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WM
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 Posted: Thu Oct 09, 2008 8:30 am Post subject: Re: Binary Tree and Pairs of Nodes |
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On 8 Okt., 22:21, LauLuna <laureanol...@yahoo.es> wrote:
[quote]
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
And so on.
Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But it is clear, that every infinite path that exists
in the infinite binary tree has one node mapped onto it.
Of course we could also have selected the path pi-3 or any other one
to start with. There is no path existing (!) in the tree being left
out by
the mapping.
Concerning the proof you were kind enough to reproduce I must say I
find it puzzling. Anytime you map a node on a path you say 'no matter
which one', so giving no criterium and defining no bijective function
between the set of paths and the set of nodes. All you say is: 'take a
node in turn and map it on a path; when the nodes are exhausted,
you>ll have exhausted the paths too'; but this is just equivalent to
postulating the equicardinality of both sets without defining a
bijection to prove it.
[/quote]
The bijection is revealed by the fact that all paths are connected.
Every path must start at the root, and in particular it must cross the
border of the domain already investigated. See the figure above.
You can treat the argument as a mathematical game: "Conquering the
binary tree." In the state depicted by the figure above I have "paid"
three nodes and "conquered" three complete paths. The aim of the game
is to prove that the complete tree can be conquered by paying not more
nodes than available in the conquered domain. And that is obviously
possible!
Every path of the tree that is different from those already mapped by
nodes, does somewhere leave the conquered domain. There I catch it.
When crossing the border, that means when it deviates from the
conquered domain, then it gets its node. Of course I can arbitrarily
select one individual path, because any other path is not another path
unless it will leave the domain meanwhile conquered.
[quote]
You say:
But it is clear, that every infinite path that exists
in the infinite binary tree has one node mapped onto it.
How could it possibly be clear when you do not even define the
intended bijection?
[/quote]
Why should I? Every path that leaves the conquered domain gets its
node. Path that do not leave the conquered domain are not different
from one of the paths already marked by a node.
[quote]
On the other hand, it>s of course your right to reject whatever is not
constructive in mathematics, but if you wish to prove an internal
inconsistency in the transfinite set theory you cannot assume
constructivity (in the usual sense); it would beg the question, since
the constructive, as usually understood, does not transcend the
countable.
[/quote]
No. The binary tree is constructive. Mathematics is constructive. In
Cantor>s list every line is different from the antidiagonal. Should
non-constructive mathematics suspect paths in the tree that cannot be
distinguished, then it s not mathematics. Same holds for the real
numbers. By the tree argument that is easy to see.
[quote]
I confess you have me wondering in what sense the infinite binary tree
is constructive; it seems easy to give a recipe to construct it but it
is not constructive in the sense that it>s the output of some
algorithm: it cannot be reduced to a unidimensional sequence.
[/quote]
No, but it can be reduced to a game. We can reduce the nodes to a
sequence. And we can prove by my game that paths can be bijected with
nodes. As there are no more reals than paths, we see that the
interpretation of Cantor>s proof as showing the existence of an aleph
larger than aleph_0 is simply wrong.
This can also be seen by proving that there are other *countable* sets
that cannot be listed (in contradiction to the definition of
countable), like the set of all diagonal numbers of constructed lists.
(See the related discussion with the misleading title: Why "meta
diagonals" are irrelevant.)
[quote]
I guess this question is important for the possibility of defining an
injection from the set of all pairs of paths into the set of all pairs
of nodes.
Let me give it a second thought.
[/quote]
Thank you for your interest.
Regards, WM |
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WM
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| Posted: Thu Oct 09, 2008 8:40 am Post subject: Re: Binary Tree and Pairs of Nodes |
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On 9 Okt., 05:39, george <gree...@email.unc.edu> wrote:
[quote]On Oct 7, 7:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
You misunderstand the binary tree.
0.
/ \
0 1
/ \ / \
0 1 0 1
...
The paths you give above can be distinguished by infinitely many
nodes.
ALL [pairs of] paths can be distinguished by infinitely many nodes.
Any two paths have some initial path-segment in common (it may
be the empty segment if their first digits are different). After
that,
EVERY node in one path is different from EVERY node in the other path;
once paths split, they never re-cross or re-intersect or re-combine.
So any one of the infinitely many nodes after the split-point on the
left path
can be paired with any one of infinitely many nodes on the split-point
after
the right path to produce a distinguishing pair. But the number of
differing
paths below the split is still uncountable, and the number of node-
pairs below
the split is still countable.
