Re: Cantor Confusion

In article <1166474763.304897.177520@xxxxxxxxxxxxxxxxxxxxxxxxxxx> "Newberry" <newberry@xxxxxxxxxx> writes:
....
That is only the case in finite trees.
It fails miserably in infinite binary trees in which no path has a last
node or last edge.

Do you agree that the number of edges is the upper bound of the number
of paths but disagree that the edges are countable?

It is the case in finite trees, but that makes it not true for infinite
trees.

I guess if the number of edges > number of paths and you accept the
diagonal argument then it follows that the edges are uncountable.

No. You can not apply the diagonal argument to the edges.

I do
not know if there is any way to prove directly that the edges are
countable.

There is. It is easy to assign natural numbers to each edge so that
all edges get assigned different natural numbers, and so there is
an injection from the set of edges to the natural numbers, and so
the set is countable.
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