How man different paths can exist in the Complete Infinite Binary Tree?

[WM]

A path in the Complete Infinite Binary Tree

0
/ \
0 1
/ \ / \
0 1 0 1
...

is an infinite sequence of nodes, 0 or 1, connected by edges / or \.

Two paths, A and B, differ, if A contains at least one node that is not in B. Vice versa this necessarily implies that B contains at least one node that is not in A.

A bunch of X paths is split into two bunches their last common node N.

What is X? How many paths can be in a bunch? Not more than there are nodes below N crossed by paths of the bunch which can further separate the paths of the bunch. Therefore every bunch can contain at most countably many paths which later can be separated.

If there are uncountably many paths in a bunch, almost all can never be separated, that means they are not observably different.

It is easy to transform this example to the case of real numbers in the interval [0, 1).

Regards, WM

[Dan Christensen]

Any particular infinite path along an infinite and complete tree can be encoded as an infinite string of binary digits (0 = left, 1 = right).

There are uncountably many such paths. The proof is a simple variation of famous Cantor's diagonal proof. Given any list of such paths, we can generate another path that is not on the list.

I hope this helps.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[burs...@gmail.com]

Nope, nodes are from N, and paths are from P(N).

[Juergen R.]

Holy Camoly! Christmas has come early this year - it is June and already we have the annual presentation of the fallacy of the binary tree by the Great Prefosser Mückenzwerg!

Once again Mückenzwerg proves that the reals are countable and you-all are to stupid to appreciate this world-shaking fact. SAD!

[Python]

Indeed! Hopefully Herr 'Venom' Frankenheim is more than 40, so The Field
medal committee won't know the dilemma of choosing between this amazing
result and the other fantastic theorems Mueckenheim has stated such as
"Any integer belongs to a finite set", "Everything belong to a
singleton" or P(N)\{2,4,6, ... } = P(N). Strange enough, but usual with
Herr Mueckenheim from Hochschule Augsburg Crank Institute, what is true
there is insignificant and what is not insignificant is false.

[Ben Bacarisse]

WM <wolfgang.m...@hs-augsburg.de> writes:

> A path in the Complete Infinite Binary Tree
>
> 0
> / \
> 0 1
> / \ / \
> 0 1 0 1
> ...
>
> is an infinite sequence of nodes, 0 or 1, connected by edges / or \.

<snip>

> A bunch of X paths is split into two bunches their last common node N.
>
> What is X?

Un-countably many. Every finite rooted path (which uniquely determines
a node) is the prefix of as many paths as are in the whole tree. It's
trivial to biject the two sets. Of course, that last past does not help
if you don't know how many paths there are in the whole tree.

> How many paths can be in a bunch? Not more than there are nodes below
> N crossed by paths of the bunch which can further separate the paths
> of the bunch.

The wording seems clumsy but it sounds correct -- there are as many
paths below N as in the whole tree. Countably many nodes have
uncountably many paths through them.

Taking / and 0 and \ as 1, a rooted path in a binary tree (including
those of finite depth) is simply the indicator function for a unique
element of P(L) where L is the set of levels -- {1, 2, ... d} for the
tree of depth d and N for the infinite tree. There is no bijection
between L and P(L) for any L. Proofs available in all good textbooks,
but not in yours.

> Therefore every bunch can contain at most countably many
> paths which later can be separated.

Every path in the tree is distinct at all times. Time has nothing to do
with it. At every node, there are un-countably many distinct paths
below it.

> If there are uncountably many paths in a bunch, almost all can never
> be separated, that means they are not observably different.

Every pair of them differ or they would be the same path! "Observably
different" is your made up term, and no doubt defines some countable
subset of paths. In mathematics what matters is whether two paths are
the same or not.

> It is easy to transform this example to the case of real numbers in
> the interval [0, 1).

Yes. This makes it clear how many paths there are in any "bunch".
Proofs of the uncountability of R abound.

--
Ben.

[Ross A. Finlayson]

There are number-theoretic proofs
of the uncountability of R, the
complete ordered field.

There are set-theoretic proofs
of the uncountability of P(N),
the powerset of ZF's ordinary N.


The sweep functions falls out of
the number-theoretic proofs and
builds a model of the interval [0,1].

In extra-ordinary ubiquitous ordinals,
the powerset function yields the
successor, order type, and anti-diagonal.


Proofs of the uncountability of R,
the complete ordered field, abound,
and sweep, the natural/unit equivalency
function, falls out of each as not being
contradicted being "onto".

[George Greene]

The title is wrong already.
"The complete infinite binary tree" IS ONE thing.
It has ONE number of paths.
The word "can" IS WRONG.
There is ONE fact of the matter about how many paths
DO
exist through this tree.

On Saturday, June 23, 2018 at 10:18:02 AM UTC-4, WM wrote:
> A path in the Complete Infinite Binary Tree
>
> 0
> / \
> 0 1
> / \ / \
> 0 1 0 1
> ...
>
> is an infinite sequence of nodes, 0 or 1, connected by edges / or \.

This is just WRONG.
You don't need the nodes AT ALL. And you certainly don't need to label the
nodes according to the direction of the edge that led to them. Except
for the root, all the nodes you have labeled 0 are labeled 0 BECAUSE they
are the LEFT child of their parent. Your tree therefore OUGHT to look like

> /
> 0
> / \
> 0 1
> ............
at the top. You actually have NO REASON for choosing 0 or 1 OR ANYTHING ELSE
as the label for the root. Of course, the improvement I have drawn here has
the problem that the leading edge doesn't connect 2 nodes, but so what?
There are many different ways of combining two things and in the case of
the binary tree, the way is THE ORDERED pair way; one of the subtrees is
always labeled "1st" or "left" or "/" and the other is always labeled "2nd"
or "right" or "\" or WHATEVER OTHER ORDERED list of 2 opposites YOU CHOOSE,
including (as you have) labeling the left "0" and the right "1".
The NAMES DON'T MATTER and the NODES DON'T MATTER.

The inifnite binary tree is the ordered pair of COPIES OF IT*SELF*.

[WM]

Am Samstag, 23. Juni 2018 18:44:23 UTC+2 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > A path in the Complete Infinite Binary Tree
> >
> > 0
> > / \
> > 0 1
> > / \ / \
> > 0 1 0 1
> > ...
> >
> > is an infinite sequence of nodes, 0 or 1, connected by edges / or \.
> <snip>
> > A bunch of X paths is split into two bunches their last common node N.
> >
> > What is X?
>
> Un-countably many. Every finite rooted path (which uniquely determines
> a node) is the prefix of as many paths as are in the whole tree. It's
> trivial to biject the two sets. Of course, that last past does not help
> if you don't know how many paths there are in the whole tree.

How many can be distinguished is easy to prove: At level n you can distinguish as many as there are nodes at level n. Since the set of nodes of all levels is countable, there cannot be a level with more than countably many nodes. This proves that not more paths can be distinguished.

>
> > Therefore every bunch can contain at most countably many
> > paths which later can be separated.
>
> Every path in the tree is distinct at all times. Time has nothing to do
> with it.

Not the time but the levels play a role.

> At every node, there are un-countably many distinct paths
> below it.

Distinct means to have different nodes at some level. This restricts the differing paths to countably many.

>
> > If there are uncountably many paths in a bunch, almost all can never
> > be separated, that means they are not observably different.
>
> Every pair of them differ or they would be the same path!

Correct. But to differ observably at level n means to occupy different nodes at level n. There are not more than countably many nodes at level n.

> "Observably
> different" is your made up term, and no doubt defines some countable
> subset of paths. In mathematics what matters is whether two paths are
> the same or not.

They differ but that cannot be observed? - Matheology!

>
> > It is easy to transform this example to the case of real numbers in
> > the interval [0, 1).
>
> Yes. This makes it clear how many paths there are in any "bunch".
> Proofs of the uncountability of R abound.

Contradictions like the above one abound too. Therefore you have to make up "differing without observably differing".

Regards, WM

[WM]

Am Samstag, 23. Juni 2018 20:28:54 UTC+2 schrieb George Greene:

The following is okay but does not change the facts.

>
> > /
> > 0
> > / \
> > 0 1
> > ............
> at the top. You actually have NO REASON for choosing 0 or 1 OR ANYTHING ELSE
> as the label for the root. Of course, the improvement I have drawn here has
> the problem that the leading edge doesn't connect 2 nodes, but so what?
> There are many different ways of combining two things and in the case of
> the binary tree, the way is THE ORDERED pair way; one of the subtrees is
> always labeled "1st" or "left" or "/" and the other is always labeled "2nd"
> or "right" or "\" or WHATEVER OTHER ORDERED list of 2 opposites YOU CHOOSE,
> including (as you have) labeling the left "0" and the right "1".
> The NAMES DON'T MATTER and the NODES DON'T MATTER.
>
> The inifnite binary tree is the ordered pair of COPIES OF IT*SELF*.

The question is how many paths can be distinguished. At level n there are 2^n nodes, and therefore 2^n paths (or bunches of paths) can be distinguished. At no level more than aleph_0 nodes exist. Therefore no more than aleph_0 paths can be distinguished in the Complete Infinite Binary Tree.

Regards, WM

[WM]

Am Samstag, 23. Juni 2018 16:57:58 UTC+2 schrieb burs...@gmail.com:
> Nope, nodes are from N, and paths are from P(N).

Nevertheless paths can only differ by nodes (or edges). At evel 3 we can distinguish 8 different bunches of paths. A level 5 we can distinguish 32 different bunches. At no level we can distinguish more than aleph_0 different bunches of paths. Whether every bunch is assumed to contain uncountably many paths is irrelevant. The belief in not distinguishable paths is matheology.