[/quote]
Then try to play my game and to loose. You have no chance.
[quote]
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
"Conquering the binary tree."[/quote]
Rule: Pay one of the nodes of the conquered part of the binary tree to
enlarge your empire.
In the state depicted by the figure above I have "paid" three nodes
(payment indicated by "o") and "conquered" three complete paths (not
completely drawn). The aim of the game is to prove that the complete
tree can be conquered by paying not more nodes than available in the
conquered domain. And that is obviously possible!
Every path of the tree that is different from those already mapped by
nodes, does somewhere leave the conquered domain. There you can catch
it. When crossing the border, that means when it deviates from the
conquered domain, then it gets its node. Of course you can arbitrarily
select one individual path, because any other path is not another path
unless it will leave the domain that you meanwhile have conquered.
Regards, WM |
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WM
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| Posted: Thu Oct 09, 2008 9:00 am Post subject: Re: Binary Tree and Pairs of Nodes |
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On 9 Okt., 05:53, george <gree...@email.unc.edu> wrote:
[quote]The diagonal argument will do quite well for refuting your conjecture
here,
[/quote]
The diagonal argument cannot work in a system that by definition
contains every real, as does the binary tree. What different path
would you like to produce?
[quote]if anybody can figure out the natural relationship for diagonalizing
the path-
pairs and the node-pairs. Every path-pair already "includes" every
node-pair
so it is not immediately obvious what to complement to "anti"-diag.
But you know it>s going to be something.
[/quote]
I know that there is no chance. I always win my game.
[quote]
For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....
o
|
/
/
/
...
The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...
o
|
o
/ \
/ \
/ \
...
The next node is mapped on another path, no matter which one, say one
always changing direction, i.e., 0.010101...
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
And so on.
Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But perhaps you have understood already, that every
infinite path that exists in the infinite binary tree has one node
mapped onto it.
You could map the root to every path if that were the only constraint.
You need to add something like ONLY ONE node gets mapped to any path.
[/quote]
No. My proof would not be invalidated when more than one node were
mapped to any path. Important is only that a node is not related to
more than one path. But that is already clear by the definition of a
mapping from nodes to paths. (Are we in sci.logic here?) Important is
further surjectivity. That is guaranteed by my game.
[quote]And if you do that, then
You CAN>T DO this, because there are MORE paths than nodes!
[/quote]
That is just the question (to be denied).There are not more distinct
paths (i.e., paths distinguishable by at least one pair of nodes) than
nodes.
Regards, WM |
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LauLuna
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| Posted: Thu Oct 09, 2008 2:49 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 8, 2:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]On 8 Okt., 13:58, David C. Ullrich <dullr...@sprynet.com> wrote:
Please stop suspecting things you do not know of - try to understand
my argument.
Sorry, I>m not going to try to understand an argument
that purports to show that there are only countably
many paths in an infinite binary tree.
But it is true, as you can see below without stretching yourself too
far, I think.
The only possible
reason for that would be to try to explain to the
author where the error is. I>ve done that - your
comment "Your statement implies mine." is
simply wrong.
You talked about a path and *every pair of nodes*. That implies at
least one pair of nodes, i.e., my statement, doesn>t it?
No point in going further since you
don>t seem to have an adequate grasp of simply logic.
Not adequate to what you may misunderstand as logic.
But thank you for the atempt. (No, I never tried to
publish the binary tree other than where it has been printed already.
I do not see any necessity to use the limited audience of a printed
medium because the audience of the internet is much broader.)
Giggle. Of course the fact that anyone can say anything
they want on the internet has nothing to do with it.
I published the binary tree already in printed form. I do not belong
to the set of people who publish every minute idea at 10 different
places.
You>re taking a very curious approach
to fixing that, making hundreds of posts containing
similar elementary errors.
There is nothing to fix. Obviously there cannot be more paths than
nodes in the binary tree.
Obvious or not, it>s not so.
It is very easy to see that there are not more paths than nodes in the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....
o
|
/
/
/
...
The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...
o
|
o
/ \
/ \
/ \
...
The next node is mapped on another path, no matter which one, say one
always changing direction, i.e., 0.010101...
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
And so on.
Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But perhaps you have understood already, that every
infinite path that exists in the infinite binary tree has one node
mapped onto it.
Of course we could also have selected the path pi-3 or any other one
to start with. There is no path existing in the tree being left out by
the mapping.
