Re: ** says: Definition: sum{i in N} i = 0

On 27 Jun., 19:43, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1182933394.549064.286...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
The Binary Tree
Let [x] = n <= x such that n + 1 > x, i.e. n is the largest integer <=
x.
Let ]x[ = n >= x such that n - 1 < x, i.e. n is the smallest integer
= x.

Level

|0 0.
| / \
|1 0 1
| / \ / \
|2 0 1 0 1
| /\ /\ /\ /\
v x

At height x the number of nodes is K(x) = 2^[x+1] - 1.
At height x the number of separated path bunches is P(x) = 2^]x[.

At level n, as indicated above, the number of paths in the finite tree
that ends at level n is Card(P( {x in N\/{0}: x < n}))

Thus the number in the infinite tree is Card(P(N))



In height x the quotient P(x)/K(x) = (2^]x[) / (2^[x+1] - 1)

In the whole tree we can estimate
P/K = lim_{x --> oo} (2^]x[) / (2^[x+1] - 1) < 2.

The ratio only holds for finite trees, In an infinite tree there is a
different path for each different subset of N, so the number of paths
equals the number of subsets of N, which is greater than the number of
elements of N.



Would you agree that this result is correct although there is not a
limit in the usual sense.

I would agree that there is one path for each subset of N.

Or would you prefer to say that because of
the quotient is alternating between the epsilon surrounding of 1/2 and
the epsilon surrounding of 1 the true result is 2^aleph_0?

Since I see the reality of a bijection between the set of paths ad the
set of subsets of N, I say the result must be larger than Card(N)

That means
P/K = lim_{x --> oo} (2^]x[) / (2^[x+1] - 1) > 2
or set theory is selfcontradictory.

Thank you.

Regards, WM

.

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