> On 31 Mai, 19:27, Virgil <virg...@nowhere.com> wrote: > > > > Any set in a sane set theory has all members equally "available".
On Jun 1, 6:54 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > Then a complete set with linear order should have a last element.
"should" is NOT a logical notion. NEITHER is "complete", as YOU use it, SINCE YOU FAIL TO DEFINE IT.
It is, moreover, fairly comical that you say "set with linear order" as though that were somehow ONE thing. ANY set can be put into ANY order. ANY ordering on ANY superset of ANY set can be RESTRICTED to a smaller set. The point is that the set and the ordering SIMPLY HAVE NOTHING TO DO WITH each other. If a set is infinite then YOU MAY ORDER IT ANY WAY YOU LIKE. Insisting that the order be linear DOES NOT AFFECT the question of whether there is a FIRST OR a last element! Since the reverse/converse of any linear order IS ALSO A LINEAR order, if it had to have a last element then it would also have to have a first one. The set of the integers HAS NEITHER.