Regards, WM
[/quote]
Your game is just a countable sequence of choices, a countable choice
function. We have no guarantee that such a device will 'finally' give
us the entire tree.
The question is not whether we can play your game and lose, the
question is whether there is a winning strategy, that is, a definable
pairing function. There is none. Try to construct it and you>ll find
that paths are shaped in more ways than any finitely expressible
criterium could take into account.
I think your idea is that whenever a new path parts we have an
available node in the 'conquered domain' to match it with, so that "in
the limit" we have a sufficient supply of nodes to tag all paths in
the tree. You are assuming that whatever holds for all finite initial
portions of the tree holds for the whole tree. The assumption is
unwarranted.
So far I can only concede that the apparently constructive nature of
the binary tree is a little baffling to sheer intuition. It seems that
we can construct it step by step just as we can generate the naturals;
the tree seems to be the outcome of a countable sequence of acts, so
that a countable choice function would give it all to us.
But this need not be so, for if we try to define a particular
bijection between the set of all paths and the set of all nodes, we
necessarily fail. This shows that the passage to the limit doesn>t
work.
Perhaps (only perhaps) it could be argued that the complete infinite
binary tree is unintelligible since it can obviously be constructed
step by step by a mathematician following finite instructions
(provided with a potentially infinite supply of paper etc.) and yet it
cannot be taken to be the outcome in the limit of such an operation
because 'somewhere' in its development the tree jumps into the
uncountable.
Or equivalently: maybe, if the tree is conceivable at all, it can only
be conceived of as the potential output of such a task and at the same
time it cannot be conceived of as such output since the potential
output of any conceivable algorithm is countable. Then (maybe) we must
infer that it is not conceivable.
As far as I can see, this would (perhaps) amount to refuting the
existence of the set of all reals.
Best. |
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WM
Guest
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| Posted: Thu Oct 09, 2008 7:20 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:
[quote]o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
And so on.
Your game is just a countable sequence of choices, a countable choice
function. We have no guarantee that such a device will 'finally' give
us the entire tree.
[/quote]
Of course there is no "finally". All we can say is that "iff there was
an infinite countable set of nodes" and "iff infinite sets could be
worked through completely", then every node would get its turn. But if
every node of the tree has been used to mark one path, how should any
further path remain? A path without nodes? Not connected to the root
node? Certainly, if set theory was right, then such paths must exist.
[quote]
The question is not whether we can play your game and lose, the
question is whether there is a winning strategy, that is, a definable
pairing function. There is none. Try to construct it and you>ll find
that paths are shaped in more ways than any finitely expressible
criterium could take into account.
[/quote]
Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.
[quote]
I think your idea is that whenever a new path parts we have an
available node in the 'conquered domain' to match it with,
[/quote]
That is obvious.
[quote]so that "in
the limit" we have a sufficient supply of nodes to tag all paths in
the tree. You are assuming that whatever holds for all finite initial
portions of the tree holds for the whole tree. The assumption is
unwarranted.
[/quote]
The assumption that all elements of an infinite set can be used is
false, of course. But it is Cantor>s assumption and is the basic axiom
of set theory. Therefore I use it here. I do not need anything else
than the axiom of infinity and a counting of the nodes like
0
1
23
4567
....
I do not talk about a limit. I use the "fact" of set theory that all
elements of an infinite set can be used.
[quote]
So far I can only concede that the apparently constructive nature of
the binary tree is a little baffling to sheer intuition.
[/quote]
Sorry, my game is not intuition. It has fixed rules. Only the choice
of paths is not fixed because that is not of interest.
[quote]It seems that
we can construct it step by step just as we can generate the naturals;
the tree seems to be the outcome of a countable sequence of acts, so
that a countable choice function would give it all to us.
But this need not be so, for if we try to define a particular
bijection between the set of all paths and the set of all nodes, we
necessarily fail. This shows that the passage to the limit doesn>t
work.
[/quote]
This shows that there is no set of all reals. It is impossible to
define more than a countable set of reals. There are no uncountable
sets at all. In the parallel thread with the misleading name the
discussion has come to a striking argument:
It is impossible to put all anti-diagonals of all constructed lists in
a list. (This list has an anti-diagonal that is not in it.) But the
set of all those anti-diagonals is countable (if all are formed
according to the same rule). Therefore the non-existence in a sequence
does not show uncountability.
