Cantor and the binary tree
muec...@rz.fh-augsburg.de
Cantor and the binary tree.
If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
= 1
then all the real numbers of the interval [0,1] are realized as
infinite paths of the binary tree
.
0 1
0 1 0 1
..................
too (read from top to bottom). Each number is given by a path
stretching over infinitely many nodes (bits). All nodes (bits) of the
tree are countable. The paths are not, according to Cantor's famous
diagonal proof.
But we find that, up to line number n, there are -1 + 2^(n+1) nodes
whereas 2^n different paths arrive at and 2^(n+1) different paths
spring off from line number n. Hence, in the enumerated domain, there
is at most one more path than nodes. After leaving any finite line
number n (if it is reasonable to make such a distinction) we can no
longer apply these formulae. But we know that any new branching
increases the number of paths by 1 and, by definition, the number of
nodes by 1 too (because any branching is a node). Therefore, the number
of paths always equals that of the nodes + 1. It is simply impossible
to assume that one of these numbers becomes uncountably infinite while
the other remains countably infinite.
Regards, WM
David C. Ullrich
>
>Cantor and the binary tree.
>
>If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
>= 1
>
>then all the real numbers of the interval [0,1] are realized as
>infinite paths of the binary tree
> .
> 0 1
>0 1 0 1
>..................
>
>too (read from top to bottom). Each number is given by a path
>stretching over infinitely many nodes (bits). All nodes (bits) of the
>tree are countable. The paths are not, according to Cantor's famous
>diagonal proof.
>
>But we find that, up to line number n, there are -1 + 2^(n+1) nodes
>whereas 2^n different paths arrive at and 2^(n+1) different paths
>spring off from line number n. Hence, in the enumerated domain, there
>is at most one more path than nodes. After leaving any finite line
>number n (if it is reasonable to make such a distinction) we can no
>longer apply these formulae. But we know that any new branching
>increases the number of paths by 1 and, by definition, the number of
>nodes by 1 too (because any branching is a node). Therefore, the number
>of paths always equals that of the nodes + 1.
All that is true for any finite binary tree, yes.
>It is simply impossible
>to assume that one of these numbers becomes uncountably infinite while
>the other remains countably infinite.
Hint: infinite sets are not the same as finite sets. What you say
is "impossible to assume" is not something we _assume_, but it's
true. And very easy to prove.
>Regards, WM
************************
David C. Ullrich
Dik T. Winter
> If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
> = 1
You may note that that is *not* an infinite sum...
> .
> 0 1
> 0 1 0 1
> ..................
...
> But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> whereas 2^n different paths arrive at and 2^(n+1) different paths
> spring off from line number n. Hence, in the enumerated domain, there
> is at most one more path than nodes. After leaving any finite line
> number n (if it is reasonable to make such a distinction) we can no
> longer apply these formulae. But we know that any new branching
> increases the number of paths by 1 and, by definition, the number of
> nodes by 1 too (because any branching is a node). Therefore, the number
> of paths always equals that of the nodes + 1. It is simply impossible
> to assume that one of these numbers becomes uncountably infinite while
> the other remains countably infinite.
Also we find that up to line n, summing up to that node along the path
gives a value for k/(2^n) for some integer k. Therefore a number along
a node is always equal to a rational with a denominator that is a power
of two. It is simply impossible to assume that one of these numbers
becomes 1/3.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Robert Kolker
> of paths always equals that of the nodes + 1. It is simply impossible
> to assume that one of these numbers becomes uncountably infinite while
> the other remains countably infinite.
Wrong. 2^(aleph_0) > aleph_0.
List all the infinite binary sequences with a bijection to the integers.
Now flip the n-th digit of the n-th sequence in the list. This cannot
occur anywhere in the list. Contradiction. Such a bijection to the
integers does not exist.
Bob Kolker
Robin Chapman
> muec...@rz.fh-augsburg.de wrote:
>
>> of paths always equals that of the nodes + 1. It is simply impossible
>> to assume that one of these numbers becomes uncountably infinite while
>> the other remains countably infinite.
"becomes"? Muck's fuzzy metaphors are sabotaging him again.
The fact is that the nodes in this tree form a countable
set and the paths form an uncountable set. "becoming"
has nowt to do with that.
>
> Wrong. 2^(aleph_0) > aleph_0.
>
> List all the infinite binary sequences with a bijection to the integers.
> Now flip the n-th digit of the n-th sequence in the list. This cannot
> occur anywhere in the list. Contradiction. Such a bijection to the
> integers does not exist.
One can hardly imagine a simpler mathematical proof. Alas, it's still
beyond the limits of Muck's feeble intellect :-(
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_
Virgil
muec...@rz.fh-augsburg.de wrote:
> Cantor and the binary tree.
>
> If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
> = 1
>
> then all the real numbers of the interval [0,1] are realized as
> infinite paths of the binary tree
> .
> 0 1
> 0 1 0 1
> ..................
>
> too (read from top to bottom). Each number is given by a path
> stretching over infinitely many nodes (bits). All nodes (bits) of the
> tree are countable. The paths are not, according to Cantor's famous
> diagonal proof.
So far so good.
>
> But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> whereas 2^n different paths arrive at and 2^(n+1) different paths
> spring off from line number n. Hence, in the enumerated domain, there
> is at most one more path than nodes. After leaving any finite line
> number n (if it is reasonable to make such a distinction) we can no
> longer apply these formulae. But we know that any new branching
> increases the number of paths by 1 and, by definition, the number of
> nodes by 1 too (because any branching is a node). Therefore, the number
> of paths always equals that of the nodes + 1. It is simply impossible
> to assume that one of these numbers becomes uncountably infinite while
> the other remains countably infinite.
One does not assume it, one proves it, as I have done several times.
Here is an outline of that proof:
Every node is represented by a terminating binary (starting at "." and
terminating at the node itself in the tree above) which is like a subset
of the rationals, which are countable.
Every unending path is represented by a non-terminating binary (also
starting at "." but never ending), which surject onto the real interval
[0,1], and are thus as uncountable as the reals.
That WM choses to reject proofs that show him wrong does not invalidate
such proofs.
muec...@rz.fh-augsburg.de
I can imagine that you are ready to prove that in 1-2-3-4-... there are
more dashes than numbers. But it is nevertheless inacceptable.
In my tree the question is not whether it is an infinite set or not,
but whether every node is member of a line with finite enumeration.
If no branching is possible without a new node (because a branching is
a node), then "the infinite" does not at all help you. *That* is very
muec...@rz.fh-augsburg.de
Dik T. Winter wrote:
> In article <1116939502.8...@o13g2000cwo.googlegroups.com>
muec...@rz.fh-augsburg.de writes:
> > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n =
0.111...
> > = 1
>
> You may note that that is *not* an infinite sum...
>
> > .
> > 0 1
> > 0 1 0 1
> > ..................
Any path is an infinite sequence of bits which by multiplying with 2^-n
and summing up establishes an infinite series representing a real
number. Every combination of countably many bits is realized by
definition.
> ...
> > But we find that, up to line number n, there are -1 + 2^(n+1)
nodes
> > whereas 2^n different paths arrive at and 2^(n+1) different paths
> > spring off from line number n. Hence, in the enumerated domain,
there
> > is at most one more path than nodes. After leaving any finite line
> > number n (if it is reasonable to make such a distinction) we can
no
> > longer apply these formulae. But we know that any new branching
> > increases the number of paths by 1 and, by definition, the number
of
> > nodes by 1 too (because any branching is a node). Therefore, the
number
> > of paths always equals that of the nodes + 1. It is simply
impossible
> > to assume that one of these numbers becomes uncountably infinite
while
> > the other remains countably infinite.
>
> Also we find that up to line n, summing up to that node along the
path
> gives a value for k/(2^n) for some integer k.
It is forbidden to stop there, by definition. If you want to realize a
terminating rational with n bits, then you must follow, from line n+1
on, the path with infinitely many zero-bits.
Therefore a number along
> a node is always equal to a rational with a denominator that is a
power
> of two. It is simply impossible to assume that one of these numbers
> becomes 1/3.
Why should 0.010101... not exist in that tree? Every path is infinite
by definition as is 0.010101..., by definition.
Regards, WM
muec...@rz.fh-augsburg.de
Robert Kolker wrote:
> muec...@rz.fh-augsburg.de wrote:
>
> > of paths always equals that of the nodes + 1. It is simply
impossible
> > to assume that one of these numbers becomes uncountably infinite
while
> > the other remains countably infinite.
>
> Wrong. 2^(aleph_0) > aleph_0.
That is one result. Obviously it is not consistent, because another
proof leads to the opposite result. Or is it forbidden to consider
other proofs? Is it forbidden to think but only to repeat the old
stuff. Point out where my proof is in error.
>
> List all the infinite binary sequences with a bijection to the
integers.
It is not possible to list them all. Therefore I constructed the tree.
It is not a list.
Regards, WM
muec...@rz.fh-augsburg.de
Robin Chapman wrote:
> >> of paths always equals that of the nodes + 1. It is simply
impossible
> >> to assume that one of these numbers becomes uncountably infinite
while
> >> the other remains countably infinite.
>
> "becomes"?
If you prefer "is", you may use it. That does not matter. It is
obviously impossible that the set of paths is uncountable when the set
of nodes is countable, because every pair of paths springs off from one
node, while one path leads to that node. Try to find an error n the
arguing, not in the result.
> One can hardly imagine a simpler mathematical proof. Alas, it's still
> beyond the limits of Chaps
Regards, WM
Tony Orlow
root node), and the paths are half of that number. There are no infinite paths
in a tree with finite nodes/branches. Patently impossible, and clearly
ridiculous. But, what else is new?
--
Smiles,
Tony
Ron Sperber
> Robert Kolker wrote:
>
>
>>muec...@rz.fh-augsburg.de wrote:
>>
>>
>>>of paths always equals that of the nodes + 1. It is simply impossible
>>>to assume that one of these numbers becomes uncountably infinite while
>>>the other remains countably infinite.
>
>
> "becomes"? Muck's fuzzy metaphors are sabotaging him again.
> The fact is that the nodes in this tree form a countable
> set and the paths form an uncountable set. "becoming"
> has nowt to do with that.
>
>
>>Wrong. 2^(aleph_0) > aleph_0.
>>
>>List all the infinite binary sequences with a bijection to the integers.
>>Now flip the n-th digit of the n-th sequence in the list. This cannot
>>occur anywhere in the list. Contradiction. Such a bijection to the
>>integers does not exist.
