Date: Mar 26, 2013 3:54 AM
Subject: Re: Matheology § 224

On 26 Mrz., 03:10, "Ross A. Finlayson" <>

> Two different paths, finite or infinite, have at least one node not in
> common.

If the path of 1/pi is not in the Binary Tree, it is impossible to
find that path in the Binary Tree, notwithstanding the fact that for
every rational approximation q of 1/pi, there is a better one q' (with
a node belonging to 1/pi but not to q). The result remains that q' is
not 1/pi.
> And they have at most countably many nodes not in common, there are
> only countably many nodes.

That is provable.
> So, are there uncountably many nodes?



> Because, there's a rational for
> each node, and the rationals are countable.  There are only countably
> many different paths, from a path.  The paths are rooted.
> Seems rather muddled.

Why. It is obvious that only countably many different steps will
construct the Binary Tree and that no step will be able to distinguish
more than one path from the body already constructed.
> Infinite sequences are written out as their specification.  No, there
> are not infinite digital resources to emit each (element of an
> infinite sequence), but, a rule to emit each suffices for many
> purposes.

I agree. There are rules. But there are ounly countably many such
rules. Set theorists accept unnameable numbers, but the Binary Tree
shows that even this excuse is not really an excuse.

Regards, WM