept it as a proved result despite the possibilities of verifying it
at formal level or not, and also maintaining that Cantor's proof is
very clear and valid, and so we deduce that we have a genuine paradox
that resulted from assuming having completed infinity, and thus we
must reject having completed infinity. That's what WM is saying
(3) To consider countability of the finite initial segments FALSE,
i.e. to say that we have uncountably many finite initial segments of
reals and as well we have uncountably many reals. This clearly
preserves congruity of the argument, but it requires justification,
and the justification can be based on the principle of "parameter free
definability of sets", since the alleged bijection between the finite
initial segments of the reals and the set N of all naturals is NOT
parameter free definable, then this bijection does not exist, and it
is false to say that it is. This claim only accepts infinite sets to
exist if there is a parameter free formula after which membership of
those sets is determined, so if there is non then it doesn't accept
the existence sets that are not parameter free definable.
(4) To consider uncountability of the reals to be FALSE by
interpreting the universal quantifier in Cantor's argument over
functions to range over *elements* of the universe of discourse, and
thus it doesn't cover all available functions (which are subsets of
the universe of discourse), so there is a bijection between the reals
and the naturals that is simply missing from being among the
*elements* of the universe of discourse. So the reality of that matter
is that the set of all reals is countable but discourse misses the
necessary bijection. This is the interpretation of having a countable
model of a theory that proves the existence of uncountably many reals.
It is a consequence of Skolem's arguments. The justification for that
is that if we go second order and consider the universal quantifier
over functions in Cantor's argument to range over all subsets of the
universe of discourse, then we are at second order logic which is not
known to support a proof system and so it can barely be seen as a kind
of logic or something that we prove things after. So we are better
with at first order. If we use Henkin semantics to explain the second
order quantifier then we'll end up by the same argument of skolem. So
Henkin semantics for second order or first order semantics both ensure
having a countable model of any theory, and since proof theory by
ordinal analysis depends on constructiveness within countable limits,
then we only need to stipulate the existence of countable models since
it is those the ones that are both provable and also they provide a
reductionist approach reminiscent of Ockam's razor, that is if we can
do the job with less so why demand more. And since there is no logical
argument to force us to adhere to uncountability without assuming it
first, so why adhere to it?
It is interesting to know the responses to all those options.
Zuhair
fom
Dec 26, 2012, 2:44:17 PM12/26/12
to
On 12/26/2012 5:29 AM, Zuhair wrote:
> On Dec 26, 12:21 pm, Virgil <vir...@ligriv.com> wrote:
>>
>> For any finite set of such strings, finite initial segments suffice to
>> distinguish all of them from each oterhbut for at least some infinite
>> set, no finite set of finite initial segments suffices.
>>
> Yes but a countable set of them suffices! no?
>
No.
> On Dec 26, 12:21 pm, Virgil <vir...@ligriv.com> wrote:
>>
>> For any finite set of such strings, finite initial segments suffice to
>> distinguish all of them from each oterhbut for at least some infinite
>> set, no finite set of finite initial segments suffices.
>>
> Yes but a countable set of them suffices! no?
>
The diagonal argument generates a distinct logical
type given a particular assumption.
A completion of an incomplete dense topological space
populates the namespace of the new logical type.
The completed infinity of eventually constant infinite
strings is required to separate the completed infinity
of arbitrary strings.
Again, THE COMPLETED INFINITY OF EVENTUALLY CONSTANT
INFINITE STRINGS is required to separate the completed
infinity of arbitrary strings. These are not the
finite strings. The finite strings collect the
infinite strings into the open sets of the
topology.
This is what is meant by a dense set. Think of
the fact that there is a rational between every
two irrationals and an irrational between every
two rationals. The order relation of the rationals
grounds the order relation of the reals, and hence
gives an identity criterion for the reals in terms
of antisymmetry. But, this is not an existence
criterion.
A completed infinity is admitted as a base
assumption of the original argument.
A completed infinity is required to substantiate
any named particular of the new logical type.
There is no countable-to-uncountable magic until
one claims that there is a correspondence --
only type distinctions.
========================
These constructions trace back to the incommensurables
of Grecian geometry. The counter-intuitiveness
arises from geometric intuitions associated with
invariance under rigid motion and the measurement
of area.
The reason I mention the measurement of area is
because of an example from a statistics book I
once read. Area is a non-linear function of
linear measurements. As such, the random errors
of linear measurements can produce systematic
error in the subsequent calculation of area.
This does not happen with linear functions.
This example speaks to the general treatment
of differentials in calculus. Modern
differentials use the little-oh concept to
deal with small linear errors in the neighborhood
of a definite value.
dx=delta(x)
dy=f'(x)dx
delta(y)=f(x+delta(x))-f(x)
delta(y)=f'(x)h + o(h) where o(h)->0 as h->0
df=f'(x)h
Whereas it once seemed that the calculus
depended on the structure of the real
numbers, that dependency is less apparent
now. At the time Cantor and Dedekind did
their investigations, infinitesimals were
still actively being researched. Continuity
seemed to depend upon the geometric completeness
of the real line, but there had been a strong
desire on the part of late nineteenth century
mathematicians to get away from geometric
reasoning. As various algebraic systems
different from naturals, rationals, or reals
had been shown to have arithmetic relations,
the notion of "number" had been expanded.
Thus Cantor and Dedekind had little inhibition
against introducing logical types having the same
relations as the numbers with which everyone
had been using and then calling them "numbers."
But, if you need something intuitive
to ground what Cantor and Dedekind
were doing, look to the fundamental
theorem of algebra which asserts that
the field of complex numbers is
algebraically closed. The proof
depends upon the fact that every
positive real number has a real
positive square root and that every
polynomial of odd degree over the
real numbers has a root in the real
numbers. These assertions depend in
the first instance on the existence
of irrational numbers and in the
second instance on the intermediate
value theorem (meaning no "gaps" like
with the rational numbers).
So, in order for (x^2-2)=0 to have
a solution, there has to be something
that is not a rational number but
behaves like a number.
All known proofs of the fundamental
theorem of algebra depend on real
analysis (which is not the same
as calculus or approximation
theory).
Example on area referred to above:
http://books.google.com/books?id=BQDP5W4xAXUC&pg=PA69&lpg=PA69&dq=statistics+bias++%22area+of+a+circle%22&source=bl&ots=ZXi5Mdiu3Y&sig=Cxnp7akTywOLw2Vo0-ZoOsWJz4A&hl=en&sa=X&ei=ljnbUI-eGaLe2AWMnoHgBQ&ved=0CFkQ6AEwBg#v=onepage&q=statistics%20bias%20%20%22area%20of%20a%20circle%22&f=false
Virgil
Dec 26, 2012, 3:14:21 PM12/26/12
to
In article
<96a45cc2-099b-4ebc...@a8g2000vby.googlegroups.com>,
As one identifies a path as a particular set of digits or nodes, how is
asking you to identify your paths in any way overlooking them?
specify at least infinitely many of its nodes as being particular to
that path before it can b separated from any other path, but you never
have done this with any of your paths, so there is no telling which
paths you have.
You have also not proved that your set of paths is countable, which can
only be done by proving that they can be listed, and most easily done by
providing such a list.
> But according
> to Cantor's argument, you should be able to distinguish the diagonal
> by nodes only.
Only if one has a list of all the paths one wishes to distinguish it
from.
Since you claim that you set of paths can b isted, lets see you actually
do it.
infinite binary sequences) which includes every possible innfinite
binary sequnce.
To prove your set is countable you must prove it to be listable, since
that is the same thing, but any listing of such infinite binary
sequences is provably incomplete, a fact know to all here but denied by
WM.
--
<96a45cc2-099b-4ebc...@a8g2000vby.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Dez., 10:12, Virgil <vir...@ligriv.com> wrote:
>
> > > �Try to find and
> > > identify by nodes only one further path.
> >
> > AS soon as you tell us what paths you have included in your tree,
>
> So you do no longer claim that the Cantor-argument works by digits or
> nodes?
As usual, you draw unwarranted conclusions from everything.
> On 26 Dez., 10:12, Virgil <vir...@ligriv.com> wrote:
>
> > > �Try to find and
> > > identify by nodes only one further path.
> >
> > AS soon as you tell us what paths you have included in your tree,
>
> So you do no longer claim that the Cantor-argument works by digits or
> nodes?
As one identifies a path as a particular set of digits or nodes, how is
asking you to identify your paths in any way overlooking them?
>
> > we
> > will be able to tell you at least some of the ones that you have left
> > out. But until you show us your paths we have no way of knowing which
> > ones are yours and which ones are not yours.
>
> Although I told you all the nodes of my paths (namely all nodes of the
> Binary Tree) and no nodes of your paths (namely the empty set of
> nodes), you cannot distinguish my paths and your paths.
A path being a set of nodes, to distinguish it from other path one must
> > we
> > will be able to tell you at least some of the ones that you have left
> > out. But until you show us your paths we have no way of knowing which
> > ones are yours and which ones are not yours.
>
> Although I told you all the nodes of my paths (namely all nodes of the
> Binary Tree) and no nodes of your paths (namely the empty set of
> nodes), you cannot distinguish my paths and your paths.
specify at least infinitely many of its nodes as being particular to
that path before it can b separated from any other path, but you never
have done this with any of your paths, so there is no telling which
paths you have.
You have also not proved that your set of paths is countable, which can
only be done by proving that they can be listed, and most easily done by
providing such a list.
> But according
> to Cantor's argument, you should be able to distinguish the diagonal
> by nodes only.
from.
Since you claim that you set of paths can b isted, lets see you actually
do it.
>
> > > Then your claim may be
> > > considered by rational and sober thinkers.
> >
> > It already has been considered by rational and sober thinkers
>
> like you
>
> > and
> > accepted by them and it is only drunk 'tinkers' like yourself who object.
>
> Object to what?
To your claim that you have a countable set of paths (in the form of
> > > Then your claim may be
> > > considered by rational and sober thinkers.
> >
> > It already has been considered by rational and sober thinkers
>
> like you
>
> > and
> > accepted by them and it is only drunk 'tinkers' like yourself who object.
>
> Object to what?
infinite binary sequences) which includes every possible innfinite
binary sequnce.
To prove your set is countable you must prove it to be listable, since
that is the same thing, but any listing of such infinite binary
sequences is provably incomplete, a fact know to all here but denied by
WM.
--
Virgil
Dec 26, 2012, 3:26:28 PM12/26/12
to
In article
<d7e06e32-533c-4b90...@a6g2000vbh.googlegroups.com>,
numbers quite effectively,
> You need always a finite definition like 0.111 and so in infinity.
>
> You cannot use a finite sequence of digits alone to define a number,
> because something unexpected could follow. And you cannot an infinite
> sequence of digits to define a number, because the definition would
> never get ready. All you can do is to use a finite definition like 1/3
> or 0.111... or 0.25 and the rest is silence or zeros.
>
>
> > For any finite set of such strings, finite initial segments suffice to
> > distinguish all of them from each other but for at least some infinite
> Therefore even the Cantor list must
> be infinite.
> But without finite definition, the Cantor-list is as
> undefined as its diagonal.
It is those who claim that the set of digit strings is countable who
must create that allegedly impossible listing to prove that that set
satisfies the definition of countability.
Any set whose members cannot be 'listed' is, by definition, uncountable.
If WM chooses to say that the set of real numbers, or the set of endless
binary strings cannot be listed, then it is WM who is arguing that those
sets are uncountable.
--
<d7e06e32-533c-4b90...@a6g2000vbh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Dez., 10:21, Virgil <vir...@ligriv.com> wrote:
>
> > Note, however, that there is no finite initial segment of any one
> > infinite binary sequence that distinguishes it from ALL others.
>
> Note that there is no sequence of digits that alone defines a number.
Actually there are lots of finite sequences of digits that define
> On 26 Dez., 10:21, Virgil <vir...@ligriv.com> wrote:
>
> > Note, however, that there is no finite initial segment of any one
> > infinite binary sequence that distinguishes it from ALL others.
>
> Note that there is no sequence of digits that alone defines a number.
numbers quite effectively,
> You need always a finite definition like 0.111 and so in infinity.
>
> You cannot use a finite sequence of digits alone to define a number,
> because something unexpected could follow. And you cannot an infinite
> sequence of digits to define a number, because the definition would
> never get ready. All you can do is to use a finite definition like 1/3
> or 0.111... or 0.25 and the rest is silence or zeros.
>
>
> > For any finite set of such strings, finite initial segments suffice to
> > set, no finite set of finite initial segments suffices.
>
> And no infinite set is available.
At least not in WMytheology.
>
> And no infinite set is available.
> Therefore even the Cantor list must
> be infinite.
> But without finite definition, the Cantor-list is as
> undefined as its diagonal.
must create that allegedly impossible listing to prove that that set
satisfies the definition of countability.
Any set whose members cannot be 'listed' is, by definition, uncountable.
If WM chooses to say that the set of real numbers, or the set of endless
binary strings cannot be listed, then it is WM who is arguing that those
sets are uncountable.
> >
> > What drives WM is shear orneryness.
>
> What is that?
The necessary quality for membership in your WMytheology.
> > What drives WM is shear orneryness.
>
> What is that?
--
Virgil
Dec 26, 2012, 3:34:09 PM12/26/12
to
In article
<4ff1889f-baa9-43d7...@a8g2000vby.googlegroups.com>,
at least show that they can be listed, since the very definition of
countability of a set requires that N can be surjected to it.
> Therefore not more than countably many
> different things can be distinguished by nodes.
Until you can prove listability, you have not proved countability.
> Where is any necessity
> for "intuition" in this clear mathematical argument?
You "intuit" but do not prove, that your sets satisfy the formal
definition of countability.
Or does WM claim a definition of countability other than the standard
one?
