On 9 Jun., 20:58, Virgil <virg...@nowhere.com> wrote: > In article > <13ee1342-16b3-47ed-8cb4-48900dab1...@o36g2000vbi.googlegroups.com>, > > > > > > MoeBlee <jazzm...@hotmail.com> wrote: > > On Jun 9, 10:36 am, MoeBlee <jazzm...@hotmail.com> wrote: > > > > Or do you mean: > > > > S is potentially infinite <-> there exists an ordering R on X such > > > that there is no R-maximal member of X > > > Correction (and incorporating Virgil's remark about the empty set): > > > S is potentially infinite <-> (S is non-empty & there exists a linear > > ordering R on S such that there is no R-maximal member of S). > > > But, isn't that equivalent with 'infinite' (i.e. 'not finite', i.e., > > not equinumerous with any natural number') anyway? > > It is! So if that is WM's definition, it is the same as at least one > definition of actual infiniteness.
My definition is: The union of finite segments is a finite segment. There is always a last segment. But it can be surpassed.
The idea that the infinite union of finite segments1 1,2 1,2,3 ... results in an infinite segment is false.
You can verify this by forming an infinite union of finite segments 1 1 1 ...
If the infinite union of finite segments always yields an infinte segment, why then is the result of this infinite union not an infinite segment?
Briefly: The set of all natural numbers that exists has always a last element, but potentially infinite sets are not static. They can grow (and they can shrink).