In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 10 Apr., 08:59, William Hughes <wpihug...@gmail.com> wrote: > > On Apr 10, 7:52 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 9 Apr., 20:15, William Hughes <wpihug...@gmail.com> wrote: > > > > <snip> > > > > > > No line is an infinite > > > > sequence of finite unions. > > > > > But if the list contains infinitely many (more than any finite number > > > of) FISONs, then it contains infinitely many (more than any finite > > > number of) unions, doesn't it? > > > > Yes, things that are elements are infinite in number. > > Numbers *are* their number. 10 positive naturals include the number > 10. n tpositive naturals include the number n.
The FIRST ten do, but not every ten, as WM apparently claims.
> > > > Good. Then we are agreed that it makes perfect > > sense to say that any one line (and all its predecessors) > > can be removed, but the collection of all lines > > cannot be removed > > By a single move. When applying induction, i.e., when n is removed, > also n+1 is removed, every finite line is removed. Then no finite > lines remain - only actually infinite lines remain. > > > Thus, the fact that there is no line (along with > > all its predecessors) that cannot be removed > > is not a contradiction. > > > > Next argument.- > > The old question that you heve refused to answer: The Binary Tree, > when constructed by all its FISONs, does it contain more than all its > FISONs?
It depends on what is allowed in that construction.
If one allows arbitrary unions of nested sets of fisons (where of each pair one is a subset of the other) then yes, since the union of any maximal such set of nested FISONs is a path, and every path is such a union.
Actually, one could even redefine a path to be a maximal nested set of FISONs, in which case as soon as all those FISONs exist, by any form of construction, so automatically does the COMPLETE INFINITE BINARY TREE, with its uncountably many paths. --