(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 branches which is still countable. For branch points we have branches.

But for infinitely many branch points, which is the case for tetrations, they have a branch point at every integer , we have many branches. And we know that already is uncountable.

On the other hand this is only under the assumption that branches dont coincide. E.g. consider . 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 , 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.