So much of WM's WMytheology is non-standard in actual mathematics, that
I would not be surprised if WM has his own private definition of
countability, that he hise the same way he hides the paths in his aleged
but unproven Complete Infinite Binary Tree.
--
<4ff1889f-baa9-43d7...@a8g2000vby.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Dez., 09:43, Zuhair <zaljo...@gmail.com> wrote:
>
> > I want to note that I'm not claiming to have paradox in the formal
> > sense, but there is a kind of extreme counter-intuitiveness involved
>
> There is nothing depending on any intuition. The CIBT can be
> constructed in countably many steps, adding one node in every step and
> never removing anything. That means thare are not more than countable
> many different configurations.
In order to PROVE only countably many, one must be able to list them, or
> On 26 Dez., 09:43, Zuhair <zaljo...@gmail.com> wrote:
>
> > I want to note that I'm not claiming to have paradox in the formal
> > sense, but there is a kind of extreme counter-intuitiveness involved
>
> There is nothing depending on any intuition. The CIBT can be
> constructed in countably many steps, adding one node in every step and
> never removing anything. That means thare are not more than countable
> many different configurations.
at least show that they can be listed, since the very definition of
countability of a set requires that N can be surjected to it.
> Therefore not more than countably many
> different things can be distinguished by nodes.
> Where is any necessity
> for "intuition" in this clear mathematical argument?
definition of countability.
Or does WM claim a definition of countability other than the standard
one?
So much of WM's WMytheology is non-standard in actual mathematics, that
I would not be surprised if WM has his own private definition of
countability, that he hise the same way he hides the paths in his aleged
but unproven Complete Infinite Binary Tree.
--
Virgil
Dec 26, 2012, 3:46:39 PM12/26/12
to
In article
<638ebf80-7340-44b9...@r14g2000vbe.googlegroups.com>,
> > distinguish all of them from each other but for at least some infinite
identified by its terminal node, if any set of finite initial segments
does it, that set certainly must be countable.
--
<638ebf80-7340-44b9...@r14g2000vbe.googlegroups.com>,
> > set, no finite set of finite initial segments suffices.
> >
> Yes but a countable set of them suffices! no?
Since there are only countably many finite initial segments, each one
> >
> Yes but a countable set of them suffices! no?
identified by its terminal node, if any set of finite initial segments
does it, that set certainly must be countable.
fom
Dec 26, 2012, 4:03:49 PM12/26/12
to
On 12/26/2012 2:34 PM, Virgil wrote:
> In article
> <4ff1889f-baa9-43d7...@a8g2000vby.googlegroups.com>,
> WM <muec...@rz.fh-augsburg.de> wrote:
>
>
> Or does WM claim a definition of countability other than the standard
> one?
>
He does. He stated it.
> In article
> <4ff1889f-baa9-43d7...@a8g2000vby.googlegroups.com>,
> WM <muec...@rz.fh-augsburg.de> wrote:
>
>
> Or does WM claim a definition of countability other than the standard
> one?
>
A set is countable iff it is a subset of a countable set.
Infinite sets are, therefore, if they exist, uncountable.
Claims of non-existence, of course, are just as metaphysical
as claims of existence. WM is an empiricist.
fom
Dec 26, 2012, 4:47:05 PM12/26/12
to
On 12/26/2012 9:49 AM, WM wrote:
> http://planetmath.org/?op=getobj&from=objects&id=12607
The example is really cute.
I just finished reading through a
great deal of Leibniz. He portrayed
his logic as being oppositely oriented
from that of scholastic thinking. So,
the hierarchy of genera in the two
logics is opposed.
Scholastic logic is extensional, but
Leibniz logic is intensional. Curiously,
he also attributes the principle of
identity of indiscernibles to Thomas
Aquinas. The summary statement was
"an individual is the lowest species"
This may clearly be interpreted along
the lines of Cantor's intersection
theorem with vanishing diameters.
Set theory, itself (along with the
Fregean "extension of a concept"),
is based on extensionality.
So, while I will take a little more
time to weigh your example, it does
reflect precisely a mismatch in
the historical development.
> http://planetmath.org/?op=getobj&from=objects&id=12607
The example is really cute.
I just finished reading through a
great deal of Leibniz. He portrayed
his logic as being oppositely oriented
from that of scholastic thinking. So,
the hierarchy of genera in the two
logics is opposed.
Scholastic logic is extensional, but
Leibniz logic is intensional. Curiously,
he also attributes the principle of
identity of indiscernibles to Thomas
Aquinas. The summary statement was
"an individual is the lowest species"
This may clearly be interpreted along
the lines of Cantor's intersection
theorem with vanishing diameters.
Set theory, itself (along with the
Fregean "extension of a concept"),
is based on extensionality.
So, while I will take a little more
time to weigh your example, it does
reflect precisely a mismatch in
the historical development.
Virgil
Dec 26, 2012, 6:03:53 PM12/26/12
to
In article
<c3b5cefe-1a77-479c...@g6g2000vbk.googlegroups.com>,
digits or nodes, Cantor did no such thing.
> And I showed
The only thing you evern sow is your own ignorance.
> not to believe in
> No set of finite subsets of |N exists, that was uncountable.
Do you mean that now there are some which have now become uncountable?
> Only a subset containing uncountably
> many infinite subsets is uncountable.
Right! And in ZFC, for example, such sets must exist.
--
<c3b5cefe-1a77-479c...@g6g2000vbk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Dez., 13:24, gus gassmann <g...@nospam.com> wrote:
> > On 26/12/2012 7:29 AM, Zuhair wrote:
> >
> >
> >
> > >> Depends on the level of distinguishability at issue.
> >
> > >> For any finite set of such strings, finite initial segments suffice to
> > >> distinguish all of them from each oterhbut for at least some infinite
> > >> set, no finite set of finite initial segments suffices.
> >
> > > Yes but a countable set of them suffices! no?
> >
> > Of course. And how many such countable sets are there? Cantor showed
> > that there are uncountably many.
>
> Cantor showed that by digits or nodes.
Since while sets may have members they need not ever contain either
> On 26 Dez., 13:24, gus gassmann <g...@nospam.com> wrote:
> > On 26/12/2012 7:29 AM, Zuhair wrote:
> >
> >
> >
> > >> Depends on the level of distinguishability at issue.
> >
> > >> For any finite set of such strings, finite initial segments suffice to
> > >> distinguish all of them from each oterhbut for at least some infinite
> > >> set, no finite set of finite initial segments suffices.
> >
> > > Yes but a countable set of them suffices! no?
> >
> > Of course. And how many such countable sets are there? Cantor showed
> > that there are uncountably many.
>
> Cantor showed that by digits or nodes.
digits or nodes, Cantor did no such thing.
> And I showed
The only thing you evern sow is your own ignorance.
> >
> > There are at least two counter-intuitive notions when dealing with
> > infinities: There is an infinite set, each of whose elements are finite
> > (viz. the sequence of initial segments {{1}, {1,2}, {1,2,3}, ...}; and
> > the set of all countable subsets of a countable set is uncountable. The
> > only thing this shows is that intuition is sometimes insufficient to
> > grasp complex things.
>
> No, it shows that there is no nonsense great enough for matheologians
Like WM
> > There are at least two counter-intuitive notions when dealing with
> > infinities: There is an infinite set, each of whose elements are finite
> > (viz. the sequence of initial segments {{1}, {1,2}, {1,2,3}, ...}; and
> > the set of all countable subsets of a countable set is uncountable. The
> > only thing this shows is that intuition is sometimes insufficient to
> > grasp complex things.
>
> No, it shows that there is no nonsense great enough for matheologians
> not to believe in
> No set of finite subsets of |N exists, that was uncountable.
> Only a subset containing uncountably
> many infinite subsets is uncountable.
>
> Cantor's and Hessenberg's "proofs" simply show that infinity is never
> finished and a complete infinite set is not part of sober thinking.
Thinking about infiniteness certainly makes WM extremely unsober.
> Cantor's and Hessenberg's "proofs" simply show that infinity is never
> finished and a complete infinite set is not part of sober thinking.
--
George Greene
Dec 26, 2012, 6:45:58 PM12/26/12
to
On Dec 26, 2:14 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> (3) To consider countability of the finite initial segments FALSE,
is NOT something you should be posting to sci.math. YOUR reputation
>
> (3) To consider countability of the finite initial segments FALSE,
is at stake.
> This clearly
> preserves congruity of the argument, but it requires justification,
> and the justification can be based on the principle
with N.
What is or isn't parameter-free has NOTHING to do with it.
George Greene
Dec 26, 2012, 6:50:13 PM12/26/12
to
On Dec 26, 4:03 pm, fom <fomJ...@nyms.net> wrote:
> A set is countable iff it is a subset of a countable set.
>
> Infinite sets are, therefore, if they exist, uncountable.
You wouldn't know logic if it spit in your eye.
> A set is countable iff it is a subset of a countable set.
>
> Infinite sets are, therefore, if they exist, uncountable.
Your argument above IS UNSOUND.
THERE EXISTS (at least one) infinite countable set. Since it is a
subset OF ITSELF,
it does NOT follow from your argument above that THIS argument fails.
More to the point, the successor OF ANY countable set is countable --
here's another iff:
a set is countable iff its successor is countable. EVERY SET HAS a
successor.
Moreover,
A set is countable iff its union with a countable disjoint subset is
countable.
This does NOT preclude the set of all even naturals, the set of all
odd naturals,
and the set of all naturals ALL being countable (which is good, since
of course THEY ALL ARE).
George Greene
Dec 26, 2012, 6:51:59 PM12/26/12
to
On Dec 26, 4:03 pm, fom <fomJ...@nyms.net> wrote:
> > Or does WM claim a definition of countability other than the standard
> > one?
>
> He does. He stated it.
>
> A set is countable iff it is a subset of a countable set.
This IS NOT all that different from the traditional definition.
> > Or does WM claim a definition of countability other than the standard
> > one?
>
> He does. He stated it.
>
> A set is countable iff it is a subset of a countable set.
This is TRUE.
This IS A CONSEQUENCE of the traditional definition. The problem is,
this definition DOES NOT imply that infinite sets are uncountable.
If there is an infinite countable set, then, obviously, both itself
and
many of its supersets (e.g. its successor) WOULD ALSO be countable.
AND THEY ARE. And this definition does NOT preclude that.
George Greene
Dec 26, 2012, 6:57:22 PM12/26/12
to
On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
> The infinite paths that are eventually constant form a
> dense set. That is, they are just like the rational numbers
> in respect to the real numbers.
I did in fact already explain (perhaps it was in another thread; this
discussion seems to jump around randomly from the messages under one
first-one to the messages under another) that the paths that are
eventually all-left
or all-right all represent rational numbers with a denominator that is
a power of 2, and that the
rational number they represent is (in the all-left case) the one
represented by the last (right-child) node
in the other direction. In the all-right case, AS I ALREADY
EXPLAINED, the represented rational is
the one that is the right sibling of the last node in the path that is
a left-child (this is assuming
left-branch is 0 and right-branch is 1; Zuhair may have a problem
seeing it correctly because he
may still be talking about labeling the nodes instead of the edges;
each EDGE is labeled 0 or 1 and
each node is labeled with the number it represents, which is just the
finite path ending at it). All
the paths ending with an infinite-unidirectional tail REPRESENT THE
SAME RATIONAL AS SOME FINITE path.
If the tail is all left then it is just the finite path you get by
chopping OFF the tail (infinite-all-left
tails are just IRRELEVANT since they mean all ZEROs).
> The infinite paths that are eventually constant form a
> dense set. That is, they are just like the rational numbers
> in respect to the real numbers.
I did in fact already explain (perhaps it was in another thread; this
discussion seems to jump around randomly from the messages under one
first-one to the messages under another) that the paths that are
eventually all-left
or all-right all represent rational numbers with a denominator that is
a power of 2, and that the
rational number they represent is (in the all-left case) the one
represented by the last (right-child) node
in the other direction. In the all-right case, AS I ALREADY
EXPLAINED, the represented rational is
the one that is the right sibling of the last node in the path that is
a left-child (this is assuming
left-branch is 0 and right-branch is 1; Zuhair may have a problem
seeing it correctly because he
may still be talking about labeling the nodes instead of the edges;
each EDGE is labeled 0 or 1 and
each node is labeled with the number it represents, which is just the
finite path ending at it). All
the paths ending with an infinite-unidirectional tail REPRESENT THE
SAME RATIONAL AS SOME FINITE path.
If the tail is all left then it is just the finite path you get by
chopping OFF the tail (infinite-all-left
tails are just IRRELEVANT since they mean all ZEROs).
George Greene
Dec 26, 2012, 7:00:39 PM12/26/12
to
On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
> There are a great many assumptions in these matters --
> one being that a completed infinity is always a
> presupposition of
It is FOR DAMN sure that the presupposition of a completed infinity
IS A PRESUPPOSITION OF the mere EXISTENCE of *THE*COMPLETE* infinite
binary
tree! GIVEN THAT WM HAS STARTED BY SWALLOWING THE ELEPHANT, he does
NOT get
to COMPLAIN about indigestion with the GNAT!
That completed infinities exist in a relevant way is something that WM
has not only already CONCEDED, but has HIMSELF STIPULATED AND
POSTULATED,
GIVEN that HE is the one talking about "THE *COMPLETE*INFINITE* Binary
Tree"!!
Qualms about the possible non-existence of "completed infinities"
therefore
SIMPLY HAVE NO PLACE WHATSOEVER IN *THIS* discussion.
Each and every one of the infinite paths he is trying to count
*IS*A*COMPLETED* infinity!!
Wherefore WE'LL have NO more of THAT nonsense!
Zuhair
Dec 26, 2012, 11:47:50 PM12/26/12
to
reals is not parameter free definable, that's why if one holds that No
set is not parameter free definable, then this bijection won't exist.
