Re: Cantor Confusion

Newberry <newberry@xxxxxxxxxx> wrote:

Virgil wrote:
In article <1167094506.703211.116860@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Newberry" <newberry@xxxxxxxxxx> wrote:

David Marcus wrote:
Newberry wrote:
paths edges
level 1: 2 = 2^1 2
level 2: 4 = 2^2 6
level 3: 8 = 2^3 14

level n: 2^n (not sure what the formula is)

Does the ratio edges/paths converge to 2 for n --> infinity?

Yes, as WM is fond of repeating ad nauseum.

It certainly makes it highly couterintutive that there are more paths
then edges although I do not know if it generates a flat contradiction.

Yes, it is counterintuitive (depending on your intuition). No, there is
no contradiction.

Just because a system avoids a contradiction of the type P & ~P does
not mean that it is justified. For example an omega-inconsistent system
may not produce any P & ~P and would be still unacceptable. Similarly a
system in which we can prove that

#edges = 2 * #paths

and at the same time that the cardinality of paths is greater than the
number of edges is unacceptable.

If you wish to argue that because the ratio of paths to edges (or nodes)
in finite trees is bounded that it must by some sort of limit argument
remain bounded for infinite trees, then you must allow the same argument
for the ratio of paths to terminal nodes.

What exactly is wrong with the limit I (or rather WM, got to give him
credit) have calculated?

It has nothing to do with the infinite case.

Is it true that at each level 2*2^n - 2)/2^n?

There is no finite level n for an infinite tree, and
you cannot just toss transfinite cardinals into an equation
willy nilly and expect it to make sense.

Is it true that lim{n-->oo} (2*2^n - 2)/2^n = 2?

Yes, but what does that have to do with anything?
Explain why you think that limit is relevant?

Is is true that the cardinality of the index n is the same as the
cardinality of the edges in an infinite path?

What index n? There is no aleph_0 level. You are apparently
just trolling at this point, as all of this has been explained
to you and you have simply ignored the explanations.

Do you think that an infinite tree has zero leaf nodes?
Do you agree that the limit of the ratio of nodes to leaf
nodes is 2 as the tree gets large?

I guess you do think that 2*0 = oo afterall.

Stephen


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