>
>
> One can hardly imagine a simpler mathematical proof. Alas, it's still
> beyond the limits of Muck's feeble intellect :-(
>
It simply boggles my mind that this simple proof gives so many people
such fits that they refuse to accept it. I continue to be sadly shocked
by the number of posts on sci.math daily refuting Cantor's proof. Of
course they are always fuzzy on details, but that's to be expected since
they can't actually disprove it.
Tony Orlow
to be referring to the "proof" of uncountability of the reals. it really only
proves that there are more reals than naturals, and that one can generate more
strings than the number of symbols each string contains, provided you have a
set of more than one symbol to choose from. By traversing the list diagonally
and assuming you have covered the list, you are assuming that it is square in
the sense that there are as many digits in each number as numbers in the list.
But, digital number systems don't work that way. If you have a number system of
base S, L digits will allow you to represent S^L numbers, so if you have N
digits, you will have 10^N numbers in decimal, 2^N numbers in binary, in your
list. It is infinitely longer than it is wide, and therefore cannot be
completely traversed diagonally. The number generated in the antidiagonal is
simply one of the 2^N-N numbers below the diagonal of traversal.
The subsequent conclusion that the reals are not "countable" rests on the
notion that all countably infinite sets are the same size, which is an
assumption that I reject for many reasons, and which has no justification
besides "oo=oo=oo".
Sure, the proof looks simple. It's a little too simple, and the critical
thought aimed at it is too.
--
Smiles,
Tony
Robert Kolker
>
> The subsequent conclusion that the reals are not "countable" rests on the
> notion that all countably infinite sets are the same size, which is an
> assumption that I reject for many reasons, and which has no justification
> besides "oo=oo=oo".
Will you stop this sloppy nonsense and listen. Let two sets A, B be in
one to one correspondence with the integers N by the mappings f and g
respecitive then f : N ->A, g : N ->B. since f and b are bijections
f_inverse maps A to N and the composition (g(f_inverse)) maps A onto B
in a 1-1 fashion which means A is equivalent to B. QED.
Now go away and learn some mathematics.
Bob Kolker
Tony Orlow
functions into your comparison. It's a mistake to ignore them.
--
Smiles,
Tony
Robert Kolker
>
> That's all very well and good, if you specify f anf g and figure those
> functions into your comparison. It's a mistake to ignore them.
Are you capable of following a proof? Even a three line proof?
Bob Kolker
Proginoskes
David C. Ullrich wrote:
> On 24 May 2005 05:58:22 -0700, muec...@rz.fh-augsburg.de
> wrote:
> [...]
> Hint: infinite sets are not the same as finite sets.
Yes. Intuition goes out the window with infinity, which is why we do
mathematics instead. 8-)
One (easy?) example of how infinite sets are different from finite sets
is that if you have an infinite set S, you can put it in 1-1
correspondence with a proper subset T (a subset which is not equal to
S).
For instance, let S = N, T = {2, 3, 4,...}. One possible 1-1
correspondence is
f(n) = n + 1; that is, match up
1 <--> 2
2 <--> 3
3 <--> 4
...
This uses up all the elements of S (on the left-hand side) and T (on
the right-hand side).
This is impossible to do if S is finite.
--- Christopher Heckman
Ron Sperber
here, countably infinite) has a very precise definition. That is a set X
is countably infinite if there exists a bijection f:N->X where N is the
set of natural numbers.
Here is a theorem that can be proved from this definiton. Let X,Y be
countably infinite sets. Then there is a bijection h:X->Y.
Proof.
Let f:N->X and g:N->Y be bijections that show that X and Y are
countably infinite. Since f and g are bijections there are bijections
f^-1:X->N, g^-1:Y->N.
Then gf^-1:X->Y and fg^-1:Y->X are functions. and since gf^-1fg^-1:Y->Y
is the identity and fg^-1gf^-1:X->X is the identity gf^-1 and fg^-1 are
bijections, thus h=gf^-1 works.
Thus your nonsense about different sized countable sets is just
that...nonsense. any 2 countably infinite sets have a bijection between
them which is the only useful measure of "size" for infinite sets.
Virgil
muec...@rz.fh-augsburg.de wrote:
> David C. Ullrich wrote:
> > Hint: infinite sets are not the same as finite sets. What you say
> > is "impossible to assume" is not something we _assume_, but it's
> > true. And very easy to prove.
>
> I can imagine that you are ready to prove that in 1-2-3-4-... there are
> more dashes than numbers. But it is nevertheless inacceptable.
>
> In my tree the question is not whether it is an infinite set or not,
> but whether every node is member of a line with finite enumeration.
>
> If no branching is possible without a new node (because a branching is
> a node), then "the infinite" does not at all help you. *That* is very
> easy to prove.
How is it that all those things that WM declares so easy to prove still
remain unproven?
Virgil
muec...@rz.fh-augsburg.de wrote:
> >
> > > .
> > > 0 1
> > > 0 1 0 1
> > > ..................
>
> Any path is an infinite sequence of bits which by multiplying with 2^-n
> and summing up establishes an infinite series representing a real
> number. Every combination of countably many bits is realized by
> definition.
But such an "infinite" binary tree is not a list, so this has nothing to
say about the validity of Cantor's theorem.
Virgil
muec...@rz.fh-augsburg.de wrote:
> Robert Kolker wrote:
> > muec...@rz.fh-augsburg.de wrote:
> >
> > > of paths always equals that of the nodes + 1. It is simply
> impossible
> > > to assume that one of these numbers becomes uncountably infinite
> while
> > > the other remains countably infinite.
> >
> > Wrong. 2^(aleph_0) > aleph_0.
>
> That is one result. Obviously it is not consistent, because another
> proof leads to the opposite result.
Then show us that "proof". Unproven claims butter no parsips.
Or is it forbidden to consider
> other proofs?
It is forbidden to claim that they exist without providing reasonable
evidence of their existence.
> Is it forbidden to think but only to repeat the old
> stuff. Point out where my proof is in error.
As your "proof" is not in evidence, only an unsupported claim of one,
the "error" is that proof is absent.
> >
> > List all the infinite binary sequences with a bijection to the
> integers.
>
> It is not possible to list them all.
Perhaps the light is finally beginning to dawn in WM's mind! That
unlistability is exactly the point!
> Therefore I constructed the tree.
> It is not a list.
And that is why it is irrelevant in deciding whether the reals can be
listed.
Virgil
muec...@rz.fh-augsburg.de wrote:
> Robin Chapman wrote:
>
>
> > >> of paths always equals that of the nodes + 1. It is simply
> impossible
> > >> to assume that one of these numbers becomes uncountably infinite
> while
> > >> the other remains countably infinite.
> >
> > "becomes"?
>
> If you prefer "is", you may use it. That does not matter. It is
> obviously impossible that the set of paths is uncountable when the set
> of nodes is countable, because every pair of paths springs off from one
> node, while one path leads to that node. Try to find an error n the
> arguing, not in the result.
For finite paths (having a root node and a leaf or terminal node) one is
there can only be countably many, but it is those paths starting at the
root which procede through infinitely many nodes in this infinite tree,
and never have a terminal or leaf node, that are under consideration.
The set of these unbounded paths can mapped bijectively to P(N), whereas
the set of nodes can be mapped bijectively to N.
>
> > One can hardly imagine a simpler mathematical proof. Alas, it's still
> > beyond the limits of Chaps
As all such false proofs should be.
Dik T. Winter
> Dik T. Winter wrote:
> > In article <1116939502.8...@o13g2000cwo.googlegroups.com>
> muec...@rz.fh-augsburg.de writes:
> > > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n =
> > > 0.111... = 1
> >
> > You may note that that is *not* an infinite sum...
Note what I said here. It is not a sum of infinitely many terms.
> > > .
> > > 0 1
> > > 0 1 0 1
> > > ..................
>
> Any path is an infinite sequence of bits which by multiplying with 2^-n
> and summing up establishes an infinite series representing a real
> number. Every combination of countably many bits is realized by
> definition.
Mathematics does *not* define summing up infinitely many terms. It uses
limits in this case.
> > > But we find that, up to line number n, there are -1 + 2^(n+1)
> > > nodes whereas 2^n different paths arrive at and 2^(n+1) different
> > > paths spring off from line number n. Hence, in the enumerated domain,
> > > there is at most one more path than nodes. After leaving any finite
> > > line number n (if it is reasonable to make such a distinction) we can
> > > no longer apply these formulae. But we know that any new branching
> > > increases the number of paths by 1 and, by definition, the number
> > > of nodes by 1 too (because any branching is a node). Therefore, the
> > > number of paths always equals that of the nodes + 1. It is simply
> > > impossible to assume that one of these numbers becomes uncountably
> > > infinite while the other remains countably infinite.
> >
> > Also we find that up to line n, summing up to that node along the
> > path gives a value for k/(2^n) for some integer k.
>
> It is forbidden to stop there, by definition. If you want to realize a
> terminating rational with n bits, then you must follow, from line n+1
> on, the path with infinitely many zero-bits.
Why is that forbidden? If I want to stop at a node I can not follow an
infinite path. But if you want to attach a node at the end of the
infinite path, you may be in for a surprise. In that case the number
of nodes is also uncountable. They are the real numbers. It is the
set of nodes (or paths) that are a finite distance away from the origin
that is countable.
> > Therefore a number along
> > a node is always equal to a rational with a denominator that is a
> > power of two. It is simply impossible to assume that one of these
> > numbers becomes 1/3.
>
> Why should 0.010101... not exist in that tree? Every path is infinite
> by definition as is 0.010101..., by definition.
The path does exist, but there is no node at the end. Or if you wish
there is a node at the and (as J.H. Conway does with his surreal numbers),
the number of nodes is uncountable.
Martin Shobe
>
>Robin Chapman wrote:
>
>
>> >> of paths always equals that of the nodes + 1. It is simply
>impossible
>> >> to assume that one of these numbers becomes uncountably infinite
>while
>> >> the other remains countably infinite.
>>
>> "becomes"?
>
>If you prefer "is", you may use it. That does not matter. It is
>obviously impossible that the set of paths is uncountable when the set
>of nodes is countable, because every pair of paths springs off from one
>node, while one path leads to that node. Try to find an error n the
>arguing, not in the result.
Sure thing. The error is that "It is simply impossible to assume that
one of these numbers becomes uncountably infinite while the other
remains countably infinite" does not follow from the previous
statements in your proof. The relationships between nodes and paths
you supplied require that the tree be finite. Attempting to use those
relationships on infinite trees (without first proving that the
relationships hold) is simply a fallicy.