[quote]
Perhaps (only perhaps) it could be argued that the complete infinite
binary tree is unintelligible since it can obviously be constructed
step by step by a mathematician following finite instructions
(provided with a potentially infinite supply of paper etc.) and yet it
cannot be taken to be the outcome in the limit of such an operation
because 'somewhere' in its development the tree jumps into the
uncountable.
[/quote]
No. See above. "Uncountable" is simply a misinterpretation.
[quote]
Or equivalently: maybe, if the tree is conceivable at all, it can only
be conceived of as the potential output of such a task and at the same
time it cannot be conceived of as such output since the potential
output of any conceivable algorithm is countable. Then (maybe) we must
infer that it is not conceivable.
[/quote]
It is conceivable and it is countable (obviously the nodes are). It
simply does never end, as every infinite set. There is no finished
infinity. But that is not directly under discussion. What the tree
shows is the falsity of the concept of uncountability.
[quote]
As far as I can see, this would (perhaps) amount to refuting the
existence of the set of all reals.
[/quote]
The tree does it. For instance, if I was forced to give an algorithm
for my game, then I could say: Put the next node on the path that
afterwards always turns left, i.e., on a terminating rational of the
form
0.xxx...xxx000...
with x either 0 or 1.
Obviously the whole tree will be exhausted by this procedure (I
conquer the whole tree). So no path remains. But I could also say: Put
the next node on the path that afterwards always turns right, like
0.xxx...xxx111...
Obviously I could find (countably) many such algorithms. Each one
would exhaust the binary tree and let me win my game.
How can that be? There are not all reals in that tree! But a tree is
merely a picture for the binary, or decimal, representation of the
reals (of [0. 1]). Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals. The
number of infinite sequences is countable and is much smaller than
expected. And it cannot be decided which form those infinite sequences
have. They do not exist on the Platonist shelf, waiting to be taken by
the mathematicians. They have to be created like the paths in my tree.
Superfluous to say that the number of such creations is not
uncountable.
Regards, WM |
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WM
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| Posted: Fri Oct 10, 2008 7:50 am Post subject: Re: Binary Tree and Pairs of Nodes |
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On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
[quote]In article
871188cc-f053-4ab5-870f-0b57ce13e...@n33g2000pri.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \ ...
And so on.
Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.
Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.
But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".
[/quote]
Not according to the rule of my game. Consider the figure above. Node
x buys you one of the path leaving the path
0.111... at position x. You may say that there are many paths
deviating from 0.111..., and you are right. But if one of them
deviates from said path, it will get get another node. There is no
escape.
[quote]o
|
o
/ \
o \
/ \ \
/ / \
/ \ x
/ / / \
... /[/quote]
\
With every won path you get infinitely many nodes. Each of them helps
you to get another path, one and only one path per node. In this game
you have the ratio of paths to nodes P/N = 0 at every step. The
infinite binary tree covers every step, but not more.
Similarly the usual decimal expansion of real numbers covers every
digit, but not more. There is no last digit of pi. There is every
digits of pi - unless one looks to close, but that is not in question
here. Set theorists insistuing of uncountably many reals must insist
that there are more than every digit of pi. I am not inclined to
discuss about such folly.
[quote]
But if every node of the tree has been used to mark one path
Unless that node is a terminal (leaf) node, it marks more than one path,
[/quote]
Try to understand the rule of the game (there is only one). Try to
understand that there is no leaf node.
Regards, WM |
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Virgil
Guest
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| Posted: Fri Oct 10, 2008 7:50 am Post subject: Re: Binary Tree and Pairs of Nodes |
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In article
<871188cc-f053-4ab5-870f-0b57ce13e72e@n33g2000pri.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \ ...
And so on.
Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.
Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.
[/quote]
But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".
[quote]But if every node of the tree has been used to mark one path
[/quote]
Unless that node is a terminal (leaf) node, it marks more than one path,
and in any infinite complete binary tree there are no terminal (leaf)
nodes at all.
Thus WM>s picture is, as usual, flawed.
[quote]Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.
[/quote]
That is an illogical argument that carefully misrepresents the essential
nature of infinite complete binary trees.
[quote]
This shows that there is no set of all reals. It is impossible to
define more than a countable set of reals.
[/quote]
But not impossible to imagine them, wad as nothing mathematical is any
more or less that a figment of imagination, they are there.
[quote]There are no uncountable
sets at all.
[/quote]
Outside of the imagination, there are no sets at all, but at least in my
imagination, and the imaginations of many mathematicians, there are
plenty of uncountable sets
[quote]
It is conceivable and it is countable (obviously the nodes are). It
simply does never end, as every infinite set. There is no finished
infinity. But that is not directly under discussion. What the tree
shows is the falsity of the concept of uncountability.