Regards, WM

[burs...@gmail.com]

bunch <> branch

You are taking the haystack for the blade of grass

What do you want to count, the blades of grass or something else.

[WM]

Am Sonntag, 24. Juni 2018 12:52:59 UTC+2 schrieb burs...@gmail.com:
> Am Sonntag, 24. Juni 2018 11:37:23 UTC+2 schrieb WM:
> > Am Samstag, 23. Juni 2018 16:57:58 UTC+2 schrieb burs...@gmail.com:
> > > Nope, nodes are from N, and paths are from P(N).
> >
> > Nevertheless paths can only differ by nodes (or edges). At evel 3 we can distinguish 8 different bunches of paths. A level 5 we can distinguish 32 different bunches. At no level we can distinguish more than aleph_0 different bunches of paths. Whether every bunch is assumed to contain uncountably many paths is irrelevant. The belief in not distinguishable paths is matheology.
> >
> You are taking the haystack for the blade of grass
>
> What do you want to count, the blades of grass or something else.

Because mathematics has to be precise. Otherwise it falls to pieces. And further it is very easy once you have lost the blinkers.

In order to distinguish a path A from another path P, you need a node of P that is not in A. The level, where this node can be found, does not matter. But of course, the level must have a finite index n, because there are no levels with infinite index omega or larger.

In order to distinguish A from n paths P1, P2, ..., Pn we need n nodes. It cannot be excluded that one node a of A is sufficient to distinguish A from all paths P1, P2, ..., Pn, but these are not n different paths unless there are nodes that distinguish P1 from P2 to Pn, and P2 from P3 to Pn and so on. In total n - 1 nodes are required to distinguish n paths P1, P2, ..., Pn.

In order to distinguish a further path from the former, there is at least one other node necessary. However, at most a countable set of nodes is available.

Regards, WM

[burs...@gmail.com]

You cannot distinguish only countable many blades of grass.
If there were only countable many blades of grass,
then you could find a bijective mapping.

But this is impossible. See Cantors theorem. You have
lost your blinkers. You think the bunch is the branch.
You think they heystack is the blade of grass.

I a separator is not what you think. A separate, when
you mean by separator, is only a node. Its the beginning
of a whole subtree, which has again uncountable

many branches.

[burs...@gmail.com]

Since you can continue this process ad infinitum,
doesn't mean you have counted anything.

You have told us already a dozen time that after
finitely many natural numbers, there are still
infinitely many natural numbers.

Now here we have the case after finitely many
nodes, there are still uncountable many
infinite branches.

Its exactly the same what you were telling us all
the time with matural numbers, only what follows
is not countable infinite but uncountable infinite.

After finitely many nodes, there are still
uncountable many infinite branches. Its exactly
the same as your Tristam Shandy or otherwise

triviality, from Augsburg Crank institute,
streetnumber 123, in Volkswagen Omlette avenue.

[burs...@gmail.com]

Here is some verbal nonsense, just for the fun
of it: "since the nodes are an unfinished infinity
they cannot count the branches."

Since I neither know what a finished infinity
or an unfinished infinity should mean, please
don't buy stockmarket or build rockets based

on the above sentence. Its just fun sentence.

[exflaso....@gmail.com]

On Sunday, June 24, 2018 at 10:48:00 AM UTC-4, WM wrote:
> In order to distinguish A from n paths P1, P2, ..., Pn we need n nodes. It cannot be excluded that one node a of A is sufficient to distinguish A from all paths P1, P2, ..., Pn, but these are not n different paths unless there are nodes that distinguish P1 from P2 to Pn, and P2 from P3 to Pn and so on. In total n - 1 nodes are required to distinguish n paths P1, P2, ..., Pn.

You're looking at this backwards. You're describing how many nodes are needed to distinguish a countable collection of N paths (N-1). That doesn't tell you how many paths are distinguished by a countable collection of N nodes (2^N).

For example, while 5 paths need only 4 nodes to be distinguished, those 4 nodes can distinguish up to 16 total paths. The problem you're describing here falls far short of the problem you initially presented.

EFQ

[exflaso....@gmail.com]

On Sunday, June 24, 2018 at 11:17:10 AM UTC-4, exflaso....@gmail.com wrote:
> On Sunday, June 24, 2018 at 10:48:00 AM UTC-4, WM wrote:
> > In order to distinguish A from n paths P1, P2, ..., Pn we need n nodes. It cannot be excluded that one node a of A is sufficient to distinguish A from all paths P1, P2, ..., Pn, but these are not n different paths unless there are nodes that distinguish P1 from P2 to Pn, and P2 from P3 to Pn and so on. In total n - 1 nodes are required to distinguish n paths P1, P2, ..., Pn.
>
> You're looking at this backwards. You're describing how many nodes are needed to distinguish a countable collection of N paths (N-1). That doesn't tell you how many paths are distinguished by a countable collection of N nodes (2^N).

Also, your N-1 is wrong. Consider the following paths:

P1 = 100̄
P2 = 010̄
P3 = 110̄
P4 = 000̄

These are clearly four distinct paths, but only two total nodes differ between them, so you don't need three nodes to distinguish them.

So, not only are you looking at the problem backwards, but you got it wrong in the direction you did you look at.

EFQ

[Ben Bacarisse]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Samstag, 23. Juni 2018 18:44:23 UTC+2 schrieb Ben Bacarisse:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > A path in the Complete Infinite Binary Tree
>> >
>> > 0
>> > / \
>> > 0 1
>> > / \ / \
>> > 0 1 0 1
>> > ...
>> >
>> > is an infinite sequence of nodes, 0 or 1, connected by edges / or \.
>> <snip>
>> > A bunch of X paths is split into two bunches their last common node N.
>> >
>> > What is X?
>>
>> Un-countably many. Every finite rooted path (which uniquely determines
>> a node) is the prefix of as many paths as are in the whole tree. It's
>> trivial to biject the two sets. Of course, that last past does not help
>> if you don't know how many paths there are in the whole tree.
>
> How many can be distinguished is easy to prove: At level n you can
> distinguish as many as there are nodes at level n. Since the set of
> nodes of all levels is countable, there cannot be a level with more
> than countably many nodes. This proves that not more paths can be
> distinguished.

No. At every level n none of the infinite paths are "distinguished".
At every level n, only a finite set of finite paths are "distinguished".
You can't learn much about the infinite paths with this argument.

However, considering levels can help. The rooted paths in the tree of
level n correspond to the 2^n elements of 2^n P({1, 2, ..., n}). The
rooted paths of the infinite binary tree correspnd to the elements of
P(N). To know how many there are, you need to know what 2^|N| is.

>> At every node, there are un-countably many distinct paths
>> below it.
>
> Distinct means to have different nodes at some level.

Yes. At level n, only a finite set of 2^n paths can be distinguished.

> This restricts the differing paths to countably many.

I means there are 2^|N| paths in the infinite binary tree.

>> > If there are uncountably many paths in a bunch, almost all can never
>> > be separated, that means they are not observably different.
>>
>> Every pair of them differ or they would be the same path!
>
> Correct. But to differ observably at level n means to occupy different
> nodes at level n. There are not more than countably many nodes at
> level n.

For every pair of infinite paths p1 and p2 there is a least n at which
p1(n) != p2(n).

And the set of 2^(n-1) nodes at level n divides the infinite paths into
2^(n-1) sets of uncountably many paths.

--
Ben.

[WM]

Am Montag, 25. Juni 2018 03:15:07 UTC+2 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Samstag, 23. Juni 2018 18:44:23 UTC+2 schrieb Ben Bacarisse:
> >> WM <wolfgang.m...@hs-augsburg.de> writes:
> >>
> >> > A path in the Complete Infinite Binary Tree
> >> >
> >> > 0
> >> > / \
> >> > 0 1
> >> > / \ / \
> >> > 0 1 0 1
> >> > ...

> At every level n none of the infinite paths are "distinguished".
> At every level n, only a finite set of finite paths are "distinguished".

So it is. These finite paths belong to bunches of paths of infinite paths. But at countably many levels only countably bunches of paths can be distinguished. Nowhere uncountably many bunches of paths (or paths) can be distinguished.

> You can't learn much about the infinite paths with this argument.

We can learn more than enough: At no level more than countably bunches of paths (or even paths themselves) can be distinguished.

>
> However, considering levels can help. The rooted paths in the tree of
> level n correspond to the 2^n elements of 2^n P({1, 2, ..., n}). The
> rooted paths of the infinite binary tree correspnd to the elements of
> P(N). To know how many there are, you need to know what 2^|N| is.

We know that all nodes are countable and that there is no level |N|.

>

> Yes. At level n, only a finite set of 2^n paths can be distinguished.
>
> > This restricts the differing paths to countably many.
>
> I means there are 2^|N| paths in the infinite binary tree.
>

No, there is no level |N|.

> > Correct. But to differ observably at level n means to occupy different
> > nodes at level n. There are not more than countably many nodes at
> > level n.
>
> For every pair of infinite paths p1 and p2 there is a least n at which
> p1(n) != p2(n).

Then there are less than uncountably many paths.