As an example of a property that holds for finite trees that doesn't
hold for infinite ones, count the number of leaf nodes. In the
infinite binary tree being discussed, there are zero. Every finite
binary tree has at least one.
Martin
Virgil
Martin Shobe <msh...@sbcglobal.net> wrote:
Nice point! Mind if I use it?
muec...@rz.fh-augsburg.de
Dik T. Winter wrote:
> In article <1116939502.8...@o13g2000cwo.googlegroups.com> muec...@rz.fh-augsburg.de writes:
> > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
> > = 1
>
> You may note that that is *not* an infinite sum...
It is defined as an infinite tree.
>
> > .
> > 0 1
> > 0 1 0 1
> > ..................
> ...
> Also we find that up to line n, summing up to that node along the path
> gives a value for k/(2^n) for some integer k. Therefore a number along
> a node is always equal to a rational with a denominator that is a power
> of two. It is simply impossible to assume that one of these numbers
> becomes 1/3.
What is the difference to an infinite decimal expansion? A number
0.333... is always equal to a rational with a denominator that is a
power of ten. It is simply impossible to assume that this number
becomes 1/3. Or is there any occult advantage of the decimal system
over the binary system?
Regards, WM
muec...@rz.fh-augsburg.de
Virgil wrote:
> In article <1116939502.8...@o13g2000cwo.googlegroups.com>,
> muec...@rz.fh-augsburg.de wrote:
>
> > Cantor and the binary tree.
> >
> > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
> > = 1
> >
> > then all the real numbers of the interval [0,1] are realized as
> > infinite paths of the binary tree
> > .
> > 0 1
> > 0 1 0 1
> > ..................
> >
> > too (read from top to bottom). Each number is given by a path
> > stretching over infinitely many nodes (bits). All nodes (bits) of the
> > tree are countable. The paths are not, according to Cantor's famous
> > diagonal proof.
>
> So far so good.
> One does not assume it, one proves it, as I have done several times.
>
> Here is an outline of that proof:
>
> Every node is represented by a terminating binary (starting at "." and
> terminating at the node itself in the tree above) which is like a subset
> of the rationals, which are countable.
>
> Every unending path is represented by a non-terminating binary (also
> starting at "." but never ending), which surject onto the real interval
> [0,1], and are thus as uncountable as the reals.
>
> That WM choses to reject proofs that show him wrong does not invalidate
> such proofs.
Cantor's proof is questioned, hence I do not accept it as an argument.
Please find out at which step my arguing fails. As you seem to accept
the first points, you may start at point 3.
1) Each number of (0,1) is given by a path stretching over infinitely
many nodes (bits).
2) All nodes (bits) of the tree belong to a countable se.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
Regards, WM
muec...@rz.fh-augsburg.de
Ron Sperber wrote:
line number n
0 0.
1 0 1
2 0 1 0 1
... ..................
> It simply boggles my mind that this simple proof gives so many people
> such fits that they refuse to accept it. I continue to be sadly shocked
> by the number of posts on sci.math daily refuting Cantor's proof. Of
> course they are always fuzzy on details, but that's to be expected since
> they can't actually disprove it.
You are talking about Cantor's proof. You could say the same about
mine. I do not say, here, that Canor's proof is wrong. All I say, here,
is that another proof leads to another result. The reason is that set
theory is inconsistent. If you do not agree, then point out that step
of my proof, which is invalid.
1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1, from 1 coming in to 2
muec...@rz.fh-augsburg.de
Are you capable to follow a five lines proof without referring to "Big
Brother" Cantor? Consistency of set theory is questioned, hence I do
not accept Cantor's proof as an argument.
line number n
0 0.
1 0 1
2 0 1 0 1
... ..................
1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
Please point out which step is wrong.
Regards, WM
muec...@rz.fh-augsburg.de
Virgil wrote:
> > If no branching is possible without a new node (because a branching is
> > a node), then "the infinite" does not at all help you. *That* is very
> > easy to prove.
>
> How is it that all those things that WM declares so easy to prove still
> remain unproven?
It is very easily proven that a branching is impossible without a node,
because, by definition, a branching _is_ a node.
Regards, WM
muec...@rz.fh-augsburg.de
The tree is not a list. Therefore I choose it. Nevertheless it has to
say much about infinite sets and Cantor's theorem, namely that the set
of nodes and the set of paths are equivalent sets. The set of nodes is
countable and the set of paths is equivalent to the set of reals in
(0,1). That's quite a lot of information, isn't it?
Regards, WM
muec...@rz.fh-augsburg.de
Virgil wrote:
> > If you prefer "is", you may use it. That does not matter. It is
> > obviously impossible that the set of paths is uncountable when the set
> > of nodes is countable, because every pair of paths springs off from one
> > node, while one path leads to that node. Try to find an error n the
> > arguing, not in the result.
>
> For finite paths
There are no finite paths in my tree. Terminating rationals are
completed by strings of zeros. For that sake enough zeros are in the
tree.
>
> The set of these unbounded paths can mapped bijectively to P(N), whereas
> the set of nodes can be mapped bijectively to N.
Consistency of set theory is questioned, hence I do not accept Cantor's
proof as an argument.
line number n
0 0.
1 0 1
2 0 1 0 1
... ..................
1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
Please point out which of these simple steps is (are) wrong.
Regards, WM
muec...@rz.fh-augsburg.de
Martin Shobe wrote:
> >If you prefer "is", you may use it. That does not matter. It is
> >obviously impossible that the set of paths is uncountable when the set
> >of nodes is countable, because every pair of paths springs off from one
> >node, while one path leads to that node. Try to find an error n the
> >arguing, not in the result.
>
> Sure thing. The error is that "It is simply impossible to assume that
> one of these numbers becomes uncountably infinite while the other
> remains countably infinite" does not follow from the previous
> statements in your proof.
Then point out, please, which step is wrong.
1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
The relationships between nodes and paths
> you supplied require that the tree be finite.
If you assert that, you should give a number n as an upper bound. Would
you assert that
0.
1
1
1
...
must be finite?
Would you assert that
0.
0 1
0 1
0 1
...
must be finite?
Which branching must terminate the tree? How many nodes are admitted?
Attempting to use those
> relationships on infinite trees (without first proving that the
> relationships hold) is simply a fallicy.
To prove that a branching is a node is easy, because it is defined so.
More is not necessary.
Regards, WM
muec...@rz.fh-augsburg.de
Dik T. Winter wrote:
> In article <1116958479.5...@g47g2000cwa.googlegroups.com> muec...@rz.fh-augsburg.de writes:
> > Dik T. Winter wrote:
> > > In article <1116939502.8...@o13g2000cwo.googlegroups.com>
> > muec...@rz.fh-augsburg.de writes:
> > > > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n =
> > > > 0.111... = 1
> > >
> > > You may note that that is *not* an infinite sum...
>
> Note what I said here. It is not a sum of infinitely many terms.
>
> > > > .
> > > > 0 1
> > > > 0 1 0 1
> > > > ..................
> >
> > Any path is an infinite sequence of bits which by multiplying with 2^-n
> > and summing up establishes an infinite series representing a real
> > number. Every combination of countably many bits is realized by
> > definition.
>
> Mathematics does *not* define summing up infinitely many terms. It uses
> limits in this case.
Does mathematics allow me to write
0.
1
1
1
...
?
> > Why should 0.010101... not exist in that tree? Every path is infinite
> > by definition as is 0.010101..., by definition.
>
> The path does exist, but there is no node at the end. Or if you wish
> there is a node at the and (as J.H. Conway does with his surreal numbers),
> the number of nodes is uncountable.
There can be no node at the end, because there is no end. We do not
need any node at the end in order to show that the set of nodes is
equivalent to that of paths. It is shown by the following steps. Please
point out which step is wrong.
1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
Regards, WM
Ross A. Finlayson
It's interesting that you note that there is one more path than node,
where the number of nodes is considered to be countable and paths
uncountable.
In the model of ubiquitous ordinals, the powerset is order type is
successor and f(x)=x+1 is the mapping between set and powerset for
which the missing element is the empty set.
The vacuity of the empty set is misleading. The empty set is an
element of any powerset by the usual definition, and regardless of
whether that particular set thus vacuously or trivially satisfies a
given expression of a regularity axiom, that it otherwise does not is
not obvious.
It is basically about that one element, the successor, that induction
guarantees to exist, towards the inductive impasse, or from it.
About set theory, and its consistency, there's a difference between
saying quantification over sets implies a universal set and thus the
regularity axiom of ZF is inconsistent and saying any set theory is
inconsistent.
As far as the tree construction goes, you might notice that the binary
representation of the tree that would represent an integer is as well
infinite, except that some of the 1 nodes would be leaf nodes, or
rather, each node has four child nodes, two each of one and zero where
one of each of those is a leaf and the other not.
About having a pair of nodes for each zero or one decision, with a leaf
node indicating the implicit infinitely trailing zeros (or ones) of a
rational that has a denominator that is a power of two or one of the
integers, it is basically the same thing as an internal node with
implicitly following afterwards the right or left branch of each
descendent in the balanced binary tree.
The tree to represent natural integers via integral moduli is still no
different than the treeto represent the reals via integral moduli .
While that may be so, it might be more convenient to have a distinction
between the leaf and non-leaf nodes to indicate where the path ends.
I think that's an interesting way to consider the construction of the
real numbers, as binary or decimal or equivalently (in the sense of
definitions) Cauchy/Dedekind or continued fractions, but as a point-set
there are also some considerations of the continuity of the real number
line that transcend (in the sense of transcendental numbers) those
definitions. That deserves more explanation.
I am glad to see you move from your ultrafinitist viewpoint to instead
one that acknowledges the infinite. However, if you proceed that way,
then your viewpoint will make obsolete your papers, which I admit to
you I don't think are very strong. Please be careful and take your
time on the argument to not expose weaknesses that are readily
exploited. Furthermore I wish you good luck and recommend my words on
these matters.
These infinite things are not finite: they're unending, there's always
one more.
Ross F.
Dik T. Winter
There is no difference. It is *your* reasoning where you draw conclusions
for the infinite tree from properties of the finite trees. My mimicking
it simply shows that such reasoning is not necessarily valid and should
be proven.
Dik T. Winter
> Dik T. Winter wrote:
> > In article <1116958479.5...@g47g2000cwa.googlegroups.com> muec...@rz.fh-augsburg.de writes:
> > > Dik T. Winter wrote:
> > > > In article <1116939502.8...@o13g2000cwo.googlegroups.com>
> > > muec...@rz.fh-augsburg.de writes:
> > > > > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n =
> > > > > 0.111... = 1
> > > >
> > > > You may note that that is *not* an infinite sum...
