On Monday, December 9, 2013 4:24:26 PM UTC-4, Zeit Geist wrote: > On Monday, December 9, 2013 10:59:11 AM UTC-7, wpih...@gmail.com wrote: > > > To understand where WM is coming from one must note > > > > > > that for WM the statement > > > > > > for all n in N, p(n) is not true > > > > > > does not imply > > > > > > there is no n in N such that p(n) is true, > > > > > > but only that > > > > > > one cannot find an n in N such that p(n) is true. > > > > > > > Yes, but he refuses to describe the meaning of "one cannot find". He uses the word "taken", which also has an unclear meaning. > > > > His argument against the Infinite rests on Philosophicial grounds. > > However, he insists that assuming the Existence of an Infinite Set leads to a Logical Contradiction. > > But, when showing this purported Contradiction, he slips in an Implict Assumption of Anti-Infinity. > > Then, a Contradiction results. Wow! I imagine that. > > > > > > > So for WM: > > > > > > If L is a constructable list of constructable 0/1 lists > > > then there is a constructable 0/1 list, y, such that one cannot > > > find a n in N with y the nth element of L. > > > > > > > So, y could "really" be on L, but we can't "find" it's n? > > Let me know if I interpreted this right?
Yes, (however, as I note, only WM seems to believe that this is possible). > > > > However, WM actually claims the we can't find All digits of y, >and hence y can Not be shown Unequal to All members on the L.
Careful, y here is a finite algorithm, not an infinite string of digits. It is possible to show by induction that y is not equal to the nth member of L (also a finite algorithm) for any n in N. WM agrees to this point. However, WM does not think that this means there is no n in N such that y is the nth member of L. WM takes Red Queen lessons.