Regards, WM

[WM]

Am Sonntag, 24. Juni 2018 17:17:10 UTC+2 schrieb exflaso....@gmail.com:
> On Sunday, June 24, 2018 at 10:48:00 AM UTC-4, WM wrote:
> > In order to distinguish A from n paths P1, P2, ..., Pn we need n nodes. It cannot be excluded that one node a of A is sufficient to distinguish A from all paths P1, P2, ..., Pn, but these are not n different paths unless there are nodes that distinguish P1 from P2 to Pn, and P2 from P3 to Pn and so on. In total n - 1 nodes are required to distinguish n paths P1, P2, ..., Pn.
>
> You're looking at this backwards. You're describing how many nodes are needed to distinguish a countable collection of N paths (N-1). That doesn't tell you how many paths are distinguished by a countable collection of N nodes (2^N).
>
> For example, while 5 paths need only 4 nodes to be distinguished, those 4 nodes can distinguish up to 16 total paths.

No. Give an example. Fail.

Regards, WM

[WM]

Am Sonntag, 24. Juni 2018 17:31:04 UTC+2 schrieb exflaso....@gmail.com:
> On Sunday, June 24, 2018 at 11:17:10 AM UTC-4, exflaso....@gmail.com wrote:
> > On Sunday, June 24, 2018 at 10:48:00 AM UTC-4, WM wrote:
> > > In order to distinguish A from n paths P1, P2, ..., Pn we need n nodes. It cannot be excluded that one node a of A is sufficient to distinguish A from all paths P1, P2, ..., Pn, but these are not n different paths unless there are nodes that distinguish P1 from P2 to Pn, and P2 from P3 to Pn and so on. In total n - 1 nodes are required to distinguish n paths P1, P2, ..., Pn.
> >
> > You're looking at this backwards. You're describing how many nodes are needed to distinguish a countable collection of N paths (N-1). That doesn't tell you how many paths are distinguished by a countable collection of N nodes (2^N).
>
> Also, your N-1 is wrong. Consider the following paths:
>
> P1 = 100̄
> P2 = 010̄
> P3 = 110̄
> P4 = 000̄
>
> These are clearly four distinct paths,

No. All paths start with the root node. Try again.

Regards, WM

[WM]

Am Sonntag, 24. Juni 2018 16:55:12 UTC+2 schrieb burs...@gmail.com:

> You cannot distinguish only countable many blades of grass.
> If there were only countable many blades of grass,
> then you could find a bijective mapping.

If such mapping made sense. It does not however. Just this is shown by the Binary Tree.

>
> But this is impossible. See Cantors theorem.

It has been contradicted. See the undisputedly countable set of nodes and the fact that n-1 nodes are required to distinguish n infinite paths.

In fact there cannot be more paths distinguished than there are nodes on one level. Their numbers grow but are always finite. A sequence with always finite terms can have limit aleph_0 or omega, never 2^aleph_0.

> I a separator is not what you think. A separate, when
> you mean by separator, is only a node. Its the beginning
> of a whole subtree, which has again

only countably many separators.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 03:15:07 UTC+2 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:


> Yes. At level n, only a finite set of 2^n paths can be distinguished.

By 2^n nodes. And what is the limit of this sequence of finite terms 2^n? It is aleph_0. It is not 2^aleph_0. Therefore even in the limit only aleph_0 paths can be distinguished.

>
> > This restricts the differing paths to countably many.
>
> I means there are 2^|N| paths in the infinite binary tree.

Chuckle. There are 2^|N| nodes? Try again.

Regards, WM

[burs...@gmail.com]

Nope 2^B is the cardinal of P(A), when B is the
cardinal of A. This is called cardinal exponentiation,

and it gives you uncountable many branches.
http://mathworld.wolfram.com/CardinalExponentiation.html

[WM]

Am Montag, 25. Juni 2018 11:47:13 UTC+2 schrieb burs...@gmail.com:
> Nope 2^B is the cardinal of P(A), when B is the
> cardinal of A. This is called cardinal exponentiation,

The sequence (2^n) has limit aleph_0. Otherwise the set of unit fractions and of natural numbers would be uncountable:
(1/2), (1/3, 1/4), (1/5, 1/6, 1/7, 1/8), ...

This is the same sequence as the nodes by level: 1, 2, 4, ...
Try again.

Regards, WM

[burs...@gmail.com]

Nope 2^B is the cardinal of P(A), when B is the
cardinal of A. This is called cardinal exponentiation,

and it gives you uncountable many branches.
http://mathworld.wolfram.com/CardinalExponentiation.html

And as you can easily read, the cardinality 2^B
is not defined with the help of some limit.
Limits don't play the same role for cardinals as
in the definition of ordinals. They are much sharper.

Also ordinal exponentiation is different from cardinal
exponentiation. Words such as uncountable refer
to cardinal and not ordinal.

[WM]

Am Montag, 25. Juni 2018 12:11:59 UTC+2 schrieb burs...@gmail.com:


> And as you can easily read, the cardinality 2^B
> is not defined with the help of some limit.

But the number of nodes N as a function N(n) of the level number n has a limit. It is omega.

> Limits don't play the same role for cardinals as
> in the definition of ordinals. They are much sharper.

The Binary Tree can distinguish as many paths at level n as it has nodes at level n. The number of nodes is countable. Hence not more than countably many infinite paths can be distinguished in the Complete Infinite Binary Tree.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 03:15:07 UTC+2 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:


> > Distinct means to have different nodes at some level.
>
> Yes. At level n, only a finite set of 2^n paths can be distinguished.

What is the limit of the sequence 1, 2, 4, 8, ... ? Is it uncountable? Then there are uncountably many unit fractions

(1/2), (1/3, 1/4), (1/5, 1/6, 1/7, 1/8), ...

and uncountably many natural numbers.

Hence there is no level with more than aleph_0 nodes and therefore not more than aleph_0 distinct paths.

Regards, WM

[exflaso....@gmail.com]

Pick whatever root node R you want:

P1 = R100̄
P2 = R010̄
P3 = R110̄
P4 = R000̄

EFQ

[WM]

Am Montag, 25. Juni 2018 15:39:25 UTC+2 schrieb exflaso....@gmail.com:


> Pick whatever root node R you want:
>
> P1 = R100̄
> P2 = R010̄
> P3 = R110̄
> P4 = R000̄
>

> These are clearly four distinct paths, but only two total nodes differ between them, so you don't need three nodes to distinguish them.

The following diagram may show your mistake:

R

/ \
0 1
/ \ / \
0 1 0 1

/\ /\ /\ /\
0 10 1 0 1 0 1

Your four paths have four different nodes at the last shown level. (There are forur different nodes with value 0.)

Regards, WM

[burs...@gmail.com]

Nope. The number of nodes has not the limit omega.
The number of nodes is not the same as the set of
nodes. The number of nodes diverges:

lim n->oo #(nodes on level n) = lim n->oo 2^n = oo

What you mean is that the set of nodes on each
level together form the set of all nodes. Well
this is obviously true:

union n in N (nodes on level n) = all nodes

[exflaso....@gmail.com]

On Monday, June 25, 2018 at 10:00:59 AM UTC-4, WM wrote:
> Am Montag, 25. Juni 2018 15:39:25 UTC+2 schrieb exflaso....@gmail.com:
>
> > Pick whatever root node R you want:
> >
> > P1 = R100̄
> > P2 = R010̄
> > P3 = R110̄
> > P4 = R000̄
> >
> > These are clearly four distinct paths, but only two total nodes differ between them, so you don't need three nodes to distinguish them.
>
> The following diagram may show your mistake:

Oh yes, I see. This still doesn't change the fact that N-1 and 2^N are different. You're answering your original question with the wrong answer.

EFQ

[burs...@gmail.com]

The easiest is to view the nodes as a set:

{t1, t2, t3, ...}

With the rule that t2n is the left successor of
tn and t2n+1 is the right successor of t2n.

You can check yourself that this works:

t4 ...
/
t2\
/ t5 ...
t1
\ t6 ...
t3/
\
t7 ...

This numbering of nodes, has also the
advantage, that the levels are easily
identifiable.

The level n, is just the nodes:

t2^(n-1) .. t2^n-1 : Nodes from level n

[burs...@gmail.com]

Corr.: Typo

With the rule that t2n is the left successor of

tn and t2n+1 is the right successor of tn.

[WM]

Am Montag, 25. Juni 2018 16:04:01 UTC+2 schrieb burs...@gmail.com:

> Nope. The number of nodes has not the limit omega.
> The number of nodes is not the same as the set of
> nodes.

The set has cardinality aleph_0. Therefore every level has not more than aleph_0 nodes.

n-1 nodes are required to distinguish n infinite paths.
All nodes can be taken at the same level. Therefore it is impossible to distinguish more than aleph_0 paths.

If more paths are sassumed to "exist", they cannot be distinguished by nodes. They clearly belong to matheology.

> The number of nodes diverges:
>
> lim n->oo #(nodes on level n) = lim n->oo 2^n = oo

In set theory this number is called aleph_0. (oo means potential infinity.)

>
> What you mean is that the set of nodes on each
> level together form the set of all nodes. Well
> this is obviously true:
>
> union n in N (nodes on level n) = all nodes

The set of nodes is countable:

0
1 2
3 4 5 6
...

That is all that counts for the argument. No level has more than aleph_0 different paths passing through it.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 16:06:02 UTC+2 schrieb exflaso....@gmail.com:


> Oh yes, I see. This still doesn't change the fact that N-1 and 2^N are different.

n-1 nodes are required to distinguish n infinite paths.

All nodes can be taken at the same level (like the 4 nodes in the example above). There is no level having more than aleph_0 nodes (since all nodes of the Binary Tree are countable). Therefore it is impossible to distinguish more than aleph_0 paths.

If more paths are sassumed to "exist", they cannot be distinguished by nodes. They clearly belong to matheology.