> >
> > Note what I said here. It is not a sum of infinitely many terms.
> > > Any path is an infinite sequence of bits which by multiplying with 2^-n
> > > and summing up establishes an infinite series representing a real
> > > number. Every combination of countably many bits is realized by
> > > definition.
> >
> > Mathematics does *not* define summing up infinitely many terms. It uses
> > limits in this case.
>
> Does mathematics allow me to write
Mathematics allows you a lot of things. Even saying 2 = 0. But it
does *not* definine a lot of things, like adding infinitely many terms.
> > > Why should 0.010101... not exist in that tree? Every path is infinite
> > > by definition as is 0.010101..., by definition.
> >
> > The path does exist, but there is no node at the end. Or if you wish
> > there is a node at the and (as J.H. Conway does with his surreal numbers),
> > the number of nodes is uncountable.
>
> There can be no node at the end, because there is no end. We do not
> need any node at the end in order to show that the set of nodes is
> equivalent to that of paths. It is shown by the following steps. Please
> point out which step is wrong.
>
> 1) Each real number of (0,1) is given by a path stretching over
> infinitely many nodes (bits).
Yup.
> 2) All nodes (bits) of the tree belong to a countable set.
Yup.
> 3) A node can only exist within a path.
Yup.
> 4) Any node increases the number of paths by 1 from 1 coming in, to 2
> going out. 2 - 1 = 1.
Eh? Now you are talking about an ever increasing finite tree, not about
an infinite tree. If you are talking about an infinite tree it may be
allowable, but because the number of paths is infinite, so it stays the
same when you add 1 to it.
> 5) Any node increases the number of nodes by 1.
Same remark here.
So what does that prove about infinite trees?
Robert Kolker
>
> Robert Kolker wrote:
>
>>Tony Orlow (aeo6) wrote:
>>
>>>That's all very well and good, if you specify f anf g and figure those
>>>functions into your comparison. It's a mistake to ignore them.
>>
>>Are you capable of following a proof? Even a three line proof?
>
>
> Are you capable to follow a five lines proof without referring to "Big
> Brother" Cantor? Consistency of set theory is questioned, hence I do
> not accept Cantor's proof as an argument.
Bullshit! I used only the definition of bijection. A set is countably
infinite if and only if it can be mapped onto the set of integers by a
bijection. That is what countably infinite -means-! The rest was trivial
inversion of one of the bijections and composition of bijections which
is a bijection.
Bob Kolker
Tony Orlow
mapping functions and then disregarding them at infinity. Are you capable of
critically considering your thought process?
--
Smiles,
Tony
Tony Orlow
>
>
> David C. Ullrich wrote:
> > On 24 May 2005 05:58:22 -0700, muec...@rz.fh-augsburg.de
> > wrote:
> > [...]
> > Hint: infinite sets are not the same as finite sets.
>
> Yes. Intuition goes out the window with infinity, which is why we do
> mathematics instead. 8-)
with your math.
>
> One (easy?) example of how infinite sets are different from finite sets
> is that if you have an infinite set S, you can put it in 1-1
> correspondence with a proper subset T (a subset which is not equal to
> S).
infinity, and are a poor tool for dealing with it.
>
> For instance, let S = N, T = {2, 3, 4,...}. One possible 1-1
> correspondence is
> f(n) = n + 1; that is, match up
> 1 <--> 2
> 2 <--> 3
> 3 <--> 4
> ...
intuitions at x=N, since taking away the first element of the naturals would
intuitively leave you with the naturals, minus 1. This is what I mean by taking
the mapping functions into account.
>
> This uses up all the elements of S (on the left-hand side) and T (on
> the right-hand side).
>
> This is impossible to do if S is finite.
what you are doing halfway through your calculation
>
> --- Christopher Heckman
>
>
--
Smiles,
Tony
Tony Orlow
are diagonal, and derive the non-results that you get, if you like. You can
believe that the even half of the naturals is as large as the entire set, or
that a sparse set like the naturals is the same size as a dense set like the
rationals. That's your choice. I am well aware of how cardinality "works", so
don't assume ignorance on my part. In my analysis, spurred by revulsion on the
part of my intuition, there are several incosistencies in this method, not the
least of which is the application of mapping functions without regard to their
relationship to relative set size. So, your proof using bijection isn't
considered a proof by me.
--
Smiles,
Tony
Tony Orlow
little axiom cave whenever questioned. He has demonstrated im more ways and
with more patience than you deserve, that there is a basic contradiction in
your assumptions of infinite sets of finite naturals and binary tree with an
infinite number of nodes but only finite branches, or infinite branches with
only elements that are a finite distance from the root. Those kinds of
illogical conjectures should easily be dispensed with. Unfortunately, it's like
trying to teach a rabid dog to swim.
--
Smiles,
Tony
Tony Orlow
"list"? Hmmmmm. Try again.
--
Smiles,
Tony
Robert Kolker
>
> Quite, thank you. I have pointed out innumerable times the problem with using
> mapping functions and then disregarding them at infinity. Are you capable of
> critically considering your thought process?
The mappings were defined for the -entirety- of the sets. There is not
"at infinity". There is no limiting process in the context of proof.
Yoda says: There is no "at infinity". That is why you fail, Tony Orlow.
Bob Kolker
Tony Orlow
--
Smiles,
Tony
Tony Orlow
> In article <1116959066.6...@o13g2000cwo.googlegroups.com>,
> muec...@rz.fh-augsburg.de wrote:
>
> > Robin Chapman wrote:
> >
> >
> > > >> of paths always equals that of the nodes + 1. It is simply
> > impossible
> > > >> to assume that one of these numbers becomes uncountably infinite
> > while
> > > >> the other remains countably infinite.
> > >
> > > "becomes"?
> >
> > If you prefer "is", you may use it. That does not matter. It is
> > obviously impossible that the set of paths is uncountable when the set
> > of nodes is countable, because every pair of paths springs off from one
> > node, while one path leads to that node. Try to find an error n the
> > arguing, not in the result.
>
> For finite paths (having a root node and a leaf or terminal node) one is
> there can only be countably many, but it is those paths starting at the
> root which procede through infinitely many nodes in this infinite tree,
> and never have a terminal or leaf node, that are under consideration.
>
> The set of these unbounded paths can mapped bijectively to P(N), whereas
> the set of nodes can be mapped bijectively to N.
Then bijections are foolish at infinity. Regardless of finitude or lack
thereof, the number of paths NEVER surpasses the number of nodes. It is simply
not possible, despite your pseudo-logical system.
> >
> > > One can hardly imagine a simpler mathematical proof. Alas, it's still
> > > beyond the limits of Chaps
>
> As all such false proofs should be.
>
--
Smiles,
Tony
Tony Orlow
pertinent, and it is a mystery to me that those with the Cantorian disease
cannot dispel their demon long enough to grasp it. I see exactly what you are
saying, and it's correct.
--
Smiles,
Tony
Tony Orlow
reasonable to conclude that adding 1 to an infinite set creates an infinite,
but finitely larger, set. oo+1<>oo, if by oo we mean the same infinite number
on both sides of the inequality. This colllapsing of functions at infinity is
an invalid method based on unfounded assumptions. WM is correct in generalizing
the realtionships between numbers of nodes, branches and paths. Thos
relationships are maintained at infinity. I am sure WM can form a nice
inductive proof to this effect.
>
> > 5) Any node increases the number of nodes by 1.
>
> Same remark here.
>
> So what does that prove about infinite trees?
>
--
Smiles,
Tony
Virgil
muec...@rz.fh-augsburg.de wrote:
A number
> 0.333... is always equal to a rational with a denominator that is a
> power of ten. It is simply impossible to assume that this number
> becomes 1/3. Or is there any occult advantage of the decimal system
> over the binary system?
That "number" does not have to BECOME 1/3, it already is 1/3.
If one regards 0.333... not as a number but only as a sequence of
partial sums, it is true that none of those partial sums is exactly 1/3,
but the mathematical standard for interpreting a repeating decimal, like
0.333..., is that it represents that NUMBER which is the limit of that
sequence of partial sums. And that number, by every reasonable analysis,
is exactly and precisely 1/3.
WM must be off his meds again.
Virgil
muec...@rz.fh-augsburg.de wrote:
WM conflates bounded paths, having terminal of leaf nodes with unbounded
unending paths which have no terminal or leaf nodes.
1) Each number of (0,1) is given by an unending path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ending paths by 1 from 1 coming in,
to 2> going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.
All unending paths in an unending binary tree contain infinitely many
nodes.
The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end).
Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.
There are moreof the non-terminating than of the terminating.
So
Robert Kolker
>
>
> That "number" does not have to BECOME 1/3, it already is 1/3.
The temporal fallacy is one of the biggest imediments to understanding
limits.
Bob Kolker
Virgil
muec...@rz.fh-augsburg.de wrote:
> Ron Sperber wrote:
>
> line number n
> 0 0.
> 1 0 1
> 2 0 1 0 1
> ... ..................
>
> > It simply boggles my mind that this simple proof gives so many people
> > such fits that they refuse to accept it. I continue to be sadly shocked
> > by the number of posts on sci.math daily refuting Cantor's proof. Of
> > course they are always fuzzy on details, but that's to be expected since
> > they can't actually disprove it.
>
> You are talking about Cantor's proof. You could say the same about
> mine. I do not say, here, that Canor's proof is wrong. All I say, here,
> is that another proof leads to another result.
Not if one analyses it correctly.
Each leaf node (end node of a finite path) corresponds to a terminating
binary proper fraction.
Each unending path, having no leaf node, corresponds to a
non-terminating binary proper fraction.
There is no one-to-one correspondence between the set of terminating and
the set of non-teminating, despite WM's weaselings.
Robert Kolker
>
> Because you people refuse to listen to basic logic, and instead retreat to your
> little axiom cave whenever questioned.
THe little axiom cave is consistent. Furthermore it has produced the
mathematics on which physics is based, which in turn has produced the
engineer art which has produced the computer with which you habitually
spray your nonsense over UseNet. Nothing succeeds like success. And
nothing fails like lack of production. Use the Force!
What have your produced either mathematically or in physics with your
Vast Insights?
Bob Kolker
Virgil
muec...@rz.fh-augsburg.de wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1, but have absolutely
nothing to do with the number of unending paths.
All unending paths in an unending binary tree contain infinitely many
nodes.
The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end).
Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.
There are moreof the non-terminating than of the terminating.
So WM is wrong yet again.
Virgil
muec...@rz.fh-augsburg.de wrote:
I am not impressed by declarations that something is easy to prove.