Zuhair
fom
Dec 27, 2012, 3:16:16 AM12/27/12
to
On 12/26/2012 5:57 PM, George Greene wrote:
> On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
>> The infinite paths that are eventually constant form a
>> dense set. That is, they are just like the rational numbers
>> in respect to the real numbers.
>
> I did in fact already explain
<snip>
> On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
>> The infinite paths that are eventually constant form a
>> dense set. That is, they are just like the rational numbers
>> in respect to the real numbers.
>
> I did in fact already explain
You did. It seems not to have stuck.
fom
Dec 27, 2012, 3:25:35 AM12/27/12
to
well. But, you are correct. I should have stipulated that the
notion of infinity involved here would have to be potential
infinity so that countable infinity is the same as a round square.
fom
Dec 27, 2012, 3:33:31 AM12/27/12
to
On 12/26/2012 6:00 PM, George Greene wrote:
> On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
>> There are a great many assumptions in these matters --
>> one being that a completed infinity is always a
>> presupposition of
>
> It is FOR DAMN sure that the presupposition of a completed infinity
> IS A PRESUPPOSITION OF the mere EXISTENCE of *THE*COMPLETE* infinite
> binary
> tree! GIVEN THAT WM HAS STARTED BY SWALLOWING THE ELEPHANT, he does
> NOT get
> to COMPLAIN about indigestion with the GNAT!
> That completed infinities exist in a relevant way is something that WM
> has not only already CONCEDED, but has HIMSELF STIPULATED AND
> POSTULATED,
<snip>
> On Dec 25, 11:47 pm, fom <fomJ...@nyms.net> wrote:
>> There are a great many assumptions in these matters --
>> one being that a completed infinity is always a
>> presupposition of
>
> It is FOR DAMN sure that the presupposition of a completed infinity
> IS A PRESUPPOSITION OF the mere EXISTENCE of *THE*COMPLETE* infinite
> binary
> tree! GIVEN THAT WM HAS STARTED BY SWALLOWING THE ELEPHANT, he does
> NOT get
> to COMPLAIN about indigestion with the GNAT!
> That completed infinities exist in a relevant way is something that WM
> has not only already CONCEDED, but has HIMSELF STIPULATED AND
> POSTULATED,
You are so good at finding those historical points
I will not argue.
But, I did ask WM certain questions that, by the
end, led to him rejecting any notion of actual
infinity. The claim was that he held a position
not unlike Abraham Robinson. The difference,
however, is that WM accepts infinity only to
the extent that the extended real number system
can be formed. Robinson rejected any actuality
of infinity but felt mathematicians should
work as if it were. Giving him the benefit
of the doubt, one must conclude he is a lazy
[...] since he should be talking about constructive
mathematics rather than harassing people
using standard mathematical definitions.
fom
Dec 27, 2012, 3:38:20 AM12/27/12
to
On 12/26/2012 5:50 PM, George Greene wrote:
> On Dec 26, 4:03 pm, fom <fomJ...@nyms.net> wrote:
>
>> A set is countable iff it is a subset of a countable set.
>>
>> Infinite sets are, therefore, if they exist, uncountable.
>
> You wouldn't know logic if it spit in your eye.
> Your argument above IS UNSOUND.
Mistaken.
> On Dec 26, 4:03 pm, fom <fomJ...@nyms.net> wrote:
>
>> A set is countable iff it is a subset of a countable set.
>>
>> Infinite sets are, therefore, if they exist, uncountable.
>
> You wouldn't know logic if it spit in your eye.
> Your argument above IS UNSOUND.
What I should have said is that a countable infinity
is like a round square in WM's philosophy.
WM is a finitist on this point.
I have a hard time thinking like a finitist and I
was trying to explain someone else's position.
This one really pissed you off since you
responded twice.
gus gassmann
Dec 27, 2012, 6:31:21 AM12/27/12
to
On 26/12/2012 5:03 PM, fom wrote:
> On 12/26/2012 2:34 PM, Virgil wrote:
>> Or does WM claim a definition of countability other than the standard
>> one?
>>
>
> He does. He stated it.
>
> A set is countable iff it is a subset of a countable set.
This definition is, of course, circular, and should convince anyone who
> On 12/26/2012 2:34 PM, Virgil wrote:
>> Or does WM claim a definition of countability other than the standard
>> one?
>>
>
> He does. He stated it.
>
> A set is countable iff it is a subset of a countable set.
reads it that the writer of this "definition" does not understand the
first points of mathematics.
> Infinite sets are, therefore, if they exist, uncountable.
> Claims of non-existence, of course, are just as metaphysical
> as claims of existence. WM is an empiricist.
Sam Sung
Dec 27, 2012, 6:35:06 AM12/27/12
to
gus gassmann schrieb:
The idiot WM thinks that every set must be constructed and while
this idiot "believes" one can construct an countable but infinite
set he does not believe one can construct an uncountable infinite
set from a countable infinite set ;)
this idiot "believes" one can construct an countable but infinite
set he does not believe one can construct an uncountable infinite
set from a countable infinite set ;)
WM
Dec 27, 2012, 7:08:13 AM12/27/12
to
On 26 Dez., 20:14, Zuhair <zaljo...@gmail.com> wrote:
> On Dec 26, 6:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
> > Cantor's and Hessenberg's "proofs" simply show that infinity is never
> > finished and a complete infinite set is not part of sober thinking.
>
> > Regards, WM
>
> To make the discussion fruitful, lets take all possibilities available
> and see what is the response to each.
>
> (1) To say that the formal proof of Cantor is clear and exact in
> formal terms, but the distinguishability argument is clear on
> intuitive level but has not been verified in formal terms, so
> accordingly we have the option of saying that Infinity do not copy
> intuitions derived from the finite world, and deem the result as just
> counter-intuitive but not paradoxical. I think this is the standard
> approach.
>
> (2) To say that the distinguishability argument is so clear and to
> accept it as a proved result despite the possibilities of verifying it
> at formal level or not, and also maintaining that Cantor's proof is
> very clear and valid, and so we deduce that we have a genuine paradox
> that resulted from assuming having completed infinity, and thus we
> must reject having completed infinity. That's what WM is saying
Yes, but it would not be correct to call it a paradox (i.e., something
> On Dec 26, 6:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
> > Cantor's and Hessenberg's "proofs" simply show that infinity is never
> > finished and a complete infinite set is not part of sober thinking.
>
> > Regards, WM
>
> To make the discussion fruitful, lets take all possibilities available
> and see what is the response to each.
>
> (1) To say that the formal proof of Cantor is clear and exact in
> formal terms, but the distinguishability argument is clear on
> intuitive level but has not been verified in formal terms, so
> accordingly we have the option of saying that Infinity do not copy
> intuitions derived from the finite world, and deem the result as just
> counter-intuitive but not paradoxical. I think this is the standard
> approach.
>
> (2) To say that the distinguishability argument is so clear and to
> accept it as a proved result despite the possibilities of verifying it
> at formal level or not, and also maintaining that Cantor's proof is
> very clear and valid, and so we deduce that we have a genuine paradox
> that resulted from assuming having completed infinity, and thus we
> must reject having completed infinity. That's what WM is saying
contrary to intuition like the relativistic twin paradox) but an
antinomy, because both results contradicting each othe can be obtained
formally.
>
> (3) To consider countability of the finite initial segments FALSE,
> i.e. to say that we have uncountably many finite initial segments of
> reals and as well we have uncountably many reals. This clearly
> preserves congruity of the argument, but it requires justification,
> and the justification can be based on the principle of "parameter free
> definability of sets", since the alleged bijection between the finite
> initial segments of the reals and the set N of all naturals is NOT
> parameter free definable, then this bijection does not exist, and it
> is false to say that it is. This claim only accepts infinite sets to
> exist if there is a parameter free formula after which membership of
> those sets is determined, so if there is non then it doesn't accept
> the existence sets that are not parameter free definable.
Here is a parameter free enumeration of all finite initial segments of
> (3) To consider countability of the finite initial segments FALSE,
> i.e. to say that we have uncountably many finite initial segments of
> reals and as well we have uncountably many reals. This clearly
> preserves congruity of the argument, but it requires justification,
> and the justification can be based on the principle of "parameter free
> definability of sets", since the alleged bijection between the finite
> initial segments of the reals and the set N of all naturals is NOT
> parameter free definable, then this bijection does not exist, and it
> is false to say that it is. This claim only accepts infinite sets to
> exist if there is a parameter free formula after which membership of
> those sets is determined, so if there is non then it doesn't accept
> the existence sets that are not parameter free definable.
the paths of the Binary Tree:
0
1, 2
3, 4, 5, 6
7, ...
Regards, WM
WM
Dec 27, 2012, 7:11:41 AM12/27/12
to
On 26 Dez., 21:14, Virgil <vir...@ligriv.com> wrote:
> As usual, you draw unwarranted conclusions from everything.
>
> As one identifies a path as a particular set of digits or nodes, how is
> asking you to identify your paths in any way overlooking them?
Since I have constructed or covered all paths that can be defined by
> As usual, you draw unwarranted conclusions from everything.
>
> As one identifies a path as a particular set of digits or nodes, how is
> asking you to identify your paths in any way overlooking them?
nodes. There is no node that you could name that could define a path
that is not in the CIBT that I constructed.
> You have also not proved that your set of paths is countable, which can
> only be done by proving that they can be listed, and most easily done by
> providing such a list.
of all finite paths extending from the root node to a given node. You
can append them by whatever tail you want, because all these tails are
also covered by my construction. There is no node left out!
Regards, WM
WM
Dec 27, 2012, 7:14:19 AM12/27/12
to
On 27 Dez., 00:57, George Greene <gree...@email.unc.edu> wrote:
> In the all-right case, AS I ALREADY
> EXPLAINED, the represented rational is
> the one that is the right sibling of the last node in the path that is
> a left-child (this is assuming
> left-branch is 0 and right-branch is 1;
But what you have not yet explained is how we can distinguish further
> In the all-right case, AS I ALREADY
> EXPLAINED, the represented rational is
> the one that is the right sibling of the last node in the path that is
> a left-child (this is assuming
> left-branch is 0 and right-branch is 1;
uncountably many paths by the set of countably many nodes or edges
that entirely belong to the countable set of paths used for
constructing or covering the CIBT.
Regards, WM
WM
Dec 27, 2012, 7:16:32 AM12/27/12
to
On 27 Dez., 09:33, fom <fomJ...@nyms.net> wrote:
>
> But, I did ask WM certain questions that, by the
> end, led to him rejecting any notion of actual
> infinity.
Yes, but that is the result of the contradictions that I found after
>
> But, I did ask WM certain questions that, by the
> end, led to him rejecting any notion of actual
> infinity.
assuming actual infinity like in the Comple Infinite Binary Tree.
> The claim was that he held a position
> not unlike Abraham Robinson. The difference,
> however, is that WM accepts infinity only to
> the extent that the extended real number system
> can be formed.
approximations we can do mathematics that fits all practical purposes
(FAPP, as the late John Bell put it (with respect to quantum
mechanics, not mathematics)).
Regards, WM
WM
Dec 27, 2012, 9:21:24 AM12/27/12
to
they define more than countably many infinite paths of the CIBT by
means of nodes? Remember, Cantor is said to have proved uncountability
my means of *digits at finite indices* - by asserting that there must
be some more reals than can be counted. But that is the way how his
great followers, like Virgil and you an Samsung, put the argument
presently.
Regards, WM
Sam Sung
Dec 27, 2012, 9:27:14 AM12/27/12
to
The idiot WM:
> ...
Piss off already, idiot. The world does neither need nor want you asshole.
> ...
Piss off already, idiot. The world does neither need nor want you asshole.
forbi...@gmail.com
Dec 27, 2012, 11:10:33 AM12/27/12
to
On Thursday, December 27, 2012 6:21:24 AM UTC-8, WM wrote:
> On 27 Dez., 12:31, gus gassmann <g...@nospam.com> wrote:
>
> > On 26/12/2012 5:03 PM, fom wrote:
>
> >
>
> > > WM is an empiricist.
>
> >
>
> > WM is a fool. An arrogant, blustering, uneducable fool.
>
>
>
> Then try to educate at least the other members of this group: How can
>
> they define more than countably many infinite paths of the CIBT by
>
> means of nodes?
Why should they? You tell me the nodes passed through by the square
> On 27 Dez., 12:31, gus gassmann <g...@nospam.com> wrote:
>
> > On 26/12/2012 5:03 PM, fom wrote:
>
> >
>
> > > WM is an empiricist.
>
> >
>
> > WM is a fool. An arrogant, blustering, uneducable fool.
>
>
>
> Then try to educate at least the other members of this group: How can
>
> they define more than countably many infinite paths of the CIBT by
>
> means of nodes?
root of 2. You're the one proposing the method. When you're done with
the square root of 2 then please move on to pi.
> Remember, Cantor is said to have proved uncountability
>
> my means of *digits at finite indices* - by asserting that there must
>
> be some more reals than can be counted.
of the rationals. He provided a method by which to prove any proposed
enumerated set of reals in the interval [0,1) was incomplete.
Assertion and proof aren't the same.
fom
Dec 27, 2012, 11:57:13 AM12/27/12
to
On 12/27/2012 5:31 AM, gus gassmann wrote:
> On 26/12/2012 5:03 PM, fom wrote:
>
>> On 12/26/2012 2:34 PM, Virgil wrote:
>
>>> Or does WM claim a definition of countability other than the standard
>>> one?
>>>
>>
>> He does. He stated it.
>>
>> A set is countable iff it is a subset of a countable set.
>
> This definition is, of course, circular, and should convince anyone who
> reads it that the writer of this "definition" does not understand the
> first points of mathematics.