Regards, WM

[exflaso....@gmail.com]

On Monday, June 25, 2018 at 10:20:09 AM UTC-4, WM wrote:
> Am Montag, 25. Juni 2018 16:06:02 UTC+2 schrieb exflaso....@gmail.com:
>
>
> > Oh yes, I see. This still doesn't change the fact that N-1 and 2^N are different.
>
> n-1 nodes are required to distinguish n infinite paths.

You're still thinking of it backwards. The question isn't how many nodes are required for N paths; it's how many paths are produced by N nodes.

EFQ

[WM]

Am Montag, 25. Juni 2018 16:12:02 UTC+2 schrieb burs...@gmail.com:

> You can check yourself that this works:
>
> t4 ...
> /
> t2\
> / t5 ...
> t1
> \ t6 ...
> t3/
> \
> t7 ...
>

I know. But the countability of the set of nodes is not disputed.

> This numbering of nodes, has also the
> advantage, that the levels are easily
> identifiable.

The important fact is that no level has more than a finite number of nodes and even in the limit there are at most aleph_0 nodes.

It remains one of the greatest mysteries of mankind for me, how an intelligtent person with say IQ > 90 can claim that there are uncountably many infinite paths in the Binary Tree. More houses than bricks. Unbelievable perversion!

Cantor really has spoilt generations.

Regards, WM

[WM]

The question that I have asked and answered is this: How many infinite paths at most can be distinguished in the Binary Tree? That's mathematics: Estimating upper and lower bounds by clever approaches.

Regards, WM

[exflaso....@gmail.com]

On Monday, June 25, 2018 at 10:58:37 AM UTC-4, WM wrote:
> Am Montag, 25. Juni 2018 16:30:34 UTC+2 schrieb exflaso....@gmail.com:
> > On Monday, June 25, 2018 at 10:20:09 AM UTC-4, WM wrote:
> > > Am Montag, 25. Juni 2018 16:06:02 UTC+2 schrieb exflaso....@gmail.com:
> > >
> > > > Oh yes, I see. This still doesn't change the fact that N-1 and 2^N are different.
> > >
> > > n-1 nodes are required to distinguish n infinite paths.
> >
> > You're still thinking of it backwards. The question isn't how many nodes are required for N paths; it's how many paths are produced by N nodes.
>
> The question that I have asked and answered is this: How many infinite paths at most can be distinguished in the Binary Tree?

That's what you asked, but then you keep answering it by saying "N-1 nodes are required to distinguish N paths". That's an answer for a different question.

EFQ

[burs...@gmail.com]

But nodes and branches are not the same. You
see this also with this numbering. Lets repeat
what is the the nodes:

{t1, t2, t3, ...} = T

t4 ...
/
t2\
/ t5 ...
t1
\ t6 ...
t3/
\
t7 ...

Now the following things are clear:
- There are countable many nodes
- There is a relationship between a node
and its two successors
- We gave the set of nodes the name T, so
that we can more easily communicate

Now lets have also a look at branches. Take for
example the left most branch B:

{t1, t2, t4, ...} = B

...
/
t4
/
t2
/
t1

Now the following things are clear:
- The left most branch is not a node, its a set of nodes
- In this set of nodes, only one of the successors is important
- If we want we can give a branch also a name,
such as B, to facilitate communication

Now if we put all the nodes and a branch side by side,
we see that B is a subset of T:

{t1, t2, t3, t4, t5, t6, t7, t8, ...} = T

{t1, t2, t4, t8, ...} = B

We can generalize this and say more generally what a
branch is, namely a brach is a subset B of T:

{n[1], n[2], n[3], ...} = B

Such that:

n[k+1]=t2n if n[k]=tn

-- or --

n[k+1]=t2n+1 if n[k]=tn

Now how big is the set of all branches?

[burs...@gmail.com]

There is a simple diagonal argument that shows
that the set of all branches is uncountable.
Assume there were a countable list of all branches:

B1 = {a1,a2,a3,...}
B2 = {b1,b2,b3,...}
B3 = {c1,c2,c3,...}
...

Now we can construct a diagonal:

D = {a1',
b2',
c3',
...}

For the diagonal itself, just choose left
when the corresponding diagonal element was
right successor. And choose right when the
corresponding diagonal elememt was left.

This way the diagonal branch will be different
to all branches in the list. Since left successors
have even numbering t2n and are differnet from
right sucessors which have an odd numbering t2n+1.

Q.E.D.

[WM]

Am Montag, 25. Juni 2018 17:56:25 UTC+2 schrieb burs...@gmail.com:
> There is a simple diagonal argument that shows
> that the set of all branches is uncountable.
> Assume there were a countable list of all branches:
>
> B1 = {a1,a2,a3,...}
> B2 = {b1,b2,b3,...}
> B3 = {c1,c2,c3,...}
> ...
>
> Now we can construct a diagonal:
>
> D = {a1',
> b2',
> c3',
> ...}
>
> For the diagonal itself, just choose left
> when the corresponding diagonal element was
> right successor.

For every right successor there is also the left successor in the tree.

> And choose right when the
> corresponding diagonal elememt was left.

For every left successor there is also the right successor in the tree.

>
> This way the diagonal branch will be different
> to all branches in the list.

But not different from all paths in the Binary Tree. The reason is that all possible paths are in the Binary Tree by definition because it is the Complete Infinite Binary Tree.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 17:50:34 UTC+2 schrieb burs...@gmail.com:

> Now if we put all the nodes and a branch

you mean a path

> side by side,

>
> Now how big is the set of all branches?

Not larger than aleph_0.

Regards, WM

[WM]

No. That's a partial answer. The final answer is aleph_0 nodes are required to to distinguish aleph_0 paths. And more nodes are not available. Therefore more paths cannot be distinguished.

Regards, WM

[burs...@gmail.com]

No, I don't mean a path. A path is from node to node.
For example this set {t1,t2,t3} is a path:

t4
/
t2
/
t1

But a branch is a infinite path that starts in
the root. So branches and paths are not the same.

There are uncountable many branches. But there
are only countable many finite paths. Each
finite path is uniquely determined by the start
and end node, namely:

{a1, .., an}

Its then easy to see that there are only countable
many finite paths. But on the other hand there
are uncountable many branches.

[burs...@gmail.com]

Corr.:

No, I don't mean a path. A path is from node to node.

For example this set {t1,t2,t4} is a path:

[j4n bur53]

You might enjoy this paper for some terminology:

Non-separable tree-like Banach spaces
and Rosenthal's ℓ1-theoremm Costas Poulios
(Submitted on 2 Oct 2012)
https://arxiv.org/abs/1210.0792

Besides some terminology you also find some notation.

Nodes and branches are explained on PDF 2,
in section 2. The basic construction.

burs...@gmail.com schrieb:

[exflaso....@gmail.com]

Again, you are looking at it in the wrong direction.

You have to start with how many nodes you have, not how many paths. You're question-begging. "X is needed for Y" doesn't tell you how big Y is when X is given.

EFQ

[WM]

Am Montag, 25. Juni 2018 18:54:19 UTC+2 schrieb burs...@gmail.com:

> No, I don't mean a path. A path is from node to node.

A path is an infinite set of nodes.

> For example this set {t1,t2,t3} is a path:
>
> t4
> /
> t2
> /
> t1
>

That is a finite path.

> But a branch is a infinite path that starts in
> the root.

Every infinite path starts in the root and is called path.

> So branches and paths are not the same.

But path is what you mean.

>
> There are uncountable many branches.

Are you really so much blinded that you cannot see the restriction? The number of paths that can be distinguished at level n is given by the number of nodes of level n. No level has more than aleph_0 nodes.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 19:00:08 UTC+2 schrieb j4n bur53:
> You might enjoy this paper for some terminology:
>
> Non-separable tree-like Banach spaces
> and Rosenthal's ℓ1-theoremm Costas Poulios
>

No. The terminology is clear, for instance given here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 280.

Regards, WM

[burs...@gmail.com]

You cannot say something is infinite as a definition,
and then apply the term to something finite.

Doesn't make much sense.

Am Montag, 25. Juni 2018 19:01:47 UTC+2 schrieb WM:
> A path is an infinite set of nodes.

[Ross A. Finlayson]

On Monday, June 25, 2018 at 2:26:33 AM UTC-7, WM wrote:
> Am Sonntag, 24. Juni 2018 16:55:12 UTC+2 schrieb burs...@gmail.com:
> > You cannot distinguish only countable many blades of grass.
> > If there were only countable many blades of grass,
> > then you could find a bijective mapping.
>
> If such mapping made sense. It does not however. Just this is shown by the Binary Tree.
> >
> > But this is impossible. See Cantors theorem.
>
> It has been contradicted. See the undisputedly countable set of nodes and the fact that n-1 nodes are required to distinguish n infinite paths.
>
> In fact there cannot be more paths distinguished than there are nodes on one level. Their numbers grow but are always finite. A sequence with always finite terms can have limit aleph_0 or omega, never 2^aleph_0.
>
>
> > I a separator is not what you think. A separate, when
> > you mean by separator, is only a node. Its the beginning
> > of a whole subtree, which has again
>
> only countably many separators.
>
> Regards, WM

WM, the path is infinite,
not just an unbounded
collection of finite paths.

If you don't accept this fact,
you don't have anything to say
about the infinite paths,
not "no", just "nothing".

[WM]

Am Montag, 25. Juni 2018 19:00:49 UTC+2 schrieb exflaso....@gmail.com:

> You have to start with how many nodes you have,

I do. There are countably many nodes. That means aleph_0 is the cardinality of the set of nodes. Every level has a finite number of nodes. But the limit level has aleph_0 nodes.