I am only impressed by presentation of the proof itself.
Virgil
muec...@rz.fh-augsburg.de wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
Virgil
muec...@rz.fh-augsburg.de wrote:
> Virgil wrote:
> > In article <1116958479.5...@g47g2000cwa.googlegroups.com>,
> > muec...@rz.fh-augsburg.de wrote:
> >
> >
> > > >
> > > > > .
> > > > > 0 1
> > > > > 0 1 0 1
> > > > > ..................
> > >
> > > Any path is an infinite sequence of bits which by multiplying with 2^-n
> > > and summing up establishes an infinite series representing a real
> > > number. Every combination of countably many bits is realized by
> > > definition.
> >
> > But such an "infinite" binary tree is not a list, so this has nothing to
> > say about the validity of Cantor's theorem.
>
> The tree is not a list. Therefore I choose it. Nevertheless it has to
> say much about infinite sets and Cantor's theorem, namely that the set
> of nodes and the set of paths are equivalent sets.
The set of leaf nodes and the set of finite or terminating paths are
equivalent, but infinite or non-terminating paths have no leaf nodes,
so that there is no equivalence.
> The set of nodes is
> countable and the set of paths is equivalent to the set of reals in
> (0,1). That's quite a lot of information, isn't it?
GIGO!
Virgil
muec...@rz.fh-augsburg.de wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
Virgil
muec...@rz.fh-augsburg.de wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> Robert Kolker said:
> > Tony Orlow (aeo6) wrote:
> > >
> > > That's all very well and good, if you specify f anf g and figure those
> > > functions into your comparison. It's a mistake to ignore them.
> >
> > Are you capable of following a proof? Even a three line proof?
> >
> > Bob Kolker
> >
> >
> Quite, thank you. I have pointed out innumerable times the problem with using
> mapping functions and then disregarding them at infinity.
Since none of those mapping functions ever REACH infinity, there is
nothing to disregard.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> Proginoskes said:
> >
> >
> > David C. Ullrich wrote:
> > > On 24 May 2005 05:58:22 -0700, muec...@rz.fh-augsburg.de
> > > wrote:
> > > [...]
> > > Hint: infinite sets are not the same as finite sets.
> >
> > Yes. Intuition goes out the window with infinity, which is why we do
> > mathematics instead. 8-)
> If your math throws your intuition out the window, perhaps there's a problem
> with your math.
There definitely is a problem with TO's intuition. :-)
> >
> > One (easy?) example of how infinite sets are different from finite sets
> > is that if you have an infinite set S, you can put it in 1-1
> > correspondence with a proper subset T (a subset which is not equal to
> > S).
> Yes, well, that pretty much just shows that bijections go out the window at
> infinity, and are a poor tool for dealing with it.
Only to one who relies on intuition despite any amount of evidence that
his intuition is flawed. ;-)
> >
> > For instance, let S = N, T = {2, 3, 4,...}. One possible 1-1
> > correspondence is
> > f(n) = n + 1; that is, match up
> > 1 <--> 2
> > 2 <--> 3
> > 3 <--> 4
> > ...
> Yes, and it is no coincidence that the inverse function, x-1, fits
> our intuitions at x=N, since taking away the first element of the
> naturals would intuitively leave you with the naturals, minus 1. This
> is what I mean by taking the mapping functions into account.
Which is, as required, a proper subset of N.
So that N one of the sets which allows bijection to a proper subset.
We could, of course, replace each instance of "infinite" with "allows
bijection to a proper subset". but abbreviation is what definitions are
for.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> Virgil said:
> > In article <1116958024.0...@g43g2000cwa.googlegroups.com>,
> > muec...@rz.fh-augsburg.de wrote:
> >
> > > David C. Ullrich wrote:
> >
> > > > Hint: infinite sets are not the same as finite sets. What you say
> > > > is "impossible to assume" is not something we _assume_, but it's
> > > > true. And very easy to prove.
> > >
> > > I can imagine that you are ready to prove that in 1-2-3-4-... there are
> > > more dashes than numbers. But it is nevertheless inacceptable.
> > >
> > > In my tree the question is not whether it is an infinite set or not,
> > > but whether every node is member of a line with finite enumeration.
> > >
> > > If no branching is possible without a new node (because a branching is
> > > a node), then "the infinite" does not at all help you. *That* is very
> > > easy to prove.
> >
> > How is it that all those things that WM declares so easy to prove still
> > remain unproven?
> >
> Because you people refuse to listen to basic logic,
On the contrary. Basic logic proceeds from assumptions to conclusions.
We merely formalize our assumptions into axiom systems.
> and instead retreat to
> your
> little axiom cave whenever questioned.
With an axiom system, one knows what one is dealing with. Without one,
no one knows what is going on. Mathematicians have learned this lesson
the hard way. It is a lesson they will not easily forget, and a
principle that they will not readily abandon. Especially for such a
mish-mash of conflicting intuitions as TO wishes to heap on them.
> He has demonstrated im more ways and
> with more patience than you deserve
What may satisfy non-mathematicians as adequate demonstrations of what
they already believe will not always satisfy more skeptical
mathematicians of what they do not believe.
When one preaches to the choir, one does not convert many skeptics.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
Such a "traverse is still not a list.
One can list rationals, and it has been done many times, but it is more
common to inject the rationals into the naturals to show that
Card(Q) <= Card(N) and then inject the naturals into the rationals
showing Card(N) <= Card(Q), allowing us to conclude Card(Q) = Card(N)
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1, but have absolutely
nothing to do with the number of unending paths.
All unending paths in an unending binary tree contain infinitely many
nodes.
The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end).
Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.
There are moreof the non-terminating than of the terminating.
So WM is wrong yet again.
And TO too!
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
The pseudo-logic is all TO's. He is conflating terminating paths with
non-terminating paths.
The number of paths having leaf nodes (terminal nodes) equals the number
of leaf nodes in any tree, but in a tree where there are unbounded paths
(which do not have any leaf nodes), counting leaves does not count paths.
.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1, but have absolutely
nothing to do with the number of unending paths.
All unending paths in an unending binary tree contain infinitely many
nodes.
The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end with a leaf
node).
Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.
There are more non-terminating paths than of terminating paths in a
complete unbounded binary tree in which no path has a leaf node.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
WM's "proof" disproved"
WM conflates bounded paths, having terminal or leaf nodes with unbounded
unending paths which have no terminal or leaf nodes, but contain
infinitely many intermediate nodes.
1) Each number of (0,1) is given by an UNENDING path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of ENDING paths, having terminal or
leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1, but have absolutely
nothing to do with the number of unending paths.
All unending paths in an unending binary tree contain infinitely many
nodes.
The number of leaf nodes exactly equals the number of ending or finite
paths in any finite binary tree (in which all paths end with a leaf
node).
Considering the binary tree whose root is "." and each branch is
indicated by a "0" or a "1", each leaf node, and therefore each path, is
represented by a terminating binary fraction , but each unending path is
represented by a non-terminating binary fraction.
There are more non-terminating paths than of terminating paths in a
complete unbounded binary tree in which no path has a leaf node.
So WM is wrong yet again. And TO too!
Tony Orlow
>
> Bob Kolker
>
>
You don't even understand what I am saying, so why bother disagreeing?
Says Yoda: Your breath hold you not when water your lungs are full of.
--
Smiles,
Tony
Tony Orlow
you have to perform all your partial sums? Isn't a limit something that never
gets there? Can you just "jump" to infinity, and declare that infinite set of
partial sums equal to some fraction? It almost sounds like you're coming to
your senses! It's about time!
--
Smiles,
Tony
Tony Orlow
a path. Actually, you need to add two nodes for each path, but that is
irrelevant. They are finitely proportional, so if one is finite, the other is,
and if one is infinite, so is the other. Infinite trees have infinite nodes and
paths. Finite tree only have finite numbers of nodes and paths. It's a fact.
--
Smiles,
Tony
Tony Orlow
in finite vs. infinite "time". It's just a metaphor for measure in general.
--
Smiles,
Tony
Tony Orlow
tree there are almost surely more infinite paths than finite ones, but that is
irrelevant. WM is talking about paths vs nodes and branches, because you are
making some bizarre unfounded claim about the paths being uncountable and the
nodes countable, when there are half as many paths as nodes. This kind of
result is what brings upon the Cantorians all their woes in the form of us
"idiots".
--
Smiles,
Tony
Tony Orlow
Only a precise and intuitively unobjectionable generalization of set size
measure from finite to infinite sets. That's all. Cardinality has nothing to do
with physics, though other areas of set theory are essential. Cardinality
measure is the target here, Bob. I am not Albert, attacking things just for
being axiomatic. Axioms are excellent logical atoms, if they are justified. In
your cave, that pile of bones goes rather nicely with the drips from the
ceiling, but you should get out more often, and find some new ideas for your
decor.
>
> Bob Kolker
>
--
Smiles,
Tony
Robert Kolker
>
>
> Only a precise and intuitively unobjectionable generalization of set size
> measure from finite to infinite sets. That's all.
You have proved no theorems, so you have no applications to anything
beyond your strange definitions. Do you know how to prove a theorem?
Bob Kolker
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> Virgil said:
> > In article <1117019416....@o13g2000cwo.googlegroups.com>,
> > muec...@rz.fh-augsburg.de wrote:
> >
> > A number
> > > 0.333... is always equal to a rational with a denominator that is a
> > > power of ten. It is simply impossible to assume that this number
> > > becomes 1/3. Or is there any occult advantage of the decimal system
> > > over the binary system?
> >
> > That "number" does not have to BECOME 1/3, it already is 1/3.
> >
> > If one regards 0.333... not as a number but only as a sequence of
> > partial sums, it is true that none of those partial sums is exactly 1/3,
> > but the mathematical standard for interpreting a repeating decimal, like
> > 0.333..., is that it represents that NUMBER which is the limit of that
> > sequence of partial sums. And that number, by every reasonable analysis,
> > is exactly and precisely 1/3.
> >
> > WM must be off his meds again.
> >
> But, Virgil, how do you know that, when you can never get to infinity?
Infinity is not a place, it is meerely a lack of any 'finite' boundary.
> Don't you have to perform all your partial sums?
Not if you can prove a limit exists, or doesn't exist, without doing so.
> Isn't a limit something that never
> gets there?
Wrong way round. The limit doesn't 'get' anywhere.
The sequence of partial sums, if convergent, 'gets' closer to that limit.
> Can you just "jump" to infinity, and declare that infinite set of
> partial sums equal to some fraction?
No! And it is nonsense to speak of an infinite set of values being equal
to some number.