>
>> Infinite sets are, therefore, if they exist, uncountable.
>
> And this is a non-sequitur.
This is just wrong, as George has pointed out.
> On 26/12/2012 5:03 PM, fom wrote:
>
>> On 12/26/2012 2:34 PM, Virgil wrote:
>
>>> Or does WM claim a definition of countability other than the standard
>>> one?
>>>
>>
>> He does. He stated it.
>>
>> A set is countable iff it is a subset of a countable set.
>
> This definition is, of course, circular, and should convince anyone who
> reads it that the writer of this "definition" does not understand the
> first points of mathematics.
>
>> Infinite sets are, therefore, if they exist, uncountable.
>
> And this is a non-sequitur.
What I was thinking but not saying is that a countable
infinity is like a round square for WM
Zuhair
Dec 27, 2012, 1:04:10 PM12/27/12
to
On Dec 27, 3:08 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> Here is a parameter free enumeration of all finite initial segments of
> the paths of the Binary Tree:
>
> 0
> 1, 2
> 3, 4, 5, 6
> 7, ...
>
> Regards, WM
Show me a parameter free formula "phi(y)" where the alleged
>
> Here is a parameter free enumeration of all finite initial segments of
> the paths of the Binary Tree:
>
> 0
> 1, 2
> 3, 4, 5, 6
> 7, ...
>
> Regards, WM
enumeration you've just depicted above is defined after i.e. suppose
your enumeration is denoted as "En" then show me that: For all y. y in
En iff phi(y).
Remember parameter free formula phi(y) means a formula in which ONLY y
occurs free. If you show that then I'd agree with you. If you don't
show that, then you didn't prove that your alleged enumeration is
parameter free. AND please spare me any responses that gives a
different definition for the term "parameter free definable" that you
have in your mind since simply it is not relevant to the "parameter
free definable" concept that I'm speaking about.
Zuhair
Virgil
Dec 27, 2012, 3:49:37 PM12/27/12
to
In article
<41e51766-079f-42ae...@d4g2000vbw.googlegroups.com>,
your paths, and to validate your claim, you must show us a listing of
all of them so that we may see that they are countable.
But you dare not do that because any allegedly complete listing of your
allegedly complete set of paths will show its own incompleteness.
evidence, which is WM's usual method: Makeing all sorts of irrelevant
claims (which WM does not even then prove), to throw dust in everyone's
eyes.
> It is the set
> of all finite paths extending from the root node to a given node.
In every Complete Infinite Binary Tree, every finite path ends at the
root node of of another Complete Infinite Binary Tree. So for every
finite path, there are uncountably many extensions of it in every
Complete Infinite Binary Tree.
You
> can append them by whatever tail you want, because all these tails are
> also covered by my construction.
Since there are of necessity uncountably many infinite tails to every
finite head, one still has uncountability, at least outside
Wolkenmuekenheim.
Not that WM again refuses to provide any evidence of the alleged but
false 'completeness' of his set of paths, or even of its conforming to
any definition of countability.
WMytheology at its worst!
--
<41e51766-079f-42ae...@d4g2000vbw.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Dez., 21:14, Virgil <vir...@ligriv.com> wrote:
>
> > As usual, you draw unwarranted conclusions from everything.
> >
> > As one identifies a path as a particular set of digits or nodes, how is
> > asking you to identify your paths in any way overlooking �them?
>
> Since I have constructed or covered all paths that can be defined by
> nodes. There is no node that you could name that could define a path
> that is not in the CIBT that I constructed.
You claim to have done so, but have not yet shown us a single one of
> On 26 Dez., 21:14, Virgil <vir...@ligriv.com> wrote:
>
> > As usual, you draw unwarranted conclusions from everything.
> >
> > As one identifies a path as a particular set of digits or nodes, how is
> > asking you to identify your paths in any way overlooking �them?
>
> Since I have constructed or covered all paths that can be defined by
> nodes. There is no node that you could name that could define a path
> that is not in the CIBT that I constructed.
your paths, and to validate your claim, you must show us a listing of
all of them so that we may see that they are countable.
But you dare not do that because any allegedly complete listing of your
allegedly complete set of paths will show its own incompleteness.
>
> > You have also not proved that your set of paths is countable, which can
> > only be done by proving that they can be listed, and most easily done by
> > providing such a list.
>
> The paths that can be identified by nodes can be listed.
Until you actually list them, any such claim is unsupported by any
> > You have also not proved that your set of paths is countable, which can
> > only be done by proving that they can be listed, and most easily done by
> > providing such a list.
>
> The paths that can be identified by nodes can be listed.
evidence, which is WM's usual method: Makeing all sorts of irrelevant
claims (which WM does not even then prove), to throw dust in everyone's
eyes.
> It is the set
> of all finite paths extending from the root node to a given node.
root node of of another Complete Infinite Binary Tree. So for every
finite path, there are uncountably many extensions of it in every
Complete Infinite Binary Tree.
You
> can append them by whatever tail you want, because all these tails are
> also covered by my construction.
finite head, one still has uncountability, at least outside
Wolkenmuekenheim.
Not that WM again refuses to provide any evidence of the alleged but
false 'completeness' of his set of paths, or even of its conforming to
any definition of countability.
WMytheology at its worst!
--
Virgil
Dec 27, 2012, 3:57:10 PM12/27/12
to
In article
<0c4e66c5-9fcd-4a2b...@n5g2000vbk.googlegroups.com>,
Not within any common axiom system. And WM's axioms, those things he
claims but does not and cannot prove, are not truths in, for example,
ZFC, in which the axioms are at least explicit.
listing of paths or of initial segments of paths, other than possible
the root node as a one element path.
So that WM is off his rocker again!
--
<0c4e66c5-9fcd-4a2b...@n5g2000vbk.googlegroups.com>,
claims but does not and cannot prove, are not truths in, for example,
ZFC, in which the axioms are at least explicit.
> >
> > (3) To consider countability of the finite initial segments FALSE,
> > i.e. to say that we have uncountably many finite initial segments of
> > reals and as well we have uncountably many reals. This clearly
> > preserves congruity of the argument, but it requires justification,
> > and the justification can be based on the principle of "parameter free
> > definability of sets", since the alleged bijection between the finite
> > initial segments of the reals and the set N of all naturals is NOT
> > parameter free definable, then this bijection does not exist, and it
> > is false to say that it is. This claim only accepts infinite sets to
> > exist if there is a parameter free formula after which membership of
> > those sets is determined, so if there is non then it doesn't accept
> > the existence sets that are not parameter free definable.
>
> Here is a parameter free enumeration of all finite initial segments of
> the paths of the Binary Tree:
>
> 0
> 1, 2
> 3, 4, 5, 6
> 7, ...
>
> Regards, WM
While that may be an incomplete listing of nodes, it is certainly not a
> > (3) To consider countability of the finite initial segments FALSE,
> > i.e. to say that we have uncountably many finite initial segments of
> > reals and as well we have uncountably many reals. This clearly
> > preserves congruity of the argument, but it requires justification,
> > and the justification can be based on the principle of "parameter free
> > definability of sets", since the alleged bijection between the finite
> > initial segments of the reals and the set N of all naturals is NOT
> > parameter free definable, then this bijection does not exist, and it
> > is false to say that it is. This claim only accepts infinite sets to
> > exist if there is a parameter free formula after which membership of
> > those sets is determined, so if there is non then it doesn't accept
> > the existence sets that are not parameter free definable.
>
> Here is a parameter free enumeration of all finite initial segments of
> the paths of the Binary Tree:
>
> 0
> 1, 2
> 3, 4, 5, 6
> 7, ...
>
> Regards, WM
listing of paths or of initial segments of paths, other than possible
the root node as a one element path.
So that WM is off his rocker again!
--
Virgil
Dec 27, 2012, 4:06:02 PM12/27/12
to
In article
<5a86f0ee-3284-422b...@r14g2000vbe.googlegroups.com>,
in countering that argument.
All nodes of even the most infinite of paths have finite indices, so
there is no problem in defining an infinite path by its finitely indexed
nodes.
One can even ennumerate the nodes giving each one a unique natural
number at which point each infinite path becomes identified with an
infinite subset of |N.
--
<5a86f0ee-3284-422b...@r14g2000vbe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Dez., 12:31, gus gassmann <g...@nospam.com> wrote:
> > On 26/12/2012 5:03 PM, fom wrote:
> >
> > > �WM is an empiricist.
> >
> > WM is a fool. An arrogant, blustering, uneducable fool.
>
> Then try to educate at least the other members of this group: How can
> they define more than countably many infinite paths of the CIBT by
> means of nodes? Remember, Cantor is said to have proved uncountability
> my means of *digits at finite indices* - by asserting that there must
> be some more reals than can be counted.
And Cantor did it so successfuly than no one since has been successful
> On 27 Dez., 12:31, gus gassmann <g...@nospam.com> wrote:
> > On 26/12/2012 5:03 PM, fom wrote:
> >
> > > �WM is an empiricist.
> >
> > WM is a fool. An arrogant, blustering, uneducable fool.
>
> Then try to educate at least the other members of this group: How can
> they define more than countably many infinite paths of the CIBT by
> means of nodes? Remember, Cantor is said to have proved uncountability
> my means of *digits at finite indices* - by asserting that there must
> be some more reals than can be counted.
in countering that argument.
All nodes of even the most infinite of paths have finite indices, so
there is no problem in defining an infinite path by its finitely indexed
nodes.
One can even ennumerate the nodes giving each one a unique natural
number at which point each infinite path becomes identified with an
infinite subset of |N.
--
Virgil
Dec 27, 2012, 4:08:16 PM12/27/12
to
Zuhair
Dec 28, 2012, 2:03:23 AM12/28/12
to
On Dec 28, 12:06 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <5a86f0ee-3284-422b-b036-25786304f...@r14g2000vbe.googlegroups.com>,
Zuhair
> In article
> <5a86f0ee-3284-422b-b036-25786304f...@r14g2000vbe.googlegroups.com>,
>
>
> One can even ennumerate the nodes giving each one a unique natural
> number at which point each infinite path becomes identified with an
> infinite subset of |N.
> --
with an infinite SEQUENCE from domain N.
>
> One can even ennumerate the nodes giving each one a unique natural
> number at which point each infinite path becomes identified with an
> infinite subset of |N.
> --
Zuhair
Virgil
Dec 28, 2012, 4:10:49 AM12/28/12
to
In article
<17ef0235-8b89-4a03...@r3g2000vbn.googlegroups.com>,
I said subset and I meant subset.
That every infinite subset of N can also be considered to be a
subsequence of N is irrelevant.
--
<17ef0235-8b89-4a03...@r3g2000vbn.googlegroups.com>,
That every infinite subset of N can also be considered to be a
subsequence of N is irrelevant.
--
WM
Dec 28, 2012, 12:13:22 PM12/28/12
to
On 27 Dez., 17:10, forbisga...@gmail.com wrote:
> On Thursday, December 27, 2012 6:21:24 AM UTC-8, WM wrote:
> > Then try to educate at least the other members of this group: How can
>
> > they define more than countably many infinite paths of the CIBT by
>
> > means of nodes?
>
> Why should they?
Because the Binary Tree contains uncountably many real numbers that
> On Thursday, December 27, 2012 6:21:24 AM UTC-8, WM wrote:
> > Then try to educate at least the other members of this group: How can
>
> > they define more than countably many infinite paths of the CIBT by
>
> > means of nodes?
>
> Why should they?
are distinct by nodes and can be distinguished by nodes.
> You tell me the nodes passed through by the square
> root of 2.
method will never catch an infinite real.
Consider all the finite paths of the Binary Tree listed. That is a
subset of the rationals of the unit interval. Nevertheless Cantor
"proves" that the list of all finite digit sequences will supply a
diagonal that differs from all finite digits sequences at a finite
place - which is simply impossible, as the list contains all possible
finite digit sequences.
> > Remember, Cantor is said to have proved uncountability
>
> > my means of *digits at finite indices* - by asserting that there must
>
> > be some more reals than can be counted.
>
> No, he did no such thing. He proved, not asserted, the countability
> of the rationals.
which is obviously false. From a fals assumption every nonsense can be
derived.
> He provided a method by which to prove any proposed
> enumerated set of reals in the interval [0,1) was incomplete.
proposed enumerated set of rationals was incomplete. Remember: He
proved an antidiagonal can be obtained in the Binary Tree. This
antidiagonal differes from every finite path of the Binary Tree at a
finite level. But as the Binary Tree is complete with respect to all
finite paths, Cantor's result is obviously wrong - by the definition
of the CIBT. Therefore not even a proof is required.
> Assertion and proof aren't the same.
better than an unwarranted assertion.
Regards, WM
WM
Dec 28, 2012, 12:14:24 PM12/28/12
to
explain it, because I am not interested in your interpretation.) But
it is obvious to me that Cantor enumerated the rational numbers just
like I enumerate the finite paths of the Binary Tree. And he
enumerated the digits of the diagonal in just the same way, namely
assuming the complete existence of all natural numbers.
Regards, WM
WM
Dec 28, 2012, 12:15:57 PM12/28/12
to
On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
> > It is the set
> > of all finite paths extending from the root node to a given node.
>
> In every Complete Infinite Binary Tree, every finite path ends at the
> root node of of another Complete Infinite Binary Tree. So for every
> finite path, there are uncountably many extensions of it in every
> Complete Infinite Binary Tree.
And every extension is contained in the CIBT constructed from all
> > It is the set
> > of all finite paths extending from the root node to a given node.
>
> In every Complete Infinite Binary Tree, every finite path ends at the
> root node of of another Complete Infinite Binary Tree. So for every
> finite path, there are uncountably many extensions of it in every
> Complete Infinite Binary Tree.
finite paths.
