Re: Cantor Confusion

Newberry <newberry@xxxxxxxxxx> wrote:

stephen@xxxxxxxxxx wrote:
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.

In what system can you prove that #edges = 2 * #paths?

In ZFC. Theorem: In an infinite binary tree there are twice as many
edges as there are paths.
Proof: At the level n the ratio of eges over paths is expressed by
2*2^n - 2)/2^n
For the entire tree we calculated the limit for {n-->oo}
lim{n-->oo} (2*2^n - 2)/2^n = 2. QED

That is not a proof. I can just as easily "prove" that an infinite
tree has an infinite number of leaf nodes, despite the obvious
fact that it has zero leaf nodes. You cannot just assume that "the limit"
is meaningful. Limits tell you what happens has n becomes large,
but remains finite.

Stephen

.

Quantcast