> not how many paths. You're question-begging. "X is needed for Y" doesn't tell you how big Y is when X is given.

But it tells the possible maximum of Y.

Regards, WM

[Ross A. Finlayson]

On Monday, June 25, 2018 at 2:47:13 AM UTC-7, j4n bur53 wrote:
> Nope 2^B is the cardinal of P(A), when B is the
> cardinal of A. This is called cardinal exponentiation,
>
> and it gives you uncountable many branches.
> http://mathworld.wolfram.com/CardinalExponentiation.html
>
>
> Am Montag, 25. Juni 2018 11:30:24 UTC+2 schrieb WM:
> > Am Montag, 25. Juni 2018 03:15:07 UTC+2 schrieb Ben Bacarisse:
> > > WM <wolfgang.m...@hs-augsburg.de> writes:
> >
> >
> > > Yes. At level n, only a finite set of 2^n paths can be distinguished.
> >
> > By 2^n nodes. And what is the limit of this sequence of finite terms 2^n? It is aleph_0. It is not 2^aleph_0. Therefore even in the limit only aleph_0 paths can be distinguished.
> > >
> > > > This restricts the differing paths to countably many.
> > >
> > > I means there are 2^|N| paths in the infinite binary tree.
> >
> > Chuckle. There are 2^|N| nodes? Try again.
> >
> > Regards, WM

Yeah, we all know a difference between
ordinal exponentiation and the notation
for a cardinal of a set's powerset's cardinal
as "cardinal exponentiation" with its
collision of notation.

That the rationals are like 2^w in a way
moreso than like w (omega) is a pretty clear
fact. This is where, besides cardinals for
set sizes, there are also other measures as
it were of sets relative sizes in their spaces,
besides their identification with a cardinal.

For example, a proper subset is smaller than
the set, in the set. Exactly half of the
integers are even. A set with infinitely
many proper subsets that are proper supersets
of a given subset is even more larger than
just the proper superset.

[exflaso....@gmail.com]

Nope. It tells you the minimum.

EFQ

[WM]

Am Montag, 25. Juni 2018 19:12:23 UTC+2 schrieb burs...@gmail.com:
> You cannot say something is infinite as a definition,
> and then apply the term to something finite.

A "path" is infinite.
A "finite path" is finite.

Regards, WM

[burs...@gmail.com]

There is no limit level. Each level is a natural
number. There is no natural number that would be
a limit level. The tree goes on and on.

This is the tree, it goes on and on there is no
maximum, and no limit. There is no node that would
be limit, the set T is unbounded:

{t1, t2, t3, t4, t5, t6, t7, ...} = T

Same for the levels, they go on and on. There is
no maximum, and no limit. There is no node set that
would form the final level.

{t1, /* level 1 */
t2, t3, /* level 2 */
t4, t5, t6, t7, /* level 3 */
...} = T

Also its not common to use aleph_0 in such discussion.
aleph_0 is a cardinality. If something would be the
limit, than an ordinal such as omega.

But aleph_0 to use is very uncommon. Because aleph
is a functional. Its only used when you want to
express this functional (Deutsch Aleph-Funktion):

a (some ordinal) --> aleph_a (some cardinal)
https://de.wikipedia.org/wiki/Aleph-Funktion

You can simply say countable many nodes.

[burs...@gmail.com]

Nope, if at all:

path = finite or infinite

and then:

infinite path
finite path

But better use branch as in the literature.

[WM]

When 1 cake is needed for two persons, then 3 cakes can feed a minimum of 6 persons but perhaps even 600???

Regards, WM

[burs...@gmail.com]

But the most important is, there is no
limit level. And there are even less some
limit nodes.

An infinite path is just a collection
of nodes, that happens to be infinite.
A finite path is just a collection

of nodes, that happens to be finite.
This is where the adjektive finite and
infinite comes from. The number of

nodes in the path.

If you want limit indexes for nodes,
you would need an ordinal such as omega+1
as the basis for the tree.

You find this also in the paper, so
called k-trees. But these here the
following nodes are only omega-trees:

{t1,t2,t3,...}

There are countably many nodes, and the
levels are indexed by elements from
omega, but there is no level omega.

[exflaso....@gmail.com]

"One cake is needed for two people" doesn't tell you the maximum number of people that can be fed by that one cake.

EFQ

[Ross A. Finlayson]

Conway's "surreal numbers" have a "day omega".

Number systems with a "point at infinity" or
about how the naturals are compact (eg, that
they contain a copy as an element) have various
usual notions of properties of the last row
as the tree is complete and "reaches infinity".

Zeno's arrow with infinite divisibility only
has that it arrives when they're altogether
infinitely added back up, the partitions of
the divisions.

[Juergen R.]

On Monday, June 25, 2018 at 4:19:50 PM UTC+2, WM wrote:
> Am Montag, 25. Juni 2018 16:04:01 UTC+2 schrieb burs...@gmail.com:
> > Nope. The number of nodes has not the limit omega.
> > The number of nodes is not the same as the set of
> > nodes.
>
> The set has cardinality aleph_0. Therefore every level has not more than aleph_0 nodes.

>
> n-1 nodes are required to distinguish n infinite paths.

That this is a crude elementary error has been pointed out to you many times before.

In a binary tree with 2n nodes, 2 on each "level", 4 edges between levels, 2^n finite paths are "distinguished" by 2n nodes.

Your erroneous argument, applied to subsets of a set instead of paths in a tree, would assert that a set with n elements can have no more than n subsets.

[Ross A. Finlayson]

Consider a measure space on ran(EF),
where the only measurable sets are
as of the image of n-sets, then,
their only measurable subsets are
also n-sets and there are n-many.

These aren't just subsets but subsets
maintaining a given property.

[Ben Bacarisse]

WM <wolfgang.m...@hs-augsburg.de> writes:
<snip>
>> I[t] means there are 2^|N| paths in the infinite binary tree.

>
> Chuckle. There are 2^|N| nodes? Try again.

No, I did not say that. There are |N| nodes and 2^|N| paths. 2^|N| is
the cardinality of the set of function from N to a two-element set. Not
even you can dispute the fact that the set of paths bijects with the set
of functions from N -> {0, 1}. So rather than chuckling, you might
consider apologising.

--
Ben.

[WM]

Am Montag, 25. Juni 2018 19:48:07 UTC+2 schrieb Juergen R.:
> On Monday, June 25, 2018 at 4:19:50 PM UTC+2, WM wrote:
> > Am Montag, 25. Juni 2018 16:04:01 UTC+2 schrieb burs...@gmail.com:
> > > Nope. The number of nodes has not the limit omega.
> > > The number of nodes is not the same as the set of
> > > nodes.
> >
> > The set has cardinality aleph_0. Therefore every level has not more than aleph_0 nodes.
> >
> > n-1 nodes are required to distinguish n infinite paths.
>
> That this is a crude elementary error has been pointed out to you many times before.

It is fact.

>
> In a binary tree with 2n nodes, 2 on each "level", 4 edges between levels, 2^n finite paths are "distinguished" by 2n nodes.

2^n finite paths are not distinguished from a given paths and from each other by 2n nodes.

>
> Your erroneous argument, applied to subsets of a set instead of paths in a tree, would assert that a set with n elements can have no more than n subsets.

The Binary Tree has a structure which invalidates your argument.

Fact is: At level n there are 2^n nodes which can distinguish 2^n paths. (The nodes of levels m < n are irrelevant for this argument.) Level n has 2^n nodes. The limit of the sequence 1, 2, 4, 8, ... is aleph_0. No level has more than aleph_0 nodes. Therefore not more than aleph_0 paths can be distinguished. Is it really possible to be too stupid to understand that?

Regards, WM

[WM]

But in the Binary Tree we know that n nodes can distinguish at most n-1 different paths from each other and from a given path.

Regards, WM

[WM]

Am Dienstag, 26. Juni 2018 03:04:57 UTC+2 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:
> <snip>
> >> I[t] means there are 2^|N| paths in the infinite binary tree.
> >
> > Chuckle. There are 2^|N| nodes? Try again.
>
> No, I did not say that.

Fine. It is wrong.

> There are |N| nodes and 2^|N| paths.

According to Cantor. But that is wrong. At level n there are 2^n nodes which can distinguish 2^n paths. (The nodes of levels m < n are irrelevant for this argument.) Level n has 2^n nodes. The limit of the sequence 1, 2, 4, 8, ... is aleph_0 or |N| as you prefer to say. No level has more than aleph_0 nodes. Therefore not more than aleph_0 paths can be distinguished.

> 2^|N| is
> the cardinality of the set of function from N to a two-element set.

According to Cantor.

> Not
> even you can dispute the fact that the set of paths bijects with the set
> of functions from N -> {0, 1}.

But the Binary Tree distinguishes at most aleph_0 paths. Therefore Matheology has to renounce identifying paths by nodes and reals by digits. Even infinitely many are not sufficient as the Binary Tree proves.

> So rather than chuckling, you might
> consider apologising.

No reason. Try to think better.

Regards, WM

[WM]

Am Montag, 25. Juni 2018 19:30:29 UTC+2 schrieb burs...@gmail.com:

> Nope, if at all:
>
> path = finite or infinite
>
> and then:
>
> infinite path
> finite path
>
> But better use branch as in the literature.