Regards, WM
Message has been deleted
Zuhair
Dec 28, 2012, 2:19:00 PM12/28/12
to
On Dec 28, 8:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> It not obvious to me, what you call parameter-free. (And you need not
> explain it, because I am not interested in your interpretation.)
> Regards, WM
> It not obvious to me, what you call parameter-free. (And you need not
> explain it, because I am not interested in your interpretation.)
If you are not interested in my interpretation of parameter free
definability (which is standard by the way) then why you answered to
my question by saying "Here is a parameter free enumeration...", you
must not attempt to answer a question of others before you understand
what others meant by their question. When I asked my question I of
course meant my interpretation of parameter free definability, and
actually I mentioned that explicitly, I of course didn't mean some
alternative explanation existing in your head. To me it was better if
you didn't answer my question, or at least say that the enumeration
you've mentioned to be just some enumeration but to go and say it is
parameter free and in response to my question about having a parameter
free definable bijection (in my sense of course since I'm the one who
is asking) is simply an act of non respect to the one who asked the
question, it is rude to do so.
By the way you said it is not obvious to you what I meant by parameter
free definable, while this is just the basics of definability of sets
and it is WELL known, this ignorance on your side that you've admitted
(which is good really) only reflects that you have little information
on those subjects. It was expected from one who aims to refute Cantor
to be more informed. It is expected from one who say that THOUSANDS of
mathematicians for a whole of a century were acting fools and
spreading nonsense to be someone who is well informed on such issues,
but Since you are obviously ignorant why go discuss matters as if you
are well informed? I think it is obvious now who is the acting fool
and spreading nonsense.
Regards,
Zuhair
Virgil
Dec 28, 2012, 5:04:50 PM12/28/12
to
In article
<2925a2ea-c16d-483e...@b16g2000vbh.googlegroups.com>,
Claimed but never proven.
And only true if one is allowed to chain finite paths from an infinite
set of finite paths like one chains nodes from the infinite set of
nodes, into chains of infinitely many finite paths.
>
--
<2925a2ea-c16d-483e...@b16g2000vbh.googlegroups.com>,
And only true if one is allowed to chain finite paths from an infinite
set of finite paths like one chains nodes from the infinite set of
nodes, into chains of infinitely many finite paths.
>
--
Virgil
Dec 28, 2012, 8:06:49 PM12/28/12
to
In article
<51eb8729-b50a-4136...@x20g2000vbf.googlegroups.com>,
one) which you regard as being "parameter free" and of one which you
regard as not being "parameter free" might clear the air.
I am not totally clear in my owm mind what you mean by the phrase.
Do you, for instance, regard the standard definition of countability of
a set (set S is countably if and only if there exists a surjection from
|N to S) as being parameer free or not, and why?
--
<51eb8729-b50a-4136...@x20g2000vbf.googlegroups.com>,
Zuhair <zalj...@gmail.com> wrote:
> By the way you said it is not obvious to you what I meant by parameter
> free definable
An example of a mathematical definition ( or even a non mathematical
> By the way you said it is not obvious to you what I meant by parameter
> free definable
one) which you regard as being "parameter free" and of one which you
regard as not being "parameter free" might clear the air.
I am not totally clear in my owm mind what you mean by the phrase.
Do you, for instance, regard the standard definition of countability of
a set (set S is countably if and only if there exists a surjection from
|N to S) as being parameer free or not, and why?
--
forbi...@gmail.com
Dec 28, 2012, 9:33:00 PM12/28/12
to
On Friday, December 28, 2012 9:13:22 AM UTC-8, WM wrote:
> On 27 Dez., 17:10, forbisga...@gmail.com wrote:
>
> > On Thursday, December 27, 2012 6:21:24 AM UTC-8, WM wrote:
>
>
>
> > > Then try to educate at least the other members of this group: How can
>
> >
>
> > > they define more than countably many infinite paths of the CIBT by
>
> >
>
> > > means of nodes?
>
> >
>
> > Why should they?
>
>
>
> Because the Binary Tree contains uncountably many real numbers that
>
> are distinct by nodes and can be distinguished by nodes.
>
>
>
> > You tell me the nodes passed through by the square
>
> > root of 2.
>
>
>
> Since the told set of nodes will always remain finite, also Cantor's
>
> method will never catch an infinite real.
It catches the enumerated ones. That's why most accept that
> On 27 Dez., 17:10, forbisga...@gmail.com wrote:
>
> > On Thursday, December 27, 2012 6:21:24 AM UTC-8, WM wrote:
>
>
>
> > > Then try to educate at least the other members of this group: How can
>
> >
>
> > > they define more than countably many infinite paths of the CIBT by
>
> >
>
> > > means of nodes?
>
> >
>
> > Why should they?
>
>
>
> Because the Binary Tree contains uncountably many real numbers that
>
> are distinct by nodes and can be distinguished by nodes.
>
>
>
> > You tell me the nodes passed through by the square
>
> > root of 2.
>
>
>
> Since the told set of nodes will always remain finite, also Cantor's
>
> method will never catch an infinite real.
the reals are uncountably infinte.
> Consider all the finite paths of the Binary Tree listed. That is a
>
> subset of the rationals of the unit interval.
path in the CIBT. All paths are infinite.
> Nevertheless Cantor
>
> "proves" that the list of all finite digit sequences will supply a
>
> diagonal that differs from all finite digits sequences at a finite
>
> place - which is simply impossible, as the list contains all possible
>
> finite digit sequences.
to include all of them. Every enumeration differs from the number
generated by the enumeration (if expressed in base 10 in the indexed
position after the decimal point for the indexed number.)
> > > Remember, Cantor is said to have proved uncountability
>
> >
>
> > > my means of *digits at finite indices* - by asserting that there must
>
> >
>
> > > be some more reals than can be counted.
>
> >
>
> > No, he did no such thing. He proved, not asserted, the countability
>
> > of the rationals.
>
>
>
> He "proved" it under the assumption that infinity can be finished
>
> which is obviously false. From a fals assumption every nonsense can be
>
> derived.
can index a real number. One merely needs a map between real and natural.
One can reference every index via the map.
> > He provided a method by which to prove any proposed
>
> > enumerated set of reals in the interval [0,1) was incomplete.
>
>
>
> He did more than that. He provided a method by which to prove any
>
> proposed enumerated set of rationals was incomplete.
if the reals in the interval cannot be mapped to the natural
numbers then neither can any other interval of real numbers.
> Remember: He
>
> proved an antidiagonal can be obtained in the Binary Tree. This
>
> antidiagonal differes from every finite path of the Binary Tree at a
>
> finite level.
your definition of them but each one has an infinite set of paths
passing through it and none stop at a node since there are no leaf
nodes.
> But as the Binary Tree is complete with respect to all
>
> finite paths, Cantor's result is obviously wrong - by the definition
>
> of the CIBT. Therefore not even a proof is required.
continue with 0 at every node past a given node but the path is
infinite.
Zuhair
Dec 29, 2012, 2:03:30 AM12/29/12
to
> <51eb8729-b50a-4136-8af6-2c527c93b...@x20g2000vbf.googlegroups.com>,
know about it. The problem is if some people professing big claims
like refuting Cantor or saying that THOUSANDS of mathematician for a
CENTURY long time are acting fools, and then it turns that those
people themselves don't know basic definitions? really strange!
Now we come to your example:
Is countability of a set parameter free definable or not?
The answer is YES.
Why?
Because there is a parameter free formula "phi(S)" such that
For all S. Countable(S) iff phi(S)
And what is meant by phi(S) being parameter free is that phi(S) is a
formula in which only the symbol S occurs free, i.e. all other
variable symbols in phi(S) are quantified within the formula, and of
course S is free.
Now lets explicitly examine this
take phi(S) to be the following formula
Exist f. Exist N. (for all y. y in N iff y is a finite ordinal) & f: S
--> N & f is injective.
The open expansion of this formula show that only the variable symbol
S occurs free.
QED
So for example in a theory like NF where we do have the set of all
countable sets. So this set is parameter free definable set, because
membership of this set follows satisfaction of a parameter free
formula.
Now let me give you an example of an object that cannot be definable
in a parameter free manner.
Lets take some theory that provides sufficient material to define
'definable real' as:
x is a definable real <-> iff x is a real & Exist phi. for all y. y in
x <-> phi(y)
where phi(y) is a parameter free formula (i.e. only y occurs free in
phi(y)).
Now one can prove using Cantor's argument that ANY bijection F between
the set R* of ALL definable reals and the set N of all naturals IS non
parameter free definable
The reason is that if we suppose the contrary i.e. the existence of
such a bijection that is parameter free definable, then the diagonal
defined after it would be a parameter free definable real that is not
in the set of ALL definable reals, which is a clear contradiction.
So although there is a bijection between R* and N, yet it is provable
that any such bijection is non parameter free definable!
In other words there is no parameter free formula phi(y) such that the
above bijection have all its membership determined after satisfaction
of phi(y).
Hope that is helpful and clear
Zuhair
>
> Zuhair <zaljo...@gmail.com> wrote:
> > By the way you said it is not obvious to you what I meant by parameter
> > free definable
>
> An example of a mathematical definition ( or even a non mathematical
> one) which you regard as being "parameter free" and of one which you
> regard as not being "parameter free" might clear the air.
>
> I am not totally clear in my owm mind what you mean by the phrase.
>
> Do you, for instance, regard the standard definition of countability of
> a set (set S is countably if and only if there exists a surjection from
> |N to S) as being parameer free or not, and why?
> --
This is a well known subject. There is no problem if some people don't
> Zuhair <zaljo...@gmail.com> wrote:
> > By the way you said it is not obvious to you what I meant by parameter
> > free definable
>
> An example of a mathematical definition ( or even a non mathematical
> one) which you regard as being "parameter free" and of one which you
> regard as not being "parameter free" might clear the air.
>
> I am not totally clear in my owm mind what you mean by the phrase.
>
> Do you, for instance, regard the standard definition of countability of
> a set (set S is countably if and only if there exists a surjection from
> |N to S) as being parameer free or not, and why?
> --
know about it. The problem is if some people professing big claims
like refuting Cantor or saying that THOUSANDS of mathematician for a
CENTURY long time are acting fools, and then it turns that those
people themselves don't know basic definitions? really strange!
Now we come to your example:
Is countability of a set parameter free definable or not?
The answer is YES.
Why?
Because there is a parameter free formula "phi(S)" such that
For all S. Countable(S) iff phi(S)
And what is meant by phi(S) being parameter free is that phi(S) is a
formula in which only the symbol S occurs free, i.e. all other
variable symbols in phi(S) are quantified within the formula, and of
course S is free.
Now lets explicitly examine this
take phi(S) to be the following formula
Exist f. Exist N. (for all y. y in N iff y is a finite ordinal) & f: S
--> N & f is injective.
The open expansion of this formula show that only the variable symbol
S occurs free.
QED
So for example in a theory like NF where we do have the set of all
countable sets. So this set is parameter free definable set, because
membership of this set follows satisfaction of a parameter free
formula.
Now let me give you an example of an object that cannot be definable
in a parameter free manner.
Lets take some theory that provides sufficient material to define
'definable real' as:
x is a definable real <-> iff x is a real & Exist phi. for all y. y in
x <-> phi(y)
where phi(y) is a parameter free formula (i.e. only y occurs free in
phi(y)).
Now one can prove using Cantor's argument that ANY bijection F between
the set R* of ALL definable reals and the set N of all naturals IS non
parameter free definable
The reason is that if we suppose the contrary i.e. the existence of
such a bijection that is parameter free definable, then the diagonal
defined after it would be a parameter free definable real that is not
in the set of ALL definable reals, which is a clear contradiction.
So although there is a bijection between R* and N, yet it is provable
that any such bijection is non parameter free definable!
In other words there is no parameter free formula phi(y) such that the
above bijection have all its membership determined after satisfaction
of phi(y).
Hope that is helpful and clear
Zuhair
Virgil
Dec 29, 2012, 2:21:16 AM12/29/12
to
WM
Dec 29, 2012, 12:17:58 PM12/29/12
to
On 29 Dez., 03:33, forbisga...@gmail.com wrote:
> > Consider all the finite paths of the Binary Tree listed. That is a
>
> > subset of the rationals of the unit interval.
>
> And the irrationals passing throught them. There is no finite
> path in the CIBT. All paths are infinite.
Nevertheless one can consider all finite initial segments of all paths
> > Consider all the finite paths of the Binary Tree listed. That is a
>
> > subset of the rationals of the unit interval.
>
> And the irrationals passing throught them. There is no finite
> path in the CIBT. All paths are infinite.
and abbreviate them by "finite paths".
>
> > Nevertheless Cantor
>
> > "proves" that the list of all finite digit sequences will supply a
>
> > diagonal that differs from all finite digits sequences at a finite
>
> > place - which is simply impossible, as the list contains all possible
>
> > finite digit sequences.
>
> Not true. All paths are infinite and any enumeration of them fails
> to include all of them.
sequences. And that is true.
>
> > He "proved" it under the assumption that infinity can be finished
>
> > which is obviously false. From a fals assumption every nonsense can be
>
> > derived.
>
> Again no. There is no largest natural number but every one of them
> can index a real number. One merely needs a map between real and natural.
> One can reference every index via the map.
>
> > > He provided a method by which to prove any proposed
>
> > > enumerated set of reals in the interval [0,1) was incomplete.
>
> > He did more than that. He provided a method by which to prove any
>
> > proposed enumerated set of rationals was incomplete.
>
> Yes, you are right. But let's focus on the interval
No. I just showed that even the "countable" subset of all rational
> > > He provided a method by which to prove any proposed
>
> > > enumerated set of reals in the interval [0,1) was incomplete.
>
> > He did more than that. He provided a method by which to prove any
>
> > proposed enumerated set of rationals was incomplete.
