Re: Cantor Confusion

In article <1166093706.459154.39710@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <457ecfca@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:


If the number of levels in the tree is countable,
then every path in the tree is finite, and marked by a specific edge
where the value arises. Is that not true?

If TO means by "level" of a node in a binary tree the number of branches
between it and the root node, then to have every path finite requires
that every path end in a leaf node ( a ode with no child nodes ).

That requires the number of "levels" be finite, not merely countable.

In order to allow infinitely
long strings, you must have an uncountable number of levels in the tree,
in which case 1/3 will exist with a specific edge, infinitely far from
the root node.

There are no such things as trees with uncountably many "levels", at
least according to any standard definition of trees.

And there are no such things as trees with uncountably many paths.

There are in ZFC and NBG. What axiom system does WM refer to in which
there are not?

there is a unique root node at level 0, and each level thereafter is the
successor to a previous level through a connected chain of levels down
to the root.

Therefore we have a continuous growth of patghs separations from 1 to
2^aleph0. Therefore the function 2^n must be continuous

Non sequitur.

Regards, WM
.

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