The technical term is path.
See Wikipedia or https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

Regards, WM

[burs...@gmail.com]

Nope, wikipedia says the same like I am saying.

a path in a graph is a **finite** or **infinite** sequence
https://en.wikipedia.org/wiki/Path_%28graph_theory%29

You are halucinating and confused as usual.
So what I said is correct:

path = finite or infinite

You should rally put the bottle down and
stop with your booze.

[Juergen R.]

On Tuesday, June 26, 2018 at 12:36:29 PM UTC+2, WM wrote:
[...]

> The limit of the sequence 1, 2, 4, 8, ... is aleph_0.

Prove it - first you might want to find out what "limit" means in mathematics.

> No level has more than aleph_0 nodes. Therefore not more than aleph_0 >paths can be distinguished.

Prove it - and pay close attention to the order of the quantifiers.

>Is it really possible to be too stupid to understand that?

Yes.

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 14:23:54 UTC+2 schrieb Juergen R.:


> > No level has more than aleph_0 nodes. Therefore not more than aleph_0 >paths can be distinguished.
>
> Prove it - and pay close attention to the order of the quantifiers.
>
>
> >Is it really possible to be too stupid to understand that?
>
> Yes.

At level n there are 2^n nodes which can distinguish 2^n paths. (The nodes of levels m < n are irrelevant for this argument.) The limit of the sequence 1, 2, 4, 8, ... is aleph_0. No level has more than aleph_0 nodes. Therefore not more than aleph_0 paths can be distinguished. Is it really possible to remain too stupid to understand that not even in the second attempt?

Regards, WM

[exflaso....@gmail.com]

On Tuesday, June 26, 2018 at 6:49:16 AM UTC-4, WM wrote:
> But the Binary Tree distinguishes at most aleph_0 paths.

If so, then you should be able to put them into a list. I can define an algorithm that can put the nodes into a list, by starting with the root node R as the first node on the list, then adding all the nodes from each successive level, from left to right, yielding the following list (nodes are labelled here by the finite path taken to reach them from R):

1. R
2. R0
3. R1
4. R00
5. R01
6. R10
7. R11
8. R000
etc.

I hope it's clear that this list is both countably infinite and that no node will be skipped.

So, what's your comparable algorithm for listing the paths in the tree? What is the very first path on your list? What is the tenth? The hundredth? The Nth?

For any natural number N, I can compute exactly which node is in the Nth position on my list. Can you do the same for the paths?

Loose, vague intuitions about infinity are notoriously worthless. You need a rigorous proof, not hand-waving.

EFQ

[Julio Di Egidio]

On Tuesday, 26 June 2018 12:36:29 UTC+2, WM wrote:

> Fact is: At level n there are 2^n nodes which can distinguish 2^n paths. (The
> nodes of levels m < n are irrelevant for this argument.)

Though the argument with all nodes is easier to carry through: those are two
monotone sequences of which one is *strictly* greater than the other (for all n).

Julio

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 16:49:01 UTC+2 schrieb exflaso....@gmail.com:
> On Tuesday, June 26, 2018 at 6:49:16 AM UTC-4, WM wrote:
> > But the Binary Tree distinguishes at most aleph_0 paths.
>
> If so, then you should be able to put them into a list.

Chuckle. The list argument is purest nonsense.

> Loose, vague intuitions about infinity are notoriously worthless. You need a rigorous proof, not hand-waving.

On priciple I do not prove 1 + 1 = 2 and related simple stuff. But since you seem to be interested I will outline the basics sufficient to let everyone with IQ > 80 understand.

The finite Binary Tree with n levels has 2^n finite paths. They can be distingusihed and identified by the nodes occupied in level n. It is impossible to distinguish more paths since there are no more paths. Further the nodes in levels m < n are insufficient and irrelevant for that purpose.

Finite paths in the finite Binary Tree are in 1-to-1 correspondence with path bunches of the same length in the Complete Infinite Binary Tree.

Therefore precisely 2^n different path bunches can be identified by the nodes at level n. Not less and not more. Most bunches can be identified at the largest level. We don't know it but we know that all levels together have aleph_0 nodes. And if there is a limit then it has aleph_0 nodes. So an upper stimate is aleph_0 path bunches. More cannot be identified.

Regards, WM

[exflaso....@gmail.com]

On Tuesday, June 26, 2018 at 12:14:32 PM UTC-4, transf...@gmail.com wrote:
> Am Dienstag, 26. Juni 2018 16:49:01 UTC+2 schrieb exflaso....@gmail.com:
> > On Tuesday, June 26, 2018 at 6:49:16 AM UTC-4, WM wrote:
> > > But the Binary Tree distinguishes at most aleph_0 paths.
> >
> > If so, then you should be able to put them into a list.
>
> Chuckle. The list argument is purest nonsense.

It's the surest way to prove that a set is countable.

If you aren't able to list it, then you aren't able to count it, and if you aren't able to count it, it isn't countable.

> > Loose, vague intuitions about infinity are notoriously worthless. You need a rigorous proof, not hand-waving.
>
> On priciple I do not prove 1 + 1 = 2 and related simple stuff. But since you seem to be interested I will outline the basics sufficient to let everyone with IQ > 80 understand.
>
> The finite Binary Tree with n levels has 2^n finite paths.

I didn't ask about the finite paths.

> They can be distingusihed and identified by the nodes occupied in level n. It is impossible to distinguish more paths since there are no more paths.

Of course there are: all of the infinite paths! For every finite path, there are an infinite number of infinite paths that pass through it, because every finite path corresponds to a node in the tree, and every node begins a new full copy of the entire tree with itself as the root node.

> Finite paths in the finite Binary Tree are in 1-to-1 correspondence with path bunches of the same length in the Complete Infinite Binary Tree.
>
> Therefore precisely 2^n different path bunches can be identified by the nodes at level n. Not less and not more. Most bunches can be identified at the largest level. We don't know it but we know that all levels together have aleph_0 nodes. And if there is a limit then it has aleph_0 nodes. So an upper stimate is aleph_0 path bunches. More cannot be identified.

Your "path bunches" correspond to the finite paths, because you are defining them based on specific nodes in specific levels, which mark the endpoints of the finite paths. You still don't have an algorithm for listing the infinite paths, because you haven't defined how to identify infinite paths.

You can't identify a unique infinite path based on a specific node, because there are an infinite number of unique infinite paths that pass through each node.

EFQ

[j4n bur53]

Thats new to me. That there would be only aleph_0
infinite paths. Are you sure that you are talking
about infinite paths, and

not only finite paths. Of course since there are
only countable infinite many nodes, there are also
only countable infinite many finite paths

from the root to these nodes. On the other hand
there are uncountable many infinite paths, which
can be easily shown by a diagonal argument.


transf...@gmail.com schrieb:

[Juergen R.]

On illanpaisto, ja silkavat saijat
luopoissa pirkeinä myörien ponkii:
surheisna kaikk' kirjuvat lorokaijat
ja vossut lonkaloisistansa ulos vonkii.

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 18:31:46 UTC+2 schrieb exflaso....@gmail.com:
> On Tuesday, June 26, 2018 at 12:14:32 PM UTC-4, transf...@gmail.com wrote:
> > Am Dienstag, 26. Juni 2018 16:49:01 UTC+2 schrieb exflaso....@gmail.com:
> > > On Tuesday, June 26, 2018 at 6:49:16 AM UTC-4, WM wrote:
> > > > But the Binary Tree distinguishes at most aleph_0 paths.
> > >
> > > If so, then you should be able to put them into a list.
> >
> > Chuckle. The list argument is purest nonsense.
>
> It's the surest way to prove that a set is countable.

It has been applied to show that the set of nodes is countable.
The set of paths however is restricted by the nodes to countable.

>
> If you aren't able to list it, then you aren't able to count it, and if you aren't able to count it, it isn't countable.

It is in bijection with a subset of a countable set. That is enough.

>
> > > Loose, vague intuitions about infinity are notoriously worthless. You need a rigorous proof, not hand-waving.
> >
> > On priciple I do not prove 1 + 1 = 2 and related simple stuff. But since you seem to be interested I will outline the basics sufficient to let everyone with IQ > 80 understand.
> >
> > The finite Binary Tree with n levels has 2^n finite paths.
>
> I didn't ask about the finite paths.

But they will help you to understand the infinite paths.

>
> > They can be distingusihed and identified by the nodes occupied in level n. It is impossible to distinguish more paths since there are no more paths.
>
> Of course there are:

Here we talk about finite paths.

>
> > Finite paths in the finite Binary Tree are in 1-to-1 correspondence with path bunches of the same length in the Complete Infinite Binary Tree.
> >
> > Therefore precisely 2^n different path bunches can be identified by the nodes at level n. Not less and not more. Most bunches can be identified at the largest level. We don't know it but we know that all levels together have aleph_0 nodes. And if there is a limit then it has aleph_0 nodes. So an upper stimate is aleph_0 path bunches. More cannot be identified.
>
> Your "path bunches" correspond to the finite paths, because you are defining them based on specific nodes in specific levels, which mark the endpoints of the finite paths. You still don't have an algorithm for listing the infinite paths, because you haven't defined how to identify infinite paths.

It is impossible to identify them by nodes.

>
> You can't identify a unique infinite path based on a specific node, because there are an infinite number of unique infinite paths that pass through each node.

Alas the paths of that infinite number cannot be distinguished.

Look at the diagonal argument. There the anti-diagonal is distinct from all other reals by digits at finite places. There is no digit beyond all finite palces. But in the Binary Tree you cannot distinguish the believed paths at finite places.