[/quote]
Only to those for whom such falsity is a part of their creed.
To those with open minds there is no such artificial restriction of the
imagination.
[quote]
As far as I can see, this would (perhaps) amount to refuting the
existence of the set of all reals.
The tree does it.
[/quote]
The set of all paths of any infinite complete binary tree establishes
the existence of at least one uncountable set.
[quote]For instance, if I was forced to give an algorithm
for my game, then I could say: Put the next node on the path that
afterwards always turns left, i.e., on a terminating rational of the
form 0.xxx...xxx000... with x either 0 or 1. Obviously the whole tree
will be exhausted by this procedure (I conquer the whole tree). So no
path remains. But I could also say: Put the next node on the path
that afterwards always turns right, like 0.xxx...xxx111... Obviously
I could find (countably) many such algorithms. Each one would exhaust
the binary tree and let me win my game.
[/quote]
Where in the path that forever alternates between left and right do you
place your node?
[quote]
How can that be? There are not all reals in that tree!
[/quote]
There are all infinite binary strings of left and right branchings in
such a tree, including uncountably many paths each branching infinitely
many times in both directions.
[quote]But a tree is merely a picture for the binary, or decimal,
representation of the reals (of [0. 1]).
[/quote]
Except that in a tree one does not combine two paths into one real.
[quote]Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals.
[/quote]
Then your trees are either not infinite or not complete.
[quote]The
number of infinite sequences is countable and is much smaller than
expected.
[/quote]
Only in WM>s incomplete trees, not in any actual complete infinite
binary (or decimal) tree. |
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WM
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| Posted: Fri Oct 10, 2008 8:22 am Post subject: Re: Binary Tree and Pairs of Nodes |
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On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
[quote]Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals.
Then your trees are either not infinite or not complete.
[/quote]
The infinite binary tree is as complete as the number system, i.e.,
the binary or decimal representation of real numbers can be. There is
no chance to introduce another path. There is no chance to introduce
another real. All combinations of nodes-values, all combinations of
digits are present.
The infinite path 0.111... is covered by the infinite sequence of
paths
0.1000..., 0.11000..., 0.111000..., ... (*)
Same holds for the decimal or binary representations. It is impossible
to have more than an infinite number of digits in an expansion. It is
impossible to have more than all natural indices in the expansion
0.111... = SUM{n in N} 10^-n (**)
The complete sequence (*) includes the expansion (**) - in the tree
and in the binary or decimal representation. Every position with a
natural index n shows digit 1. It is without interest what might
follow behind all positions indexed by naturals.
Regards, WM |
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george
Guest
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| Posted: Fri Oct 10, 2008 7:47 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]Anyhow, the binary tree is a crystal-clear construction that avoids
the confusing opinion of undistinguishable numbers. It does never
insert its roots into the mud of the continuum (Kontinuumssauce).
[/quote]
This is just bullshit. The INFINITE part of the binary tree IS
ALWAYS
the continuum. The part you have investigated thus far is finite;
the part you have not investigated yet IS ALWAYS UNcountabl.
[quote]Therefore all its paths are and remain countable,
[/quote]
You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"! All sometimes means EACH around here!
And no INDIVDIDUAL path can be uncountable; they are all countably
infinitely LONG! But they are uncountably WIDE -- there are
uncountably
infinitely many OF them!
I realize your native language is not English but that is simply not
an excuse.
[quote]It is very easy to see that there are not more paths than nodes in
the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....
o
|
/
/
/
...
The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...
[/quote]
You CANNOT SAY "no matter which one". It MATTERS which one.
Here is the short version of your proof that the reals are countable:
1) pick a 1st real, no matter which one.
2) pick a 2nd real, no matter which one, as long as it is different
from all the previous
ones.
3) repeat ad infinitum.
4) This clearly maps every natural to a real
5) therefore (unsound) it clearly maps EVERY real to a natural.
THIS IS ALL you have said. It is completely unsupported.
Here is a better way for you to look at this.
If every (denumerably long) path is being mapped to a node,
then that node IS AT THE END OF A FINITE path.
Since that finite path to the node IS A FINITE BIT-STRING,
it uniquely represents A NATURAL NUMBER for the node.
