On 12 Jan., 20:34, Ben Bacarisse <ben.use...@bsb.me.uk> wrote: > WM <mueck...@rz.fh-augsburg.de> writes: > > On 12 Jan., 16:24, Ben Bacarisse <ben.use...@bsb.me.uk> wrote: > >> WM <mueck...@rz.fh-augsburg.de> writes: > >> > Matheology § 191 > > >> > The complete infinite Binary Tree can be constructed by first > >> > constructing all aleph_0 finite paths and then appending to each path > >> > all aleph_0 finiteley definable tails from 000... to 111... > > >> > 0 > >> > 1, 2 > >> > 3, 4, 5, 6 > >> > 7, ... > > >> > This Binary Tree contains aleph_0 * aleph_0 = aleph_0 paths. > > >> No, it contains aleph_0 * whatever the cardinality of the set of tails > >> is. Talk about begging the question! > > > A tail can be defined by a finite word *only*. Nobody can quote an > > infinite string digit by digit - although most mathematicians believe > > that instinctively when pondering about set theory (but never when > > doing analysis). > > That's the part that was missing. Without it you were begging the > question because the set of paths and the set of tails are equinumerous. > With it, the argument about tails and finite paths is pointless -- it's > just padding around the same tired old claim.
It's only in order to easen the problem for you and others, who are comprehending very slowly.
Paths are solely defined by nodes, according to set theory. (Otherwise you could never gather uncountably many.) Now try to find a path that is missing in the Binary Tree constructed by countably many paths. You fail. Your assertion fails. But you insist in believing that old nonsense because it is so familiar. So you ask for further paths by means of finite definitions like 1/3. Therefore I have appended all possible tails to all finite paths (which would be sufficient to cover the tree but not sufficient to answer the latter question).