Regards, WM

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 18:47:36 UTC+2 schrieb j4n bur53:
> > At level n there are 2^n nodes which can distinguish 2^n paths. (The nodes of levels m < n are irrelevant for this argument.) The limit of the sequence 1, 2, 4, 8, ... is aleph_0. No level has more than aleph_0 nodes. Therefore not more than aleph_0 paths can be distinguished.

> Thats new to me. That there would be only aleph_0
> infinite paths. Are you sure that you are talking
> about infinite paths,

Of course. All paths that hit the limit level having alleph_0 nodes.

> Of course since there are
> only countable infinite many nodes, there are also
> only countable infinite many finite paths
> from the root to these nodes.

I consider all paths traversing the complete tree, but their number is reszricted to the nodes per level. That is in no case more than aleph_0.

> On the other hand
> there are uncountable many infinite paths, which
> can be easily shown by a diagonal argument.

And as easily it is contradicted by the fact that all paths have to traverse all levels and are restricted by at most aleph_0 nodes per level.

The diagonal argument distinguishes real numbers at finite places only. In the Binary Tree not more than countably many paths can be distinguished at finite places.

By the way: All anti-diagonals are defined and therefore belong to the countable set of all definable numbers.

Regards, WM

[burs...@gmail.com]

WM wrote nonsense: "All paths that hit the limit
level having aleph_0 nodes."

No there is no limit level for an infinite path.

You are halucinating and confused as usual.

An infinite path, the paths we are currently
dealing with, look as follows:

B = {a1,a2,a3,a4,...}

Where a1,a2,a3,a4,... are nodes from this set
here, namely the nodes:

T = {t1,t2,t3,t4,t5,...}

When the infinite path starts at the root, we
have a1=t1, this is a branch. A branch is
an unbounded set {a1,a2,a3,a4,...}, there is
no limit level whatever. You are hallucinating

and confused as usual. In the nodes the levels
are as follows:

T = {t1, /*level 1*/
t2,t3, /*level 2*/
t4,t5,t6,t7, /*level 3*/
...}

In a branch the levels are more immediate:

B = {a1, /*level 1*/
a2, /*level 2*/
a3, /*level 3*/
...}

In boch cases there is no limit of levels. Aleph0 is not
a sign to denote when there is no limit. You are

halucinating and confused as usual.

[exflaso....@gmail.com]

On Tuesday, June 26, 2018 at 2:39:44 PM UTC-4, transf...@gmail.com wrote:
> Am Dienstag, 26. Juni 2018 18:31:46 UTC+2 schrieb exflaso....@gmail.com:
> > On Tuesday, June 26, 2018 at 12:14:32 PM UTC-4, transf...@gmail.com wrote:
> > > Am Dienstag, 26. Juni 2018 16:49:01 UTC+2 schrieb exflaso....@gmail.com:
> > > > On Tuesday, June 26, 2018 at 6:49:16 AM UTC-4, WM wrote:
> > > > > But the Binary Tree distinguishes at most aleph_0 paths.
> > > >
> > > > If so, then you should be able to put them into a list.
> > >
> > > Chuckle. The list argument is purest nonsense.
> >
> > It's the surest way to prove that a set is countable.
>
> It has been applied to show that the set of nodes is countable.
> The set of paths however is restricted by the nodes to countable.

The set of finite paths, yes. Indeed, every node defines exactly one finite path: the path it takes to get to it from the root node. So any list of the nodes is automatically a list of the finite paths.

But the infinite paths cannot be defined by reference to any specific nodes. They must be defined in other ways, such as with infinite sequences of nodes: R000̄, R100̄, R010̄, R110̄, etc.

> > If you aren't able to list it, then you aren't able to count it, and if you aren't able to count it, it isn't countable.
>
> It is in bijection with a subset of a countable set. That is enough.

If you actually had such a bijection, you could easily provide a list, just by bijecting with the natural numbers and using their inherent ordering.

So, provide your list. If you have a bijection with the natural numbers, you should be able to tell me which path is 1st, the 10th, the 100th, and in general, the Nth.

> > > > Loose, vague intuitions about infinity are notoriously worthless. You need a rigorous proof, not hand-waving.
> > >
> > > On priciple I do not prove 1 + 1 = 2 and related simple stuff. But since you seem to be interested I will outline the basics sufficient to let everyone with IQ > 80 understand.
> > >
> > > The finite Binary Tree with n levels has 2^n finite paths.
> >
> > I didn't ask about the finite paths.
>
> But they will help you to understand the infinite paths.

I "understand" just fine. I'm not interested in loose, vague intuitions. I only want a rigorous proof, and as a natural consequence of the definition of "countable", in order for a set to be countable, it must be listable.

If it is impossible to provide such a list, then the set is not countable.

Maybe you want "countable" to mean something else, but if so, you should use a different term, because that one is already taken.

> > Your "path bunches" correspond to the finite paths, because you are defining them based on specific nodes in specific levels, which mark the endpoints of the finite paths. You still don't have an algorithm for listing the infinite paths, because you haven't defined how to identify infinite paths.
>
> It is impossible to identify them by nodes.

No, it is impossible to identify them by a finite set of nodes. As long as you allow an infinite sequence of nodes, you can identify an infinite path. For example, the sequence R0̄ (zero repeating) is the unique infinite path that traces out the left edge of the tree, while R1̄ (one repeating) is its counterpart on the right edge.

Every infinite sequence of nodes (a_i) defines a unique infinite path, and every unique infinite path is defined by an infinite sequence of nodes (a_i).

> > You can't identify a unique infinite path based on a specific node, because there are an infinite number of unique infinite paths that pass through each node.
>
> Alas the paths of that infinite number cannot be distinguished.

Sure they can:

P1 = (a_i), defined for all i in N, where a_1 = R and a_i = 0 for all i > 1
P2 = (a_i), defined for all i in N, where a_1 = R, a_2 = 1, and a_i = 0 for all i > 2

These are both clearly infinite paths (because they have one node for every natural number), and they differ at node a_2, so they are distinguished.

> Look at the diagonal argument. There the anti-diagonal is distinct from all other reals by digits at finite places.

Not just at some finite places. At ALL of them. It differs from all of them in an infinite number of places.

> There is no digit beyond all finite palces.

Yes, this is sort of fundamental to why the reals are uncountable.

> But in the Binary Tree you cannot distinguish the believed paths at finite places.

I don't know what a "believed path" is, but since the infinite paths pass through an infinite number of nodes, they too cannot be distinguished if you only refer to specific finite positions. You must refer to full infinite sequences.

EFQ

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 20:55:01 UTC+2 schrieb burs...@gmail.com:
> WM wrote nonsense: "All paths that hit the limit
> level having aleph_0 nodes."
>
> No there is no limit level for an infinite path.

Then there are only finite levels with 1, 2, 4, 8, ... nodes. All less than aleph_0.

Regards, WM

[exflaso....@gmail.com]

And there are more infinite paths than there are nodes in any particular level, so aleph_0 is at best a lower bound on the number of infinite paths. You haven't demonstrated that it is an upper bound (because you haven't defined how infinite paths are identified).

EFQ

[Juergen R.]

On Tuesday, June 26, 2018 at 4:32:59 PM UTC+2, transf...@gmail.com wrote:

Oh Lordy, Lord, great God Almighty, either deliver us from this Mad Hatter,
or endow him with wisdom. The poor fool is still confused by the notion of
"limit" in mathematics.

What is he to do when he has to learn about continuity and integrals and
suchlike if he can't understand limits?

[George Greene]

On Monday, June 25, 2018 at 4:53:00 AM UTC-4, WM wrote:
> Am Montag, 25. Juni 2018 03:15:07 UTC+2 schrieb Ben Bacarisse:
> > I means there are 2^|N| paths in the infinite binary tree.
> >
> No, there is no level |N|.

You are SUCH a dumbass -- THERE DOESN'T NEED to be any level |N| in order
for there to be 2^|N| paths through the tree. If you label the ROOT level *ZERO*,WHICH YOU *SHOULD*, since there are ZERO edges along the path from the root to the root, and the path distance from the root to the root IS ZERO, then the
number of paths down to level n is EQUAL TO THE NUMBER OF LEAVES AT level n,
since every different path ends in a different leaf. THAT number is 2^n for n
IN the tree, but infinity IS NOT A NATURAL NUMBER, so why do you even think
this follows? There is no level |N| in the tree but there are only |N| NODES in the tree. The requirement that there be 2^n paths ending at level n applies when the paths DO IN FACT END at level n, but the infinite paths BY DEFINITION do NOT END AT ALL, so THERE ARE NO end-nodes/leaves with which to complete the analogy.

That really does mean YOU DON'T GET TO USE that argument.

[Ben Bacarisse]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Dienstag, 26. Juni 2018 03:04:57 UTC+2 schrieb Ben Bacarisse:

<snip>

>> 2^|N| is
>> the cardinality of the set of function from N to a two-element set.
>
> According to Cantor.

No, I am telling you the definition of that notation. If you have your
own meaning for 2^|N| then you are just playing the typical crank game
of making statements about terms and notation you have chosen to give
private meanings to.

>> Not
>> even you can dispute the fact that the set of paths bijects with the set
>> of functions from N -> {0, 1}.
>
> But the Binary Tree distinguishes at most aleph_0 paths.

Are you afraid to agree about anything at all? Please at least confirm
that you agree that the set of paths bijects with the set of functions
from N -> {0, 1} even if you have made up your own meaning for 2^|N|.

--
Ben.