>
> Yes, you are right. But let's focus on the interval
numbers of the unit interval is not countable. And that is a
contradiction. Cantor finds an antidiagonal to the set of all finite
initial segments of the binary tree. And that is obviously a
contradiction. Therefore further discussion about intervals is
useless.
Regards, WM
WM
Dec 29, 2012, 12:22:44 PM12/29/12
to
> <2925a2ea-c16d-483e-91da-3bb0084e7...@b16g2000vbh.googlegroups.com>,
Regards, WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>
> > > > It is the set
> > > > of all finite paths extending from the root node to a given node.
>
> > > In every Complete Infinite Binary Tree, every finite path ends at the
> > > root node of of another Complete Infinite Binary Tree. So for every
> > > finite path, there are uncountably many extensions of it in every
> > > Complete Infinite Binary Tree.
>
> > And every extension is contained in the CIBT constructed from all
> > finite paths.
>
> Claimed but never proven.
Definitions need not be proven.
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>
> > > > It is the set
> > > > of all finite paths extending from the root node to a given node.
>
> > > In every Complete Infinite Binary Tree, every finite path ends at the
> > > root node of of another Complete Infinite Binary Tree. So for every
> > > finite path, there are uncountably many extensions of it in every
> > > Complete Infinite Binary Tree.
>
> > And every extension is contained in the CIBT constructed from all
> > finite paths.
>
> Claimed but never proven.
Regards, WM
WM
Dec 29, 2012, 12:22:50 PM12/29/12
to
On 28 Dez., 20:19, Zuhair <zaljo...@gmail.com> wrote:
> On Dec 28, 8:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > It not obvious to me, what you call parameter-free. (And you need not
> > explain it, because I am not interested in your interpretation.)
> > Regards, WM
>
> If you are not interested in my interpretation of parameter free
> definability (which is standard by the way) then why you answered to
> my question by saying "Here is a parameter free enumeration...",
Why? Because my enumeration is parameter free. And that's because
> On Dec 28, 8:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > It not obvious to me, what you call parameter-free. (And you need not
> > explain it, because I am not interested in your interpretation.)
> > Regards, WM
>
> If you are not interested in my interpretation of parameter free
> definability (which is standard by the way) then why you answered to
> my question by saying "Here is a parameter free enumeration...",
there is no parameter involved. You know what a parameter is? You have
found a parameter in my enumeration? No? That is why I call my
enumeration parameter free.
> By the way you said it is not obvious to you what I meant by parameter
> free definable, while this is just the basics of definability of sets
> and it is WELL known,
It was expected from one who aims to refute Cantor
> to be more informed. It is expected from one who say that THOUSANDS of
> mathematicians for a whole of a century were acting fools and
> spreading nonsense to be someone who is well informed on such issues,
> but Since you are obviously ignorant why go discuss matters as if you
> are well informed?
languages or parametric definitions without a parameter.
> I think it is obvious now who is the acting fool
> and spreading nonsense.
Regards, WM
WM
Dec 29, 2012, 12:24:07 PM12/29/12
to
On 29 Dez., 08:03, Zuhair <zaljo...@gmail.com> wrote:
> Is countability of a set parameter free definable or not?
>
> The answer is YES.
>
> Why?
Because Cantor did it. And who asserts that Cantor gave a parametric
> Is countability of a set parameter free definable or not?
>
> The answer is YES.
>
> Why?
definition in formal language shows that he is a fool.
Regards, WM
forbi...@gmail.com
Dec 29, 2012, 2:10:10 PM12/29/12
to
On Saturday, December 29, 2012 9:17:58 AM UTC-8, WM wrote:
> On 29 Dez., 03:33, forbisga...@gmail.com wrote:
> > > Consider all the finite paths of the Binary Tree listed. That is a
> > > subset of the rationals of the unit interval.
> > And the irrationals passing throught them. There is no finite
> > path in the CIBT. All paths are infinite.
>
> Nevertheless one can consider all finite initial segments of all paths
> and abbreviate them by "finite paths".
No you can't. This causes infinite paths to have the same
> On 29 Dez., 03:33, forbisga...@gmail.com wrote:
> > > Consider all the finite paths of the Binary Tree listed. That is a
> > > subset of the rationals of the unit interval.
> > And the irrationals passing throught them. There is no finite
> > path in the CIBT. All paths are infinite.
>
> Nevertheless one can consider all finite initial segments of all paths
> and abbreviate them by "finite paths".
abbreviation and that doesn't make sense because it fails to
distinguish between different numbers. You can abbriviate some
numbers by way the assertion you are talking about those whose
continuation are regular repeating sequences, for instance
.111... is an abbreviation for the same number (in binary)
as is 1 and 1.000... and 1.[0]. You can't abbreviate the number 98/99ths
as 1 though you can round it to 1. There's no abbreviation for pi
at all even though you can use the greek letter to represent it.
> > > Nevertheless Cantor
> > > "proves" that the list of all finite digit sequences will supply a
> > > diagonal that differs from all finite digits sequences at a finite
> > > place - which is simply impossible, as the list contains all possible
> > > finite digit sequences.
> >
> > Not true. All paths are infinite and any enumeration of them fails
> > to include all of them.
>
> I said all finite digit sequences and I meant all finite digit
> sequences. And that is true.
doesn't means such paths do not exist or cannot be distinguished
from those who can be so represented using the abbreviation rules
specified above. The abbreviations aren't the paths even when they
point to them.
> > > He "proved" it under the assumption that infinity can be finished
> > > which is obviously false. From a fals assumption every nonsense can be
> > > derived.
> >
> > Again no. There is no largest natural number but every one of them
> > can index a real number. One merely needs a map between real and natural.
> > One can reference every index via the map.
>
> And one can find to every reference infinitely many missing numbers.
that's what Cantor proved.
> > > > He provided a method by which to prove any proposed
> > > > enumerated set of reals in the interval [0,1) was incomplete.
> > > He did more than that. He provided a method by which to prove any
> > > proposed enumerated set of rationals was incomplete.
> >
> > Yes, you are right. But let's focus on the interval
>
> No. I just showed that even the "countable" subset of all rational
> numbers of the unit interval is not countable.
> And that is a
> contradiction. Cantor finds an antidiagonal to the set of all finite
> initial segments of the binary tree.
finite sets of reals. If you're looking to prove the set of reals
is finite you've got a long ways to go.
> And that is obviously a
> contradiction. Therefore further discussion about intervals is
> useless.
argument.
Virgil
Dec 29, 2012, 3:36:44 PM12/29/12
to
Virgil
Dec 29, 2012, 3:40:36 PM12/29/12
to
In article
<5902a2c7-147a-4db9...@gu9g2000vbb.googlegroups.com>,
>
> > By the way you said it is not obvious to you what I meant by parameter
> > free definable, while this is just the basics of definability of sets
> > and it is WELL known,
>
> among a gang of big mouths like you?
WM's mouth dwarfs everyone else's here.
--
<5902a2c7-147a-4db9...@gu9g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 28 Dez., 20:19, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 28, 8:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > It not obvious to me, what you call parameter-free. (And you need not
> > > explain it, because I am not interested in your interpretation.)
> > > Regards, WM
> >
> > If you are not interested in my interpretation of parameter free
> > definability (which is standard by the way) then why you answered to
> > my question by saying "Here is a parameter free enumeration...",
>
> Why? Because my enumeration is parameter free.
Your ennumeration is existence-free. It does not exist.
> On 28 Dez., 20:19, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 28, 8:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > It not obvious to me, what you call parameter-free. (And you need not
> > > explain it, because I am not interested in your interpretation.)
> > > Regards, WM
> >
> > If you are not interested in my interpretation of parameter free
> > definability (which is standard by the way) then why you answered to
> > my question by saying "Here is a parameter free enumeration...",
>
> Why? Because my enumeration is parameter free.
>
> > By the way you said it is not obvious to you what I meant by parameter
> > free definable, while this is just the basics of definability of sets
> > and it is WELL known,
>
> among a gang of big mouths like you?
>
> It was expected from one who aims to refute Cantor
> > to be more informed. It is expected from one who say that THOUSANDS of
> > mathematicians for a whole of a century were acting fools and
> > spreading nonsense to be someone who is well informed on such issues,
> > but Since you are obviously ignorant why go discuss matters as if you
> > are well informed?
>
> In order to refute Cantor one need not study nonsense like uncountable
> languages or parametric definitions without a parameter.
And NOT studying things is WM's modus operandi!
> It was expected from one who aims to refute Cantor
> > to be more informed. It is expected from one who say that THOUSANDS of
> > mathematicians for a whole of a century were acting fools and
> > spreading nonsense to be someone who is well informed on such issues,
> > but Since you are obviously ignorant why go discuss matters as if you
> > are well informed?
>
> In order to refute Cantor one need not study nonsense like uncountable
> languages or parametric definitions without a parameter.
--
Virgil
Dec 29, 2012, 3:42:57 PM12/29/12
to
In article
<4c775e80-209d-4953...@r13g2000vbd.googlegroups.com>,
But
As usual, WM bombs.
--
<4c775e80-209d-4953...@r13g2000vbd.googlegroups.com>,
"every extension is contained in the CIBT
constructed from all finite paths"
is a claim but not a definition.
constructed from all finite paths"
As usual, WM bombs.
--
WM
Dec 29, 2012, 5:21:35 PM12/29/12
to
On 29 Dez., 20:10, forbisga...@gmail.com wrote:
> On Saturday, December 29, 2012 9:17:58 AM UTC-8, WM wrote:
> > On 29 Dez., 03:33, forbisga...@gmail.com wrote:
> > > > Consider all the finite paths of the Binary Tree listed. That is a
> > > > subset of the rationals of the unit interval.
> > > And the irrationals passing throught them. There is no finite
> > > path in the CIBT. All paths are infinite.
>
> > Nevertheless one can consider all finite initial segments of all paths
> > and abbreviate them by "finite paths".
>
> No you can't.
I have done. That proves I can.
> On Saturday, December 29, 2012 9:17:58 AM UTC-8, WM wrote:
> > On 29 Dez., 03:33, forbisga...@gmail.com wrote:
> > > > Consider all the finite paths of the Binary Tree listed. That is a
> > > > subset of the rationals of the unit interval.
> > > And the irrationals passing throught them. There is no finite
> > > path in the CIBT. All paths are infinite.
>
> > Nevertheless one can consider all finite initial segments of all paths
> > and abbreviate them by "finite paths".
>
> No you can't.
> This causes infinite paths to have the same
> abbreviation and that doesn't make sense because it fails to
> distinguish between different numbers.
All paths are different.
>There's no abbreviation for pi
> at all even though you can use the greek letter to represent it.
finite name like pi or series of Gregory-Leibniz times 4 etc.
> > I said all finite digit sequences and I meant all finite digit
> > sequences. And that is true.
>
> Not all paths can be represented by finite digit sequences.
should they not be distinguished in the Binary Tree? (Of course no
infinite path can be defined and distinguished from its companoins
other than by a finite word. And all finite words are countable.)
> > No. I just showed that even the "countable" subset of all rational
> > numbers of the unit interval is not countable.
>
> No you didn't.
> Why do you think you did?
i.e., it happens with a finite initial segment of a path. So only my
finite paths are in question. There is nothing happening with an
infinite path.
>
> > And that is a
> > contradiction. Cantor finds an antidiagonal to the set of all finite
> > initial segments of the binary tree.
>
> His argument applies to all countably infinite sets of reals not to
> finite sets of reals.
I use the countably infinite set of all finite paths (or sequences of
> > And that is a
> > contradiction. Cantor finds an antidiagonal to the set of all finite
> > initial segments of the binary tree.
>
> His argument applies to all countably infinite sets of reals not to
> finite sets of reals.
digits). Therefore the complete diagonal cannot differ up to any
finite place from all entries of the list. But there is no other
possibility to differ at all.
Regards, WM
forbi...@gmail.com
Dec 29, 2012, 8:16:13 PM12/29/12
to
On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> Every finite path has as its the abbreviation the node where it ends.
> All paths are different.
We go round and round. There are no finite paths in the CIBT.
> Every finite path has as its the abbreviation the node where it ends.
> All paths are different.
The rest follows from your failure to accept the truth of this.
WM
Dec 30, 2012, 3:14:43 AM12/30/12
to
On 30 Dez., 02:16, forbisga...@gmail.com wrote:
> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> > Every finite path has as its the abbreviation the node where it ends.
> > All paths are different.
>
> We go round and round. There are no finite paths in the CIBT.
There are finite initial segments of all paths in the Binary Tree.
> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> > Every finite path has as its the abbreviation the node where it ends.
> > All paths are different.
>
> We go round and round. There are no finite paths in the CIBT.
I call them "finite paths".
I show that every anti-diagonal of the list of all finite digit
sequences is up to every didgit n identical to a digit sequence. So
Cantor's argument fails.
> The rest follows from your failure to accept the truth of this.
paths?
Why should anybody so blind not to understand that "finite path" is
but an abbreviation of a finite initial segment of a path? Why should
it be forbidden to consider these finite initial segments?
The reason is obvious: Cantor's proof would fail. But remember, also
Cantor's proof deals with finite initial segments only.
Regards, WM
fom
Dec 30, 2012, 4:07:40 AM12/30/12
to
On 12/30/2012 2:14 AM, WM wrote:
> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>> Every finite path has as its the abbreviation the node where it ends.
>>> All paths are different.
>>
>> We go round and round. There are no finite paths in the CIBT.
>
> There are finite initial segments of all paths in the Binary Tree.
> I call them "finite paths".
> I show that every anti-diagonal of the list of all finite digit
> sequences is up to every didgit n identical to a digit sequence. So
> Cantor's argument fails.
