In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 6 Apr., 22:30, William Hughes <wpihug...@gmail.com> wrote: > > > > > My claim is: > > > > Let D be the set of all lines, and let > > E be any one finite subset of D. > > Then D\E is not empty. > > Your claim is true. But your premise is wrong. E (the set that can be > removed without changing the union of the lines) is the set of all > lines that have a follower. This set of lines is infinite. By the > existence of the follower to all lines, all lines, i.e., E can be > removed without changing the union of all lines, namely |N.
As usual, WM's logic is ill.
If, from a well ordered set, one is only allowed to remove a member or set of members so long as at least one subsequent member remains, how does WM claim that one can ever remove all remainng members without leaving any behind?
Removing all members does exactly that forbiddeen thing, so is not allowed!.
Thus E cannot be empty.
A similar argument shows that E cannot be finite. --