One definition says a number, L, is the limit of an infinite sequence if
and only if all but finitely many values in that sequence are within any
pre-assigned positive distance of L.
More formally: L is the limit of f: N -> R if and only if
for any given positive real epsilon ,
the set {n e N: |f(n) -L| > epsilon} is finite.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> > There are more of the non-terminating than of the terminating.
> >
> > So
> >
> So what? As you pointed out, when you add a node, you add a node, a branch,
> and
> a path. Actually, you need to add two nodes for each path, but that is
> irrelevant. They are finitely proportional, so if one is finite, the other
> is,
> and if one is infinite, so is the other. Infinite trees have infinite nodes
> and
> paths. Finite tree only have finite numbers of nodes and paths. It's a fact.
But it is also a fact that the set of nodes in an infinite binary tree
bijects to the set of terminating binary proper fractions, and to N,
while that set of unending paths in that same tree for which no such
bijection exists.
TO seems to have this desire to ignore what he regards as unpleasant.
Mathematics does not allow that luxury.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
When one has a finite binary tree every maximal path ends in a leaf node.
When one has an infinite binary tree, ther is no such thing as a leaf
node, and every maximal path is unending.
So why should the bijective correpondence between leaf nodes and finite
paths in finite trees have anything to do with trees in which there are
NO leaf nodes?
That makes no more sense than to argue that there must be gaps between
real numbers because there are gaps between natural numbers.
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
Our caves are a considerable step up from the trees TO hangs from.
Alan Morgan
In a finite tree there are half as many paths as nodes. I totally agree
with this statement (okay, (nodes+1)/2. Big deal). I think you'll get
general agreement with this statement.
In an infinite tree there are as many *finite* paths as nodes. Again,
I think you can count on general agreement (although there will be
some grumbling that there are infinite numbers of both and I may be
off by a factor of 2 but let's forget about that).
I await, with some interest, your proof that, in an infinte tree, there
are half as many paths (or just as many paths or twice as many paths or
whatever) *of* *all* *kinds* as there are nodes.
When WM was talking about adding a node being the same as adding a path
he was, although he didn't know it, talking about *finite* paths. Adding
a node *is* the same as adding a *finite* path. The situation with
infinite paths is much messier.
>This kind of
>result is what brings upon the Cantorians all their woes in the form of us
>"idiots".
No, the source of the woes is anti-Cantorians insistence that infinite and
finite are pretty much the same, except that infinite is larger. Anti-
cantorians think that the integers include infinitely large values.
Cantorians think that the integers include arbitrarily large values, but
know that arbitrarily large is not infinite. Anti-cantorians think that
infinite paths are like finite paths, but longer. Cantorians actually
puzzle out what in the heck is meant by an infinitely long path and
conclude that it behaves completely differently from finite ones.
Alan
--
Defendit numerus
Ron Sperber
> Ron Sperber said:
>
>>Tony Orlow (aeo6) wrote:
>>
>>>
>>>
>>>>Robin Chapman wrote:
>>>>
>>>>
>>>>>Robert Kolker wrote:
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>>muec...@rz.fh-augsburg.de wrote:
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>>of paths always equals that of the nodes + 1. It is simply impossible
>>>>>>>to assume that one of these numbers becomes uncountably infinite while
>>>>>>>the other remains countably infinite.
>>>>>
>>>>>
>>>>>The fact is that the nodes in this tree form a countable
>>>>>set and the paths form an uncountable set. "becoming"
>>>>>has nowt to do with that.
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>>Wrong. 2^(aleph_0) > aleph_0.
>>>>>>
>>>>>>List all the infinite binary sequences with a bijection to the integers.
>>>>>>Now flip the n-th digit of the n-th sequence in the list. This cannot
>>>>>>occur anywhere in the list. Contradiction. Such a bijection to the
>>>>>>integers does not exist.
>>>>>
>>>>>
>>>>>One can hardly imagine a simpler mathematical proof. Alas, it's still
>>>>>
>>>>
>>>>It simply boggles my mind that this simple proof gives so many people
>>>>such fits that they refuse to accept it. I continue to be sadly shocked
>>>>by the number of posts on sci.math daily refuting Cantor's proof. Of
>>>>course they are always fuzzy on details, but that's to be expected since
>>>>they can't actually disprove it.
>>>>
>>>
>>>to be referring to the "proof" of uncountability of the reals. it really only
>>>proves that there are more reals than naturals, and that one can generate more
>>>strings than the number of symbols each string contains, provided you have a
>>>set of more than one symbol to choose from. By traversing the list diagonally
>>>and assuming you have covered the list, you are assuming that it is square in
>>>the sense that there are as many digits in each number as numbers in the list.
>>>But, digital number systems don't work that way. If you have a number system of
>>>base S, L digits will allow you to represent S^L numbers, so if you have N
>>>digits, you will have 10^N numbers in decimal, 2^N numbers in binary, in your
>>>list. It is infinitely longer than it is wide, and therefore cannot be
>>>completely traversed diagonally. The number generated in the antidiagonal is
>>>simply one of the 2^N-N numbers below the diagonal of traversal.
>>>
>>>The subsequent conclusion that the reals are not "countable" rests on the
>>>notion that all countably infinite sets are the same size, which is an
>>>assumption that I reject for many reasons, and which has no justification
>>>besides "oo=oo=oo".
>>>
>>>Sure, the proof looks simple. It's a little too simple, and the critical
>>>thought aimed at it is too.
>>
>>your refutations are nonsense. Being countable (or to be more precise
>>here, countably infinite) has a very precise definition. That is a set X
>>is countably infinite if there exists a bijection f:N->X where N is the
>>set of natural numbers.
>>
>>Here is a theorem that can be proved from this definiton. Let X,Y be
>>countably infinite sets. Then there is a bijection h:X->Y.
>>
>>Proof.
>>
>> Let f:N->X and g:N->Y be bijections that show that X and Y are
>>countably infinite. Since f and g are bijections there are bijections
>>f^-1:X->N, g^-1:Y->N.
>>
>>Then gf^-1:X->Y and fg^-1:Y->X are functions. and since gf^-1fg^-1:Y->Y
>>is the identity and fg^-1gf^-1:X->X is the identity gf^-1 and fg^-1 are
>>bijections, thus h=gf^-1 works.
>>
>>Thus your nonsense about different sized countable sets is just
>>that...nonsense. any 2 countably infinite sets have a bijection between
>>them which is the only useful measure of "size" for infinite sets.
>>
>
> That's all very nice, and you can draw your lines and ignore the fact they they
> are diagonal, and derive the non-results that you get, if you like. You can
> believe that the even half of the naturals is as large as the entire set, or
> that a sparse set like the naturals is the same size as a dense set like the
> rationals. That's your choice. I am well aware of how cardinality "works", so
> don't assume ignorance on my part. In my analysis, spurred by revulsion on the
> part of my intuition, there are several incosistencies in this method, not the
> least of which is the application of mapping functions without regard to their
> relationship to relative set size. So, your proof using bijection isn't
> considered a proof by me.
Ah, so basically what it boils down to is that you want everyone to
accept your own bizarre definition of "size" of a set. Well, have fun then.
Proginoskes
Tony Orlow (aeo6) wrote:
> Proginoskes said:
> >
> >
> > David C. Ullrich wrote:
> > > On 24 May 2005 05:58:22 -0700, muec...@rz.fh-augsburg.de
> > > wrote:
> > > [...]
> > > Hint: infinite sets are not the same as finite
> > > sets.
> >
> > Yes. Intuition goes out the window with infinity, which is
> > why we do mathematics instead. 8-)
>
> If your math throws your intuition out the window, perhaps
> there's a problem with your math.
It's not the math that throws the intuition out, it's infinity.
> > One (easy?) example of how infinite sets are different
> > from finite sets is that if you have an infinite set S,
> > you can put it in 1-1 correspondence with a proper
> > subset T (a subset which is not equal to S).
> Yes, well, that pretty much just shows that bijections go out the window at
> infinity, and are a poor tool for dealing with it.
But without bijections, how do you show that {3,4} and {1,2} (two
finite sets) have the same size? If you throw out bijections of
infinite sets, you also throw out bijections of finite sets, since
bijections can be used for both.
> > For instance, let S = N, T = {2, 3, 4,...}. One
> > possible 1-1 correspondence is
> > f(n) = n + 1; that is, match up
> > 1 <--> 2
> > 2 <--> 3
> > 3 <--> 4
> > ...
> Yes, and it is no coincidence that the inverse function,
> x-1, fits our intuitions at x=N,
What do you mean, the element x of N which is equal to N? N is not in
N; N contains numbers, not sets.
> since taking away the first element of the naturals would
> intuitively leave you with the naturals, minus 1. This is
> what I mean by taking the mapping functions into account.
> > This uses up all the elements of S (on the left-hand side) and T (on
> > the right-hand side).
> >
> > This is impossible to do if S is finite.
> It is possible to do for finite sets,
Okay. Let S = {1,2,3} and T = {1,2}; then T is a proper subset of S.
Give me a bijection from S to T.
> but it is not wise to do that and forget
> what you are doing halfway through your calculation
I know what I'm doing at all stages of my posts. Do you?
--- Christopher Heckman
Dik T. Winter
...
> So what? As you pointed out, when you add a node, you add a node, a branch,
> and a path. Actually, you need to add two nodes for each path, but that is
> irrelevant. They are finitely proportional, so if one is finite, the other
> is, and if one is infinite, so is the other. Infinite trees have infinite
> nodes and paths. Finite tree only have finite numbers of nodes and paths.
> It's a fact.
How do you jump from finite nodes to infinite nodes? An infinite path
never ends.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> Dik T. Winter said:
...
> > > 4) Any node increases the number of paths by 1 from 1 coming in, to 2
> > > going out. 2 - 1 = 1.
> >
> > an infinite tree. If you are talking about an infinite tree it may be
> > allowable, but because the number of paths is infinite, so it stays the
> > same when you add 1 to it.
>
> This is precisely the root mistake made by Cantorians. It is entirely
> reasonable to conclude that adding 1 to an infinite set creates an infinite,
> but finitely larger, set. oo+1<>oo, if by oo we mean the same infinite
> number on both sides of the inequality. This colllapsing of functions at
> infinity is an invalid method based on unfounded assumptions. WM is correct
> in generalizing the realtionships between numbers of nodes, branches and
> paths. Thos relationships are maintained at infinity. I am sure WM can
> form a nice inductive proof to this effect.
First off, he can't, and second, I think he disagrees with you strongly.
Like Eckard Blumschein he thinks there are only two infinities, a
potential infinity and an actual infinity.