Obviously, the nodes can be ordered:
the root is 1, the (node at the end of the)
terminating path 0 is 2, the terminating path 1 is 3,
the path 00 terminates at the node numbered 4, the path 01
termiantes at node#5,
the path 10 terminates at the node numbered 6, and the terminating
path 11 is 7, etc.
You just order them left-to-right across each row, with 0000 on the
left and 1111 on
the right. If you want to know the number corresponding to a path,
just put a 1
in front of the path-as-a-bit-string and evaluate the resulting string
as a natnum in binary.
So you say you have a mapping of every path to a node.
Well, that node has a number (from above), so you also have a mapping
from every path TO A NATURAL NUMBER.
We diagonalize this mapping by constructing the following denumerably
long
infinite path that YOU DO NOT map TO ANY natnum:
Call your mapping Y(.).
Y(p)=n, where p is some path and n is a node
in the infinite binary tree. Let the i>th bit of path p be called
p_i: if the i>th step in the path is to the left on the binary tree,
then p_i=0; if it is to the right then p_i=1.
If you want to know node n>s number n, represent the path
(from the root) to the node in binary, and replace the root with a 1
in front (left).
Evaluate that bit-string as a natnum. That is node n>s number (also
n).
It is easy to design a path, MY path, M, that PROVABLY, for EVERY n,
is NOT the path
that You map to n: Y(M)=/=n FOR ANY n.
MY PATH M IS NOT INCLUDED in your mapping, so your mapping DOES NOT
include ALL the infinite paths.
First of all, Y(M)=/=0 since no path maps to 0 in this scenario.
You do not map My path to 1, because whichEVER path YOU mapped to 1,
MY path takes THE OPPOSITE direction at step 1.
You see, your problem is, if you want to prove that there are NO MORE
infinite paths
than nodes, you have to not only map each path to 1 node, you have to
ALSO
map each NODE to at MOST ONE path. Otherwise you could map EVERY path
to 0,
which wouldn>t prove anything.
THIS MEANS YOUR MAPPING Y IS INVERTIBLE.
Y(p)=n ( you falsely allege), for every path p, but if it>s 1-1
and if you>re not leaving out any numbers, then Y has an inverse, call
it Y again
(it>ll be clear which from the type of the arguments) satisfying
Y(n)=p. In other words, you ALSO map every NODE to some PATH.
Third of all, my path does not map to 2, because whatever path
YOU mapped to 2, MY path goes THE OTHER WAY
in its 2nd step. Do you get it yet? MY PATH IS NOT IN your map.
For all i, M_i=1-[Y(n)]_i,
where Y(n) is the path that you mapped to n.
In other words,
YOU>RE STUPID. |
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george
Guest
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| Posted: Fri Oct 10, 2008 8:05 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]It is very easy to see that there are not more paths than nodes in
the
binary tree.
[/quote]
It CANNOT be "easy to see" SINCE IT IS FALSE!
NOBODY, INCLUDING YOU, HAS EVER seen this!
[quote]For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....
o
|
/
/
/
...
[/quote]
This means that any node that turns right at the 0th step will NOT map
to 0
(or to node 0; your mapping to nodes is equivalent to mapping to
natnums
since the nodes are countable, which means they can be exactly matched
1-1 with the natnums).
[quote]The next node
[/quote]
WHICH node is next?
We>ll say it>s the one on the left, the path to which is just 0.
We>ll call that node 1.
[quote]is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...
[/quote]
This means that any path that turns LEFT at the 1th (second) step will
NOT
be mapped to node 1. So MY path, just to make sure it maps to
neither 0 nor 1,
will turn right on the 0th step and left on the 1st.
[quote] o
|
o
/ \
/ \
/ \
...
The next node is mapped on another path, no matter which one, say one
always changing direction, i.e., 0.010101...
[/quote]
The next node, node 2, Has a 0 in position 2 in your example.
So MY path will have a 1 there. So so far, my path is 101.
It differs FROM ALL THREE of the example paths you have thus far
mentioned.
It differs from the path you map to 0 digit-position 0 (the first).
It differs from the
path you map to 1 at digit-position 1 (the second). It differs from
the path you
map to 2 at digit-position 2 (the third).
I CAN keep this up ad infinitum, DUMBASS.
[quote]
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...
And so on.
[/quote]
And so on, my finite path keeps growing, DISAGREEING WITH EVERY
infinite path you have proferred, at a rather early FINITE position
(the same position
as the number of the node you are now up to, in fact).
[quote]Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But it is clear, that every infinite path that exists
in the infinite binary tree has one node mapped onto it.