Every? On a countably infinite collection? How so, cheater?
> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>> Every finite path has as its the abbreviation the node where it ends.
>>> All paths are different.
>>
>> We go round and round. There are no finite paths in the CIBT.
>
> There are finite initial segments of all paths in the Binary Tree.
> I call them "finite paths".
> I show that every anti-diagonal of the list of all finite digit
> sequences is up to every didgit n identical to a digit sequence. So
> Cantor's argument fails.
>
>> The rest follows from your failure to accept the truth of this.
>
> Why should it be disallaowed to construct a CIBT from all finite
> paths?
from the theory of metric spaces.
> Why should anybody so blind not to understand that "finite path" is
> but an abbreviation of a finite initial segment of a path?
well-founded and a tree built from infinite paths
is said to be ill-founded. A coherent, intelligent
person is either speaking of one or the other, or
possibly the relation that a well-founded tree has
to an ill-founded tree.
You are denying the existence of the ill-founded
tree while trying to present the well-founded tree
as being the same thing.
> Why should
> it be forbidden to consider these finite initial segments?
>
the ill-founded tree. There are only uncountably
infinite collections of infinite paths NAMED by
initial segments.
> The reason is obvious: Cantor's proof would fail.
one name not on a given list after someone such
as yourself has claimed that every name has been
placed there. As Wittgenstein observed, it
leverages the ambiguity of the word "all." It makes
no claim beyond one name.
What Cantor defined were "fundamental sequences"
He then asserted that the logical foundation of
the real number system rested on taking these
fundamental sequences to be the logical type
from which the real number system is composed.
> But remember, also
> Cantor's proof deals with finite initial segments only.
of strings commonly believed to informatively distinguish
any distinct pair of real numbers from one another.
fom
Dec 30, 2012, 4:46:47 AM12/30/12
to
had to be named and proven to exist
WM
Dec 30, 2012, 7:12:03 AM12/30/12
to
On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
> On 12/30/2012 2:14 AM, WM wrote:
>
> > On 30 Dez., 02:16, forbisga...@gmail.com wrote:
> >> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> >>> Every finite path has as its the abbreviation the node where it ends.
> >>> All paths are different.
>
> >> We go round and round. There are no finite paths in the CIBT.
>
> > There are finite initial segments of all paths in the Binary Tree.
> > I call them "finite paths".
> > I show that every anti-diagonal of the list of all finite digit
> > sequences is up to every didgit n identical to a digit sequence. So
> > Cantor's argument fails.
>
> Every? On a countably infinite collection?
Put the countable infinite collection of all finite strings of digits
> On 12/30/2012 2:14 AM, WM wrote:
>
> > On 30 Dez., 02:16, forbisga...@gmail.com wrote:
> >> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> >>> Every finite path has as its the abbreviation the node where it ends.
> >>> All paths are different.
>
> >> We go round and round. There are no finite paths in the CIBT.
>
> > There are finite initial segments of all paths in the Binary Tree.
> > I call them "finite paths".
> > I show that every anti-diagonal of the list of all finite digit
> > sequences is up to every didgit n identical to a digit sequence. So
> > Cantor's argument fails.
>
> Every? On a countably infinite collection?
in a list. Every finite string of the anti-diagonal is as an entry in
the list. There is no chance for the anti-diagonal to differ "in the
infinite" from all entries. And there is no chance to differ in the
finite (up to a finite index) from all entries. Hence Cantor fails.
>
>
>
> >> The rest follows from your failure to accept the truth of this.
>
> > Why should it be disallaowed to construct a CIBT from all finite
> > paths?
>
> Because the definition of "completeness" is a property arising
> from the theory of metric spaces.
Stop blathering nonsense, cheater.
>
>
> >> The rest follows from your failure to accept the truth of this.
>
> > Why should it be disallaowed to construct a CIBT from all finite
> > paths?
>
> Because the definition of "completeness" is a property arising
> from the theory of metric spaces.
A Binary Tree that contains only all finite paths is complete.
>
> > Why should anybody so blind not to understand that "finite path" is
> > but an abbreviation of a finite initial segment of a path?
>
> Because a tree built from finite paths is said to be
> well-founded and a tree built from infinite paths
> is said to be ill-founded. A coherent, intelligent
> person is either speaking of one or the other, or
> possibly the relation that a well-founded tree has
> to an ill-founded tree.
Unless matheology is concerned, the CIBT is defined by its nodes and
> > Why should anybody so blind not to understand that "finite path" is
> > but an abbreviation of a finite initial segment of a path?
>
> Because a tree built from finite paths is said to be
> well-founded and a tree built from infinite paths
> is said to be ill-founded. A coherent, intelligent
> person is either speaking of one or the other, or
> possibly the relation that a well-founded tree has
> to an ill-founded tree.
its edges. And with respect to them, both your approaches are
identical.
>
> You are denying the existence of the ill-founded
> tree while trying to present the well-founded tree
> as being the same thing.
With respect to nodes and edges they are.
> You are denying the existence of the ill-founded
> tree while trying to present the well-founded tree
> as being the same thing.
Everybody who thinks different is ill in his head.
Regards, WM
WM
Dec 30, 2012, 7:13:03 AM12/30/12
to
On 30 Dez., 10:46, fom <fomJ...@nyms.net> wrote:
> On 12/29/2012 11:22 AM, WM wrote:
>
>
>
>
>
> > On 28 Dez., 23:04, Virgil <vir...@ligriv.com> wrote:
> >> In article
> >> <2925a2ea-c16d-483e-91da-3bb0084e7...@b16g2000vbh.googlegroups.com>,
>
> >> WM <mueck...@rz.fh-augsburg.de> wrote:
> >>> On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>
> >>>>> It is the set
> >>>>> of all finite paths extending from the root node to a given node.
>
> >>>> In every Complete Infinite Binary Tree, every finite path ends at the
> >>>> root node of of another Complete Infinite Binary Tree. So for every
> >>>> finite path, there are uncountably many extensions of it in every
> >>>> Complete Infinite Binary Tree.
>
> >>> And every extension is contained in the CIBT constructed from all
> >>> finite paths.
>
> >> Claimed but never proven.
>
> > Definitions need not be proven.
>
> On 12/29/2012 11:22 AM, WM wrote:
>
>
>
>
>
> > On 28 Dez., 23:04, Virgil <vir...@ligriv.com> wrote:
> >> In article
> >> <2925a2ea-c16d-483e-91da-3bb0084e7...@b16g2000vbh.googlegroups.com>,
>
> >> WM <mueck...@rz.fh-augsburg.de> wrote:
> >>> On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>
> >>>>> It is the set
> >>>>> of all finite paths extending from the root node to a given node.
>
> >>>> In every Complete Infinite Binary Tree, every finite path ends at the
> >>>> root node of of another Complete Infinite Binary Tree. So for every
> >>>> finite path, there are uncountably many extensions of it in every
> >>>> Complete Infinite Binary Tree.
>
> >>> And every extension is contained in the CIBT constructed from all
> >>> finite paths.
>
> >> Claimed but never proven.
>
> > Definitions need not be proven.
>
> But you said that everything in mathematics
> had to be named and proven to exist.
Trying to show that you are in a bad mood again?
Regards, WM
Virgil
Dec 30, 2012, 3:18:51 PM12/30/12
to
On 12/29/2012 11:22 AM, WM wrote:
> On 28 Dez., 23:04, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <2925a2ea-c16d-483e-91da-3bb0084e7...@b16g2000vbh.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>> On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>>
>>>>> It is the set
>>>>> of all finite paths extending from the root node to a given node.
>>
>>>> In every Complete Infinite Binary Tree, every finite path ends at
the
>>>> root node of of another Complete Infinite Binary Tree. So for every
>>>> finite path, there are uncountably many extensions of it in every
>>>> Complete Infinite Binary Tree.
>>
>>> And every extension is contained in the CIBT constructed from all
>>> finite paths.
>>
>> Claimed but never proven.
>
> Definitions need not be proven.
That they are instantiated must be proven, and that you have not done.
> On 28 Dez., 23:04, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <2925a2ea-c16d-483e-91da-3bb0084e7...@b16g2000vbh.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>> On 27 Dez., 21:49, Virgil <vir...@ligriv.com> wrote:
>>
>>>>> It is the set
>>>>> of all finite paths extending from the root node to a given node.
>>
>>>> In every Complete Infinite Binary Tree, every finite path ends at
the
>>>> root node of of another Complete Infinite Binary Tree. So for every
>>>> finite path, there are uncountably many extensions of it in every
>>>> Complete Infinite Binary Tree.
>>
>>> And every extension is contained in the CIBT constructed from all
>>> finite paths.
>>
>> Claimed but never proven.
>
> Definitions need not be proven.
You have not proven that there is a CIBT constructed or constructible so
as NOT to contain uncountably many paths, whereas many have proven that
any CIBT must have more than any countable set of paths in order to be a
COMPLETE INFINITE BINARY TREE.
Given any countable set of such paths, any proof that they are countable
requires that they be listable, but one can prove that they are not
listable by showing that no list of them can be complete.
And no mater how vociferously WM tries to argue otherwise, in standard
mathematics that can all be done.
--
Ross A. Finlayson
Dec 30, 2012, 3:57:05 PM12/30/12
to
notion of a relevant structure of what "uncountability" might be.
Then, to declaim that modern mathematics as "standard" proves that a
diagonal argument as Al-Jofar's for the tree holds, is well true, but,
modern mathematics is incomplete, and, as described above, a bread-
first traversal of the tree, sees that not hold.
Basically establishing a symmetry through the center of the tree, in
the lexicographic ordering of the paths, then from seeing that from
left at zero to right at one there is the inexhaustibility of the
domain as to course-of-passage, with a "completed" infinity or for
that matter, as modeled from the finite, completed symmetry: this non-
standard (or not-yet-standard) course, that is well supported by the
classical and as well in asymptotics by the modern and concrete, sees
different results from that.
Draw the line, Cantor shows is it's point-to-point, not the stippling
of the stellation, our complete ordered field as structure above the
continuum follows from simpler principles (of points that make space).
Draw the line. Split the tree. Diagonal? Where's the middle? The
middle is defined by the ends.
Regards,
Ross Finlayson
Virgil
Dec 30, 2012, 4:15:56 PM12/30/12
to
In article
<8a425f72-80f2-4aee...@vb8g2000pbb.googlegroups.com>,
--
<8a425f72-80f2-4aee...@vb8g2000pbb.googlegroups.com>,
"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:
> > requires that they be listable, but one can prove that they are not
> > listable by showing that no list of them can be complete.
> > And no mater how vociferously WM tries to argue otherwise, in standard
> > mathematics that can all be done.
> > --
>
>
> Well, not when "standard" was "pre-Cantorian"
I used only the present tense which eliminates pre-Cantorianism.
> > requires that they be listable, but one can prove that they are not
> > listable by showing that no list of them can be complete.
> > And no mater how vociferously WM tries to argue otherwise, in standard
> > mathematics that can all be done.
> > --
>
>
> Well, not when "standard" was "pre-Cantorian"
--
Ross A. Finlayson
Dec 30, 2012, 4:44:09 PM12/30/12
to
> <8a425f72-80f2-4aee-9bb9-01f1c6f12...@vb8g2000pbb.googlegroups.com>,
mathematics is incomplete, there are to be discovered true things
about its domain, not its theorems.
And no, there's no proof that ZF (as a general foundation for modern
mathematics) is consistent and complete, and there aren't that it's
consistent, either, and Goedel shows, in modern mathematics, it's not
both. And, measure theory uses countable additivity (of the non-
finite non-zero infinitesimal differential patches) for real analysis,
and concrete mathematics uses asymptotics and sometimes, regular
infinite ordinals: with no applied results solely due transfinite
cardinals, and indeed transfinite set theory is somewhat a raw,
disposed shoehorn of real analysis. And half of the integers are
even.
Let's work more on posts, and progress, than replies. Quit worrying
so much about covering your ass, as getting your head out of it.
And as described above, a breadth-first traversal, sees different
results for the tree's "anti-diagonal", as that it's simply at the end
of the traversal, end-to-end, point-to-point.
Then the question arises, diagonal of what? Where's the middle?
Where's the square.
Draw a line: without putting pencil to paper. That's mathematics.
Regards,
Ross Finlayson
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > > requires that they be listable, but one can prove that they are not
> > > listable by showing that no list of them can be complete.
> > > And no mater how vociferously WM tries to argue otherwise, in standard
> > > mathematics that can all be done.
> > > --
>
> > Well, not when "standard" was "pre-Cantorian"
>
> I used only the present tense which eliminates pre-Cantorianism.
> --
Then you shouldn't discount the future where, as we know modern
>
> > > requires that they be listable, but one can prove that they are not
> > > listable by showing that no list of them can be complete.
> > > And no mater how vociferously WM tries to argue otherwise, in standard
> > > mathematics that can all be done.
> > > --
>
> > Well, not when "standard" was "pre-Cantorian"
>
> I used only the present tense which eliminates pre-Cantorianism.
> --
mathematics is incomplete, there are to be discovered true things
about its domain, not its theorems.
And no, there's no proof that ZF (as a general foundation for modern
mathematics) is consistent and complete, and there aren't that it's
consistent, either, and Goedel shows, in modern mathematics, it's not
both. And, measure theory uses countable additivity (of the non-
finite non-zero infinitesimal differential patches) for real analysis,
and concrete mathematics uses asymptotics and sometimes, regular
infinite ordinals: with no applied results solely due transfinite
cardinals, and indeed transfinite set theory is somewhat a raw,
disposed shoehorn of real analysis. And half of the integers are
even.
Let's work more on posts, and progress, than replies. Quit worrying
so much about covering your ass, as getting your head out of it.
