Re: The complete infinite binary tree has only countably many infinite paths.

On 29 Mar, 13:04, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Sat, 28 Mar 2009 11:52:01 -0700 (PDT), LudovicoVan

<ju...@xxxxxxxxxxxxx> wrote:
On 28 Mar, 12:29, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Fri, 27 Mar 2009 13:26:20 -0700 (PDT), LudovicoVan

<ju...@xxxxxxxxxxxxx> wrote:
On 24 Mar, 11:56, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
The complete infinite binary tree has only countably many infinite
paths.

Absolutely!

not.

When the nodes are countable, how could the paths be not?

"How could they not be?" is not a proof of anything.

Indeed, it's even straightforward that there is a bijection between
the paths and the leaf nodes...

Huh? In the tree in question there are _no_ leaf nodes.

The set of "leaves" cannot be empty. You might say its cardinality is
the cardinality of the continuum, while I'd rather think of limit
points. The outcome is indeed different, which is the point in
question.

What one "might" say has little to do with the truth of the
matter. Words have _definitions_. By definition a leaf
in a tree is a node with no child nodes. Every node
in that tree has children, hence there are no leaves.

There is no formal definition of "leaf" node: you seem to mistake the
formal for the informal. I am refering to limit points, no mention of
child nodes. Your tree above is not transfinite, and so never
complete.

-LV
.

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