Re: Approaching the infinite binary tree

On 30 Sep, 05:10, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<6bf9ced4-d917-4874-8da2-997cf15b8...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
 LudovicoVan <ju...@xxxxxxxxxxxxx> wrote:
On 29 Sep, 22:18, Virgil <Vir...@xxxxxxxxx> wrote:
In article <290920081547112928%ed...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
 "G. A. Edgar" <ed...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

In article
<36da9a22-353b-4a59-b222-36c755be8...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
David R Tribble <da...@xxxxxxxxxxx> wrote:

Mariano Suarez-Alvarez
You seem to mean by this that you have a last level, the one
indexed by w.

LudovicoVan wrote:
I can drop the "informally": A tree with w levels! What is the problem
with that?

Nothing at all.

However, you have a huge problem if you are saying there
is a level that has index w.

It may make perfectly good sense, depending on your definition of
"tree".
 http://en.wikipedia.org/wiki/Tree_(set_theory)

Just one of the many problems this creates for you:
what is the index of the level immediately before level w?

Depending on your definition of "tree", it can happen that "level
immediately before" need not be defined.

Probably means something like an infinite complete binary tree as
defined inhttp://en.wikipedia.org/wiki/Binary_tree, in which case each
node other than the root must have a parent node in the "level
immediately before" it.

That article is quite irrelevant to a mathematical discussion (and
what you furtherly remark is of course the usual nonsense).

That article IS a mathematical discussion.

And my  nonsense is considerably  more sensible than you inept attempts
at sense.

The only mention to infinite trees is here:

[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??

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

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.

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?

-LV
.

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