(10/22/2009, 09:45 PM)Base-Acid Tetration Wrote: Then I would think the number of branches is still countable, though i doubt it because a[4]b 's cardinality is that of the continuum.
Potentially there uncountably many branches.
Branches are usually determined by the non-homotopic curves that lead to a point.
E.g. take the logarithm. There is one branching point at 0.
Paths are homotopic iff they wind the same number around 0.
log has a branch for each winding number which is countably many.
Now consider a function with two branch points. For each branch point we have a winding number and so we have \( \mathbb{Z}\times\mathbb{Z} \) branches which is still countable. For \( n \) branch points we have \( \mathbb{Z}^n \) branches.
But for infinitely many branch points, which is the case for tetrations, they have a branch point at every integer \( \le 2 \), we have \( \mathbb{Z}^\omega \) many branches. And we know that already \( 2^\omega \) is uncountable.
On the other hand this is only under the assumption that branches dont coincide. E.g. consider \( \sqrt{z} \). It has a branch point at 0. But only two branches.
But even if we assume that each branch point gives only two branches we still have \( 2^\omega \), i.e. uncountably many branches.
If however more branches coincide it can lessen the number of different branches.
!!!EDIT!!!: After reading mike3's posting I see what is the error in the above reasoning:
A path with finite length can not wind around infinitely many branchpoints of tetration.
It always needs to choose a finite number out the infinitely many branchpoints to wind around.
And thatswhy the number of branches must be countable.
