
ran(EF) contains ~EF(n)
Posted:
Apr 30, 2013 11:11 AM


Basically this EF is the limit of a function. Notarealfunction, it's modeled by real functions and in fact very simply n/d. Then, the properties of its range have that: it is symmetric about some midpoint in its range 1/2, for f^1(1/2), defined for even d. Just as ran(f) starts from .000... and increases in constant monotone, in reverse it starts with .111... (or for example in decimal, .999...) and decreases. Then, as there is a less than finite difference between elements of the range for consecutive elements of the domain by their natural ordering, there is in the range an element that starts with .1, .11, .111, ..., and each finite initial segment of integers as expansion. This is from seeing that the elements of ran(f) are the same elements of ran(REF) for REF the "reverse" equivalency function.
EF: n/d REF: (dn)/d
As there exists dn for each n and d, there exists here ~EF(n) = REF(n) constructively, and each element in ran(EF) is in ran(REF).
Then, here, constructively: ~EF(n) e ran(EF), quod erat demonstrandum.
There are replete properties of this function and its range that its range is R_[0,1].
Regards,
Ross Finlayson