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 21:11:46 UTC+2 schrieb exflaso....@gmail.com:
> On Tuesday, June 26, 2018 at 2:39:44 PM UTC-4, transf...@gmail.com wrote:


> > It has been applied to show that the set of nodes is countable.
> > The set of paths however is restricted by the nodes to countable.
>
> The set of finite paths, yes. Indeed, every node defines exactly one finite path: the path it takes to get to it from the root node. So any list of the nodes is automatically a list of the finite paths.
>
> But the infinite paths cannot be defined by reference to any specific nodes.

The proof is this: Try to distinguish three infinite paths from the path 111... and from each other by less than three differing nodes.

Regards, WM

[transf...@gmail.com]

Am Dienstag, 26. Juni 2018 21:11:46 UTC+2 schrieb exflaso....@gmail.com:

> They must be defined in other ways, such as with infinite sequences of nodes: R000̄, R100̄, R010̄, R110̄, etc.

Since no infinite sequence can be written, each infinite sequence has to be defined by a formula. Alas there are only countably many formulas.

Same result as in the Binary Tree: Nothing is uncountable.

"The possible combinations of finitely many letters form a countable set, and since every determined real number must be definable by a finite number of words, there can exist only countably many real numbers – in contradiction to Cantor's classical theorem and its proof." [H. Weyl: "Das Kontinuum", Veit, Leipzig (1918) p. 18]

> > Alas the paths of that infinite number cannot be distinguished.
>
> Sure they can:
>
> P1 = (a_i), defined for all i in N, where a_1 = R and a_i = 0 for all i > 1
> P2 = (a_i), defined for all i in N, where a_1 = R, a_2 = 1, and a_i = 0 for all i > 2
>
> These are both clearly infinite paths

They are clearly defined by finite formulas.


Paul Bernays, 1934-1939 Hilbert's co-author of "Grundlagen der Mathematik (Foundations of Mathematics)" said: "If we pursue the thought that each real number is defined by an arithmetical law, the idea of the totality of real numbers is no longer indispensable, and the axiom of choice is not at all evident." [P. Bernays: "On Platonism in mathematics" (1934) p. 5ff]

Kurt Schütte, 1933 Hilbert's last doctoral student, said: "If we define the real numbers in a strictly formal system, where only finite derivations and fixed symbols are permitted, then these real numbers can certainly be enumerated because the formulas and derivations on the basis of their constructive definition are countable." [K. Schütte: "Beweistheorie", Springer (1960)]

Regards, WM

[transf...@gmail.com]

Am Mittwoch, 27. Juni 2018 00:46:19 UTC+2 schrieb George Greene:
>

>
> That really does mean YOU DON'T GET TO USE that argument.

Distinguish the path 111... from three other infinite paths of the Binary Tree and also these from each other. Try to achieve an "observable difference" using less than three different nodes.

Come back, when you are done. Or try to get rid of Cantor's brain-damaging nonsense.

Regards, WM

[transf...@gmail.com]

Am Mittwoch, 27. Juni 2018 03:32:15 UTC+2 schrieb Ben Bacarisse:


> > But the Binary Tree distinguishes at most aleph_0 paths.

If you don't agree try to distinguish four different paths from the paths 111... by less than four different nodes.

Can you?

>
> Are you afraid to agree about anything at all? Please at least confirm
> that you agree that the set of paths bijects with the set of functions
> from N -> {0, 1} even if you have made up your own meaning for 2^|N|.

The set of path bijects with the set P(|N) and with the set of functions

> from N -> {0, 1}.

All these sets are not uncountable. (I avoid to say they are countable because this is a notion of same meaninglessness.)

Regards, WM

[exflaso....@gmail.com]

On Wednesday, June 27, 2018 at 8:43:16 AM UTC-4, transf...@gmail.com wrote:
> Am Dienstag, 26. Juni 2018 21:11:46 UTC+2 schrieb exflaso....@gmail.com:
>
>
> > They must be defined in other ways, such as with infinite sequences of nodes: R000̄, R100̄, R010̄, R110̄, etc.
>
> Since no infinite sequence can be written,

I just did.

> each infinite sequence has to be defined by a formula.

Everything has to be written in some notation. The symbol 1 isn't actually the concept 1. It's just a representation of it. That doesn't mean that it isn't a valid representation.

> "The possible combinations of finitely many letters form a countable set, and since every determined real number must be definable by a finite number of words, there can exist only countably many real numbers

Or rather, only countably many "determined" real numbers (whatever that means; I suspect it's intended something along the lines of "algebraic").

Clearly, then, there must also be uncountably many non-"determined" real numbers (such as the transcendentals).

> > > Alas the paths of that infinite number cannot be distinguished.
> >
> > Sure they can:
> >
> > P1 = (a_i), defined for all i in N, where a_1 = R and a_i = 0 for all i > 1
> > P2 = (a_i), defined for all i in N, where a_1 = R, a_2 = 1, and a_i = 0 for all i > 2
> >
> > These are both clearly infinite paths
>
> They are clearly defined by finite formulas.

So? The paths themselves are still infinite. The set of natural numbers N is infinite, but I only need a single symbol to notate it.

EFQ

[exflaso....@gmail.com]

I'll do even better than that. You can list out all infinite paths, and I'll still be able to give you a new infinite path that isn't on your list.

EFQ

[Juergen R.]

On Sunday, June 24, 2018 at 11:32:50 AM UTC+2, WM wrote:
> Am Samstag, 23. Juni 2018 20:28:54 UTC+2 schrieb George Greene:
>
> The following is okay but does not change the facts.
> >
> > > /
> > > 0
> > > / \
> > > 0 1
> > > ............
> > at the top. You actually have NO REASON for choosing 0 or 1 OR ANYTHING ELSE
> > as the label for the root. Of course, the improvement I have drawn here has
> > the problem that the leading edge doesn't connect 2 nodes, but so what?
> > There are many different ways of combining two things and in the case of
> > the binary tree, the way is THE ORDERED pair way; one of the subtrees is
> > always labeled "1st" or "left" or "/" and the other is always labeled "2nd"
> > or "right" or "\" or WHATEVER OTHER ORDERED list of 2 opposites YOU CHOOSE,
> > including (as you have) labeling the left "0" and the right "1".
> > The NAMES DON'T MATTER and the NODES DON'T MATTER.
> >
> > The inifnite binary tree is the ordered pair of COPIES OF IT*SELF*.
>
> The question is how many paths can be distinguished. At level n there are 2^n nodes, and therefore 2^n paths (or bunches of paths) can be distinguished. At no level more than aleph_0 nodes exist. Therefore no more than aleph_0 paths can be distinguished in the Complete Infinite Binary Tree.
>
> Regards, WM

The Walrus and the Carpenter
Were walking close at hand;
They wept like anything to see
Such quantities of sand:
If this were only cleared away,'
They said, it would be grand!'

If seven maids with seven mops
Swept it for half a year,
Do you suppose,' the Walrus said,
That they could get it clear?'
I doubt it,' said the Carpenter,
And shed a bitter tear.

[Julio Di Egidio]

On Wednesday, 27 June 2018 18:55:37 UTC+2, exflaso....@gmail.com wrote:

> I'll do even better than that. You can list out all infinite paths, and I'll
> still be able to give you a new infinite path that isn't on your list.

Just give up then, you too have no arguments.

Julio

[Ross A. Finlayson]

You say "list" but really it
is "bijection to the naturals".

Then, the anti-diagonal argument
is reading off the values of the
range of the function to form a
list, and from that construct an
anti-diagonal.

So, "countable" is "bijects with
the naturals". Making a "list"
is just the artifact of the anti-
diagonal argument.

[exflaso....@gmail.com]

On Wednesday, June 27, 2018 at 1:31:24 PM UTC-4, Ross A. Finlayson wrote:
> On Wednesday, June 27, 2018 at 9:55:37 AM UTC-7, exflaso....@gmail.com wrote:
> > On Wednesday, June 27, 2018 at 8:40:39 AM UTC-4, transf...@gmail.com wrote:
> > > Am Dienstag, 26. Juni 2018 21:11:46 UTC+2 schrieb exflaso....@gmail.com:
> > > > On Tuesday, June 26, 2018 at 2:39:44 PM UTC-4, transf...@gmail.com wrote:
> > >
> > >
> > > > > It has been applied to show that the set of nodes is countable.
> > > > > The set of paths however is restricted by the nodes to countable.
> > > >
> > > > The set of finite paths, yes. Indeed, every node defines exactly one finite path: the path it takes to get to it from the root node. So any list of the nodes is automatically a list of the finite paths.
> > > >
> > > > But the infinite paths cannot be defined by reference to any specific nodes.
> > >
> > > The proof is this: Try to distinguish three infinite paths from the path 111... and from each other by less than three differing nodes.
> >
> > I'll do even better than that. You can list out all infinite paths, and I'll still be able to give you a new infinite path that isn't on your list.
>

> You say "list" but really it
> is "bijection to the naturals".

Yes.

> Then, the anti-diagonal argument
> is reading off the values of the
> range of the function to form a
> list, and from that construct an
> anti-diagonal.

Also yes.

> So, "countable" is "bijects with
> the naturals". Making a "list"
> is just the artifact of the anti-
> diagonal argument.

Pretty much, yes.

"List" is only four letters.

EFQ

[Ross A. Finlayson]

This is about that the bijection is
a function is a bridge that's infinitely-
many wide for the countably infinite,
that all the elements maintain their
order from both the left, and the right,
and they all cross (or are mapped) at
once, in parallel, not serially.

This is about
Cantor-(Schroeder)-Bernstein
theorem, or here,
Cantor +- Schroeder +- Bernstein.