Of course we could also have selected the path pi-3 or any other one
to start with. There is no path existing (!) in the tree being left
out by
the mapping.
[/quote]
There IS SO TOO. If you give the mapping, I AM GIVING a path that IS
NOT INCLUDED
in it! |
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george
Guest
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| Posted: Fri Oct 10, 2008 8:07 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]Anyhow, the binary tree is a crystal-clear construction that avoids
the confusing opinion of undistinguishable numbers. It does never
insert its roots into the mud of the continuum (Kontinuumssauce).
[/quote]
This is just bullshit. The INFINITE part of the binary tree IS
ALWAYS
the continuum. The part you have investigated thus far is finite;
the part you have not investigated yet IS ALWAYS UNcountabl.
[quote]Therefore all its paths are and remain countable,
[/quote]
You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"! All sometimes means EACH around here!
And no INDIVDIDUAL path can be uncountable; they are all countably
infinitely LONG! But they are uncountably WIDE -- there are
uncountably
infinitely many OF them!
I realize your native language is not English but that is simply not
an excuse.
[quote]It is very easy to see that there are not more paths than nodes in
the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....
o
|
/
/
/
...
The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...
[/quote]
You CANNOT SAY "no matter which one". It MATTERS which one.
Here is the short version of your proof that the reals are countable:
1) pick a 1st real, no matter which one.
2) pick a 2nd real, no matter which one, as long as it is different
from all the previous
ones.
3) repeat ad infinitum.
4) This clearly maps every natural to a real
5) therefore (unsound) it clearly maps EVERY real to a natural.
THIS IS ALL you have said. It is completely unsupported. |
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george
Guest
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| Posted: Fri Oct 10, 2008 8:14 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
Every node in your infinite binary tree occurs at the
END OF A FINITE path.
Since that finite path to the node IS A FINITE BIT-STRING,
it uniquely represents A NATURAL NUMBER for the node.
Obviously, the nodes can be ordered:
the root is 1, the (node at the end of the)
terminating path 0 is 2, the terminating path 1 is 3,
the path 00 terminates at the node numbered 4, the path 01
termiantes at node#5,
the path 10 terminates at the node numbered 6, and the terminating
path 11 is 7, etc.
You just order them left-to-right across each row, with 0000 on the
left and 1111 on
the right. If you want to know the number corresponding to a path,
just put a 1
in front of the path-as-a-bit-string and evaluate the resulting string
as a natnum in binary.
If you want to know node n>s number n, represent the path
(from the root) to the node in binary, and replace the root with a 1
in front (left).
Evaluate that bit-string as a natnum. That is node n>s number (also
n).
So you say you have a mapping of every path to a node.
Well, that node has a number (from above), so you also have a mapping
from every path TO A NATURAL NUMBER.
We diagonalize this mapping by constructing the following denumerably
long
infinite path that YOU DO NOT map TO ANY natnum:
Call your mapping Y(.).
Y(p)=n, where p is some path and n is a node
in the infinite binary tree. Let the i>th bit of path p be called
p_i: if the i>th step in the path is to the left on the binary tree,
then p_i=0; if it is to the right then p_i=1.
It is easy to design a path, MY path, M, that PROVABLY, for EVERY n,
is NOT the path
that You map to n: Y(M)=/=n FOR ANY n.
MY PATH M IS NOT INCLUDED in your mapping, so your mapping DOES NOT
include ALL the infinite paths.
First of all, Y(M)=/=0 since no path maps to 0 in this scenario.
You do not map My path to 1, because whichEVER path YOU mapped to 1,
MY path takes THE OPPOSITE direction at step 1.
You see, your problem is, if you want to prove that there are NO MORE
infinite paths
than nodes, you have to not only map each path to 1 node, you have to
ALSO
map each NODE to at MOST ONE path. Otherwise you could map EVERY path
to 0,
which wouldn>t prove anything.
THIS MEANS YOUR MAPPING Y IS INVERTIBLE.
Y(p)=n ( you falsely allege), for every path p, but if it>s 1-1
and if you>re not leaving out any numbers, then Y has an inverse, call
it y
(it>ll be clear not only from case but from type of the arguments as
well,
which Y we mean)) satisfying
y(n)=p. In other words, you ALSO map every NODE to some PATH.
Third of all, my path does not map to 2, because whatever path
YOU mapped to 2, MY path goes THE OTHER WAY
in its 2nd step. Do you get it yet? MY PATH IS NOT IN your map.