Dik T. Winter
> Virgil said:
...
> > If one regards 0.333... not as a number but only as a sequence of
> > partial sums, it is true that none of those partial sums is exactly 1/3,
> > but the mathematical standard for interpreting a repeating decimal, like
> > 0.333..., is that it represents that NUMBER which is the limit of that
> > sequence of partial sums. And that number, by every reasonable analysis,
> > is exactly and precisely 1/3.
> But, Virgil, how do you know that, when you can never get to infinity? Don't
> you have to perform all your partial sums?
No.
> Isn't a limit something that never
> gets there?
Indeed.
> Can you just "jump" to infinity, and declare that infinite set of
> partial sums equal to some fraction?
No, apart from the use of the word "set" in this context.
A limit is defined using finite sums only.
Dik T. Winter
In a thread Tony appears to have left I asked the following:
> A question for you Tony. Let's say we have a set K which contains all
> natural numbers that in their binary expansion have a leftmost digit 1
> in an even position (and say that we start counting positions with 0).
> And to be clear, the set starts with:
> {1, 4, 5, 6, 7, 16, 17, 18, ..., 31, 64, 65, ..., 127, 256, ...}
> Are there more, less, or equally many numbers in that set as in the set
> of even numbers? And, whatever your answer is, how do we come at that
> answer?
>
> Note: lim{n -> oo} |{s in P, s <= n}| / |{t in N, t <= n}| does not exist.
Perhaps he will answer in this thread.
Tony Orlow
> In article <1117020437.3...@z14g2000cwz.googlegroups.com>,
> muec...@rz.fh-augsburg.de wrote:
>
> > Robert Kolker wrote:
> > > Tony Orlow (aeo6) wrote:
> > > >
> > > > That's all very well and good, if you specify f anf g and figure those
> > > > functions into your comparison. It's a mistake to ignore them.
> > >
> > > Are you capable of following a proof? Even a three line proof?
> >
> > Are you capable to follow a five lines proof without referring to "Big
> > Brother" Cantor? Consistency of set theory is questioned, hence I do
> > not accept Cantor's proof as an argument.
> >
> > 0 0.
> > 1 0 1
> > 2 0 1 0 1
> > ... ..................
> >
> >
> > infinitely many nodes (bits).
> > 2) All nodes (bits) of the tree belong to a countable set.
> > 3) A node can only exist within a path.
> > going out. 2 - 1 = 1.
> >
> > Please point out which step is wrong.
>
>
> WM conflates bounded paths, having terminal or leaf nodes with unbounded
> unending paths which have no terminal or leaf nodes, but contain
>
> 1) Each number of (0,1) is given by an UNENDING path stretching over
> infinitely many nodes (bits).
And yet you claim that no node is infinitely far from the root? Wrong.
besides, I thought you were objecting earlier to appending trailing
zeroes....maybe you changed your mind.
>
> 2) All nodes (bits) of the tree belong to a countable set.
>
> 3) A node can only exist within a path.
>
> 4) Any node increases the number of ENDING paths, having terminal or
You are really hung up on this lack of end. That's your problem, whether it has
to do with paths on a tree, or the "greatest finite natural". You need to look
where you can actually see something for once.
Anyway, one path is added for any TWO nodes. But do go on.....
>
> 5) Any node increases the number of nodes by 1, but have absolutely
> nothing to do with the number of unending paths.
Actually, if you have an infinite binary tree, with all paths unending, it's
rather difficult to add a node, isn't it? I suggest we would have a more
fruitful conversation talking about inserting nodes in the middle of an
infinite tree, and see what that does for us.
>
>
> All unending paths in an unending binary tree contain infinitely many
> nodes.
Most of which are infinitely far from the root.
>
> The number of leaf nodes exactly equals the number of ending or finite
> paths in any finite binary tree (in which all paths end).
And which you have conveniently dispensed with for the sake of your argument.
Typical Cantorian prestadigitation.
>
> Considering the binary tree whose root is "." and each branch is
> indicated by a "0" or a "1", each leaf node, and therefore each path, is
> represented by a terminating binary fraction , but each unending path is
> represented by a non-terminating binary fraction.
>
> There are moreof the non-terminating than of the terminating.
>
> So WM is wrong yet again.
>
And again you miss the point entirely, but will probably repeat the same
nonsense at least 6 more times, as is your tendency.
Let's put this baby to bed. See if you can poke a hole in this:
Let us consider the middle of a tree, finite or infinite, and what happens when
we insert a node. We have the following tree, where each node has two children,
2x and 2x+1 where the parent node is x:
1
2 3
4 5 6 7
8 9 10 11 12 13 14 15
Let's insert a node between 2 and 5. Sure, this will break the pattern I just
mentioned; that was just so you could make sure you knew whose children were
whose. We'll call our new node X, and say for argument that 5 will be its right
child node. This leaves X with no left child node (I put a period where one
would be):
1
2 3
4 X 6 7
8 9 . 5 12 13 14 15
10 11
Notice that we have not added a path yet, but only extended the paths that
include the branch between 2 and 5, by one branch. Every time we add a node, we
add a branch, but not necessarily a path. Not every node MUST have two
children. In fact, if one creates a binary "tree" with no left nodes, we have a
linked list which consists of a single path, even if it is infintiely long. It
is only when we add a second child to a node that we create a new path. It
takes two nodes to add a path. So, if we then add another element, Y, as the
left child of X, THEN we have produced a new path, ending in Y, a new leaf
node:
1
2 3
4 X 6 7
8 9 Y 5 12 13 14 15
10 11
So, it takes two nodes to add a path. Note that node 1 may have a parent node,
and nodes 8,9,10,11,12,13,14,15 all may have children. It doesn't matter
whether this is a finite tree, or an infinite one. It doesn't matter whether
any leaf nodes existed previously, or whether the paths are finite or infinite.
In all cases, the addition of a node adds one branch, and MAY add a path, but
no more than one additional path for every two nodes or branches. Therefore,
there are ALWAYS fewer paths than nodes, and the proposition that any binary
tree has more branches than nodes is clearly wrong.
If someone wants to tell me that one can't insert a node in the tree, I suggest
they ask their programmer friends, before wasting my time.
--
Smiles,
Tony
Tony Orlow
> In article <1117021029....@o13g2000cwo.googlegroups.com>,
> muec...@rz.fh-augsburg.de wrote:
>
> > Virgil wrote:
> > > In article <1116958479.5...@g47g2000cwa.googlegroups.com>,
> > > muec...@rz.fh-augsburg.de wrote:
> > >
> > >
> > > > >
> > > > > > .
> > > > > > 0 1
> > > > > > 0 1 0 1
> > > > > > ..................
> > > >
> > > > and summing up establishes an infinite series representing a real
> > > > number. Every combination of countably many bits is realized by
> > > > definition.
> > >
> > > But such an "infinite" binary tree is not a list, so this has nothing to
> > > say about the validity of Cantor's theorem.
> >
> > The tree is not a list. Therefore I choose it. Nevertheless it has to
> > say much about infinite sets and Cantor's theorem, namely that the set
> > of nodes and the set of paths are equivalent sets.
>
> equivalent, but infinite or non-terminating paths have no leaf nodes,
> so that there is no equivalence.
>
> > The set of nodes is
> > countable and the set of paths is equivalent to the set of reals in
> > (0,1). That's quite a lot of information, isn't it?
>
>
> WM's "proof" disproved"
>
> WM conflates bounded paths, having terminal or leaf nodes with unbounded
> unending paths which have no terminal or leaf nodes, but contain
> infinitely many intermediate nodes.
>
> 1) Each number of (0,1) is given by an UNENDING path stretching over
> infinitely many nodes (bits).
>
>
> 3) A node can only exist within a path.
>
> leaf nodes, by 1 from 1 coming in, to 2> going out. 2 - 1 = 1.
>
> nothing to do with the number of unending paths.
>
>
> nodes.
>
> paths in any finite binary tree (in which all paths end).
>
> indicated by a "0" or a "1", each leaf node, and therefore each path, is
> represented by a terminating binary fraction , but each unending path is
> represented by a non-terminating binary fraction.
>
> There are moreof the non-terminating than of the terminating.
>
> So WM is wrong yet again.
>
posted regarding insertion of nodes in ANY tree. Sorry, the prior existence of
leaf nodes is not a ruse that's available for you in this one.
--
Smiles,
Tony
Tony Orlow
were paying attention, I think you'd realize this is not related to what WM is
claiming. Look at my argument concerning insertion of a node, and respond to
that. It's the relative number of paths vs. nodes that WM is addressing, in
response to your claims that you have an infinite tree with uncountable paths
(because you're hung up on leaf nodes for some reason) and countable nodes,
when paths are always less numerous than nodes. Pay attention. It's no wonder
these discussions with you go around in circles, when you don't even know what
they're about.
--
Smiles,
Tony
Tony Orlow
Cantorians are wrong, as I have explained in detail for each one, using
established mathematics from other areas. Mathematikers declare axioms and
consider them law, but they are educated guesses. You have a system that you
think "works", but the results it derives are at odds with reality. Sorry,
cardinality of infinite sets is a sinking ship.
Real mathematicians solve problems, derive equations, and get results that are
useful. Axiomatic systems are very good for this, but axioms are always
questionable, unless universally justified in some manner, which is probably
impossible. At least SOME justification is in order.
--
Smiles,
Tony
Tony Orlow
a step up from the clouds, too? Typical Cantorian thought pattern. Very good.
--
Smiles,
Tony
Tony Orlow
I should hope so.
>
> In an infinite tree there are as many *finite* paths as nodes. Again,
> I think you can count on general agreement (although there will be
> some grumbling that there are infinite numbers of both and I may be
> off by a factor of 2 but let's forget about that).
If "there are as many *finite* paths as nodes", then what do the infinite paths
consist of? Don't they have nodes as well? Are there infinite paths, if all the
nodes are used up in the finite paths?
>
> I await, with some interest, your proof that, in an infinte tree, there
> are half as many paths (or just as many paths or twice as many paths or
> whatever) *of* *all* *kinds* as there are nodes.
Good. I got called into duty yesterday before I could finish it but it's posted
now. And, no leaf nodes or lack thereof for Virgil to chew on, redundantly
reiterating the same repetitive statement verbatim over and over.
>
> When WM was talking about adding a node being the same as adding a path
> he was, although he didn't know it, talking about *finite* paths. Adding
> a node *is* the same as adding a *finite* path. The situation with
> infinite paths is much messier.
Nope, not. It requires two nodes to make a new path, anyway. One creates a new
path when one adds a second child to a node. refer to my demonstration of node
insertion.