And as described above, a breadth-first traversal, sees different
results for the tree's "anti-diagonal", as that it's simply at the end
of the traversal, end-to-end, point-to-point.
Then the question arises, diagonal of what? Where's the middle?
Where's the square.
Draw a line: without putting pencil to paper. That's mathematics.
Regards,
Ross Finlayson
Virgil
Dec 30, 2012, 6:21:56 PM12/30/12
to
In article
<fb7fa7ab-bc25-4cb9...@i2g2000pbi.googlegroups.com>,
And, as thing stand in the present, standard mathematics supports the
Cantor diagonal argument and that every Complete Infinite Binary Tree
which really is a Complete Infinite Binary Tree instead of one of WM's
corrupted versions of one, must have uncountably many paths.
--
<fb7fa7ab-bc25-4cb9...@i2g2000pbi.googlegroups.com>,
> On Dec 30, 1:15 pm, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <8a425f72-80f2-4aee-9bb9-01f1c6f12...@vb8g2000pbb.googlegroups.com>,
> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> > > > requires that they be listable, but one can prove that they are not
> > > > listable by showing that no list of them can be complete.
> > > > And no matter how vociferously WM tries to argue otherwise, in standard
> > In article
> > <8a425f72-80f2-4aee-9bb9-01f1c6f12...@vb8g2000pbb.googlegroups.com>,
> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> > > > requires that they be listable, but one can prove that they are not
> > > > listable by showing that no list of them can be complete.
> > > > mathematics that can all be done.
> > > > --
> >
> > > Well, not when "standard" was "pre-Cantorian"
> >
> > I used only the present tense which eliminates pre-Cantorianism.
> > --
>
>
> Then you shouldn't discount the future
I don't, but neither do I pretend to predict it, the way you do.
> > > > --
> >
> > > Well, not when "standard" was "pre-Cantorian"
> >
> > I used only the present tense which eliminates pre-Cantorianism.
> > --
>
>
> Then you shouldn't discount the future
And, as thing stand in the present, standard mathematics supports the
Cantor diagonal argument and that every Complete Infinite Binary Tree
which really is a Complete Infinite Binary Tree instead of one of WM's
corrupted versions of one, must have uncountably many paths.
--
Ross A. Finlayson
Dec 30, 2012, 7:29:16 PM12/30/12
to
> <fb7fa7ab-bc25-4cb9-9b7c-a073c2fe7...@i2g2000pbi.googlegroups.com>,
"pretend": pure, abstract mathematics.
Unless you're a Platonist now.
I'm a Platonist: those things are real. Show an application solely
due transfinite cardinals.
And the universe is infinite, and it's real. And: as mathematical
object: it's its own powerset.
And, there may always be more than our "standard" mathematics, but,
existence is that, for what we know it.
Regards,
Ross Finlayson
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Dec 30, 1:15 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <8a425f72-80f2-4aee-9bb9-01f1c6f12...@vb8g2000pbb.googlegroups.com>,
> > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > > > > requires that they be listable, but one can prove that they are not
> > > > > listable by showing that no list of them can be complete.
> > > > > And no matter how vociferously WM tries to argue otherwise, in standard
> > > > > mathematics that can all be done.
> > > > > --
>
> > > > Well, not when "standard" was "pre-Cantorian"
>
> > > I used only the present tense which eliminates pre-Cantorianism.
> > > --
>
> > Then you shouldn't discount the future
>
> I don't, but neither do I pretend to predict it, the way you do.
>
> And, as thing stand in the present, standard mathematics supports the
> Cantor diagonal argument and that every Complete Infinite Binary Tree
> which really is a Complete Infinite Binary Tree instead of one of WM's
> corrupted versions of one, must have uncountably many paths.
> --
Until you find applications for transfinite cardinals, it's all
>
>
>
>
>
>
>
>
>
> > On Dec 30, 1:15 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > > <8a425f72-80f2-4aee-9bb9-01f1c6f12...@vb8g2000pbb.googlegroups.com>,
> > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > > > > requires that they be listable, but one can prove that they are not
> > > > > listable by showing that no list of them can be complete.
> > > > > And no matter how vociferously WM tries to argue otherwise, in standard
> > > > > mathematics that can all be done.
> > > > > --
>
> > > > Well, not when "standard" was "pre-Cantorian"
>
> > > I used only the present tense which eliminates pre-Cantorianism.
> > > --
>
> > Then you shouldn't discount the future
>
> I don't, but neither do I pretend to predict it, the way you do.
>
> And, as thing stand in the present, standard mathematics supports the
> Cantor diagonal argument and that every Complete Infinite Binary Tree
> which really is a Complete Infinite Binary Tree instead of one of WM's
> corrupted versions of one, must have uncountably many paths.
> --
"pretend": pure, abstract mathematics.
Unless you're a Platonist now.
I'm a Platonist: those things are real. Show an application solely
due transfinite cardinals.
And the universe is infinite, and it's real. And: as mathematical
object: it's its own powerset.
And, there may always be more than our "standard" mathematics, but,
existence is that, for what we know it.
Regards,
Ross Finlayson
Virgil
Dec 30, 2012, 9:09:34 PM12/30/12
to
In article
<0d8fdfbc-711a-44fe...@ah9g2000pbd.googlegroups.com>,
What's wrong with pure mathematics, and why do you claim it to be only
pretend?
There is nothing in your EF or other notions that is not even more
pretend than anything in my "pure" mathematics.
--
<0d8fdfbc-711a-44fe...@ah9g2000pbd.googlegroups.com>,
pretend?
There is nothing in your EF or other notions that is not even more
pretend than anything in my "pure" mathematics.
--
fom
Dec 30, 2012, 10:50:18 PM12/30/12
to
On 12/30/2012 6:12 AM, WM wrote:
> On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
>> On 12/30/2012 2:14 AM, WM wrote:
>>
>>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>>>> Every finite path has as its the abbreviation the node where it ends.
>>>>> All paths are different.
>>
>>>> We go round and round. There are no finite paths in the CIBT.
>>
>>> There are finite initial segments of all paths in the Binary Tree.
>>> I call them "finite paths".
>>> I show that every anti-diagonal of the list of all finite digit
>>> sequences is up to every didgit n identical to a digit sequence. So
>>> Cantor's argument fails.
>>
>> Every? On a countably infinite collection?
>
> Put the countable infinite collection of all finite strings of digits
> in a list.
A list implies a logical type.
> On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
>> On 12/30/2012 2:14 AM, WM wrote:
>>
>>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>>>> Every finite path has as its the abbreviation the node where it ends.
>>>>> All paths are different.
>>
>>>> We go round and round. There are no finite paths in the CIBT.
>>
>>> There are finite initial segments of all paths in the Binary Tree.
>>> I call them "finite paths".
>>> I show that every anti-diagonal of the list of all finite digit
>>> sequences is up to every didgit n identical to a digit sequence. So
>>> Cantor's argument fails.
>>
>> Every? On a countably infinite collection?
>
> Put the countable infinite collection of all finite strings of digits
> in a list.
A list implies that one list can be discerned from
another unless there is only one list.
A list is a unity in relation to those lists from which
it can be discerned.
Is this list named oo?
fom
Dec 30, 2012, 10:51:32 PM12/30/12
to
WM
Dec 31, 2012, 12:37:03 PM12/31/12
to
On 31 Dez., 04:50, fom <fomJ...@nyms.net> wrote:
> On 12/30/2012 6:12 AM, WM wrote:
>
>
>
>
>
> > On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
> >> On 12/30/2012 2:14 AM, WM wrote:
>
> >>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
> >>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> >>>>> Every finite path has as its the abbreviation the node where it ends.
> >>>>> All paths are different.
>
> >>>> We go round and round. There are no finite paths in the CIBT.
>
> >>> There are finite initial segments of all paths in the Binary Tree.
> >>> I call them "finite paths".
> >>> I show that every anti-diagonal of the list of all finite digit
> >>> sequences is up to every didgit n identical to a digit sequence. So
> >>> Cantor's argument fails.
>
> >> Every? On a countably infinite collection?
>
> > Put the countable infinite collection of all finite strings of digits
> > in a list.
>
> A list implies a logical type.
No logic of types is required for Cantor's proof.
> On 12/30/2012 6:12 AM, WM wrote:
>
>
>
>
>
> > On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
> >> On 12/30/2012 2:14 AM, WM wrote:
>
> >>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
> >>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
> >>>>> Every finite path has as its the abbreviation the node where it ends.
> >>>>> All paths are different.
>
> >>>> We go round and round. There are no finite paths in the CIBT.
>
> >>> There are finite initial segments of all paths in the Binary Tree.
> >>> I call them "finite paths".
> >>> I show that every anti-diagonal of the list of all finite digit
> >>> sequences is up to every didgit n identical to a digit sequence. So
> >>> Cantor's argument fails.
>
> >> Every? On a countably infinite collection?
>
> > Put the countable infinite collection of all finite strings of digits
> > in a list.
>
> A list implies a logical type.
>
> A list implies that one list can be discerned from
> another unless there is only one list.
Two infinite and identcal lists, if they existed without being
> A list implies that one list can be discerned from
> another unless there is only one list.
finitely defined, could never be distinguished and could never proven
to be identical.
============
> But you said that everything in mathematics
> had to be named and proven to exist.
but a name (except some points and lines in geometry). But the proof
of existence of fundamentals is not done from axioms but by the common
understanding of two or more persons of the meaning of a fundamental
word like 1 or 2 or 3 or + or divisor or ... and so on.
Regards, WM
fom
Jan 3, 2013, 5:02:34 AM1/3/13
to
On 12/31/2012 11:37 AM, WM wrote:
> On 31 Dez., 04:50, fom <fomJ...@nyms.net> wrote:
>> On 12/30/2012 6:12 AM, WM wrote:
>>
>>
>>
>>
>>
>>> On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
>>>> On 12/30/2012 2:14 AM, WM wrote:
>>
>>>>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>>>>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>>>>>> Every finite path has as its the abbreviation the node where it ends.
>>>>>>> All paths are different.
>>
>>>>>> We go round and round. There are no finite paths in the CIBT.
>>
>>>>> There are finite initial segments of all paths in the Binary Tree.
>>>>> I call them "finite paths".
>>>>> I show that every anti-diagonal of the list of all finite digit
>>>>> sequences is up to every didgit n identical to a digit sequence. So
>>>>> Cantor's argument fails.
>>
>>>> Every? On a countably infinite collection?
>>
>>> Put the countable infinite collection of all finite strings of digits
>>> in a list.
>>
>> A list implies a logical type.
>
> No logic of types is required for Cantor's proof.
I am not arguing for Cantor.
> On 31 Dez., 04:50, fom <fomJ...@nyms.net> wrote:
>> On 12/30/2012 6:12 AM, WM wrote:
>>
>>
>>
>>
>>
>>> On 30 Dez., 10:07, fom <fomJ...@nyms.net> wrote:
>>>> On 12/30/2012 2:14 AM, WM wrote:
>>
>>>>> On 30 Dez., 02:16, forbisga...@gmail.com wrote:
>>>>>> On Saturday, December 29, 2012 2:21:35 PM UTC-8, WM wrote:
>>>>>>> Every finite path has as its the abbreviation the node where it ends.
>>>>>>> All paths are different.
>>
>>>>>> We go round and round. There are no finite paths in the CIBT.
>>
>>>>> There are finite initial segments of all paths in the Binary Tree.
>>>>> I call them "finite paths".
>>>>> I show that every anti-diagonal of the list of all finite digit
>>>>> sequences is up to every didgit n identical to a digit sequence. So
>>>>> Cantor's argument fails.
>>
>>>> Every? On a countably infinite collection?
>>
>>> Put the countable infinite collection of all finite strings of digits
>>> in a list.
>>
>> A list implies a logical type.
>
> No logic of types is required for Cantor's proof.
>>
>> A list implies that one list can be discerned from
>> another unless there is only one list.
>
> Two infinite and identcal lists, if they existed without being
> finitely defined, could never be distinguished and could never proven
> to be identical.
important distinctions.
>
> ============
>
>> But you said that everything in mathematics
>> had to be named and proven to exist.
>
> Yes, everything in mathematics must have a name because it is nothing
> but a name (except some points and lines in geometry). But the proof
> of existence of fundamentals is not done from axioms but by the common
> understanding of two or more persons of the meaning of a fundamental
> word like 1 or 2 or 3 or + or divisor or ... and so on.
look at a subject called pragmatics.
There is a great deal of nonsense in mathematical logic
concerning undefined language primitives. There are
certainly such primitives in the sense of symbols used
in the transformation rules of a logical calculus.
However, this notion has been extended to whatever one
wants to put into a signature. If one reads the literature
concerning this, (Bolzano, Frege) it is clear that there
are pragmatic influences involved. This seems to be
ignored by authors who want to crowd a bunch of syntax
in the beginning of their expositions without examples.
Someone should have shot Peano for the crap he pulled
with formal proofs that no one could read.
Ross A. Finlayson
Jan 4, 2013, 11:11:42 PM1/4/13
to
On Dec 30 2012, 1:44 pm, "Ross A. Finlayson"
Rather casually: rays through ordinal points.
Draw the binary tree with the endpoints of the paths of the finite
tree to points 0, ..., 2^n-1.
Consider rays from the origin, to the ordinal points 0, ..., 2^n-1.
They are countable.
Then with rays from the origin through 0, ..., 2^w-1, they are
countable.
Are not the rays dense in the paths?
Regards,
Ross Finlayson
Draw the binary tree with the endpoints of the paths of the finite
tree to points 0, ..., 2^n-1.
Consider rays from the origin, to the ordinal points 0, ..., 2^n-1.
They are countable.
Then with rays from the origin through 0, ..., 2^w-1, they are
countable.
Are not the rays dense in the paths?
Regards,
Ross Finlayson
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