For all n, M_n=1-[y(n)]_n,
where y(n) is the path that you mapped to n.
This, unlike all your handwaving about the picture,
IS A PROOF.
And it does NOT DISAGREE with your picture in ANY way (although
it does contradict it). It BEGINS BY ASSUMING THAT YOUR PICTURE IS
CORRECT AND THAT YOU DO HAVE a list.
But it constructs, FROM YOUR mapping, a path that IS NOT IN YOUR
mapping. |
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george
Guest
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| Posted: Fri Oct 10, 2008 8:22 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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On Oct 9, 3:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.
[/quote]
You HAVE NOT DONE this!
All you have said is "Take node 1. Give it a path.
Take node 2. Give it a path, it doesn>t matter which.
Take node 3, give it a path. It doesn>t matter which".
THAT IS ALL you have said! That obviously DOES NOT NECESSARILY
use all paths! I could do THAT using ONLY paths of all 1s followed by
all 0>s!
I take node 0. I give it the all 0>s path.
I take node 1. I give it the one 1 followed by all 0>s path.
I take node 2. I give it the two 1>s followed by all 0>s path.
I can do this for ALL natnums. When I get to the end I will have
used up all the natnums, BUT NOWHERE NEAR all the paths!
In fact, NO MATTER WHAT way you assign the paths, YOU WILL NEVER
COME CLOSE to using all the paths!
The easy way to prove this is just (even though you can>t finish)
to ASSUME THAT YOU HAVE finished and given EVERY node a path!
THEN, WE CAN CONSTRUCT a path that YOU DID NOT include!
It is just the path whose nth step is THE OPPOSITE WAY from the
nth step of the path that YOU mapped to node n!
THAT path IS NOT IN YOUR map! |
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Virgil
Guest
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| Posted: Fri Oct 10, 2008 11:30 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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In article
<5f06517f-bb41-4a02-b04d-130f16f4a7f3@i76g2000hsf.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
In article
871188cc-f053-4ab5-870f-0b57ce13e...@n33g2000pri.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:
o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \ ...
And so on.
Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.
Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.
But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".
Not according to the rule of my game. Consider the figure above. Node
x buys you one of the path leaving the path
0.111... at position x. You may say that there are many paths
deviating from 0.111..., and you are right. But if one of them
deviates from said path, it will get get another node. There is no
escape.
o
|
o
/ \
o \
/ \ \
/ / \
/ \ x
/ / / \
... /
\
With every won path you get infinitely many nodes. Each of them helps
you to get another path, one and only one path per node. In this game
you have the ratio of paths to nodes P/N = 0 at every step. The
infinite binary tree covers every step, but not more.
Similarly the usual decimal expansion of real numbers covers every
digit, but not more. There is no last digit of pi. There is every
digits of pi - unless one looks to close, but that is not in question
here. Set theorists insistuing of uncountably many reals must insist
that there are more than every digit of pi. I am not inclined to
discuss about such folly.
[/quote]
Your "game" overlooks every path with infinitely many branchings in both
left and right directions, which is the vast majority of paths.
[quote]
But if every node of the tree has been used to mark one path
Unless that node is a terminal (leaf) node, it marks more than one path,
Try to understand the rule of the game (there is only one). Try to
understand that there is no leaf node.
[/quote]
It is the absence of any such leaf nodes that makes you game fail to
distinguish each path from all its neighbors. |
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Virgil
Guest
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| Posted: Fri Oct 10, 2008 11:40 pm Post subject: Re: Binary Tree and Pairs of Nodes |
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In article
<f0f8ab79-568e-4989-b712-5f5c2608091b@u27g2000pro.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
[quote]On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals.
Then your trees are either not infinite or not complete.
The infinite binary tree is as complete as the number system
[/quote]
Then in YOUR world, both are incomplete.
, i.e.,
[quote]the binary or decimal representation of real numbers can be. There is
no chance to introduce another path. There is no chance to introduce
another real. All combinations of nodes-values, all combinations of
digits are present.
[/quote]
The binary expansions for 1/3 or sqrt(2) are neither represented in your
i complete tree, as they both require infinitely many branchings in both
directions, which you game does not allow.
[quote]
The infinite path 0.111... is covered by the infinite sequence of
paths
0.1000..., 0.11000..., 0.111000..., ... (*)
[/quote]
But then WM is speaking of INFINITE SETS of his "finite" paths, not just
the individual paths themselves, and there are always more such sets of
such paths than there are such paths.
So WM loses again. |
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