>
> >This kind of
> >result is what brings upon the Cantorians all their woes in the form of us
> >"idiots".
>
> No, the source of the woes is anti-Cantorians insistence that infinite and
> finite are pretty much the same, except that infinite is larger. Anti-
> cantorians think that the integers include infinitely large values.
> Cantorians think that the integers include arbitrarily large values, but
> "know" that arbitrarily large is not infinite. Anti-cantorians think that
> infinite paths are like finite paths, but longer. Cantorians actually
> puzzle out what in the heck is meant by an infinitely long path and
> conclude that it behaves completely differently from finite ones.
>
> Alan
>
Cantorians pretend they know what they are talking about, but always seem to
frget the details of the subject matter along the way.
--
Smiles,
Tony
Tony Orlow
where I have interest, even if the idea seems foreign, I try to entertain it
and see if it works. If I object, it's usually in the form of a question, not
an insult. You see, rather than taking offense at the notion that someone may
have discovered something new, I take interest, since they may be right. I have
disvoered a lot of new things, only to discover I wasn't the first. In this
case, maybe I am. No one else seems to see these things like I do. Of course I
want to convey what I see, to test it against objections, and to get the idea
out there. If you're not interested, boohoo for you. Try coming up with an
original idea yourself. Or, think about my simple set size measure and see if
you really think it's bizarre.
I had hoped to have a web page up on this by now, but I am afraid that is going
to have to wait until this weekend. Life is busy.
--
Smiles,
Tony
Tony Orlow
>
>
> Tony Orlow (aeo6) wrote:
> > Proginoskes said:
> > >
> > >
> > > David C. Ullrich wrote:
> > > > On 24 May 2005 05:58:22 -0700, muec...@rz.fh-augsburg.de
> > > > wrote:
> > > > [...]
> > > > Hint: infinite sets are not the same as finite
> > > > sets.
> > >
> > > Yes. Intuition goes out the window with infinity, which is
> > > why we do mathematics instead. 8-)
> >
> > If your math throws your intuition out the window, perhaps
> > there's a problem with your math.
>
> It's not the math that throws the intuition out, it's infinity.
intuitions have been dealing well with infinity for 25 years. Fancy that.
>
> > > One (easy?) example of how infinite sets are different
> > > from finite sets is that if you have an infinite set S,
> > > you can put it in 1-1 correspondence with a proper
> > > subset T (a subset which is not equal to S).
>
> > Yes, well, that pretty much just shows that bijections go out the window at
> > infinity, and are a poor tool for dealing with it.
>
> But without bijections, how do you show that {3,4} and {1,2} (two
> finite sets) have the same size? If you throw out bijections of
> infinite sets, you also throw out bijections of finite sets, since
> bijections can be used for both.
sets or to infinite sets, as li=ong as they are mapped using invertible
functions. An invertible fnction can be defined for any pair of equal-sized
finite sets. For infinite sets, this is not always possible, but where it is,
the function used indicates the size of the mapped set realtive to the one it
is mapped from.
>
> > > For instance, let S = N, T = {2, 3, 4,...}. One
> > > possible 1-1 correspondence is
> > > f(n) = n + 1; that is, match up
> > > 1 <--> 2
> > > 2 <--> 3
> > > 3 <--> 4
> > > ...
>
> > Yes, and it is no coincidence that the inverse function,
> > x-1, fits our intuitions at x=N,
>
> What do you mean, the element x of N which is equal to N? N is not in
> N; N contains numbers, not sets.
read many things from context, or at least we should be able to.
>
> > since taking away the first element of the naturals would
> > intuitively leave you with the naturals, minus 1. This is
> > what I mean by taking the mapping functions into account.
>
> > > This uses up all the elements of S (on the left-hand side) and T (on
> > > the right-hand side).
> > >
> > > This is impossible to do if S is finite.
>
> > It is possible to do for finite sets,
>
> Okay. Let S = {1,2,3} and T = {1,2}; then T is a proper subset of S.
> Give me a bijection from S to T.
context snipped. No bijection there.
>
> > but it is not wise to do that and forget
> > what you are doing halfway through your calculation
>
> I know what I'm doing at all stages of my posts. Do you?
Yes, you are snipping context.
>
> --- Christopher Heckman
>
>
--
Smiles,
Tony
Tony Orlow
> In article <MPG.1cfe62b73...@newsstand.cit.cornell.edu> Tony Orlow (aeo6) <ae...@cornell.edu> writes:
> > Dik T. Winter said:
> ...
> > > > 4) Any node increases the number of paths by 1 from 1 coming in, to 2
> > > > going out. 2 - 1 = 1.
> > >
> > > Eh? Now you are talking about an ever increasing finite tree, not about
> > > an infinite tree. If you are talking about an infinite tree it may be
> > > allowable, but because the number of paths is infinite, so it stays the
> > > same when you add 1 to it.
> >
> > This is precisely the root mistake made by Cantorians. It is entirely
> > reasonable to conclude that adding 1 to an infinite set creates an infinite,
> > but finitely larger, set. oo+1<>oo, if by oo we mean the same infinite
> > number on both sides of the inequality. This colllapsing of functions at
> > infinity is an invalid method based on unfounded assumptions. WM is correct
> > in generalizing the realtionships between numbers of nodes, branches and
> > paths. Thos relationships are maintained at infinity. I am sure WM can
> > form a nice inductive proof to this effect.
>
> First off, he can't, and second, I think he disagrees with you strongly.
> Like Eckard Blumschein he thinks there are only two infinities, a
> potential infinity and an actual infinity.
>
conclusions, yes we disagree, since I am convinced that there are a wide
spectrum of distinct infinities within any level, and an infinite number of
nested levels, and dimensional extensions of infinities. I seem to be alone in
this for the most part. But, as regards WM's specific objection on this topic,
he and I agree fully.
--
Smiles,
Tony
Tony Orlow
> In article <MPG.1cfe8ee5a...@newsstand.cit.cornell.edu> Tony Orlow (aeo6) <ae...@cornell.edu> writes:
> ...
> > So what? As you pointed out, when you add a node, you add a node, a branch,
> > and a path. Actually, you need to add two nodes for each path, but that is
> > irrelevant. They are finitely proportional, so if one is finite, the other
> > is, and if one is infinite, so is the other. Infinite trees have infinite
> > nodes and paths. Finite tree only have finite numbers of nodes and paths.
> > It's a fact.
>
> How do you jump from finite nodes to infinite nodes? An infinite path
> never ends.
>
about it? If it has infinite branches, it has infinite nodes, right? Otherwise,
what connects the branches? That's not a leap, but two step logical argument.
Which of those two steps do you object to?
--
Smiles,
Tony
Robert Kolker
>
> And which you have conveniently dispensed with for the sake of your argument.
> Typical Cantorian prestadigitation.
Those are end points of finite pathes.
In the set of all countable sequences of 0 an 1 there are no end points.
Pretty soon you will tell me there is only a finite set of finite
integers. Can you see the mistake in that?
Bob Kolker
Tony Orlow
> In article <HqydnYVafLA...@comcast.com> Robert Kolker <now...@nowhere.com> writes:
> > Tony Orlow (aeo6) wrote:
> > >
> > > Only a precise and intuitively unobjectionable generalization of set size
> > > measure from finite to infinite sets. That's all.
> >
> > You have proved no theorems, so you have no applications to anything
> > beyond your strange definitions. Do you know how to prove a theorem?
>
> In a thread Tony appears to have left I asked the following:
>
> > A question for you Tony. Let's say we have a set K which contains all
> > natural numbers that in their binary expansion have a leftmost digit 1
> > in an even position (and say that we start counting positions with 0).
> > And to be clear, the set starts with:
> > {1, 4, 5, 6, 7, 16, 17, 18, ..., 31, 64, 65, ..., 127, 256, ...}
> > Are there more, less, or equally many numbers in that set as in the set
> > of even numbers? And, whatever your answer is, how do we come at that
> > answer?
> >
> > Note: lim{n -> oo} |{s in P, s <= n}| / |{t in N, t <= n}| does not exist.
>
> Perhaps he will answer in this thread.
>
>
to keep track. So, let me address this here.
First, let me say that this is not an invertible function that I can tell.
Perhaps I am wrong and you can give me a formula to compute f(n) from n, and
its inverse function. I doubt it.
However, the solution to this is rather easy using basic logic. In this subset
of the naturals we have 1 element included, then 2 excluded, then 4 included,
then 8 excluded.... For every subset included, we have a subsequent subset of
the naturals excluded which is twice as large. Therefore, of the natural
numbers, twice as many are excluded as included, and since they add up to the
entire set of naturals, we can do simple algebra and say x+2x=N =>3x=N =>
x=N/3. Since the set of evens is determined to be N/2 by the inverse function
of f(x)=2x, we can say the set of evens is a larger set than the one above by a
factor of 3/2.
That wasn't so hard, even if it didn't fit the method for invertible functions.
--
Smiles,
Tony
imagin...@despammed.com
I notice that Dik said "paths" rather than branches. But that aside,
I'd like to ask you, Tony: what are these "infinite nodes" actually
like? I presume you have passed through the twilight zone, and seen
them? Do they just look like perfectly ordinary nodes? (If so, how
would one actually be aware of passing the boundaries of Twilight?) Do
the infinite nodes stop at any point? If not, and particularly if they
don't really look any different from the finite nodes, couldn't they
all just be one happy family of plain "nodes"? Are there any
mathematical results that follow from these "infinite nodes"?
This whole discussion seems unlikely to be profitable to me, since it's
basically immense ramifications of the natural numbers all over again.
Have you abandoned the bigulosity threads?
Brian Chandler
http://imaginatorium.org
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> > 1) Each number of (0,1) is given by an UNENDING path stretching over
> > infinitely many nodes (bits).
> And yet you claim that no node is infinitely far from the root? Wrong.
When you can produce a mathematically valid proof that adding 1 to a
finite number will produce an infinite number, you may claim that there
are nodes infinitely far from the root node, but until then, it is you
who are worng
Virgil
Tony Orlow (aeo6) <ae...@cornell.edu> wrote:
> > All unending paths in an unending binary tree contain infinitely many
> > nodes.
> Most of which are infinitely far from the root.
Then how do they connect to the root node? In a path you must have each
node linked by one branch to a next node, which means that the number of
links connecting any two nodes in a path is finite.
TO is still suffering from the delusion that adding 1 to a finite number
can produce a non-finite number.
He keeps fantasizing about these allegedly infinite numbers produced by
finite processes, but cannot explain which finite step produces an
infinite result.