Re: Approaching the infinite binary tree

In article
<4ee217e3-6c4a-4e7e-af3b-e5026db36e5f@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
LudovicoVan <julio@xxxxxxxxxxxxx> wrote:

On 1 Oct, 20:51, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<db939ad0-7286-4526-8bf8-e9b6c2d0d...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
 LudovicoVan <ju...@xxxxxxxxxxxxx> wrote:
On 1 Oct, 00:35, David R Tribble <da...@xxxxxxxxxxx> wrote:
LudovicoVan wrote:
[quote]
An infinite complete binary tree is a tree with Aleph_0 levels, where
for each level d the number of existing nodes at level d is equal to
2^d. The cardinal number of the set of all nodes is Aleph_0. The
cardinal number of the set of all paths is 2^Aleph_0.
[/quote]

Anyway I am puzzled: The number of nodes is Aleph_0??

Virgil wrote:
Yes, aleph_0 is the cardinality of the set of natural numbers, which
means that the nodes can be counted.

LudovicoVan wrote:
Are you sure? One then could conclude that the set of paths must be
countable too, since it can be put in a bijection with a subset of the
nodes, namely the set of the leafs.

That might be what you'd expect intuitively, but in fact that is
not the case. The leaves (nodes) are not commensurate with
the paths.

For every node there is a finite path leading to that node.
But there are infinite paths in the complete binary tree that do
not have terminal nodes, i.e., there are endless paths having no
leaves. [*]

And who says that?

Further argument shows that there can be no bijection
between the nodes and the infinite paths. In fact, there are
far more infinite paths than finite paths in the tree.

And who says that?

Lots of us agree that the definitions say that.

Liar! It's CANTOR that says that!! No bloody "definitions" of anything
involved here.

Actually, Cantor didn't say much of anything about infinite binary tree,
though he might well have if he had been questioned about them.

The sort of complete infinite binary trees under discussions as in

http://en.wikipedia.org/wiki/Binary_tree

Quote: An infinite complete binary tree is a tree with aleph_0 levels,
where for each level d the number of existing nodes at level d is equal
to 2^d. The cardinal number of the set of all nodes is aleph_0. The
cardinal number of the set of all paths is 2^ aleph_0.

Another thing to consider is that if all the paths (including the
infinite paths) were countable, then you could arrange them in
their natural order (from left to right), and then you could show
what the successor was for any given path in the tree. But you
can't do this. For example, what is the successor of (the path next
to) the path defined as alternating left/right turns (i.e., the path
given as L,R,L,R,L,R,..., or equivalently, the binary sequence
01010101...)?

Of course I can do that, and I have already done. Below is few members
before and after the one you mention:

...
(01)0010
(01)0011
(01)0100
(01)
(01)0110
(01)0111
(01)1000
...

If (01)0010 represents a sequence endless on the left and ending in
00010

No, you just and still confuse a numeral for a number. You again
confirm that you don't even know what you are talking about.

I certainly do not know what you are talking about.

, then none of julio's seqeinces are of the correct form.

In their corrected  reverse form, between 0100(01) and 1100(01), ther
are lots of strings, including 0110(xx) for any x and y om {0,1}.

There is no problem in answering *your* question, the problems are
elsewhere, but we can't get there until we put aside the dogma.

It is your dogma, not ours, that you are being bit by.

My dogma?? My dogma is that you guys are a waste of time, with an
intellectal level of a 4 years old kid.

Based on that dogma, you should stop posting, as you can hardly expect
learn anything from us not teach us anything according to that dogma.

There
is NO problem in answering YOUR question as given above, but in YOUR
dogmatic framework of reference, which is anyway not mine.

This question has no answer, as a consequence of the fact
that the infinite paths in the tree are not countable.

You are just under a spell, so to say.

I'd indeed had expected the cardinality of the set of nodes to look
something like [2^(Aleph_0+1)]-1, maybe equal to 2^Aleph_0, but how
can it just be Aleph_0, where Aleph_0 is the cardinality of the
levels?

Because just as you can number the levels (as 1, 2, 3, 4, etc.),
you can number the nodes in all the levels (as 1, then 2, 3, then
4, 5, 6, 7, and so on). �Each level can be given a unique natural
index, and likewise each node can be given a unique natural
index.

Since each index is a natural, and there is no last index in
either case, there are Aleph_0 levels and Aleph_0 nodes.

And so what? The problem is not that the nodes are countable, it's
that your dogma imposes the paths not to be: which is simply
ridiculous.

It may seem ridiculous to julio, it is nevertheless true.

True this balls! YOUR DOGMA IMPOSES THE PATHS NOT TO BE!!! Moron and
liar.

Such incoherence!

There are many sets besides N that have cardinality Aleph_0,
and all of them can be bijected with N. The set of even naturals,
for example, where each natural k in N is mapped to a unique
even natural 2k. Another example is Galileo's set of squares,
where each natural k in N is mapped to k^2. All of these sets have
the same cardinality of Aleph_0.

Ditto.

It may seem ridiculous to julio, it it is nevertheless true.

Yeah, it's true that on this argument too there is nothing more to
add: YOUR MATH IS PROVABLY A JOKE.

You have not provided any such proofs, and have carefully avoided
presenting anything of the nature of mathematics based counter-argument.
.

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